Introduction To Condensed Matter Physics
Introduction To Condensed Matter Physics
Introduction To Condensed Matter Physics
Antonio H. Castro Neto Department of Physics Boston University February 28, 2006
Contents
1 Atoms and Molecules 1.1 Interactions in Isolated Atoms 1.2 Atomic Magnetism . . . . . . 1.2.1 Thermal Effects . . . . 1.3 Molecules . . . . . . . . . . . 1.3.1 The H+ molecule . . . 2 1.3.2 The H2 molecule . . . 1.3.3 Ionic interactions . . . 1.4 Problems . . . . . . . . . . . 7 7 11 15 16 16 20 23 23 27 28 31 33 34 39 39 41 44
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2 Crystals 2.1 Discrete Symmetries . . . . . . . . . . . . . . . . . . 2.1.1 The reciprocal lattice . . . . . . . . . . . . . . 2.1.2 The one-dimensional chain . . . . . . . . . . . 2.2 Elastic scattering . . . . . . . . . . . . . . . . . . . . 2.2.1 Experimental constraints . . . . . . . . . . . . 2.3 Defects in solids . . . . . . . . . . . . . . . . . . . . . 2.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Appendix: Dirac and Kronecker delta functions
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3 Elasticity Theory 45 3.1 Elastic properties of crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 3.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 4 Atoms in motion 4.1 The one-dimensional chain . . . . . . . . . . . 4.2 Phonons in higher dimensions . . . . . . . . . 4.3 Phonons at nite temperature . . . . . . . . . . 4.4 Inelastic scattering . . . . . . . . . . . . . . . 4.5 Problems . . . . . . . . . . . . . . . . . . . . 4.5.1 Appendix: Thermal averages for bosons 55 55 57 63 67 71 72
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5 Electrons in Solids 75 5.1 The one-dimensional chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 5.2 Dimensions higher than one . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 5.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 3
4 6 The Free Electron Gas 6.1 Elementary excitations . . . . . . . . . . . . . . . . 6.2 Free electron gas in a magnetic eld . . . . . . . . . 6.2.1 Zeeman effect and Pauli susceptibility . . . . 6.2.2 Landau levels and de Hass-van Alphen effect 6.2.3 Flux quantization . . . . . . . . . . . . . . . 6.3 Free electrons at nite temperature . . . . . . . . . . 6.3.1 Specic heat . . . . . . . . . . . . . . . . . 6.3.2 Correction to the Pauli susceptibility . . . . . 6.3.3 Landau diamagnetism . . . . . . . . . . . . 6.4 Problems . . . . . . . . . . . . . . . . . . . . . . . 7 Disordered electronic systems 7.1 Weak scattering . . . . . . . . 7.2 Strong scattering . . . . . . . 7.3 Scaling Theory of Localization 7.4 Problems . . . . . . . . . . . 8 Semiconductors 8.1 Intrinsic Semiconductors 8.2 Extrinsic Semiconductors 8.3 The p-n junction . . . . . 8.4 Problems . . . . . . . .
CONTENTS
95 98 101 102 105 110 112 116 118 119 121 123 123 127 130 133 135 137 138 141 144 145 145 147 150 151 152 154 156 157 158 160 163 165 167 167 171 172 175 176 176 177 182 184
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9 Electron-phonon interactions 9.1 Boson operators . . . . . . . . . . 9.2 Fermion operators . . . . . . . . . 9.2.1 Free electrons in a box . . 9.2.2 Free electrons in a lattice . 9.3 Electron-phonon interaction . . . 9.3.1 The optical case . . . . . . 9.3.2 The polaron problem . . . 9.3.3 The acoustic case . . . . . 9.3.4 Solitons in one dimension 9.3.5 Retarded Interactions . . . 9.3.6 Cooper pairs . . . . . . . 9.4 Problems . . . . . . . . . . . . .
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10 Magnetism 10.1 Crystal elds . . . . . . . . . . . . . . 10.2 The Heitler-London theory . . . . . . . 10.3 Coulomb interactions . . . . . . . . . . 10.4 Magnetic anisotropy . . . . . . . . . . 10.5 Localized Magnetism . . . . . . . . . . 10.5.1 Problems . . . . . . . . . . . . 10.6 Magnetic interactions in metallic alloys 10.7 The Double Exchange problem . . . . . 10.7.1 Problems . . . . . . . . . . . .
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CONTENTS
10.8 The Anderson Hamiltonian 10.9 The Kondo problem . . . . 10.9.1 Problems . . . . . 10.10Itinerant Magnetism . . . . 10.10.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 185 191 200 201 205 207 207 210 213 217 218 219 223 225 227 227 232 232 235 239 241 242 244 249 250 253 254 260
11 Magnetic phase transitions 11.1 Spontaneous symmetry breaking . . . . . . . . . . 11.1.1 Critical exponents . . . . . . . . . . . . . 11.2 Mean eld approach . . . . . . . . . . . . . . . . 11.3 Mean eld and beyond . . . . . . . . . . . . . . . 11.3.1 Interactions with innite range . . . . . . . 11.3.2 Local interactions . . . . . . . . . . . . . . 11.4 Continuous symmetries and the Goldstone theorem 11.5 Problems . . . . . . . . . . . . . . . . . . . . . . 12 Magnetic excitations 12.1 The one-dimensional Ising model 12.2 The Heisenberg-Ising Hamiltonian 12.2.1 The ferromagnetic case . . 12.2.2 The antiferromagnetic case 12.3 Problems . . . . . . . . . . . . .
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13 The Interacting Electron Gas 13.1 The Thomas-Fermi approach . . . . . . . . . . . . . . 13.2 The Random Phase Approximation: RPA . . . . . . . 13.3 Response of the one-dimensional electron gas . . . . . 13.4 The Kohn anomaly and the Peierls distortion . . . . . . 13.5 Electron-electron interactions . . . . . . . . . . . . . . 13.6 Landaus Theory of the Fermi Liquid: An Introduction 13.7 Problems . . . . . . . . . . . . . . . . . . . . . . . .
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CONTENTS
Chapter 1
where pP (pe ) is the momentum, rP (re ) the position of the proton (electron) and e is the electric charge. In quantum mechanics, momentum and position are operators that act on a Hilbert space of functions. Moreover, momentum and position are conjugated so that they obey commutation rules, namely, [xi , pj ] = i i,j h (1.2)
where we have introduced the components of the vector as r = (x1 , x2 , x3 ) and p = (p1 , p2 , p3 ) (i,j = 1 if i = j and 0). Moreover, the operators for the electron and proton commute among themselves since they are particles with different quantum numbers. The state of the system can be represented in terms of the positions of electron and proton by |rP , re , e , P where is the spin degree of freedom of each one of these particles. Protons and electrons have spin 1/2 and are called fermions. In this case can only have two possible projections on a xed axis, that is, up () or down (). Observe that although |rP , re , e , P is a legitimate state of the problem and span the whole Hilbert space, it does not represent an eigenstate of the Hamiltonian. The reason is that the momentum operator that appears in the kinetic term of (1.1) does not commute with the position operator and therefore it induces transitions between states with different positions (that is, the electron and proton move around). As usual in any problem in quantum mechanics one has to nd the basis that properly describes the system of interest. The obvious thing to do, as in classical mechanics, is to transform the Hamiltonian (1.1) 7
mp rP + me re mp + me r = rP re . P2 p2 e2 + 2M 2 r
(1.3)
where P = pP + pe is the total momentum operator, p = (mp pe me pP )/M is the relative momentum operator, M = me + mp is the total mass and = 1/me + 1/mp is the reduced mass of the system. Observe that since mp 1000me we can rewrite M mp and me with good accuracy. Hamiltonian (1.4) already represents a major simplication in regards to (1.1). First of all we observe that the center of mass motion decouples from the relative motion since R does not appear explicitly in (1.4). Notice that almost all kinetic energy of the center of mass is carried by the proton that is much heavier. It implies that we can diagonalize the problem in the basis of total momentum operator P: P|K = hK|K (1.5)
that is equivalent say that the electron-proton system moves freely in space. The non-trivial part of the problem is the solution of the relative motion. Because the potential is central the problem has a symmetry of rotations in space. Rotations in space are generated by the angular momentum operator L = r p that can be rewritten in a convenient form: Li = j,k i,j,k xj pk where i,j,k is the Levi-Civita tensor that is completely antisymmetric (that is, i,j,k = 0 if i = j or i = k or j = k and the other components are dened such that 1,2,3 = +1 as well as all other cyclic permutations, 2,3,1 = 3,1,2 = +1 and all the non-cyclic permutations are 1: 1,3,2 = 1). The angular momentum operators obey the so-called Lie algebra [Li , Lj ] = i i,j,k Lk h (1.6)
that you can easily show from denition of Li and (1.2). The fact that the angular momentum operators do not commute among themselves implies that one cannot classify the states in terms of these operators independently. Instead we use one of them, say, L3 (Lz ) and its amplitude L2 = i L2 . It is indeed trivial i to show that the Hamiltonian (1.4) commutes with these operators, [H, L2 ] = [H, L3 ] = 0 and therefore the states of the system can be classied according to the eigenvalues of these operators, namely, L2 |l, m = h2 l(l + 1)|l, m = hm|l, m (1.7)
L3 |l, m
where m = l, l + 1, l..., l 1, l in unit steps. The Schrdinger equation for the problem determines the principal quantum number n. Thus, we can dene an eigenstate of the Hamiltonian has the form |K, l, m, n, e , P so that, H|K, n, l, m, e , P = h 2 K 2 13.6(eV ) 2M n2 |K, n, l, m, e , P (1.8)
is the full solution of the problem. Observe that the solution of this problem does not involve l = 0, 1, 2..., n 1 (these are usually called s, p, d, f, ... states) , m = l, ..., +l or the spin degrees of freedom. Thus the
states of the system are degenerate in this quantum numbers (there is more than one state of the system with the same energy). Moreover, it is obvious that the ground state corresponds to occupy an electron state with n = 1 and K = 0 corresponding to a static Hydrogen atom. The eigenstates of the problem can be represented in real space by projecting |Kn, l, m, e , P , that is, K,n,l,m(R, r, , ) = R, r, , |K, l, m, n (1.9) eiKR Yl,m(, )Rn,l (r)
where Yl,m (, ) is a spherical harmonic and Rn,l (r) is the so-called radial wavefunction that extends over a distance of order of the Bohr radius, a0 = h2 /(e2 ). The shape of some of these functions is depicted in Fig.1.1.
n=1,l=0
n=2,l=1
n=2,l=0
+ +
l=0
l=1
l=2
Figure 1.1: Radial and angular dependence of some of the wavefunctions of the H atom. Things are relatively simple in the H atom because the proton only acts as an external potential. In atoms with more than one electron the situation is not so simple because electrons interact among themselves via the Coulomb interaction. Consider, for instance, the case of the He atom that has 2 electrons and 2 protons. The nucleus of the atom has mass 2mp and the full Hamiltonian of the problem reads H= p2 p2 p2 2e2 2e2 e2 P + 1 + 2 + 4mp 2me 2me |rP r1 | |rP r2 | |r1 r2 | (1.10)
where r1,2 is the position of the two electrons. This is a quite complicated problem. The wavefunction of the system is a function of the coordinates, (rP , r1 , r2 ). What we need, however, is not the complete solution of the problem but an approximate solution that provides qualitative understanding and a reasonable quantitative description. Since the mass of the proton is so much larger than the mass of the electron one expects the kinetic energy of the protons to be small compared to the other terms in the Hamiltonian, that is, we expect the electrons to move distances much larger than the protons in the same time interval. During this time interval the protons appear static for the fast electron. If the protons are static then we are back to our one-body quantum mechanics problem where the electrons move under a potential eld created by the two protons. This potential is parameterized by the distance between the protons (this implies that we can
10
solve the Schrdinger equation for each conguration of the protons). If this is the case the wavefunction of the electrons can be written as rp (r1 , r2 ), and depends on the proton coordinates as a parameter. Therefore, it is intuitive to look for a variational solution of the quantum mechanical problem with the form: (rp , r1 , r2 ) = rp (r1 , r2 )(rp ) (1.11)
where (rP ) is the nucleus wavefunction. Eq.(1.11) is known as the Born-Oppenheimer approximation. In order for this wavefunction to make any sense one imposes that rp (r1 , r2 ) is an eigenstate of HS (rP ) = p2 p2 2e2 2e2 e2 1 + 2 + 2me 2me |rP r1 | |rP r2 | |r1 r2 | (1.12)
with energy E (rP ) where labels the possible quantum numbers of (1.12). Once the eigenenergies of (1.12) are known the nucleus wavefunction (rp ) is an eigenstate of HN = p2 P + E (rP ) . 4mp (1.13)
This separation of energy scales between nucleus and electrons is natural and it will be discussed in more detail later. There is no exact analytical solution for the quantum mechanical problem of three bodies (contrary to its classical counterpart). We can understand the physics of the many-electron atom qualitatively based on what we know about the H atom and Paulis exclusion principle that states that there are no two electrons with the same quantum numbers. As you probably know this is a consequence of the Fermi-Dirac statistics obeyed by electrons. Let us go back to He atom in the Born-Oppenheimer approximation (1.12) and set rP = 0 for simplicity, that is, HS = p2 2e2 2e2 e2 p2 1 + 2 + . 2me 2me |r1 | |r2 | |r1 r2 | (1.14)
that describes two electrons moving in the Coulomb potential of a charge +2e and interacting with each other. Observe that because the electrons are practically in the same volume, the direct interaction between the electrons and the nucleus is approximately 2 times larger than the electron-electron interaction and has to be considered rst. At short distances the electron experiences the full potential of the nucleus with charge +2e while at larger distances it feels a smaller effective charge because of the other electron, that is, it sees a Coulomb potential with charge 2e e = +e. This effect only occurs because the electrons are free to move and adapt to the changes in the Coulomb potential. In a static system (that is, a problem where me ) this is not so and the electron actually experiences the full charge of the nucleus. The process in which the kinetic energy of the electrons leads to a smaller effective nuclear charge is called screening. What we are proposing is that, beyond the Hamiltonian (1.10), there is a simpler Hamiltonian or effective theory that describes the problem. For the moment being this effective theory is hidden due to our current ignorance and lack of sophistication. Nevertheless, due to the symmetry of the problem, the form of the potential does not change substantially with distance since the charge distribution is spherically symmetric. Therefore the wavefunction of the problem looks like a H-like state with a slightly different energy than the H atom. Thus, the He atom is obtained by lling up the 1s state (n = 1 and l = 0) of the H atom with electrons of opposite spin (Paulis principle). Observe, moreover, that because the rst shell is lled, the He atom is not very reactive since, as seen from far away, it looks like a neutral object to a foreign electron. Indeed, the ionization energy of the He atom is 24.6 eV instead of the 13.6 eV of the H atom (1 eV 104 K).
11
Consider now a somewhat more complicated situation: the Li atom that has charge +3e in the nucleus. Again using our effective theory we conclude that the wavefunctions are H-like and that in the ground state we can occupy the rst allowed 1s state with two electrons with opposite spins. That is, we form rst a He+ atom. Where does the second electron go? Naturally it should go to a n = 2 state that can be in the s or p state (l = 0 or l = 1, respectively). In the H atom these two states are degenerate but is this true for the Li atom? The reason for the non-degeneracy of the Li atom is related with the charge of the nucleus and the shape of the wavefunctions in the s and p states. In Fig.1.2 we show the result of the combination of the radial and angular part of the wavefunction as shown in Fig.1.1. Observe that in the s state the electron is closer to the nucleus that has charge +3e, while in the p orbital the electronic charge is distant from nucleus. Thus, the s state has lower electron-proton Coulomb energy than the p state and is occupied rst. Again, it is the preponderance of the Coulomb interaction (that in the case of Li is 3 times larger than the electron-electron interaction) that determines the ground state properties. Observe that since the rst 1s shell if lled this state is H-like and the ionization energy for this electron is 5.4 eV showing that Li is chemically reactive. Most of the atoms in the periodic table can be understood by simple arguments like these ones. The understanding of the atoms and how one describes their ground state is fundamental for the understanding of solids. The formation of a solid depends on how the protons interact with the electrons and how the electrons interact among themselves.
Figure 1.2: The shaded areas are proportional to the probability of nding the electron in each orbital.
H0 =
i=1
e 1 pi + A(ri ) 2m c
+ B g0
i=1
B Si / h
(1.15)
where ri and pi are the coordinate and momentum of the ith electron in the atom, A(r) is the vector potential (B = A), the last term is the Zeeman energy of the electron spin in a eld B, Si is the spin of the ith
12
electron, B = e /(2mc) is the Bohr magneton, and g0 2 is the g-factor. In (1.15) we have h
N
HI =
i=1
Ze2 + |ri |
N i,j=1,i=j
e2 , |ri rj |
(1.16)
describes the Coulomb energy for the interaction between the electron and the nucleus (Z is the nuclear charge) and between the electrons themselves. We assume that the magnetic eld is applied in the z direction and choose an electromagnetic gauge such that A= B (xy yx) 2 (1.17)
that is called the symmetric gauge (notice that B = Bz). Substituting (1.17) into (1.15) we get H0 =
i
p2 i + B B 2m
(Lz,i + g0 Sz,i )/ + h
i
e2 B 2 8mc2
2 x2 + yi i i
(1.18)
We can show that for practical purposes the terms in the Hamiltonian where the magnetic eld appears explicitly are small when compared to the terms that are eld independent. In this case the magnetic eld can be well treated in perturbation theory. Thus, assuming the magnetic eld to be small we nd that, up to second order in perturbation theory, the change in the ground state energy due to the magnetic eld is given by: E0 = B B 0|
i
| m|
m=0
i (Lz,i
+ g0 Sz,i )/ |0 |2 h E0 Em (1.19)
i (Lz,i
e2 B 2 8mc2
0|
We can immediately estimate the size of these terms. The rst order correction is B B n| g0 Sz,i )/ |n B B c where h h c = eB mc
(1.20)
is the cyclotron frequency. This term is of order of 104 eV ( 1 K) for a eld of 1 Tesla (T). In the third 2 h term we have e2 B 2 /(8mc2 ) n| i ri |n (e2 B 2 a2 )(8mc2 ) ( c )2 /(e2 /a0 ) that is of order of 109 0 5 K!) for a eld of 1 T. Indeed these terms are small when compared to the characteristic atomic eV ( 10 energies that are of order of 104 K. In order to calculate the perturbative shift in energy given in (1.19) one needs to know the nature of the ground state of the atom in the absence of the eld. In the absence of interactions among the electrons this is given by the energy levels of the H atom. The system is degenerate because the problem has rotational symmetry and for each state l there are 2l + 1 degenerate states corresponding to the possible projections of l, that is, to the quantum numbers m = l, ..., l. In a system with N electrons we can distribute the electrons among 2(2l + 1) states (the factor of 2 comes from the two possible spin orientations) without changing the energy of the state. The total number of possible combinations of quantum numbers is [2(2l + 1)]!/(N ![2(2l + 1) N ]!). Firstly, let us consider the problem of an atom with lled shell, that is, with N = 2l + 1. This is the case of noble gases such as He and Ar. It is clear that in the ground state we have Lz |0 = S z |0 = 0 and 2 2 2 the only term that matters is the last one in (1.19). Using that fact that x2 = yi = zi = ri /3 for the i
13
N e2 B 2 2 a 12mc2
(1.21)
(1.22)
The magnetization per unit of volume in the system for Na atoms is given by M = N e2 a2 B Na E = V B 6mc2 (1.23)
that has a negative sign implying that the response is diamagnetic, that is, in responding to an applied eld the atoms want to reduce its value ( = Na /V is the atomic density). The magnetic susceptibility is given by = M N e2 a2 = . B 6mc2 (1.24)
The situation described above is simple because we are dealing with a system with a closed shell. Most atoms have only partially lled shells. In this case the matrix elements in (1.19) do not vanish. The interaction among the electrons becomes important and the degeneracy of the orbitals is lifted by the Coulomb interaction. We observe that because the Coulomb interaction is spherically symmetric the total angular momentum, L, and the total spin, S, are constants of motion (that is, they commute with the Hamiltonian) and the states can still be classied in terms of the eigenstates of these two operators, that is, we can simplify the problem by working with the basis: |L, S, Lz , Sz . Moreover, the system is degenerate since for a xed value of L and S we can have (2L + 1)(2S + 1) states for different values of Lz and Sz corresponding to different projections of L and S. This is called a multiplet. Consider, for instance, an atom with a conguration like 4f 2 . The f orbital can comport 14 electrons. Since there are two electron they can be organized into the orbitals in 14!/(2!12!) = 91 different ways! In this case we have S = 0, 1 and L = 0, 1, 2, 3, 4, 5 and for each value of S and L there are (2S + 1)(2L + 1) degenerate states. The relevant question is: what state |S, L has the lowest energy? In order to answer this question one needs a quantitative calculation that involves the coupling of all the electrons via the Coulomb term. Observe, however, that the exclusion principle does not allow two electrons with the same spin to be in the same place in space. Thus, electrons with the same spin effectively repel each other (we will give a natural explanation for this phenomenon later when we consider the Coulomb interaction in more detail). Naturally this effect lowers the Coulomb energy. Thus, electrons in an atom tend to have all their spins aligned. This is called Hunds rst rule: in an atom the electrons maximize the total spin S. The rst rule solves the problem of the spin but not of the orbital angular momentum. Again the name of the game here is: minimize the Coulomb energy between electrons. On the one hand, in states with small L (like an s-state) the electrons spend much of their time close to the nucleus and therefore pay the energetic price of the repulsion among themselves. On the other hand, in states with large L the electrons are apart from the nucleus and feel a weaker Coulomb repulsion (they explore a larger volume of space). This gives rise to Hunds second rule: For a maximum value of S the energy is minimized by the largest value of L. The rst and second Hunds rules specify the values of L and S for which the energy is minimum but still for xed L and S we have (2L + 1)(2S + 1) degenerate states. Is this degeneracy real in an atom? The answer is no! The reason being that the orbital motion is coupled to the spin of the electron by the
14
so-called spin-orbit effect. Consider one electron moving with velocity v around a nucleus. From the point of view of the electron (that is, looking at the problem at the frame co-moving with the electron) the nucleus moves around with velocity v. Since the nucleus is charged, we can imagine the nucleus motion as a little current of charge circulating around the electron. This current generates a magnetic eld at the position of the electron that is proportional to r v L. Thus, the Zeeman energy created by this eld is also proportional to L S, that is, the spin-orbit coupling. One usually writes the Hamiltonian associated with this coupling as HSO = L S (1.25)
where depends on details of the atomic problem and can be obtained from the atomic spectra. Indeed, if one adds (1.25) to our original Hamiltonian (1.15) we see that L and S do not commute with the Hamiltonian separately. However, the total angular momentum J=L+S (1.26)
commutes with the Hamiltonian. It means that we can still classify the eigenstates of the problem in terms of J, that is, the states can be labeled as: |J, Jz , L, S . This implies that the degeneracy of the (2L+ 1)(2S + 1) states is lifted and the states split into (2J + 1) degenerate states corresponding to different values of Jz . Indeed we can rewrite (1.25) in terms of this operator as HSO = 2 J L2 S2 2 (1.27)
since the Hamiltonian commutes with J, L2 and S2 . Observe that due to (1.26) the allowed values of J go from |L S| to L + S in unit steps and the allowed valued of Jz go from J to J (that gives the degeneracy of 2J + 1). The energy associated with each one of these states is E(J, Jz , S, L) = [J(J + 1) L(L + 1) S(S + 1)] . 2 (1.28)
It is experimentally observed that > 0 for shells that are less than half lled (N 2l + 1) and < 0 for shells that are more than half lled (N 2l + 1). If > 0 it is clear from (1.28) that the energy is minimized for a given S and L for a conguration with smallest J, that is, J = |L S|. Otherwise, if < 0 the energy is minimized for the largest value of J, that is, J = L + S. This is the so-called Hunds third rule. Hunds rules can guide us to understand the response of an atom to a magnetic eld. Consider a system where J = 0. In this case we can use the Wigner-Eckart theorem and show that the rst term in the energy in (1.19) actually vanishes. We are left with the two other terms. The third term is just the diamagnetic response we studied in the case of atoms with lled shell. The second term is negative (remember that second order perturbation theory always lowers the energy of the ground state) and therefore it gives rise to a susceptibility with positive sign. This is the so-called Van Vleck paramagnetic response. If J = 0 then the rst term in (1.19) does not vanish and it is the largest contribution to the energy shift. We can rewrite this term in a more appropriate form HZ = B B (L + g0 S) = B B (J + (g0 1)S) that allows us to interpret J + (g0 1)S as the effective magnetic moment of the atom, M = B (J + (g0 1)S) (1.30) (1.29)
15
J S
that is not a constant of motion since S is not conserved. It turns out, however, that J is a constant of motion and we can think of J as being xed and that L and S precess around J as in Fig.1.3. Thus the magnetic moment is given by the component of L + g0 S parallel to J. This is the only component of the magnetic moment that contributes in rst order perturbation theory since the components of S in the direction transverse to J introduce transition between different values of Jz and give zero average when we calculate with an unperturbed state |J, Jz , L, S . The parallel component of S can be calculated from the angle between J and S, SJ = = (J S) J J2 (J2 L2 + S2 ) J 2J2
(1.31)
which, in a state |J, Jz , S, L has an expectation value: SJ = [J(J + 1) L(L + 1) + S(S + 1)] J 2J(J + 1) (1.32)
and therefore the effective magnetic moment in the direction of J is given in (1.30) M = gB J where g(J, L, S) = (g0 + 1) (g0 1) S(S + 1) L(L + 1) + 2 2 J(J + 1) (1.34) (1.33)
is the Land g-factor. The energy associated with (1.29) is E(J, Jz , S, L) = gJz B B (1.35)
where Jz = J, ..., J. This nal expression gives the splitting between the energy levels of a multi-electron atom in the lowest order approximation.
16 ( = 1/(kB T ))
J J
Z =
Jz =J
eE(J,Jz ,S,L) =
Jz =J
eg(J,L,S)Jz B B (1.36)
where in the last line we used the sum of a geometric series. The magnetization is obtained like in (1.23) and it is given by M (B) = gB JBJ (gJB B) where BJ (x) = (2J + 1) coth 2J (2J + 1)x 2J x 1 coth 2J 2J (1.38) (1.37)
is the so-called Brillouin function. Notice that for gB B 1 (that is, low temperatures) the magnetization saturates at M gB J as expected since all the moments are aligned. At high temperatures, that is, gB B 1 we nd M and gives a magnetic susceptibility (T ) = J(J + 1)(gB )2 3kB T (1.40) J(J + 1)(gB )2 B 3kB T (1.39)
that is known as the Curie susceptibility and it is important because it allows an experimentalist to extract the effective moment of the atom, J. Notice that g, B and kB are universal quantities and can be measured independently. Thus, by plotting 1 (T ) as a function of T one should get a straight line where the slope is essentially J(J + 1).
1.3 Molecules
1.3.1 The H+ molecule 2
We have seen that the physics of many-electron atom can be quite complex, but qualitatively explainable in terms of basic quantum mechanics. Let us consider a problem one notch higher in complexity, namely, the formation a molecule. Many of the concepts that lead to the understanding of the formation of molecules are useful in order to understand the stability of matter. To get a qualitative understanding of the problem let us consider the problem of two protons and one electron, that is, the H+ molecule. The Hamiltonian of this problem can be written as 2 H= p2 + p2 p2 e 2 + 1 e2 2me 2mp 1 1 1 + |re R1 | |re R2 | |R1 R2 | (1.41)
where re is the position of the electron and R1 and R2 are the positions of the two protons. The rst three terms are the kinetic energies of each one of these particles and the last three their interaction energy.
1.3. MOLECULES
17
This is a quite complicated problem. The wavefunction of the system is a function of all the coordinates, (re , R1 , R2 ). In order to simplify the problem we make use of the Born-Oppenheimer approximation and assume that the protons are static during the time of motion of the electron. Suppose that initially the protons are innitely apart. For simplicity assume R1 = 0 and R2 = R. In this case the problem has two solutions that are degenerate with each other, that is, either the electron is bound to proton 1 or to proton 2 with the same energy E0 . The wavefunction of the electron, 0 , is well localized in each proton (it is a H-like wavefunction). Let us consider a simplication of the problem that disregards all the other states of the problem except the states where the electron is localized in one of the two protons as in Fig.1.4. Let 1| ( 2|) be the state of the electron bound to proton 1(2). If R one has H0 |1 where 1|2 = 0 and 1|1 = 2|2 = 1.
H0 |2
= E0 |1
= E0 |2
(1.42)
|1>
|2>
R
Figure 1.4: Two states retained in the two-level system approximation of H+ . 2 Equation (1.42) can be rewritten as H0 = E0 (|1 1| + |2 2|) . (1.43)
As the protons get closer to each other there is a nite probability that the electron hops from proton 1 to proton 2 and vice-versa. This is the so-called quantum tunneling and is depicted in Fig.1.5. The tunneling depends on the amount of overlap between the wavefunction of the electron in the two different protons. There is an energy scale associated with the tunneling that we are going to call t. t is a function of R, vanishing when R , and becomes large when R 0. Thus, in order to incorporate tunneling into the problem one has to add a perturbation that mixes the two states. This perturbation we call HT and it has to be such that HT 1 2 (1.44)
2 and of course HT I since if we hop the electron twice it has to return to the same atom. It is obvious that the tunneling Hamiltonian must have the form:
HT = t (|1 2| + |2 1|) .
(1.45)
18
Any eigenstate of H = H0 + HT has to be a linear combination of states 1| and 2|. This problem can be studied by rewriting the Hamiltonian in matrix form. The matrix elements are: 1|H|1 1|H|2 that can be rewritten in matrix form: [H] = E0 t t E0 . (1.47) = = 2|H|2 = E0 2|H|1 = t (1.46)
Observe that in terms of the matrix formulation the states are represented by vectors [1 ] = 1 0 0 1 (1.48)
(1.49)
H| = E| .
(1.50)
From basic quantum mechanics the most general solution of this problem has to be a linear combination of base states, that is, | = a|1 + b|2 . Normalization of the wavefunction (or probability) insures that | = 1 = |a|2 + |b|2 . In the language of (1.47) we have [] = a b (1.51)
and the Schrdinger equation becomes a simple eigenvalue problem [H][] = E[] ([H] E[I])[] = 0 where [I] = 1 0 0 1 (1.53) (1.52)
is the identity matrix. It is a simple exercise to show that eigenstates of the problem are: H|A H|B = (E0 + t)|A = (E0 t)|B . (1.54)
1.3. MOLECULES
where |A |B = = 1 (|1 |2 ) = 2 1 (|1 + |2 ) = 2 1 2 1 2 1 1 1 1 .
19
(1.55)
These states are called anti-bonding and bonding, respectively. Observe that the bonding state is the ground state of the problem. In the representation of position, these two states can be easily written as R, r|B R, r|A = B,R (r) = 0 (r) + 0 (r R) = A,R (r) = 0 (r) 0 (r R)
(1.56)
where 0 is the ground state wavefunction of the H atom (n=0,l=0,m=0 (r)). Observe that the anti-bonding state has a node in the middle position between the protons (A,R (r = R/2) = 0) while the bonding state is always nite. Thus in the bonding state the amount of charge" between the protons is larger (the probability of nding the electron in the middle of the protons is higher) than in the anti-bonding state. That is the reason its energy is lowest: the electrons act as a glue between protons. In the anti-bonding case the electronic charge is mostly centered around the two protons that are therefore "shielded" from each other. Thus, the anti-bonding state is unstable. A plot of square of the wavefunction for each one of these states is shown on Fig. 1.6.
V(r)
+
r
Figure 1.5: Potential energy for H+ molecule showing the overlap of the electron wavefunction in the two 2 different protons. Another way to understand this problem is to realize that at nite R we can expand the potential term in (1.41) in powers of r/R and it is clear that the rst term is of order r/R2 . Since this term is small, we can perform perturbation theory. The rst order term cancels due to symmetry 0 |r|0 = 0. The second order term has the form E = n=0 | |V |n |2 /(E0 En ). This term is always negative if E0 is the ground state. From the previous argument it is clear that |t| = E > 0. Thus, in this way, we relate our original problem with the two-level system calculation. On the one hand, the energy of the ground state (in this case a bonding state) has to decrease as we decrease the distance between atoms. On the other hand, the rst excited state (the anti-bonding state) increases in energy as we decrease the distance between the protons (since t 0 as R ). When the distance between atoms goes to zero the energy has to diverge since the Coulomb energy term 1/R diverges. Thus we conclude that there must be a minimum of the ground state energy at some distance R0 (see Fig.1.7). In a rst approximation the energy close to the minimum is parabolic with R and is quantized in units of 0 E0 . Moreover, since this corresponds to the potential h
2 kme E0 (R R0 )2 2 2 h
(1.57)
where k is a constant of order unit. Thus, by direct substitution in (1.13) we see that the protons undergo harmonic motion around the equilibrium position with a frequency 0 me /mp E0 / . h
(a)
(b)
Figure 1.6: Contour plot for the square of the wavefunction for: (a) Bonding state; (b) Anti-bonding state. At this point it is interesting to check the validity of the Born-Oppenheimer approximation. By direct substitution of (1.11) and (1.13) into the Schrdinger equation we nd that the have neglected terms of the form 2 1 1 /mp . Since variations in R1 produces variations in r we can write this term as h approximately |pe ||pp |/(mp ). Thus, in order for our approximation to be valid we have to require p2 /mp p |pe ||pp |/mp or |pp | |pe |. The momentum of the electron in a bound state of the H-atom is approximately pe me E0 while for the protons undergoing harmonic motion we have pp mp me /mp E0 . Thus the Born-Oppenheimer approximation is reasonable when
1/4 me m1/4 p
(1.58)
that is a good approximation in most cases ((me /mp )1/4 0.16). Thus we have shown that within these approximations the protons undergo harmonic motion and within a period of oscillation of the protons the electron can be found in a bonding state. This is an example of a covalent bond. The term covalent implies that the electron is equally shared by the protons.
1.3. MOLECULES
E
21
Antibonding state
R0 R
Bonding state
Figure 1.7: Energy of the H+ system as a function of the distance between the protons R. 2 negative. Thus, second order perturbation theory produces an effective interaction that behaves like HV dW = 1 2 R6 (1.59)
where 1,2 are the polarizabilities of the atoms. This is a short ranged interaction if compared with the bare Coulomb interaction and furthermore, it is attractive. This is known as the Van der Waals interaction. Our reasoning regarding the ground state of the problem is similar to the one in the problem of the H+ 2 system. Within the Born-Oppenheimer approximation we consider the wavefunction of the problem with two electrons and two protons separated by a xed distance R. We have various possibilities among the various arrangements of the electrons around the protons. We can have one electron around each proton or we can have the two electrons around one proton. In this last situation the Coulomb repulsion between the electrons is large. Therefore, the ground state has to be such that one has on average one electron per proton. This is the so-called Heitler-London approximation. Thus, we have two states that are degenerated when the protons are innitely apart, that is, with the electrons in their original position or with the electrons exchanged between the protons. These two states are depicted on Fig.1.8.
|1>
|2>
R
Figure 1.8: Two states retained in the two-level system approximation of H2 . The situation described here is completely equivalent to the H+ system and as one imagines the protons 2 approaching each other they can exchange their electrons. Let r1,2 be the position of each electron. The
(1.60)
where , are coefcients and i (rj ) is the state of electron j on atom i. If the distance between the atoms is large any linear combination of the type (1.60) is a solution of the problem. As the distance between H atoms is reduced, the electrons from each atom can tunnel from one proton to another. Using the twolevel system approach we nd that the states that matter are the bonding and anti-bonding states that can be written as, A (r1 , r2 ) = NA (1 (r1 )2 (r2 ) 1 (r2 )2 (r1 )) ,
(1.61)
where NA,B are normalization constants. Thus, in principle, we have found the solution of the problem which is entirely analogous with the H+ molecule. 2 Nevertheless, there is a possible inconsistency with our solution. The Pauli principle requires that when we exchange the electrons the wavefunction has to change sign (indeed, one must have, (r1 , r2 ) = (r2 , r1 )) and therefore the wavefunction of two fermions has to vanish at equal position ((r1 , r1 ) = 0). Thus, state B (that we concluded to be the ground state) has the wrong statistics. What is wrong with this picture? What is wrong is that we forgot the electron spin. We can still have a wavefunction that is symmetric with respect to the coordinates if we have a spin part that is anti-symmetric. Thus, from the four states available for the two electron problem, the only spin states that matter are: s (s1 , s2 ) = |s1 =, s2 = |s1 =, s2 = ,
(1.62)
that are singlet and triplet combinations of the spins, respectively. Thus, accordingly to the Pauli principle the states allowed are s (r1 , s1 ; r2 , s2 ) = B (r1 , r2 )s (s1 , s2 ) t (r1 , s1 ; r2 , s2 ) = A (r1 , r2 )t (s1 , s2 ) . In this particular case the singlet state must be the ground state of the problem. It is clear that the Pauli principle has strong consequences here. The tunneling of electrons between atoms favors the bonding state but if the electrons have the same spin projection this state not allowed. Thus atoms with the same spin projection repel each other. As we bring two H atoms together we are trying to form a He atom. If the spins were the same the electronic conguration should be 1s1 2s1 by the Pauli principle. If we had the atoms with electrons with opposite spin then we would get 1s2 that is lower in energy. For the anti-bonding state we can have the electrons with the same spin but in this case the atom is not stable (as shown in Fig.1.7). Thus, the ground state of the H2 atom has to be a singlet state. This discussion implies that the Pauli principle acts as an interaction between the electrons, that is, there is a quantum mechanical repulsion between the atoms with the same spin. There is no classical analogue to this effect. Moreover, at shorter distances the Coulomb repulsion between the electronic clouds becomes large and this effect also increases the repulsion. It is hard to calculate the combined effect of all interactions from rst principles and one usually uses the a phenomenological approach and introduce a repulsive term of the form 1/R12 that is short ranged and is called Lennard-Jones potential so that the atom has a minimum at some R0 as before. Thus the energy of the molecule again looks very much like in Fig.1.7. (1.63)
1.4. PROBLEMS
23
1.4 Problems
1. Verify the following identities related with the Levi-Civita tensor: (1) i,j,k l,m,n = i,l j,m k,n + i,m j,n k,l + i,n j,l k,m i,n j,m k,l i,m j,l k,n i,l j,n k,m ; (2)
k i,j,k k,l,m
Use the denition of Li and (1.2) and the properties of the Levi-Civita tensor to prove (1.6). 2. The ionization energy of the oxygen is 13.6 eV and it is lower than the energy of its neighbors on the periodic table (the ionization energy of N is 14.5 eV and of F is 17.4 eV). Explain qualitatively why this is so in terms of the interaction between the electrons. Hint: Start the problem by thinking what happens to Boron (B) and go up in the atomic number. 3. Argon (A) is a noble gas and has lled shell with a conguration 3 p6 . The next atom in the periodic table is Potassium (K) that has the conguration of lowest energy with 4 s1 instead of 3 d1 ! Provide an argument that explains this observation. 4. Show that equation (1.18) is correct. 5. Show that the total angular momentum dened in (1.26) commutes with the Hamiltonian in the presence of spin-orbit coupling. 6. Use Hunds rules to nd the conguration of lowest energy for an atom in which the last incomplete shell has a conguration: 1) d8 ; 2) f9 . What is the value of the total magnetic moment in each one of these congurations. 7. What is the condition for J = 0 in terms of N and l? 8. Use the Wigner-Eckart theorem to show that for J = 0 the rst term in (1.19) vanishes. 9. Use the algebra of angular momentum (Clebsh-Gordan coefcients) and prove (1.33) and (1.34). 10. Prove equation (1.35). 11. Prove equation (1.36).
13. Two localized spins 1/2 interact via an exchange mechanism that is described by a Hamiltonian H = JS1 S2 . (1.64)
i) What are all the possible congurations of the two spins in the basis of S1 , S1Z , S2 , S2Z ? ii) What are the energies and their respective eigenstates of the system when J > 0? iii) What are the energies and their respective eigenstates of the system when J < 0? iv) Suppose a magnetic eld, B, is applied to the system so that we have to add the Zeeman energy to the Hamiltonian in (1.64): HB = B B (S1Z + S2Z ) (1.65)
where B > 0 is the effective Bohr magneton. Make a plot of the energy of the states you found on item 2) as a function of magnetic eld. What is the state with lowest energy when B ? What is its physical meaning?
14. Consider the Schrdinger equation (1.50). Assume that | = a|1 + b|2 and calculate a and b by direct substitution.
15. Solve the eigenvalue problem of equation (1.52) in matrix form and show that the solution can be written in terms of (1.54) and (1.55).
(1.66)
where R > 0 is the relative coordinate of the nuclei with mass M, m is the mass of the electron and r is the coordinate of the electron relative to the center of mass. Using the Born-Oppenheimer approximation, that is, assuming (r, R) = R (r)(R) (1.67)
nd, by direct substitution of (1.67) into (1.66) what terms are neglected. Assuming that the energy has a minimum close to R0 show that the nuclei oscillate with a frequency m/M . From this result show that we can use the Born-Oppenheimer approximation when (m/M )1/4 1. 17. Show that the dominant interaction between two H atoms has a dipole form at large values of the separation between them (call it R).
18. Now consider a molecule made out of three atoms as shown of Fig.1.9.
1.4. PROBLEMS
25
t R R t R
Figure 1.9: Three atom molecule as a triangle of side R and hopping energy t. In this case the states of the electrons localized in each one of the atoms can be written as: 1 1 = 0 0 0 2 = 1 0 0 3 = 0 . 1 (i) If tunneling between the atoms is allowed show that the Hamiltonian is written as 0 1 1 H = t 1 0 1 . 1 1 0 (ii) Diagonalize the Hamiltonian and show that the eigenvalue problem gives the following energies 2t and t with eigenvectors, 1 1 G = 1 3 1 1 1 E1 = 0 2 1 1 1 E2 = 1 (1.68) 2 0 respectively. Observe that E1 and E2 are not orthogonal to each other and one has to orthogonalize them. Use the Gram-Schmidt method and nd an orthogonal basis.
Note: Observe that the degeneracy of the problem was lifted by the tunneling. However, two of the
26
Chapter 2
Crystals
We have seen that by sharing or exchanging electrons stable molecules of atoms can be formed. Depending on external conditions such as temperature or pressure, as atoms or molecules get closer to each other, a solid can be formed. Solids are highly symmetric structures that can be found in many different shapes. The shapes depend on the type of orbitals that participate in the binding between atoms. As an illustration, let us consider the example of the famous high temperature superconductors (HTC) that layered materials formed by planes of CuO2 . The atoms in these planes are arranged in squares as in Fig.2.1. An isolated O atom has an electronic conguration 1s2 2s2 2p4 while Cu has 1s2 2s2 2p6 3s2 3p6 3d10 4s1 . Thus, by stealing two electrons the O atom can close its p-shell and acquire the same conguration of Ne. We say that O has valence 2. Therefore, in the CuO2 planes we have Cu+2 and O2 conguration. The O atom has a close p shell and the Cu is in a 3d9 conguration. It means that there is a place for a single electron in the d shell of Cu (there is one unpaired electron). It is very reasonable to imagine that the bond between O and Cu is done by a hybridization (or mixing) of the p orbital of the O with the d orbital of Cu. Since these orbitals have an anisotropic structure and are oriented 90 degrees in respect to each other we expect a square lattice such the one in Fig.2.1(a). The orbitals overlap like in Fig.2.1(b). It is interesting to note that this simple orbital structure is probably responsible for the remarkable properties of these materials. A striking property of a crystalline solid is its symmetry. Imagine yourself walking over a lattice of atoms which show a periodic structure such as the one in Fig.2.1(a). You immediately note that as you move from atom to atom you see exactly the same structure. Moreover, as you look at the system from some specic angles it looks the same. Thus, one expects based on this observation that in perfect crystals the physical properties are the same at each point of the lattice, that is, the physical properties of the system are invariant under symmetry operations. These symmetry operations can be mathematically dened and help us to predict many different properties of crystals. In terms of quantum mechanics symmetries can be expressed by the fact that there are quantum mechanical operators O that generate them. For instance, the operator that generates translations by an amount R is
h OT = eiPR/
(2.1)
where P is the momentum operator. It is very simple to show that this is true. Suppose we apply this operator to a wavefunction (r) and suppose the R = r is an innitesimal quantity. Then, OT (r) (r) + r (r) (r + r) (2.2)
where we have expanded the exponential and used that P = i . h For systems with translation symmetry described by a Hamiltonian H, the operator OT must be a con27
28
CHAPTER 2. CRYSTALS
stant, that is, it must commute with the Hamiltonian: [H, OT ] = 0. In this way, we know from the fundamentals of quantum mechanics that the wavefunctions of the Hamiltonian can be classied according to the eigenstates of the operator OT . This operator can be diagonalized straightforwardly since its eigenstates are the momentum eigenstates,
h OT |k = eikR/ |k
(2.3)
Thus, even when we do not know how to calculate exactly the eigenstates of the Hamiltonian (and we dont in most cases), we know that the momentum is a good quantum number and therefore the wavefunctions can be labeled by the momentum. This is a major advantage since we can make predictions based on simple calculations. Translation is a simple example of a continous symmetry. There are many other symmetries that can be expressed in terms of operators as well.
(a)
Cu O porbital
dorbital
(b)
Figure 2.1: (a) Spatial structure of a CuO2 plane; (b) Atomic orbitals involved in the binding.
where a1,2,3 are three independent vectors and n1,2,3 three arbitrary integers (in two dimensions we have, of course, only two of them). A periodic array of points is called a lattice if given R on the array then R = R + T is also on the array. Thus by choosing different sets of integers we can generate the whole lattice. For each lattice point we can assign different atoms. In this case we have a basis. Let R be the coordinate of these atoms with respect to a lattice point. Here, = 1, 2, ..., Nb , where Nb is the number of
29
atoms in a basis. A crystal structure is combination of a lattice plus a basis, that is, any distance between two atoms on a lattice can be written as T + R . Observe that there is no unique way to dene a lattice but it is common to dene the primitive quantities as the most economic way to describe the crystal. We call the primitive translation vectors as the smallest a1,2,3 that still allow the denition of a lattice. An example of a two-dimensional lattice is shown on Fig.2.2.
a2 a1
Figure 2.2: Example of a two-dimensional lattice with a basis. We also dene what is called as the unit cell as a certain volume that ll out the space when translated by all possible T. It is clear that this denition is not unique. We can dene a primitive unit cell that is the one with the smallest volume or the Wigner-Seitz unit cell which is obtained by linking the nearest neighboring atoms together and then cutting these lines in the middle by planes (see Fig.2.3). Observe that if ai are the primitive vectors then the volume of the primitive unit cell is simply V0 = |a1 (a2 a3 )| . (2.5)
(a)
(b)
Figure 2.3: Conventional unit cell; (b) Wigner-Seitz cell. Together with the translation symmetry the point symmetries dene what is known as the Bravais lattices. To each symmetry we have an operator which changes the coordinates of the system around a point on the lattice with an axis through it. The principal axis is the axis with the highest symmetry (that is, the one with the largest number of symmetry operations). The symmetry operations are: (i) Identity (R R); (ii) Inversion (R R); (iii) Rotations, Cn , by an angle of 2/n where n is an integer; (iv) Reection by a plane; (v) Improper rotations which are combinations of a rotation and a reection through a plane perpendicular to a principal axis. These operations can be easily identied by inspection. In two dimensions there are only ve types of Bravais lattices that can be obtained from the operations above. They are shown in Fig.2.4. It is easy to show that in two dimensions it is not possible to have a Bravais lattice with 5 fold symmetry (that is, it is not possible to ll out the plane with pentagons) and there is no Bravais lattice with rotations higher than six-fold symmetry. Thus, we are only left with the Bravais lattices of Fig.2.4: (1) Oblique (which is symmetric only under C2 ); (2) Rectangular (which is symmetric
30
CHAPTER 2. CRYSTALS
under C2 and has two reection planes); (3) Rectangular face centered; (4) Square (which is symmetric under C4 and three reection planes); (5) Hexagonal (which is symmetric under C6 , C3 , and six reection planes).
Oblique
Rectangular
Square
Hexagonal
Figure 2.4: Possible Bravais lattices in two-dimensions. Observe that the hexagonal lattice is the most symmetric of all Bravais lattices in two dimensions (that is, is invariant under the largest number of symmetry operations). Moreover, the hexagonal lattice is special because it is close packed, that is, it is a lattice that has the densest packing of hard spheres. Indeed, in the hexagonal lattice we can densely ll the lattice by placing spheres of radius a/2 (where a is the lattice spacing) in the center of a triangular lattice as in Fig.2.5.
Figure 2.5: Close packing structure for the hexagonal lattice. In three dimensions there are fourteen Bravais lattices in seven different types of structures. One of the most important is the cubic structure that has three Bravais lattices: Simple Cubic (SC); Body Centered Cubic (BCC); Face Centered Cubic (FCC). These Bravais lattices can be again classied by the symmetry operations described before and are shown in Fig.2.6.
31
TriclinicP
MonoclinicP MonoclinicB
OrthorhombicP
OrthorhombicF TetragonalP
HexagonalP
TrigonalR
CubicP
CubicI
CubicF
(r) =
i=1
(r Ri )
(2.6)
(observe that if we integrate the above expression in all space we obtain N which is the total number of atoms). Observe that from our previous discussion we can rewrite for a crystal Ri = T + R . Therefore, for a crystal, one can rewrite the density as (r) =
T
(2.7)
(r T R ) .
(2.8)
Observe therefore that if we have Ns unit cells with Nb atoms in a basis we must have N = Ns Nb . The translation symmetry of the crystal requires that (r) = (r + T) (2.9)
for all T as one can easily see if we use (2.8) and notice that the sum of two lattice vectors is another lattice vector. Observe that this property has strong consequences if we write the Fourier series of the density: (r) =
q
(q)eiqr
(2.10)
CHAPTER 2. CRYSTALS
(2.11)
where n is an integer. It is clear from the denition of T in (2.4) that we can always dene G by G = m1 b1 + m2 b2 + m3 b3 where m1,2,3 are three arbitrary integer and b1,2,3 three independent vectors such that bi aj = 2ij ii = 1 and ij = 0 if i = j is a Kronecker delta. It is simple to show that b1 = 2 a2 a3 a1 (a2 a3 ) (2.14) (2.13) (2.12)
and the other vectors are just cyclic permutations of the above relation. By the same token one has a1 = 2 b2 b3 . b1 (b2 b3 ) (2.15)
Observe that these new vectors span a new lattice. This lattice is called reciprocal lattice. Everything we said before in regards to Bravais lattices is also valid for reciprocal lattices. In particular, the Wigner-Seitz cell of the reciprocal lattice is called Brillouin zone. Moreover, observe that (2.11) denes planes in the real space such that each plane is perpendicular to G (see Fig.2.7). In order to see that this is true remember that the density is written as: (r) =
G
G eiGr ,
(2.16)
thus, for each G the density is given by a plane wave of the form cos(G r) in the direction of G. Each maximum of the wave corresponds to a plane of atoms (since the density is maximum at these planes) where the empty space between the planes corresponds to a minimum in the wave. Since the wavelength of this wave is 2/|G| and the distance between planes perpendicular to G is d we must have d= 2 . |G| (2.17)
Another way to see this relation if through (2.11). Consider two points in neighboring planes, say Rn and Rn1 , such that |Rn Rn1 | = d is the distance between planes. Then from (2.11) one has |G| = 2 . d (2.18)
Thus G labels an innite number of parallel planes in real space. The classication of these planes in particularly useful in crystallography. Since we saw that G = hb1 + kb2 + lb3 we can label planes by the set of number [hkl] which are called Miller indices. For instance, for a set of planes in the x direction we have (100) and for a set of planes in the x + y direction we would have (110), etc. Observe that the planes (200) and (100) = ( 100), for instance, are parallel to the planes (100). In this way it is very simple to think at the planes in real space as labeled by reciprocal lattice vectors. We have seen that the density can be written in terms of a Fourier series of reciprocal lattice vectors
33
Figure 2.7: Series of planes and its associated reciprocal lattice vector. G, as in (2.16), which guarantees the periodic properties required by symmetry. The Fourier components G can be calculated by the inverse transform. This is done by multiplying (2.16) by eiKr where K is a reciprocal lattice vector and integrating in the volume of the unit cell, V0 . Observe therefore that we are left with the integral dd rei(GK)r = V0 G,K .
V0
(2.19)
In order to prove that this is indeed true we rst remember that the sum of two reciprocal lattice vectors is another reciprocal lattice vector that can be written in the form (2.12). The result is trivial if G = K. Moreover, observe that the integrand has the periodicity of the lattice and its integral over a unit cell cannot depend on the choice of cell. In particular, it cannot change if we translate the unit cell by an arbitrary vector R. Therefore, from this result we have G = 1 V0 dd reiGr (r) .
V0
(2.20)
(x) =
j=0
(x ja)
(2.21)
where a is the lattice spacing. Eq. (2.21) has a simple Fourier series (x) =
k
eikx nk
(2.22)
and because the system has discrete translational symmetry by a, that is, n(x + a) = n(x) (in (2.21) the shift by a is just equivalent to a renumbering of the lattice sites!), we must have eika = 1 2j kj = a
(2.23)
Thus, the values allowed by symmetry form a lattice with lattice spacing 2/a which is the reciprocal lattice. The reciprocal lattice vectors are simply G = 2n . Also it is very simple to show that (2.19) and (2.20) are a
CHAPTER 2. CRYSTALS
e2ij a j
(2.24)
where the sum runs over all integers (positive and negative). Multiply both sides of the above equation by x e2im a and integrate from 0 to a
a 0
dxe2im a (x) =
j
j
0
dxe2i(jm) a .
(2.25)
dxe2i(jm) a = a
e2i(jm) 1 =0 2i(j m)
(2.26)
dxe2i(jm) a = am,j
(2.27)
which is the one-dimensional version of (2.19). Going back to (2.25) one nds j = 1 a
a 0
dxe2ij a (x)
(2.28)
x N1 N 1 2 3
35
k = 2/. The atoms in the crystal absorb the light and re-emit it in spherical waves as in Fig.2.9. These waves interfere with each other and an observer located at R observes a scattered wave with wave-vector k . Here we consider only the elastic scattering by the crystal so that no energy is lost in the scattering of the light with the crystal, that is, k = k = 2/ . (2.29)
Consider an atom located at a position r in the crystal. The amplitude of the electric eld of incident light at that position is E(r) = E0 eikr (2.30)
where E0 is a constant vector. The atom at r re-emits the light so that the contribution of this atom to the observed light at R is E(R) E(r)eik r where r = R r (see Fig.2.9). Thus, E(R) E0 eir(kk ) eik R . The contribution from the whole crystal is obtained by integrating (2.32) over the entire volume: E(R) E0 eik R
(2.31)
(2.32)
dV (r) eirk
crystal
(2.33)
where k = k k and (r) is the density of atoms. Observe that the rst factor in (2.33) is just a phase factor which is not important since it does not include the superposition of all the elds created by all atoms. As we have seen, the crystal has a periodic structure that needs to be taken into account. Since the crystal is composed by N unit cells we do not have to integrate over all crystal, we just have to integrate over one cell and sum over all the other cells. This means that we can rewrite r = T + r where r describes the position of each atom in the unit cell. Thus, we can rewrite (whole) as E(R) E0 E0 dV (T + r )ei(T+r )k
T Cell
eiTk
T Cell
dV (r )eir k
(2.34)
where we have used the translational symmetry of the problem, that is, (T + r ) = (r ). Let us now focus on the rst term of (2.34). For an arbitrary k the sums of the exponential is only different from zero if T k = 2n (2.35)
where n is an integer. This condition means that in order to have a constructive interference we have to require that k = G , (2.36)
is a reciprocal lattice vector. Thus, (2.36) tells us is that there is only scattering if k = k G, that
36
CHAPTER 2. CRYSTALS
k k
Origin
Figure 2.9: Geometry for scattering of light by a crystal.
this, the incoming and the out-going plane waves can only differ by a reciprocal lattice vector. This has profound consequences. In the elastic scattering the energy of the incident and scattered beam is the same and therefore |k | = |k|. Thus, as a consequence we have, 2|k| cos = |G| (k )2 = (k)2 + (G)2 2k G (2.37)
where is the angle between k and G. But we know from (2.18) that associated with each vector G we have a set of planes such that |G| = 2/d and given a plane wave we also have |k| = 2/ where is the wavelength of the beam. If instead of we use the angle between the incident beam and the plane (see Fig.2.10) the above equation can be written as 2 d sin = which is the Bragg law for elastic scattering. Another way to compute the same result is based on the use of basic quantum mechanics by calculatingthe differential cross-section for scattering of a plane wave with wave-vector k into another wave-vector k . This is given in the Born approximation by 2 d2 | k|V |k |2 d h (2.39) (2.38)
where V is the scattering potential. In a condensed systems (gas, liquid or solid) this potential is the sum of the atomic potential of each atom. Thus we can write
N
V (r) =
i=1
Vi (r Ri )
(2.40)
where Ri is the position of each atom. Using the fact that this potential is periodic we can write V (r) =
G
VG eiGr
(2.41)
37
dd rei(qG)r (2.42)
(2)d
G
where q = k k . What this last equation tells us is that there is only scattering if k = k G, that this, the incoming and the out-going plane waves can only differ by a reciprocal lattice vector in complete agreement with our previous discussion.
k G n+1 n d n1
Figure 2.10: Bragg reection through a set of planes. We can now rewrite the differential cross-section as d2 1 | k|V |k |2 = d (2)d |VG |2 q,G , (2.43)
where we used the results of Appendix (2.5.1) for the substitution of the Dirac delta function for the Kronecker delta. This equation is quite interesting because it tells us that the intensity of the reection at q = G is proportional to |VG |2 . Thus, the Bragg condition is not sufcient in order to see elastic scattering at some vector G but also it is required that the potential has a nite Fourier component at this wavevector. We can get even more insight into (2.43) if we rewrite the Fourier components of the potential in terms of the potential created by isolated atoms. In this case, within the unit cell, we write V (r) =
U (r R )
(2.44)
where U is the strength of the potential for a particular atom . Using (2.20) we nd VG = 1 V0 U eiGR
(2.45)
which depends only on the atoms on a unit cell. In this case (2.43) becomes 1 d2 2 d V0
U U eiG(R R ) G , U U F (R R )
1 = V0
(2.46)
38 where F (R ) = 1 V0 eiGR .
G
CHAPTER 2. CRYSTALS
(2.47)
This result implies that the scattering depends on the local structure of the atoms on the unit cell.
These equations simply considerably if we assume that the atoms are the same. In this case one can write V (r) = U0
(r R )
(2.48)
(2.49)
(2.50)
is the Fourier transform of the density. It is straightforward to conclude from the above equations that d2 |U0 |2 S(q) d where S(q) = (2)d
G
(2.51)
|G |2 q,G
(2.52)
is called the static structure factor. Observe that a scattering at some vector q = G is only possible if G = 0. This factor is the same that appears in equation (2.34). Thus, even if the scattering is allowed the positions of the atoms on the unit cell determines if there is any intensity for that particular scattering.
The scattering in a disordered system such as a glass (or liquid very viscous uid) can also be immediately obtained from these equations. Since the system is disordered the unit cell is the volume of the system itself. Moreover, the sum in (2.50) is a sum of random phases that leads to destructive interference if the phases vary wildly. Thus, from (2.50), G = 1 V eiGR = nG,0
(2.53)
where n = N/V is the average density of the system. By direct substitution of this expression in (2.52) we nd S(q) = (2)d n2 q,0 shows that in this case the system only has forward scattering. (2.54)
39
For 1 one needs E 104 eV which is the X-ray part of the spectrum. Another possibility is scattering by electron waves. In this case the scattering process is due to the electron-electron interaction in the system. The energy of an electron is given by E= h2 h 2 k2 = . 2me 2me 2 (2.56)
For 1 one needs E 100 eV. However, the most precise measurement of crystal lattices is done by neutron scattering. Neutrons have no electric charge and therefore are insensitive to charge degrees of freedom in the solid. They interact with the magnetic moments in the solid (nuclear and/or electronic). The energy has the same form as above for electrons but since its mass is a thousand times larger the relevant energies are a thousand times smaller, that is, E 0.1 eV. Although we can have different probes it is easy to see that each one of them measures the system at different scales of energy. Thus, the choice of probe depends strongly on what kind of energy scale one wants to probe. In condensed matter physics one is usually interested in energies of the order of a few meV which is the energy scale of the neutrons. Neutron scattering is also particularly important because it probes directly the magnetic excitations. Electron scattering is a complex probe because the scattered electron tends to interact strongly with the other electrons in the system (this is called multiple scattering) and our simple Born approximation formula is not valid. We really need a probe that interacts weakly with the system of interest. X-rays have an energy that is usually orders of magnitude larger than the energy scales of interest. It turns out, however, that they are excellent in order to measure static properties such as, S(q), via elastic scattering.
40
CHAPTER 2. CRYSTALS
In order to understand the entropic nature of such a defect consider a perfect lattice made out of N atoms and M vacancies which are randomly distributed (we are considering that there is no clustering of these vacancies and that the energy of the vacancies do not depend whether there are other vacancies in its immediate neighborhood). In this case it is clear that the system has many equivalent congurations. The number of these congurations is simply (N + M )!/(N !M !). Thus, the total entropy at zero temperature is: S = kB ln (N + M )! N !M ! . (2.57)
Here we are going to consider the case where N, M 1 but M/N is nite. Thus we can use the Stirling approximation ln(N !) N ln(N ) N . Now consider the free energy of the system which is given by F (M ) = U (M ) T S(M ) (2.59) (2.58)
where U is the internal energy of the system with M vacancies. In the case under consideration the energy of create M vacancies is just M times the energy to create a single vacancy since we are disregarding vacancy-vacancy interactions. In this case we have U (M ) = 0 M (2.60)
where 0 is the energy required to create a single vacancy. For a xed value of N the equilibrium is attained when the free energy is a minimum with respect to variations of M . Thus, we have to minimize F (M ) 0 M kB T ln (N + M )N +M NNMM (2.61)
(2.62)
2.4. PROBLEMS
and the derivative becomes F = 0 kB T ln M and thus 0 kB T ln or M (T ) = e Ns
0 kB T
41
Ns M M
(2.63)
Ns M M
(2.64)
(2.65)
+1
which shows that there is always a nite number of vacancies in the crystal. Observe that when kB T 0 we have M (T ) Ns /2. However, in normal materials 0 is of order of 1 eV (10, 000K) and therefore one usually has the opposite limit, that is, kB T 0 , where M (T )/Ns e
k
0 BT
(2.66)
Figure 2.12: Interstitials in a crystal. Another type of defect is the interstitial which is similar to a vacancy but the atom, instead of leaving the bulk of the crystal, moves to an intermediary position between other atoms. This is shown on Fig.2.12 and it is called a Frenkel defect. This type of defects are more common in ionic crystal where a positively charged ion can move in between negatively charged ions in a crystalline matrix. In this case charge neutrality requires that an equal number of negative and positive defects are generated. It is also possible to replace a negative ion by an electron localized at the defect. In this case the electron is localized in a quantum well. Int his case the absorption of light in the crystal changes due to the existence of vacancies since the electrons in a quantum well absorb light at different frequencies than in a perfect crystal. It implies that the crystal changes its color. For this reason this kind of defects are called color centers. An example is depicted on Fig.2.13.
2.4 Problems
1. Prove that for n = 5 it is not possible to dene a smallest vector a1 that generates the lattice. The proof of the non-existence of Bravais lattices for n 7 is analogous.
42
+ + + + + + + + + + + + + + + + + + + + + + + + + + +
CHAPTER 2. CRYSTALS
+ + + + +
Figure 2.13: Two types of color centers in an ionic crystal. 2. Prove that the Wigner-Seitz unit cell of a planar hexagonal lattice of lattice spacing 2R has an area of 2 3R2 . Show that the fractional area occupied by the spheres relative to this unit cell area is 0.907. 3. Prove that the vectors bi can be written as in (2.14). 4. Prove that (2.15) is indeed correct. 5. Prove that the volume of the Brillouin zone, VB = (2)3 /V0 where V0 is the unit cell volume. 6. Show that by translating the unit cell to another cell V0 in (2.19) the integral is zero if G = K. 7. We have seen that a reciprocal lattice vector labels a series of planes (hkl). Show that a equivalent way to label the planes in real space consists of two steps: 1) nd the intercept of the plane with the axes in terms of the vectors a1 , a2 , a3 ; 2)take the reciprocal of these numbers and then reduce to three integers having the same ratio, usually the smallest three integers. The result, that is, (hkl) is the index of the plane. 8. The primitive vectors for a bcc lattice are a1 = a2 = a3 = a (x + y + z) 2 a (x y + z) 2 a (x + y z) . 2
(2.67)
Find the reciprocal lattice vectors. What lattice does it form? Make a drawing of the two lattices. 9. The primitive vectors for a fcc lattice are a1 = a2 = a3 = a (y + z) 2 a (x + z) 2 a (x + y) . 2
(2.68)
Find the reciprocal lattice vectors. What lattice does it form? Make a drawing of the two lattices.
2.4. PROBLEMS
43
10. In this problem we are going to study the effect of the second factor in (2.34), the static form factor. We write it as: SG =
Cell
dV n(r)eirG
(2.69)
since k = G is the Bragg condition. When there are Nb atoms in the unit cell we can write
Nb
n(r) =
j=1
nj (r rj )
(2.70)
SG =
j=1
fj eirj G
(2.71)
where fj =
Cell
dV nj (r)eirG
(2.72)
is called atomic form factor which depends only on the kind of atom is participating on the lattice formation. Consider now that rj can be written in terms of the primitive vectors as rj = xj a1 + yj a2 + zj a3 where xj , yj , zj are any real number. Using that G =
Nb 3 i=1 mi bi
SG =
j=1
11. Calculate SG in previous problem for a fcc lattice with one atom per unit cell assuming that the fcc lattice can be thought of a simple cubic lattice with a basis given by the vectors (0, 0, 0), (0, 1/2, 1/2), (1/2, 0, 1/2), (1/2, 1/2, 0). (ii) Assume that fj = f and show that if m1 , m2 , m3 are all even or all odd we have SG = 4f . (iii) Show that if one of the mj is even and the other two are odd then SG = 0. (iv) Show that if one of the mj is odd and the other two are even then SG = 0. (v) Compare the scattered reections that you would get from a fcc lattice without a basis and the ones you got from assuming that the fcc lattice can be seen as a simple cubic lattice with a basis. Do you get identical results? 12. Consider a two dimensional crystal in a square lattice with two different types of atoms with different cross-sections (a two dimensional version of NaCl). What would be a result of a neutron scattering experiment in such a system? What are the allowed values of momentum for scattering? 13. Consider an ionic crystal with positive and negative charges. Assuming that the energy required to create an interstitial with positive (negative) charge requires an energy p (n ) and that the system has charge neutrality show that the number of positive and negative interstitials is np = nn = Np Nn e
p +n 2
CHAPTER 2. CRYSTALS
where i = x, y, z. Thus a Dirac delta can be written as (k) = (kx )(ky )(kz ) = (2nx /L)(2ny /L)(2nz /L) V = k,0 (2)d
(2.76)
where k,0 is a Kronecker delta and V = L3 is the volume of the quantization box. In this way one can replace sums by integrals, V (2)d dd q . (2.77)
Another way to rewrite this expression is due to the result of Problem 5: if N is the number of primitive cells and VB is their volume we have immediately, N VB dd q . (2.78)
Chapter 3
Elasticity Theory
Point defects are not the only type of defects that occur in crystals. There are extended defects that are responsible for very important mechanical properties of crystalline systems. Line or plane defects in crystals are possible under application of pressure or stress. When a deformation due to stress is reversible we call it a elastic deformation, otherwise, when it is not reversible, we call it a plastic deformation. In order to understand the difference between plastic and elastic deformations consider the simple toy model proposed by Frenkel for the shear of a perfect crystal as shown in Fig.3.1. Suppose a stress, , is applied to a plane of atoms which is displaced by another plane of atoms by an amount x. Under Hooks law the stress, , is linearly related to the displacement by the shear modulus G (x) G x d (3.1)
where d is the distance between planes. Of course this equation is valid for small displacements. Suppose we apply a strong shear stress so that all the atoms move one lattice spacing a. In this case we are back to the original situation with the atoms in their equilibrium position in zero stress. This implies that the stress must be a periodic function of the lattice spacing. For simplicity let us consider a harmonic variation with position: (x) = G a sin 2d 2x a (3.2)
that turns into (3.1) in the limit of x a/(2). Observe that (3.2) shows that the maximum stress occurs for x = a/4 where c = (a/4) = G a . 2d (3.3)
If the stress applied is larger than c the upper plane in Fig.3.1 has to move freely and never return to the original unstressed situation. This would imply plastic ow of the crystal. Since a d 1 the critical stress is approximately G/6. It turns out that this prediction is completely at odds with the experiments. For instance for a single crystal of Al we nd that experimentally c G/60, 000! Thus, the conclusion is that plastic deformations occur at a much smaller stress than the theory predict and they must be generated not by elastic deformations of the crystal but by defects. These defects are called dislocations. Plastic deformations in crystals occur when planes of atoms slide over each other due the presence of dislocations. The simplest types of dislocations are edge dislocations and screw dislocations. In a edge dislocation there is a mismatch of planes as if an extra planes of atoms have been inserted in the crystal while in a screw dislocation there is a mismatch between planes of atoms as shown in Fig.3.2. 45
46
x d a
(a)
Lower Plane
(b)
Upper Plane
47
where i, j = 1, 2, 3, ..., d corresponding to x, y, z.... Observe that i uj can be seen as a matrix, or more formally, a tensor with d d components. From this argument we conclude that the free energy has to be written as F = F [i uj ]. Another symmetry of the problem is inversion symmetry, that is the free energy has to be invariant under u u. It implies that linear terms are not allowed and therefore F = F [(i uj )2 ]. Now let us consider another important symmetry which is the symmetry of rigid rotation of the lattice by an angle . Consider, for instance, a plane of atoms labeled by a reciprocal lattice vector G which is rotated by this angle. Along this plane all the atoms are displaced by u(r) = |r| as shown in Fig.3.3. Another way to write this displacement is to consider the vector oriented anti-clockwise in the rotation direction. It is clear that u(r) = r . This relation can be easily inverted with the help of differential calculus to 1 = u(r) 4 (3.6) (3.5)
which expresses the angle of rotation in terms of the displacement. Since the whole crystal is rigidly rotated the free energy cannot depend of , that is, the system is invariant under rigid rotations. Observe that in (3.6) the displacements appear in combinations of the form i uj j ui which is the anti-symmetric part of the tensor i uj . Thus, due to the rotation symmetry the free energy can only depend on the symmetric part of the tensor i uj , that is, uij = 1 (i uj + j ui ) 2 (3.7)
which is called the strain tensor. In summary, due to rigid translations, inversions and rotations the free energy due to elastic deformations of a crystal can only be a function of uij ukl .
u(r) r
Figure 3.3: Rigid rotation of an atomic plane. It is very simple to understand the physical meaning of each component of the strain tensor. Consider for instance a compression or dilation of the system along the direction x. If Lx is the size of the crystal in this direction then x = Lx Lx (3.8)
gives the relative compression in that direction (obviously for compression x < 0 and for dilation x > 0).
48
In this case the volume of the whole crystal is changed by V . By the same token the volume of the unit cell, V0 = |a1 (a2 a3 )| is changed by V0 because ai changes under compression. It turns out, however, that since V = N V0 where N is the total number of atoms we must have V V0 = . V V0 (3.9)
This relation is valid because we are considering the crystal in the absence of vacancies and interstitials. In the presence of defects (3.9) has to be modied. Moreover, since the reciprocal lattice vectors are tied to the direct lattice vectors by ai bj = 2ij it is clear that a compression in real space will lead to a dilation in reciprocal space and vice-versa. Consider the set of N planes separated by a distance d in the x direction before the application of pressure. In this case we have Lx = N d. After pressure is applied the distance between planes becomes d so that L = N d . Therefore x d d L Lx x = x . = Lx d (3.10)
We label the displacement of the nth plane of the crystal by un . Let us assume that the rst plane is xed before and after pressure is applied so that u0 = 0. It is simple to see in Fig. 3.4 that second plane is displaced by u1 = d d = x d, the second plane is displaced by u2 = 2d 2d = 2x d and so on. In general un = nx d . If we label the position of each plane by x = nd we see that this last expression can be rewritten as u(x) = x x x u(x) = x . (3.12) (3.11)
u1
u2
u3
u4
In a more general way we can consider a set of planes labeled by the reciprocal lattice vector G. We know that |G| = 2/d where d is the distance between planes. After compression or dilation we must have
49
(3.13)
Moreover, in this case, the displacement of the atoms due to the compression or dilation can be written as G u(r) = (G G ) r = x G x , and therefore u = x x leading to: uxx = x , (3.16) (3.15)
(3.14)
and all other derivatives vanish. Thus, it is clear that uii measures the lattice compression or dilation in the direction of the vector xi . The total volume of the system is V = Lx Ly Lz and therefore we must have V = Lx Ly Lz + Lx Ly Lz + Lx Ly Lz . Thus, it is simple to show that V = V uii
i
(3.17)
that is, the relative change in the volume of the solid is given by the trace of the strain tensor. Equation (3.17) has to be interpreted carefully. Observe that uii (r) is a local function of the position and of course V is a global change in the system. Thus (3.17) is really only valid in a perfect crystal without defects like vacancies where changes in volume can be non-uniform. In general the diagonal elements of the strain tensor are related to the local changes in the unit cell volume such that, V0 (r) = V0 (r) uii (r)
i
(3.18)
which is a local relation. Of course, for ordered systems equation (3.9) is valid and (3.17) is identical to (3.18). We can always decompose the distortion of a solid in terms of volume changes and shear (that never affects the unit cell volume). We have seen that volume changes have to do with the diagonal elements of the strain tensor. The off-diagonal components of the strain tensor represent the shear distortion of the system as you can easily show. Since shear does not change the volume of the crystal it is represented by a traceless tensor sij = uij ij Notice that indeed
i sii k
ukk . d
(3.19)
= 0 since
i ii
= d.
From now on we are going to consider only small distortions of the crystal and from the symmetry
(3.20)
where Cijkl are the so-called elastic constants of the crystal which in principle can have d4 components. Now observe that uij = uji and therefore we must have the following symmetries: Cijkl = Cjikl = Cijlk = Cjilk = Cklij . (3.21)
In addition the free energy has extra symmetries that depend on the point symmetries of the crystal itself. A highly symmetric crystal has less independent elastic constants than less symmetric crystals. For instance, a three dimensional cubic crystal has 3 independent elastic constants while a triclinic crystal has 21 independent elastic constants. Moreover, it is very simple to estimate the order of magnitude of the elastic constants if we remember that uij is a dimensionless quantity and therefore Cijkl has dimensions of energy divided by lengthd , that is, energy density. The energy here is just the biding energy per atom of the solid which is of order of a few electron volts while the length is of order of the lattice spacing, that is, a few angstroms. Things simplify considerably in isotropic solids where compression and shear stress are independent of the direction they are applied. It must be clear that in this case there are only two elastic constants: one associated with compressions and dilations (the so-called bulk modulus, B) and another associated with shear distortions (the so-called shear modulus, G). Since we have seen that compressions have to do with uii and shear are related to sij the free energy has to be written as 2 1 dr B s2 . (3.22) uii + 2G F = ij 2
i ij
In order to see that this expression has indeed the form of (3.20) we use (3.19) and rewrite (3.22) as 2 1 2G F = uii + 2G u2 dr B (3.23) ij 2 d
i i,j
Cijkl =
2G d
ij kl + G (il jk + ik jl )
(3.24)
that has the symmetry properties as required. Of course the expression for the free energy as given by (3.20) is useful if we can relate the internal distortions (the strain) with the external agents such as pressure or strain. In order to do that one has to calculate the work done on the system by an external force. Consider an external force F applied to an element of volume V . If the interaction between the atoms is short ranged this force is transmitted by the neighboring volumes through the surface S that surrounds the volume V . The total force in the volume in the ith direction is Fi =
V
drfi
(3.25)
where f is the force per unit of volume of the undistorted solid. Since the force applied is a local function
51
j ij
(3.26)
dSj ij
(3.27)
where we used Gauss theorem. Observe that the stress tensor ij gives the force per unit of area in the direction i exerted by the surrounding medium on a volume element across its surface oriented on the direction j. Consider a solid surrounded by an isotropic uid at pressure P . This solid will experience an stress given by the hydrostatic pressure which is P . In this case the stress tensor is simply ij = P ij . Suppose uniaxial pressure T is applied along the x axis of a crystal. In this case it is obvious that xx = T (3.29) (3.28)
and all other components are zero. In the more generic case consider the work done by a force density f which displaces the volume elements of a crystal by u(r). This work is simply W = P V = = = dr
i,j
dr
(3.30)
where we have integrated by parts and neglected the surface terms. The change in the free energy due to the changes in the strain eld is the negative of the work done by the internal forces, thus, F = W = Observe that this leads to the important relation: ij = which relates the strain tensor with the stress tensor. F uij (3.32) dr
i,j
ij uij
(3.31)
ull ij + 2Gsij
(3.33)
52
that we can invert to write the strain in terms of the stress. Firstly we observe that for i = j we have ij = 2Gsij = 2Guij where we have used (3.19). Secondly, for i = j we have ii = 2Guii + B and therefore
i ii
(3.34)
2G d
ull
l
(3.35)
= dB
uij = ij
(3.36)
This last equation is very useful. Consider a solid subject to a hydrostatic pressure P . Then, from (3.36) one has uii =
i
1 dB
ii =
P . B
(3.37)
(3.38)
which is the usual thermodynamic denition of the bulk modulus as expected. Suppose that uniaxial stress T is applied in the z direction. Again we have uzz = T d 1 d1 + dB 2G T 1 1 = d 2G dB
uxx = uyy
(3.39)
Observe that the change in the size of the system along z always follows the applied force while in the transverse directions it will depend on the ration dB/(2G). The Young modulus, Y , of the system is dened as Y Y and the Poissons ration is dened as = = uxx uzz dB 2G . d(d 1)B + 2G = = T uzz 2d2 GB 2G + d(d 1)B (3.40)
(3.41)
3.2. PROBLEMS
53
3.2 Problems
1. Write down the elastic free energy for a three dimensional crystal with cubic symmetry. Show that the elastic tensor Cijkl has only 3 independent components, namely, C1111 , C1122 and C1212 (1, 2, 3 refers to x, y, z, respectively). Calculate: 1)the bulk modulus; 2)the Poisson ratio for stresses along one of the symmetry axes in terms of the elastic constants.
54
Chapter 4
Atoms in motion
In the previous section we studied the problem of static deformation of a crystal. As we have seen previously a crystalline structure each time we take an atom from its position we have to pay an energetic price which for elastic deformations. For small deformations the energy is a simple quadratic function of the displacement. It is known from basic quantum mechanics that even at zero temperature motion does not to cease to exist entirely due to quantum uctuations. The basic example of this effect is the harmonic oscillator problem which is described by a potential V (x) = m 2 x2 /2 where m is the oscillator mass and its oscillation frequency. In quantum mechanics the ground state of this problem has nite energy /2 and therefore even h in the ground state the oscillator is not static but uctuates with amplitude A /(m) (since V (A) h h /2). In classical mechanics the lowest energy state have zero energy and therefore A = 0 (indeed, when h 0 we recover the classical case). Therefore, in order to understand the behavior of solids at very low temperatures one has to take into consideration the kinetic energy of the oscillations in the solid.
(un un+1 )2 2
(4.1)
since we are assuming that only the nearest neighbor atoms are coupled. If U was everything we would have it is clear that the conguration of lowest energy has un = 0 for all n. But in quantum mechanics we have to include the kinetic energy of the atoms which is given by K=
n
p2 n 2m
(4.2)
where pn is the momentum of the nth atom. The quantization condition for this problem is that the displacement and the momentum are canonically conjugated and therefore have well dened commutation 55
(4.3)
As we all know from basic quantum mechanics the fact that two operators do not commute imply the Heisenberg uncertainty principle, that is, pn un > , which implies that even we are certain that the h oscillator is at certain position un from equilibrium we lose completely the information about its momentum. This is exactly what causes the harmonic oscillator to have a ground state with nite amplitude. There are many ways to study the Hamiltonian H = K + U . Here we are going to study the problem via the equations of motion by using the Heisenberg representation for the problem. In this representation the operators evolve in time accordingly to: i h un t pn i h t = [un , H] = [pn , H] (4.4)
which can be easily calculated by using (4.3) and the Hamiltonian: un t pn t = pn M (4.5)
which can be rewritten by substituting the rst equation into the second one: 2 un = (un+1 2un + un1 ) 2 t M (4.6)
which is the equation for the time evolution of the operator un . This equation is simply a simple second order linear differential equation and can be solved by assuming un (t) has a simple harmonic form, that is, un (t) = un eit which leads to ( 2 + 2/M ) un /M (un+1 + un1 ) = 0 . (4.7)
Notice that the above equation relates the displacement at nth atom with the displacement at n + 1 and n 1 atoms. The solution to this problem is given again by a simple harmonic solution: un = ueikna where u is a constant. Direct substitution of (4.8) into (4.7) requires that (k) = 2 |sin(ka/2)| . M (4.9) (4.8)
Which shows that there is a one to one correspondence between the frequency of oscillation and k the wavenumber of the oscillation. Observe that the periodicity of the chain requires that uN +1 = u1 which, by (4.8) requires that eikN a = 1 2m k(m) = Na
(4.10)
57
where m is an integer. Therefore that the wave-numbers are quantized in units of 2/(N a) and these give the allowed quantum states of the problem. Notice, however, that the total number of states in the problem has to be conserved. In the absence of interactions between the atoms there are N allowed states in the problem corresponding to the rotation of the chain by N 2/N . We have to end up with the same number of states in momentum space, as well. If N is even the allowed values of m are m = 0, 1, 2, ..., (N/2 1), N/2 and for N odd, m = 0, 1, 2, ..., (N 1)/2. . Thus, we see that k dened above varies between (/(2a)(1 1/N ) k /(2a). In the limit of a macroscopic number of atoms, N , the distance between the allowed states shrinks to zero and the wave-numbers form a continuum in the interval /(2a) < k /(2a). This is exactly the Brillouin zone for a one dimensional system, as expected from the periodicity of the problem. Finally, we should point out that in the limit of very long wavelengths, that is, when k 1/a we can expand (4.9) as (k) a |k| M (4.11)
and we see that the frequency of oscillation is linear with the wavenumber. This relationship is identical to the dispersion of photons: (k) = c|k| where c is the light velocity. Like photons the oscillations we are discussing propagate through the solid which a characteristic velocity cs = a M which is called the sound velocity. Indeed, from basic quantum mechanics we know that the group velocity of a wave v(k) is given by v(k) = dEk /dk. Remember that in quantum mechanics there is no distinction between waves h (or oscillations) and particles. Indeed we could say that we have discovered a new particle which has been named acoustic phonon which propagates with the sound velocity. At longer wave-lengths k /(2a) the phonon does not propagate since (k) 2 M does not depend on k and therefore its group velocity vanishes. Observe that phonons are an effect of the interaction between atoms (when 0 we nd (k) = 0) and they do not exist outside of the many-body system. This is the major difference between condensed matter physics and high energy physics where particles exist in the vacuum.
M a
n1
u n+1
(a)
M1
M2
(b)
Figure 4.1: One-dimensional lattice: (a) one atom per basis; (b) two atoms per basis.
58
Since the atoms undergo harmonic oscillation the potential energy associated with the displacement has the generic form U= 1 2
, u, (R)K, (R R )u, (R )
(4.13)
R,R , ,
, where K, with , = x, y, z and , = 1, ..., Nb is a matrix dNb dNb which tells us how the energy of the system change when displace an atom in some unit cell at position R in the direction , relative to another atom in the unit cell located at R in the direction . Observe that this matrix can only depend on the relative position of the atoms. This is an important property as we are going to see. In the one dimensional problem discussed previously where we have just one atom in the unit cell and just one direction to move them this matrix has a very simple form which you can directly check, namely,
K(n m) = (2nm,0 nm,1 nm,1 . The kinetic energy of the atoms is simply K=
R,
(4.14)
P2 (R) 2M
(4.15)
where M is the mass of each atom in the basis and P the momentum of each atom. The momentum and the displacement are canonically conjugated, that is,
h u (R), P (R ) = i , , R,R .
(4.16)
The commutation relation leads to a quantum problem we want to solve. Notice that the Hamiltonian of the system H = K + U is rather complex and will change depending on the type of lattice we are working on. In order to simplify the problem one has to look for generic properties which are common to all Bravais lattices. There are two of these properties: (1) the harmonic interaction between atoms only depends on the relative distance between the atoms; (2) all Bravais lattices have inversion symmetry. Because K only depends on the relative distance between the atoms it is convenient to Fourier transform the displacement operators: 1 u (R) = N eikR u (k)
k
(4.17)
where k is the wave-vector of the problem. If, like in the one dimensional case we studied previously, we impose periodic boundary conditions in all directions, that is u (x + Nx ax , y, z) = u (x, y + Ny ay , z) = u (x, y, z + Nz az ) = u (x, y, z), where N is the number of atoms in each direction (observe that the total number of atoms is N = Nx Ny Nz ) and a is the lattice spacing in each direction, we will have quantized values for the components of k: k = where N /2 < n N /2 when N . 2n . N a (4.18)
59
1 2
(4.19)
(4.20)
P (k) P (k) . 2M
(4.21)
Notice that (4.19) is diagonal in momentum space. Thus part of our problem is solved and we just have to diagonalize the problem in the discrete indices = x, y, z and = 1, ..., Nb . We would like to nd a transformation of coordinates such that (4.19) is diagonal. Therefore there must be a unitary transformation U (U 1 = U T ) between u and a new set of coordinates q , that is, u (k) =
, , U, q (k)
(4.22)
such that U 1
, , , 2 K , U , = M , , , . ,
(4.23)
, , ,
This problem involves dNb dNb matrices and therefore we will nd dNb values of , . Instead of working with two indices, and it is common to use a short notation and introduce a single index s = 1, ..., dNb for the eigenvalues s . To each eigenvalue one has a corresponding eigenvector. Each eigenvector is related to a different polarization vector es (k). From (4.22) we see that each polarization vector corresponds to a row (or column) of the matrix U. Thus (4.22) can be rewritten as u (k) =
s
(4.24)
where the polarization vectors are orthogonal. The orthogonality condition is simply given by the condition that U U T = I, which can be written as
dNb
e (k)e (k) = , , s, s,
s=1
(4.25)
or conversely,
Nb =1
(4.26)
60
It is clear that after the diagonalization that the Hamiltonian of the system is written H=
s k
(4.27)
where ps is the momentum canonically conjugated to qs . Thus we have reduced the problem to decoupled harmonic oscillators as one would expect from the beginning since we used the fact that the forces among the atoms are harmonic. Therefore, the excitation spectrum of the system is given by Es (k) = hs (k) ns (k) + 1 2 (4.28)
to each mode of this spectrum we associate a elementary excitation that we call the phonon. In condensed matter physics we are interested in obtaining these elementary excitations. The problem of vibrating atoms is particularly simple since it involves only harmonic forces. Let us now go back to the problem of obtaining the frequencies through eq. (4.23). Multiplying this equation by the left by U one obtains K M 2I U = 0 which only has a non-trivial solution (U = 0) if
2 det K(k) M s (k)I = 0
(4.29)
(4.30)
which is the equation that denes the eigenmodes and it involves the diagonalization of a dNb dNb matrix and therefore, for each value of s, the eigenvectors have dimension dNb . Each one of these eigenvectors represent the polarization of the phonon waves in the solid. In order to proceed we need more information about K(k). Since we are not dealing with a specic lattice we have to use the most general properties of these systems, that is, we have to use their symmetries. The rst important symmetry which is true for any crystal is the inversion symmetry, that is, K(R) = K(R) . This property implies, from denition (4.20), that K(k) = K(R) cos (k R) (4.32) (4.31)
which is an even function of the momentum, that is, K(k) = K(k). Another important property of the system is that if we translate all the atoms by the same xed but arbitrary amount, say u (R) = R0 , then the energy of the system has to be the same. This is called Galilean invariance. Thus, from (4.13), one nds R0 K, (R R ) R0 = 0 (4.33)
R,R ,,
R,,
61
(4.35)
Let us consider, as an illustration, the case of a basis with just one type of atom (Nb = 1). It implies that we will have d modes in the system only. In this case (4.35) becomes K(k = 0) = 0. Thus, using (4.32) one can write K(k) = K(R) [cos (k R) 1]
R
(4.36)
(4.37)
where nk is the unit vector on the direction of k. The matrix C(nk ) = 1 2M (nk R)2 K(R) (4.38)
has no information on the amplitude of k but only its direction. Then, by direct substitution in (4.30), we obtain s (k) = cs (nk )k where det C(nk ) c2 (nk )I = 0 s (4.40) (4.39)
determine the phonon velocity cs (nk ) in the direction of k. Phonons which have a linear dispersion with momentum, such the ones described by (4.39) are called acoustical phonons. Our approximation in (4.37) requires that the sum in (4.38) to converge. We can estimate this sum by replacing it by an integral over R and we see that C dRRd1 R2 K(R) only converges if K behaves at least like 1/Rd+2 at large distances which is usually the case in solids. For atoms with more than one atom in the basis the matrix K does not necessarily vanish at k = 0 although the sum of the components of the matrix must vanish as explicit in (4.35). However, the form (4.32) is still valid. It implies that when k = 0 we are going to have solutions such that the frequency s (k = 0) = 0. These modes can be obtained directly from (4.30) by setting k = 0,
2 det K(k = 0) M O,s I = 0 .
(4.41)
These modes which are dispersionless at very small k are called optical phonons. We have concluded that in a lattice with a basis we must have d branches of acoustical phonons and d(Nb 1) branches of optical phonons. The typical dispersion for the phonon modes is shown on Fig.4.2. We have to plot the dispersion only in the unit cell of the reciprocal lattice, that is, on the Brillouin zone, because of the periodicity of the
62
system. Moreover, from the fact that the crystal has inversion symmetry (4.32) we know that K(k) = K(k) and therefore we only need to know half of the Brillouin zone.
s(k)
Optical
Acoustical
G/2
Figure 4.2: Typical dispersions for a three-dimensional with two atoms in the basis.
We have shown that for atoms interacting through harmonic forces the problem reduces to a set of decoupled harmonic operators as given in the Hamiltonian (4.27). We can therefore use all the technology we have for decoupled harmonic oscillators for this problem. In particular we are interested in the operators that create or destroy phonons in the system. The annihilation and creation operators for harmonic modes are dened as, as (k) = a (k) = s M s (k) 2 h M s (k) 2 h qs (k) + i ps (k) M s (k) i ps (k) M s (k) (4.42)
qs (k)
respectively. Notice that the displacements can now be written with help of (4.17), (4.24) and (4.42) as u (R) =
s k
es, (k)eikR
(4.43)
Notice that e (k) = es, (k). s, In terms of these operators the Hamiltonian (4.27) can be rewritten as H=
s,k
1 2
(4.44)
Thus the states of the system can be labeled by the occupation in each phonon state, |ns (k) and the energy
63
(4.46)
The ground state of the system is therefore the empty state, |0 which is dened by as (k)|0 = 0 (4.47)
since there are no phonons to be destroyed. The excitations of the problem are therefore obtained by applying creation operators a to |0 .
n|eH |n =
eEn
n
(4.48)
where tr is the trace of the operator and the sum is extended over all eigenstates En and = 1/(kB T ). The free energy, F , of a given statistical mechanical problem is related to the partition function via Z = eF 1 F = = ln(Z) . Observe that the mean energy density of the problem is giving by E = = 1 tr HeH V tr [eH ] 1 Z 1 n En eEn = . En V V Z ne
(4.49)
(4.50)
When a given system is made out of independent parts the free energy is an additive quantity, that is, F = i Fi , and thus from (4.49) the partition function is a product of the partition function of the parts, Z = i Zi . From the phonon problem we have shown that the system decouples into a set of harmonic oscillators which are labeled by s and k. Thus, the partition function reads,
Z=
s,k ns (k)=0
h e s (k)(ns (k)+1/2)
(4.51)
where we have to sum over all possible occupations. The sum over the occupations is just a geometric series and one gets Z=
s,k h e s (k)/2 = h e s (k) 1
s,k
1 2 sinh( s (k)/2) h
(4.52)
h s (k) ns (k) +
1 2
(4.53)
where ns (k) = 1
h e s (k)
(4.54)
is the mean number of phonons on a given eigenstate at some temperature T . This is called the Bose-Einstein h distribution function.h The term E0 = s,k s (k)/2 is the ground state of the problem (take the limit of T 0 in (4.53). This energy is not a observable quantity. We can however observe how a system exchange energy with a heat bath at some temperature T . The quantity that characterizes this exchange of energy is the specic heat at constant volume, CV . The specic heat measures how the energy changes as we vary the temperature of the system. Since temperature is related to the number of excited states, the specic heat is essentially a counting of the number of available states in the system. This quantity is given by CV = 1 1 E = V T T V h s (k) s (k) h e . (4.55)
s,k
In order to calculate the thermodynamic quantities we dene the concept of density of states. The density of states is the number of states with a given energy E. Mathematically it is given by, N (E) = For phonons the density of states is just N (E) = 1 V (E s (k)) . h (4.57) 1 V (E En ) . (4.56)
s,k
dE
0
N (E)E eE 1
(4.58)
which can be proved by direct substitution. Thus, all the thermodynamic function can be obtained directly from the knowledge of the density of states. In general, the density of states is a very complicated function of the energy due to the shape of the phonon dispersion relation. We have seen, however, that we have d acoustic modes and d(Nb 1) optical modes. Moreover, these modes are characterized by the fact that the acoustical modes have a dispersion proportional to k while the optical modes are dispersionless at small k. What is the effect of temperature on the phonons? At low temperatures only acoustical modes are excited since to excite an optical mode one has to pay an energy 0 . Therefore for kB T 0 we do not expect to excite any optical mode. The h h acoustical modes, however, can always be excited at low temperatures since the dispersion vanishes with k, that is, it is always possible to nd a mode k such that kB T ck. Indeed, these two types of excitations h are the most basic ones in condensed matter physics: gapless and gapfull. Most of physical properties of solids can be understood on the basis of this classication. Without doing any calculation we can predict the behavior of the thermodynamic functions quite well.
65
(a)
kx
n(E)
(b)
E/(k BT)
Figure 4.3: (a) Phase space for phonons in two dimensions; (b) Shape of the Bose-Einstein distribution.
For an acoustical mode at some given temperature T the number of modes available is given by the spherical volume in k-space of radius k at energy kB T (see Fig.4.3)(a)). In d dimensions this volume is proportional to T d . Naturally the number of states available at this temperature is proportional to T d . Thus the specic heat has to behave like T d . The energy of the states is of order kB T . Therefore the thermal energy of the system, E, (E = E E0 ) has to be proportional to T d+1 (in accordance with d ). This argument is valid as far as the Bose-Einstein occupation (4.55) the specic heat is proportional to T number does not give an important contribution. This requires low temperatures (see Fig.4.3(b)). At high temperatures the occupation number gives a contribution of order kB T /E and therefore the mean energy in (4.58) is just E kB T dEN (E) which is linear in the temperature. The specic heat, by its turn, has to be temperature independent. This is known as the Dulong-Petit law. The integral dE N (E) is essentially the total number of states per unit of volume in the system which is dN/V 1/ad where a is the lattice spacing. Observe that besides the thermal energy kB T the only other quantities which can appear in the expression of the energy is the phonon velocity c, the Planck constant and the lattice spacing a. The only h quantity with dimensions of energy that we can construct from these constants is c/a. Since we know that h d+1 the only form allowed for the mean energy per unit of the mean thermal energy is proportional to (kB T ) volume is E = (kB T )d+1 d ( c/a)d h h c/a kB T (4.59)
where d (x) is a dimensionless universal function such that for x 1 it is constant and when x 1 it behaves like xd1 accordingly to the Dulong-Petit law. Eq.(4.59) is called a scaling form of the mean energy because it was only based on dimensional analysis and the study of trivial limits of the problem. This scaling form of the energy is possible because acoustic modes are critical, that is, the frequency and momentum scale in a well dened way. In this case we can dened what is called a dynamical exponent, z. This can be understood in the following way: suppose the lattice constant of the problem is changed by a constant b, that is, a ba. Since the momentum is proportional to 1/a then it has to change as k k/b. But the frequency is also proportional to the momentum and therefore /b. Thus the ratio
66
k/ is unchanged by any change in the scale of the problem. In this case we say that z = 1. In a system where the frequency does not depend linearly with momentum a change in the lattice spacing can lead to a change in the frequency such that /bz and thus the invariant ratio is not k/ but kz / which denes a generic dynamical exponent. The physical reason for this is due to the relativistic invariance of the acoustic problem, that is, the system of acoustic phonons is invariant under a Lorentz transformation.
ky
kD
kx
Figure 4.4: Geometry of the Debye approximation in two dimensions: the square Brillouin zone is replaced by a circle of radius kD . Now that we know the result of the calculation let us consider how a serious calculation would lead to the form (4.59). We approximate the actual frequency by the acoustic relation (4.39). We have to remember, however, that the dispersion is a periodic function of momentum and by replacing the exact dispersion by a linear one we have disregarded wavevectors close to the zone boundary (k G/2). We have therefore to cut-off the integrals at some maximum vector, kD . In order to estimate this wavevector we replace the original unit cell by a spherical one with radius kD (see Fig.4.4). The number of states in the original cell is N , the number of atoms in the crystal. Moreover, the volume in k-space per wave vector is (2)d /V and therefore the number of states is (2)d N/V = (2)d n. In order to preserve the number of states for the spherical problem we require that (2)d n = kD = dd k (kD k) d(2)d n Sd
1/d
(4.60)
where Sd = d is the total solid angle in d dimensions (S1 = 2, S2 = 2 and S3 = 4) and (x) = 1 if x > 0 and vanishes otherwise is a step function. This is called the Debye model. Observe that kD is proportional to 1/a and therefore acts as a high energy cut-off for the model. For this model the density of states for acoustical modes is written as ND (E) =
s
=
s
(4.61)
67
(4.62)
where kB D = hD = hckD c/a is the Debye energy and is the scale of energy that should appear in h (4.59). Moreover, 1 1 = d c d d 1 Sd cd (nk ) s (4.63)
gives an estimate of the average phonon velocity. Notice that the density of states is nite only for energies smaller than the Debye energy. By substitution of (4.62) into (4.58) one nds (4.59) with d (x) = dSd (2)d
x
dz
0
zd . ez 1
(4.64)
The characteristic energy scale in solids is dened by the Debye temperature. The sound velocity is of order of 104 m/s in most solids and since the lattice spacing is of order of the typical Debye temperatures are of order of hundreds of Kelvin. The optical modes can be treated in a much simpler way since they are dispersionless and have a gap for excitation of energy 0 . It is clear from (4.58) that in this case the physical properties are going to h be completely determined by the Bose-Einstein factor which is exponentially small at low temperatures. Thus one expect all the thermal properties of optical modes to decay exponentially with temperature, that h is, e 0 /(kB T ) . Since we have Nb (d 1) optical branches in the spectrum we can write that the density of states is simply NE (E) = Nb (d 1)Ns (E 0 ) h V (4.65)
where the prefactor is chosen in such way that the total number of states is correct. Thus, using (4.58), one nds E = (d 1) 0 nh h e 0 1 (4.66)
where we have used that N = Ns Nb . At low temperatures, as expected, the mean energy vanishes exponentially and at high temperatures one recovers the Dulong-Petit result. This simple model for optical phonons is called Einstein model.
68
EF at some distant time t . Observe that the energy must be conserved in the scattering and therefore EI EF = E , (4.67)
where E is the change in energy of the probe during scattering. The scattering is inelastic because the probe does not conserve its energy during scattering. The rate of transition is simply given by Fermis golden rule T (q, E) = = 2 h 2 h (E EI + EF ) | p , F |V |p, I |2 (E EI + EF ) dd reiqr F |V (r)|I |2 (4.68)
1 | V
t |F> p
EF
Figure 4.5: Scattering process. We will assume as before that the interaction of the probe and the solid is short ranged and can be written as V (r) = V0
R
V (r) = V0
R
where we have used (4.12) and for simplicity of notation we assume one atom per unit cell. Substitution in (4.68) leads to T (q, E) = 2V02 V 2 h (E EI + EF ) | eiqR F |eiqu(R) |I |2 . (4.70)
The cross-section of the crystal is proportional to the transition rate and this allows us to dene the so called
69
(4.71)
which can be Fourier transformed in the frequency domain, S(q, ) = where S(q, t) = = = = 1 N 1 N 1 N 1 N
h ei(EF EI )t/ F R,R
dt it e S(q, t) 2
(4.72)
F R,R
F R,R
R,R
(4.73)
Where we have used that: (1) |I and |F are eigenstates of H0 ; (2) the operator identity U eA U 1 = 1 h h eU AU ; (3) changed from the Schrdinger representation to the Heisenberg representation, that is, eiH0 t/ AeiH0 t/ = A(t); and (4) that the nal states form a complete set F |F F | = 1. We will assume further that initially the solid is an equilibrium situation at temperature T so that I|...|I becomes a thermal average. Notice that the dynamical form factor only depends on the properties of the crystal since the probe is already out of the problem. Notice that we can write (4.73) in a simpler form if we use (2.50) S(q, t) = N I|(q, 0)(q, t)|I (4.74)
is the so-called density-density correlation function. This result is a natural result since the probe is coupled to the ion density in (4.69) and therefore the response of the solid occurs by creating density uctuations. In order to calculate the thermal average in (4.73) we are going to use a result from the appendix which says that if the ground state of the problem is the one for harmonic oscillators and the operators A and B are linear combination of destruction and creation operators we have eA eB = e 2
1
A2 +2AB+B 2
(4.75)
where in our case A = iq u(R ) and B = iq u(R, t). Thus we have eiqu(R ) eiqu(R,t)
= e2 = e2
1
q q C, (RR ,t)
(4.76)
is the phonon-phonon correlation function which depends only on the relative position of the ions. Thus,
eiqR e
q q C, (R,t)
(4.78)
Most of the phenomena in physics can be expressed in terms of correlation function such as (4.77). This is one important example since most of the excitations in solids (even electronic ones) have bosonic character. The calculation of the correlation function (4.77) is very simple because the u operators are just linear combinations of creation and annihilation operators. We can easily show using (4.43) and the commutation relations between the creation and annihilation operators that C, (R, t) =
s
Notice that the complete evaluation of S(q, ) is very complicated. One can however look at some simple limits. Let us study the limit of elastic scattering, that is, when the energy is conserved in the scattering. In this case we have to take the limit of 0 in (4.72). This is the so-called static limit and is equivalent of taking the limit of long times t in (4.72) since the exponential term oscillates strongly in this limit and the integral is dominated by small values of the frequency. In this limit the exponential terms in (4.79) also oscillate strongly and give vanishing contribution. The only term left is limt = dd k
s
1 2
(2)d
|q es (k)|2
which is the so-called Debye-Waller factor. The physical meaning of this factor is clear if we calculate the dynamical form factor from (4.78) and (4.72): S (0) (q, ) = e2W (T ) ()N q,G (4.81)
which is the result of elastic scattering obtained before except by the Debye-Waller factor which tells us that virtual transitions induced by the collision of the probe with the crystal at nite temperatures decreases the intensity of the scattering. We have been able, therefore, to reproduce the results of elastic scattering. The next correction to elastic scattering is obtained by retaining the time dependence on correlation function (4.77). In order to do that we replace the exponent in (4.78) by S(q, t) =
R
eiqR exp
1 2
=
R
iqR 2W (T )
n=0
1 2
q q C, (R, t)
,
(4.82)
4.5. PROBLEMS
71
It is possible to show that each power in n in (4.82) is due to the scattering of the probe to n phonons. The rst order correction to the static term produces a correction in the dynamic form factor which is given by S (1) (q, ) = e2W (T )
s
|q es (q)|2
ns (q) ( s (q))]
observe that the rst term increases the number of phonons by one and therefore is related with the emission of one phonon by the probe. The second process is associated with the absorption of one phonon by the probe. The functions guarantee the conservation of energy for each one of these process. Other corrections can be calculated immediately from (4.82). An important property of (4.84) is that the scattering happens when = E/ = s (q). Thus, by varying the energy of the probe one can map the dispersion relation of h the phonons.
4.5 Problems
1. Calculate the commutation relations between u (k) and P (k) starting from (4.16). 2. The group velocity of an excitation is dened as the velocity at which a wave-packet made out of a superposition of plane waves propagate and is given by vg = d(k) . dk
What is the group velocity of an acoustical wave when k 0? What is the group velocity of an optical wave when k 0? 3. Calculate the phonon dispersion and polarizations for an one-dimensional lattice with two atoms with masses M1 and M2 as in Fig.4.1(b). 4. What is the sound velocity for the problem in which the chain has two different masses? 5. Assume that a chain is made out of 3 different atoms with different masses. What can you say about the spectrum of the system? Where are gaps located? Hint: no calculations needed. 6. Using the commutation relations between qs (k) and ps (k) show that the creation and annihilation operators obey the commutation relations (4.45). 7. Here we are going to consider the long wavelength theory of acoustic phonons. Dene two new elds, s (r) = s (r) = 1 V 1 V ps (k)eikr
k
qs (k)eikr
k
(4.85)
h which obey canonical commutation relations [s (r), s (r )] = i (r r )s,s . By linearizing the phonon dispersion, s (k) = cs k show that Hamiltonian density associated with the acoustic case can be written in terms of these elds as H= 2 (r) c2 s + s (s (r))2 2 2 (4.86)
72
8. Prove that the equation (4.64) is correct and obey the scaling properties. Evaluate the low temperature specic heat for the Debye model in d = 1, 2, 3. 9. Calculate the specic heat for the Einstein model for phonons with frequency 0 . 10. Show that the eld theory associated with the Einstein model is given by the following Hamiltonian density: H= 11. Prove eq.(4.79). 12. Calculate the Debye-Waller factor in 1, 2 and 3 dimensions at zero temperature for acoustic phonons. What is the physical meaning of your results? This result is related to the famous Mermim-Wagner theorem. 13. Prove that eq.(4.84) is correct.
2 2 (r) aM 0 2 s + s (r) . 2 2
(4.88)
we want to evaluate an average of the form eA eB . First we use an operator identity which is valid when the commutator [A, B] is a c-number: eA eB = eA+B e[A,B]/2 . Notice that A + B can now be written as A + B = a + a where = A + B = A + B . (4.92) (4.91) (4.90)
4.5. PROBLEMS
Thus we are interested in calculating ea+a
73
= e/2 ea ea = e/2 ea ea
(4.93)
(4.94)
= tr eH ea ea /Z = tr ea eH ea
/Z
= tr eH eH ea eH ea = eH ea eH e
a
/Z (4.95)
The Hamiltonian for the problem is simply H = h(a a + 1/2) and if we use the operator identity A U 1 = eU AU 1 and Ue exa a aexa we obtain from (4.95): ea ea
a
= ex a
(4.96)
ee
a h
ea
h = f (e )
(4.97)
(4.98)
h f (e2 ) = e(1+e
+e2 ) h h
h f (e3 )
(4.99)
and so on. If we do this process an innite number of times we nd f () = e/(1e where we have used the result of a geometric series f (0) = ea and therefore f () = e/(1e
) h ) h
(4.100)
n 0 x
=
n
h e n n|ea |n = 1
(4.101)
(4.102)
1
h e
(4.103)
(4.104)
With this result at hand we can rewrite ea+a and nally eA eB = exp = exp 1 [A B A B + (A + B )(A + B )(1 + 2 )] n 2 1 2 A + 2AB + B 2 2
n = e 2 (1+2 )
(4.105)
(4.106)
which can be obtained directly from the denition of A and B, the commutation relation [a, a ] = 1, the 2 = 0. denition n = a a and that a2 = a
Chapter 5
Electrons in Solids
We have seen that the difference in the mass of the electrons and ions allow us to separate their time (or energy) scales. This introduces a major simplication since we can treat electrons and ions separately. We have investigated the problem of the coupling between ions and showed that the ion problem has elementary excitations which are called phonons. The short range character of the interaction between ions and the fact that they form a lattice makes their theoretical treatment quite simple. The same thing does not happen with electrons. This can be seen clearly from the experimental fact that electronic properties of solids can vary wildly, that is, we can have metals, insulators, magnetic systems, superconductors, etc, with properties which depend only on the way the electrons interact among themselves. In this chapter we are going to study the interaction between the electrons and the atoms in a crystal. We are going to explore the fact that in a crystal the potential felt by the electrons has to be periodic due to the periodic arrangement of the atoms. We are going to consider rst the case of a one-dimensional chain of atoms since the mathematical treatment is quite simple and it is relevant for the experimental case of one-dimensional systems such as organic conductors, polymers and other low dimensional systems. The treatment of the higher dimensional problem is completely analogous and we will generalize the onedimensional results to higher dimensions.
(5.1)
where p is the momentum operator. In order to understand why this operator generates translations consider the transformation
h h R(l)xR1 (l) = eipl/ xeipl/ = x + l
(5.2)
75
76
as you can easily shown by using the commutation relations [, p] = i . In the absence of the atoms the x h Hamiltonian of the problem has to commute with R(l) for any value of l by translation (rotation) invariance.
t
1
N N1
2 3
Figure 5.1: A ring of atoms with radius R and hopping energy t. In the presence of the atoms this is not true because the electron feels the potential of each atom. The potential V (x) felt by the electron has to be periodic, that is, if x is the coordinate along the ring one must have V (x + a) = V (x) where a = 2R/N is the lattice spacing. Thus, although the Hamiltonian of the problem does not commute with R(l) for any value of l it has to commute for l = a. Thus, the problem does not have a continuous symmetry but a discrete one. Since the Hamiltonian has to commute with R(a) one knows that H and R(a) share the same eigenvectors. Thus, if we nd the eigenvectors of R(a) we will also nd the eigenvectors of H. Moreover, because R(a) is written like (5.1) its eigenvalues are related to the eigenvalues of the momentum operator, that is, p|k = hk|k . In this case the eigenvalue problem for R(a) reads R(a)|k = eika |k . (5.4) (5.3)
We would like, however, to rewrite the states |k in terms of the states |n of the electron localize on atom n. As we used previously this state is essentially a vector with N entries where only the nth entry is 1 and all the others are zero: 0 0 0 ... (5.5) |n = 0 . 1 0 ... 0
77
We know from basic quantum mechanics that if we have a complete set of orthogonal states any state in the Hilbert space has to be written as a linear combination of these states, in other words, the states |n span the Hilbert space ( n |n n| = 1) |k =
n
|n n|k =
ck,n |n
(5.6)
where ck,n are unknown coefcients. We further know that R(a) has to move the electron one unit in the lattice and therefore R(a)|n R
1
(a)|n
= |n + 1
= |n 1 .
(5.7)
ck,n R(a)|n =
n
ck,n |n + 1 =
ck,n1 |n .
(5.8)
Using (5.4) together with (5.6) and the fact that the states |n are orthogonal to each other ( n|m = n,m ) we see that ck,n = ck,n1 eika (5.9)
which gives a recursion relation for each ck,n starting from the previous one. Suppose we start from ck,1 . In this case it is very simple to see that ck,n = ck,1 eikan and thus the expansion (5.6) becomes |k = ck,1 eikan |n (5.11) (5.10)
and from the fact that |k has to be normalized ( k|k = 1) we nd 1 |ck,1 | = . N We have therefore solved the eigenvalue problem in (5.4). (5.12)
Because the system has the periodicity generated by R(a) it means that the energy of the system does not change when we translate the system by a, that is, the Hamiltonian commutes with R(a): [H, R(a)] = 0. From basic quantum mechanics we know that it implies that the eigenstates of R(a) are also eigenstates of H. This result, however, does not tell us what is the energy of these states. In order to know that we have to use the Hamiltonian itself. Now, let us go back to our original problem of the tunneling of electrons between different atoms. The Hamiltonian can be constructed in exactly the same way as for the case of the other
1 0 0 0 1 0
Observe that if we apply H to a vector like (5.5) we immediately obtain H|n = t (|n + 1 + |n 1 ) .
(5.13)
(5.14)
The meaning is rather obvious: the Hamiltonian has only a hoping term which moves the electron between nearest neighbor sites on the chain. Since we know that |k is an eigenstate of the system let us calculate its energy. In order to do that we just have to apply the Hamiltonian the state: H|k t = N t = N eikan (|n + 1 + |n 1 ) eika(n1) + eika(n+1) |n (5.15)
(5.16)
(5.17)
which gives the spectrum of the problem and is shown as a function of k in Fig.5.2. Observe that the state of lowest energy has k = 0 with energy 2t. From (5.11) this state is simply given by 1 |n = N 1 1 1 ... 1 1 1 ... 1 (5.18)
1 |k = 0 = N
which shows that the probability of nding the electron in any atoms is the same and given by 1/N , that is, the electron is spread uniformly over the entire chain! That is the way the kinetic energy is maximized. Observe that we started with N states and therefore we have to end with N states. It means that there are only N allowed values of k! To nd out what are these values we have to remember that the ring is periodic
79
E 2
k -3 -2 -1 1 2 3
-1
-2
Figure 5.2: Spectrum for an electron moving in a one-dimensional chain as a function of k. and therefore |n + N = |n . Thus, from (5.10) we must have eikaN = 1 which implies that the values of k are quantized and given by km = 2m Na (5.20) (5.19)
where m = 0, 1, ..., (N 1)/2, N/2 for N even and m = 0, 1, ..., (N 1)/2 for N odd produces the required N states. Observe that the available states go from k = /a(1 1/N ) to k = /a. In the limit of N the "distance" between two states which is just 2/(N a) goes to zero and we have /a < k /a which is called the Brillouin zone. Notice that the tunneling between atoms has broken the degeneracy of the original N states into N non-degenerate states for each value of k. Thus if the distance between the atoms is large enough so that t 0 no tunneling takes place and the electron is stuck in one of the atoms. No motion can occur and the system is an insulator that is highly degenerate. As t increases when we approach the atoms the tunneling grows and the degeneracy is broken as shown in Fig.5.3. The argument here is not only valid for the ground state but for any state of the atom. For instance, we could have electrons coming from other orbitals and going from atom to atom. Thus, each orbital gives rise to what is called a band of states which essentially have a dispersion given by (5.17). We also say that each band has a nite bandwidth, W , which is the energy difference between the highest and lowest energy state which for (5.17) is just W = 4t. Because the electron is spread over the entire lattice it is very easy to move it around. This is true if the band is not full. This is simple to understand: from the Pauli principle we can only have one electron for each value of k. Suppose we put 1 electron on a lattice of N atoms. The state of lowest energy has k = 0 (compare this state with the states we found for the H2 and the molecule with 3 atoms!). Once this state is occupied we can put another electron on this state but with opposite spin. The next state to be occupied is the closest to k = 0 since, according to (5.17) (see Fig.5.2) the energy grows with k. Thus, if
80
E N
N N W N
1/a
Figure 5.3: Energy of the system as a function of the distance between atoms. we put Ne electrons in the system they have to occupy symmetrically the states around k = 0. Obviously the band will be full when Ne = 2N . In this case there are no more states available! The next state in the system corresponds to an atomic state with larger energy, that is, there is a gap, , between bands which in the atomic picture is just the energy between the discrete states. Thus, in order to make an excitation in the system we have to excite the electron over the gap and this costs energy. Thus, at low temperatures no conduction can happen and the system is a band insulator. In real space this is also simple to understand: when Ne = 2N the electrons cannot hop anymore to the same orbital since it is fully occupied (Pauli principle) and therefore it has to hop to another orbital at higher energy, as shown in Fig.5.4. The scenario that we have been describing here is known as tight binding approximation because we assume that the wavefunction on each atom is very close to an atomic wavefunction. This is only true is the tunneling t is small. From Fig. 5.3 one sees that when R decreases the bands can start to overlap and the gap between bands disappear. This implies that the electrons just feel a very weak potential created by the lattice. In this case one can just treat the periodic potential created by the lattice as a perturbation. Let us consider this case in more detail. In the absence of a periodic potential the electron undergoes free motion in a circle of length L. The Schrdinger equation for this case is simply h 2 d2 (x) = E(x) 2m dx2 (5.21)
1 eikx L 2 2 h k . 2m
(5.22)
Observe that the energy is a continuous function of k in complete contrast with the tight binding picture and
81
t (a)
(b)
Figure 5.4: Motion of electron in a periodic array of atoms: (a) band less than lled; (b) lled band.
it is shown on Fig.5.5(a). The question that remains to be answered is what happens when we introduce the periodic potential of the lattice. Observe that because the potential is periodic we have V (x) = V (x + a) and if one expands V (x) in Fourier series V (x) =
k
Vk eikx
(5.23)
where n is an integer. Thus, even without computing the effects of the potential we already know that the momentum is only dened in the Brillouin zone, that is, /a < k /a. Thus, one has to shift the portions of the energy Ek inside of the Brillouin zone as shown in Fig.5.5(b). This is the effect of the pure periodicity of the problem even if V (x) 0! When V (x) is nite but small we have to take into account the effect of the potential in the Schrdinger equation h 2 d2 + V (x) E (x) = EE (x) . 2m dx2 (5.25)
Let us treat the problem in perturbation theory. Since we know that the problem when V (x) = 0 is solved in terms of plane waves we look for a problem of (5.25) which is a linear combination of plane waves 1 E (x) = L ck,E eikx
k
(5.26)
where we would like to know the values of ck,E . Because of the periodic boundary conditions, that is,
82
Ek
(a)
k
/a /a
Ek
(b)
k
/a /a
Figure 5.5: Electron dispersion: (a) in the absence of a periodic potential; (b) in the presence of a periodic potential.
where n = 0, 1, ..., N/2. We have now to normalize the wavefunction over all space
L/2 L/2
dx|(x)|2 = 1
(5.28)
dxei(kp)x = 1 .
(5.29)
In order to perform the integral one has to remember that k and p are given by (5.27) which implies that the above integral has the form:
L/2 L/2
dxei(pk)x = 2
(5.30)
Observe that the above integral is zero unless n = m (that is, p = k) in which case we write
L/2 L/2
dxei(pk)x = Lp,k
(5.31)
where p,k is the so-called Kronecker delta which has the properties that p,k = 1 if p = k and p, k = 0 if
83
(5.32)
gives the normalization of the wavefunction. If we substitute (5.26) into (5.25) one nds h 2 k2 ck,E eikx + 2m Vk eikx cp,E eipx =
k,p k
Eck,E eikx
(5.33)
in order to solve this equation we multiply the equation above by eipx and integrate over x. We are going to use the following useful integral (5.31) to obtain
0 (E Ep )cp,E =
Vpp cp ,E
p
(5.34)
where Vkp = 1 L
L/2 L/2
dxV (x)ei(kp)x .
(5.35)
Observe that (5.34) is a matrix equation where Vk,p are the matrix elements of the potential and the collection of ck,E for xed E form a vector. In order to see that let us remember that because of the boundary conditions kn = 2n/L with n = 0, 1, ..., N/2 there are N allowed values of k. Thus, let us number km with m = 1, ..., N . We will use m to number the eigenstate (which is equivalent to use k in the Brillouin zone). That is, we can rewrite (5.34) as E Ek1 0 ... 0 ck1 ,E ck2 ,E 0 E Ek2 ... 0 ... ... ... ... ... ck ,E 0 ... E EkN1 0 N1 0 ... 0 E EkN ckN ,E V1,1 V1,2 ... V1,N ck1 ,E V2,1 V2,2 ... V2,N ck2 ,E ... = ... ... ... ... (5.36) VN 1,1 ck ,E ... VN 1,N 1 VN 1,N N1 VN,1 ... VN,N 1 VN,N ckN ,E which is a N N diagonalization problem exactly like in the tight binding problem we solved earlier.
We do not wish at this point to solve (5.34) in its full form but for small values of V , that is, we would like to solve the problem as a power series expansion of the potential. Observe that for V = 0 the solution is straightforward, namely, for a given ith eigenvalue we have: if E = Eki then cki ,E = 1; if E = Eki we have cki ,E = 0. Another way to rewrite this result is to dene a wave number k such that: E= h 2 k2 2m (5.37)
Vpp cp ,k
p
(5.38)
and therefore the solution for V = 0 is: if p = k we have ck,k = 1 and if p = k cp,k = 0. The result can be summarized as c0 = p,k . p,k
(1) (1)
(5.39)
(1)
0 The next order correction is obtained if we assume E = Ek + Ek and cp,k = c0 + cp,k with Ek and p,k
cp,k given in rst order in V and substitute on (5.38): h 2 2 (1) (1) (k p2 ) + Ek (p,k + cp,k ) = 2m Ek p,k +
(1) 2
(1)
Vpp (p ,k + cp ,k )
p
(1)
(5.40)
where we have kept only terms to rst order in V . The solution of this equation is straightforward again: if k = p we nd Ek = V0 and if k = p we nd cp,k =
(1) (1)
(5.41)
Vpk h 2 2 2 2m (k p )
(5.42)
which is the rst order correction. Notice that ck,k = 0! Observe that Vkk=0 = 1 L
L/2 L/2
(1)
dxV (x) = V
(5.43)
is just the mean value of the potential in the system. Thus, we conclude that rst correction to the energy is a simple shift in the overall energy of the system. We observe that the wavefunction of the problem at this order in perturbation theory can be written as Vk,p 1 eipx . (5.44) k (x) eikx + 0 E0 Ep L k
p=k (1) (2)
0 The second order correction proceeds in the same way, that is, we write E = Ek + Ek + Ek and (2) (1) ck,p = c0 + ck,p + ck,p and substitute on (5.38). The rst order terms cancel because of (5.40) and the k,p second order equation becomes (2) (1) (1)
Ek p,k + Ek cp,k +
h 2 2 (2) (k p2 )cp,k = 2m
Vpp cp ,k
p
(1)
(5.45)
85
Vkp cp ,k =
p
(1)
h 2 2 2m (k
|Vp k |2
(p )2 )
(5.46)
where we used (5.42) and the fact that Vp = Vp . If k = p then, from (5.45) we have
cp,k =
p
(2)
Vpp Vp k h 2 2 2 2m (k (p ) )
V0 Vpk h 2 2 2 2m (k p )
(5.47)
which proportional to V 2 . In order to calculate the change in the energy we have to calculate the matrix element Vk,p. From (5.35) and (5.23) one has Vk,p = =
n
1 L
L/2
Vn
n L/2
Vn k,p+2n/a
Thus, Ek =
n=0 (2)
|Vn |2 . 0 0 Ek Ek2n/a
(5.49)
Observe that if k = n/a the denominator in the above equation vanishes and perturbation theory fails! Thus, for this particular points something else has to happen and one cannot use non-degenerate perturbation theory. For the other values of k the result (5.49) is ne since no divergences occur. Thus, let us go back to the original equation (5.34) and try to understand what is going on. For simplicity we are going to assume that V (x) = V0 cos(2x/a). This implies that we have V0 = 0 and V1 = V1 = U0 /2. Substituting (5.48) into (5.34) one nds
0 (Ek E)ck,E +
Vn ck2n/a,E = 0 .
n
(5.50)
But because of our choice of V (x) this equation simplies to U0 h k2 E ck,E + c + ck2/a,E = 0 2m 2 k+2/a,E which again is an innite matrix problem. The rst three terms read U0 h (k + 2/a)2 E ck+2/a,E + c + ck,E = 0 2m 2 k+4/a,E h k2 U0 E ck,E + c + ck2/a,E = 0 2m 2 k+2/a,E h (k 2/a)2 U0 E ck2/a,E + c + ck,E = 0 2m 2 k4/a,E (5.51)
(5.52)
What do we do now? Well our perturbation theory failed for k = /(2a). Thus we will retain in this matrix problem only the values of k and k 2/a and forget about all the other Fourier components. In this case
Ek U0 /2
U0 /2 h (k2/a)2 2m
Ek
ck,E ck2/a,E
= 0.
(5.53)
h (k 2m
2/a)2
h k2 2m
2 2 + U0
(5.54)
E/a, = |U0 |/2 that is, there is a gap in the spectrum with magnitude /a = E/a,+ E/a, = |U0 |
(5.55)
(5.56)
which implies that the two branches in Fig.5.5(b) are now split and there is a gap between them as shown in Fig.5.6.
Ek
U0
/a /a
U0 k
Figure 5.6: Effect of a periodic potential on a free electron system. Moreover, by direct substitution of (5.55) into the matrix equation (5.53) we nd ck,+ = ck/a,+ ck, = ck/a, which implies from (5.26) that +,k (x) cos(x/a) ,k (x) sin(x/a) (5.57)
(5.58)
which have a very simple meaning if we consider a plot of the potential felt by the atoms and the shape of these function as in Fig.5.7. The ground state wavefunction G = has its maximum at the position of the ions while the next excited state E = + has its maximum in between the ions. Since the charge of the electron is opposite to the ions the energy is minimized when the electrons have most of the charge on the top of the positive ions as in Fig.5.7(a). Thus, the appearance of gaps in the spectrum of a free electron system are related to the breaking of the translational symmetry of the system (which now has only discrete translational symmetry). It does not matter from what limit we start, that is, electrons localized on atoms or free electrons. When the hybridiza-
87
x (a)
a V(x)
| |2
E
x (b)
Figure 5.7: Wavefunctions for an electron in a periodic potential of the ions: (a) ground state; (b) rst excited state .
tion between atoms is small the system behaves more like isolated atoms and the bands reect the quantum nature of the atomic states. As the hybridization grows the system behaves more like a free electron gas. Observe that the electron plays a dual role in this dance: it binds atoms together and also gives rise to charge conduction. Depending on the type of system and on the type of orbital that participates on the formation of a solid the properties of the solid can change dramatically as one sees from the simples arguments we have given. In Fig.5.8 we show how the free electron solution compares with the solution close to the Brillouin zone boundary k = G/2 = /a. We see that the free electron solution works very well close to the bottom of the band but fails to describe the system at the Brillouin zone boundary.
where due to the symmetry we have V (r) = V (r + T) where T = of this symmetry we can write V (r) =
G
VG eiGr
(5.60)
88
E 1
0.8
0.6
0.4
0.2
Figure 5.8: Solution of the problem close to k = G/2 and comparison with the free electron result (dashed line).
where G is a reciprocal lattice vector (notice that T G = 2n which is the Bragg law). The general solution of (5.59) can be written as E (r) =
k
k,E eikr
(5.61)
where the Fourier components obey the equation (by direct substitution of (5.60) and (5.61) into (5.59)) h k2 E k,E + 2m which is the higher dimensional analogue of (5.34). Let us consider now the general properties of (5.62). (5.62) can be thought (as in the case of (5.36)) as a matrix equation for the coefcients kG,E . For a given eigenvalue E the coefcients k,E have to be linear combinations of other coefcients at wavevectors shifted by G. Thus, if we dene again E = h2 k2 /(2m) then we must have accordingly to (5.61): k (r) =
G
VG kG,E = 0
G
(5.62)
kG ei(kG)r .
(5.63)
We can rewrite this wavefunction in a more interesting way, k (r) = eikr uk (r) where uk (r) =
G
(5.64)
kG eiGr .
(5.65)
89
Observe that uk (r) = uk (r + T) is a periodic function. Thus, we have shown that in a periodic lattice the wavefunction can be written in a very special form which is extended over the whole lattice. This is called Blochs theorem. The result of this demonstration is that the wavefunction can be dened in the Brillouin zone of the material since we can reach any k outside of the zone by translating by a reciprocal lattice vector, that is, k+G (r) = k (r). Moreover, the energy is also a periodic function of the reciprocal lattice since we can always shift the momentum in (5.62) by a reciprocal lattice vector, Ek = Ek+G . This result is completely general and does not depend on the particular form of the potential! In the limit where the potential is weak, as we have seen previously, we can try perturbation theory. As before we will nd that the perturbation theory fails at k = G/2 which is the border of the Brillouin zone. This failure is again related with the opening of a gap in the spectrum due to the Bragg scattering, that is, the energy of the system is only slightly modied in for most wave-vector except when 0 = 0 kG k where perturbation theory gives a divergent contribution. The origin of this divergence can be understood immediately since the condition for the divergence is equivalent to |k| = |k G| (5.66)
which is the Bragg law (see (2.38)). This result makes a lot of sense since electrons are as good waves as any other microscopic particles. Thus when the electron wave-vector is such that the Bragg law (5.66) is obeyed the electron undergoes coherent Bragg diffraction. Observe that this condition is equivalent to say that k has to lie on the zone boundary in order to undergo Bragg diffraction and this is the point where perturbation theory fails and one has to use the matrix approach described earlier. In this case we obtain the formation of a gap at these points, that is, the gap is a straight consequence of Bragg diffraction. Consider for instance the one dimensional case studied before. Condition (5.66) implies that perturbation theory fails for k = G/2 = n/a and one has a sudden jump in the energy at this point. A graphical way to depict this situation is shown in Fig.5.9 where we see that the only substantial difference between the free electron case and the actual case occurs whenever a gap opens at the Bragg points. In order to obtain the bands discussed in previous approach one has just to fold back the points back to the Brillouin zone as in Fig.5.8.
E
k
|U 2/a |
|U /a|
3 /a 2 /a /a 0 /a 2 /a 3 /a
Figure 5.9: Modication of the free electron dispersion (dashed line) due to the Bragg reection of the electrons. This construction can be carried out to higher dimensions where the points where gaps open are still given by the Bragg condition (5.66). Consider for instance the problem of a two dimensional square lattice
90
such as in Fig.5.10. As the momentum varies as shown in Fig.5.10(a) the gaps open as in Fig.5.10(b).
ky
k /a a /a /a /a 2/a b c
kx (a)
Ek
(b) k
a b c
Figure 5.10: (a) Brillouin zone in k-space (shaded area) and neighboring zones; (b) Electron dispersion along the direction of k. At this point one natural question comes out: how are these results related with the tight binding calculation we have done previously? The answer to that is very straightforward. In the tight binding limit the electrons are tightly bounded to the atoms and therefore the potential is strong. However, accordingly to Blochs theorem, (5.64), we should be able to solve the problem in terms of periodic functions. The wavefunction of the problem has to obey Blocks theorem which is a statement of the periodicity of the system, nothing more. Observe that from (5.64) and (5.65) the wavefunction of the problem must obey k (r + T) = eikT k (r) . (5.67)
We would like to construct the wavefunction from localized orbitals instead of the plane waves which are so natural for the weak potential problem. These localized orbitals are the solution of the atomic Schrdinger equation: h 2 2 + VA (r) A (r) = EA A (r) 2m (5.68)
where VA (r) is the atomic potential and EA the binding energy of the electron. Let us consider the problem where the atoms are far apart so the tunneling is small. We would like to construct the wavefunction from localized orbitals like the ones given in (5.68) which has this property. Remember that in the case of molecules this can be done by taking linear combinations of localized orbitals (which we called bonding and anti-bonding). A similar idea here is to choose the linear combination 1 k (r) = N eikT A (r T) . (5.69)
Observe that this wavefunction obeys (5.67) Moreover, in this approximation each band carries the atomic
91
Observe that (5.69) is a solution of the atomic equation (5.68) but instead we would like to solve the actual Schrdinger equation (5.59). It is clear that this is not the actual ground state of the problem. However, in the limit of innite separation between the atoms it is an eigenstate. Moreover, the spectrum of the problem can be written from (5.59) by using the orthogonality of the wavefunctions as Ek =
dd r k (r)
h 2 2 + V (r) k (r) 2m
(5.70)
which, as we said before, is periodic on the reciprocal lattice, that is, Ek+G = Ek . Since the energy is periodic on the reciprocal lattice it can be always expand as a Fourier series of the direct lattice (remember that the direct lattice is the reciprocal of the reciprocal lattice!) that is, Ek =
T
tT eikT .
(5.71)
In order to calculate the coefcients tT we are going to use (5.70). If we substitute (5.69) into (5.70) we nd Ek =
T,T
eik(TT ) N eik(TT ) N
dd r A (r T )
=
T,T
dd r A (r + T T )
=
T
eikT
dd r A (r + T)
h 2 2 + V (r) A (r) 2m
where we have used the periodicity of the potential U (r + T) = U (r). Thus we have shown that (5.71) is correct and that tT =
dd r A (r + T)
h 2 2 + V (r) A (r) 2m
(5.73)
where the integral runs through the unit cell. We now rewrite V (r) = VA (r) + V (r) where V (r) = V (r) VA (r). In the limit where the distance of the atoms is very large we expect U 0 and therefore can be treated as a perturbation. In this case, using (5.68) one gets tT = EA
dd r A (r + T)A (r) + dd r A (r + T)V (r)A (r) .
(5.74)
Observe now that the atomic wavefunction decays exponentially with distance (A (r) er/ao where a0 is the Bohr radius) and therefore the overlap integrals above will be very small for T = 0. In this case it is a good approximation to use the on-site (T = 0) and nearest neighbor contributions since the integrals will decay exponentially with distance. This is known as the tight binding approximation. In this approximation we write tT EA T,0 tA T,
(5.75)
where represent the nearest neighbors sites and tA is the overlap given in (5.74) for nearest neighbors.
(5.76)
Observe that corrections to the tight binding approximation can be computed directly from (5.73). As an example consider the one-dimensional case. In this case we have just two nearest neighbors which correspond to = a where a is the lattice spacing. In this case the energy becomes, Ek = EA 2tA cos(ka) (5.77)
and has the shape shown in Fig.5.2. For a cubic system in d dimensions the tight binding model predicts a dispersion
d
Ek = EA 2tA
cos(ki a)
i=1
(5.78)
which has its minimum value at EA 2dtA and its maximum at EA + 2dtA . Thus, we say that the system has a bandwidth W = 4dtA . In the atomic limit tA 0 and the bandwidth vanishes while as the tunneling between atoms increases the bandwidth increases proportionally to the hopping tA as in Fig.5.3. Observe that gaps that we have computed in the free electron gas just represent the difference in energy between the atomic levels that decreases as the tunneling between atoms increases. When the energy levels start to overlap we are back to the nearly free electron gas. Notice that at long wavelengths ka 1 we can approximate the problem by expanding the exponential in (5.76) to second order in ka, Ek EA ZtA + tA 2 (k )2 (5.79)
where Z is the number of nearest neighbor sites (Z = 2d for a cubic system) and we have used that k = 0 by inversion symmetry. We can now use that
d
=
i=1
ani
(5.80)
where a is the lattice spacing and ni is the unit vector along the main crystal axis. In this case we rewrite (5.79) as
d
i=1
h 2 2 ki 2m A
(5.81)
h 2 2tA a2
(5.82)
is the so-called effective mass of the electrons. Observe that this mass can be very different from the free electron mass since its physical meaning is completely different of that one. The effective mass measures the static dragging of the lattice by the electron motion since the electron interacts with the atomic cores. This
93
can be easily understood if we take a semi-classical approach. As we have seen the electrons are described by a dispersion k . From this dispersion we can dene the velocity by vk = 1 k h (5.83)
which can be checked to be correct for the free electron case where k = h2 k2 /(2m) and vk = hk/m. Suppose an electrical eld E is applied to the system. This electric eld will cause the electrons to accelerate. It turns out that the electrons deep inside the Fermi sea cannot accelerate since they would have to make transitions to already occupied states. This is forbidden by Paulis principle. The electrons at the Fermi surface, however, can be accelerated freely. If we imagine this electrons to behave like wavepackets (that is, like classical particles) then the acceleration is given by dvk dt = 1 d h dt 1 h
d i,j=1 d i=1
k ni ki (5.84)
2 k dkj ni ki kj dt
which is valid for k kF and where we used the chain rule and ni is the unit vector in the i-direction. Observe that the momentum of the electron is simply k and therefore by Newtons equation of motion in h an external eld we have h which substituted in (5.84) leads to e dvk = 2 dt h
d i,j=1
dk = eE dt
(5.85)
2 k Ej ni . ki kj
(5.86)
Observe that (5.84) is very similar to the classical equation dv eE = . dt m This similarity allows us to dene the effective mass tensor m1 = ij 1 2 k . h 2 ki kj (5.88) (5.87)
Observe that for a cubic system with dispersion given by (5.78) in the limit where kF a 1 we obtain (5.82) as expected.
In this picture the difference between metals and band insulators is very simple to understand. At band llings smaller than one there is in average less than two electron per site as in Fig.5.4(a). Thus the electrons can move freely since electrons with opposite spin can hop from site to site. When the band becomes full (that is we have two electrons per site) the Pauli principle does not allow motion as shown in Fig.5.4(b). In order to move an electron one has to excite the electron to an unlled atomic state and therefore one has to pay an energy price which is the gap energy.
94
5.3 Problems
1. Prove eq.(5.62). 2. Show that (5.69) obeys (5.67). 3. Show that a band insulator always has an even number of electrons per atom. Also argue that the opposite is not true, that is, solids with an even number of electrons per atom can be metallic. 4. In the free electron gas the mass of the electron is just m. In the tight binding case the mass of the electron is dened as 1 d2 Ek 1 = 2 . m h dk2 (1) What is the mass of the electron as a function of the momentum? Why is this mass different from m? (2) What is the mass of the electron when the band is empty? What happens to this mass when the tunneling between atoms goes to zero? Give a physical interpretation for your result. 5. In these notes we have used the so-called reduced zone scheme in which the momentum is plotted only in the rst Brillouin zone /a < k /a. Another possible scheme is to plot the energymomentum relation over all momentum space in which case we do not fold the energy in the Brillouin zone. Starting from the free electron problem (parabolic band) make a qualitative plot of the energy of a particle in a periodic potential as a function of k from 4/a to 4/a. What are the positions of the gaps in momentum space? What are the values of the gaps in terms of the Fourier components of the periodic potential? Hint: no calculations needed. 6. Solve exactly the matrix problem in (1.52) for the case where ck4pi/a,E = 0. Make a plot of the energy bands you get from the solution. What is the value of the gaps in the spectrum? 7. Show that for a cubic system we have: m1 = ij cos(ki a) i,j . m
Chapter 6
where i = x, y, z where we have written the size Li = Ni a. Observe that maximum allowed value of ni is Ni /2 since at this point we reach the boundary of the Brillouin zone, Gi /2 = /a for the cubic lattice. Thus, the number of states available is N = Nx Ny Nz which is the number of primitive cells. If we take into account the spin we have 2N independent states in each energy band. If the atom contributes with just one electron the band is half-lled with electrons. If each atom contributes with two electrons the band is completely lled. The state of lowest energy is obtained by lling up the spectrum starting with the k = 0 as in Fig.6.1(a). For each k we can have two electrons with opposite spins. In momentum space this is equivalent of lling up all the states symmetrically up to the largest value which depend on the total number of electrons as in Fig.6.1(b). We are interested in the thermodynamic limit, that is, the limit that N and L but = N/Ld is nite. In this limit the distance between the states goes to zero and therefore one lls densely the momentum space. It is clear that the number of electrons in the system can be written as N =2
k
(kF k)
(6.2)
where kF is the largest possible momentum and the factor 2 comes from the two spin projections. In the thermodynamic limit we use our old trick Ld (2)d 95 dd k (6.3)
96
Ek
(a)
k kx
2/L
(b)
ky
Figure 6.1: (a) Ground state of a free Fermi gas as a function of energy; (b) Ground state in momentum space showing the lines of equal energy (circles).
dkkd1 (6.4)
(6.5)
that depends only on the electron density and the dimensionality of the system. Associated with the Fermi momentum there is the so-called Fermi energy which is the energy of the highest occupied state, EF = h 2 2 kF . 2m (6.6)
Thus, in this case, the nite size spectrum of Fig.6.1 becomes the one in Fig.6.2. The points in momentum space, kF , such that kF = EF (6.7)
denes the Fermi surface. Observe that the system is highly degenerated since we have a large number of states for a given energy. The magnitude of this degeneracy is again dened in terms of a density of states which is dened exactly like in the phonon problem, that is, Eq.(4.56). For the free electron problem one
97
Ek EF (a) kF kx (b) kF ky kF k
Figure 6.2: (a) Ground state of the free Fermi gas; (b) Ground state in momentum space.
has N (E) = = = 2 V (E k ) = 2
0
dd k (E k ) (2)d h 2 k2 2m = Sd (2)d 2m h 2
d/2
dkkd1 E
d2 2
d2 2
E EF
(6.8)
where we have used (6.5) and (6.6). The concept of a Fermi surface is one of the most important concepts in the physics of electrons in d solids. In a solid of density the volume occupied by one electron is in average Sd r0 /d and therefore the mean distance between electrons is of order of r0 = d Sd
1/d
(6.9)
Since we are talking about atomic scales it is convenient to use the Bohr radius, a0 = h2 /(me2 ), as a distance scale. Thus, the number, rs = r0 a0 (6.10)
gives us an idea of the density of the solid relative to the atomic volume of the atom. In this units the Fermi
me4 1 . 2 2 rs h 2
(6.11)
Observe that apart from the geometric factors the Fermi energy is proportional to the ionization energy of the Hydrogen atom EI = me4 /(2 2 ) 13.6 eV. In most metals rs is of order of one or smaller (the distance h between electrons is of the order of the distance between atoms). In this case it is easy to see that the Fermi energy is of order of eV, that is, of order of 10, 000 K. The ground state energy can now be easily calculated: E0 =
k
h 2 k2 (kF k) 2m (6.12)
E0 N
d EF . d+2
Ek EF
(a) kF kx (b)
k k q
kF k
ky
Figure 6.3: (a) Particle-hole excitation across the Fermi surface; (b) Particle-hole excitation in momentum space. The lowest energy excitation corresponds to take an electron with momentum k < kF and transfer it to a state with momentum k > kF . In this case a hole is left behind (see Fig.6.3). The momentum of this particle-hole excitation is q = k k (6.13)
99
(6.14)
At some temperature T all particle-hole pairs such that (q, k) kB T can be excited. At low temperh atures only a region of energy kB T can be excited around the Fermi surface. This is clear contrast with the phonon problem where any phonon state can be excited with any number of phonons. The fact that the phase space for excitations in the electronic system is much smaller than in the phonon problem is a pure consequence of Paulis principle since one cannot occupy states that are already occupied. Obviously the excitation with lowest energy has |k| = kF and q 0. In this limit the excitation energy reduces to (q|| ) vF q|| where q|| is the component of q in the direction of kF and vF = h kF m (6.16) (6.15)
is the Fermi velocity. The striking discovery here is that the dispersion relation for particle-hole excitations is very similar to the one of acoustic phonons (see 4.39)). Observe that what we actually have done is the linearization of the dispersion of the electrons close to kF as shown in Fig.6.4. The fact that at low energies particle-hole excitations behave as acoustic phonons is not a mere coincidence. It turns out that the elementary excitations of Fermi systems have bosonic character. We will come back to this interesting issue later but now we are going to consider another important excitation of Fermi systems.
EF
K BT
k kF kF
Figure 6.4: Linearization of the electron dispersion close to the Fermi surface.
So far we have considered the situation where we only have excitations with small q, that is, when q kF . However, we can also have low energy excitations with large momentum q. Consider the case where k is close to the Fermi surface but k = k + q is on the opposite side of the Fermi sphere. In this case we can make an excitation with q 2kF that also has a low energy excitation since (q = 2kF , k h kF ) 2 /(2m)(2(2kF )kF + (2kF )2 ) 0. Thus, 2kF excitations with low energy are clearly possible h (later we will see that these excitations are rather important). Thus, the excitation spectrum of the electron
h 2 (2kF q + q 2 ) 2m h 2 (2kF q + q 2 ) 2m
(6.17)
which correspond, respectively, to the small momentum and large momentum processes. Dimensionality also plays a role here. Notice that in one dimension the Fermi sea becomes a line that runs from kF to +kF and there are no excitations with zero energy at nite momentum since the only possible excitations have either q = 0 or q = 2kF (see Fig.6.5). In higher dimensions there is a macroscopic number of excitations at low energies because the Fermi surface is a continuum. It is easy to see that in one-dimension it is not possible to generate any particle-hole excitation with energy smaller than h min = + (q) = h h 2 (2kF q q 2 ) 2m (6.18)
while this is possible in 2 and 3 dimensions. If one draws the curves generated by (6.17) and (6.18) in the q diagram we get the situation shown in Fig.6.6.
q~0
(a) (b)
kF
q ~ 2 kF
+k F
k F
+k F
Figure 6.5: Fermi points in one dimension and the possible particle hole excitations: (a) small momentum transfer; (b) 2kF momentum transfer.
q
0
q
0
(a)
2k F
(b)
2k F
Figure 6.6: (a)Particle-hole continuum in 1D; (b)Particle-hole continuum in 2D and 3D. Let us go back to our original problem of electrons on a box of size L in three dimensions. Suppose we displace the whole electron gas through a distance x with respect to background of charge produced by the ions. In this case we will have an excess of positive charge on one side of the box and an excess of negative charge on the opposite side of the box (see Fig.6.7). This effect will create an electric eld across the box. By Gauss law the electric eld is simply E = 4 where is the charge density on the wall of the box. This density is simply = ex. Of course this eld will act on the electrons in such a
101
way to move the displaced charge density back to its original place so charge neutrality is obtained again. Mathematically this come from the fact that the force applied on the charge on the opposite side of the box is F = N eE = N e4 = 4e2 N x. Thus the equation of motion for the displaced density is (N m) d2 x dt2 d2 x dt2 = 4e2 N x(t)
2 = p x(t)
(6.19)
where
2 p =
4e2 m
(6.20)
is the so-called plasmon frequency of the electron system. Observe that (6.19) predicts that the displaced charge undergoes harmonic motion with the plasmon frequency. Since the plasmon frequency is associated with the displacement of the whole electron system it has a characteristic wave-vector q = 0. Thus this simple argument shows that the electron gas, besides the particle-hole excitations described by (6.15), also has a collective mode called the plasmon that has a dispersion (q) p (6.21)
at small wavelengths. Again, we have a similarity between the optical phonon problem and the electron gas. These similarities do not stop here. Observe that the plasmon is a high energy excitation since we need temperatures of the order of p in order to excite this mode. The plasmon is a massive excitation. Using h the equations given before we can show that
3/2 h p 41.7rs
(6.22)
in eV. Thus, for metallic densities the plasma frequency is of order of the order of 40 eV or 40, 000 K. Thus, in order to excite a plasmon in a metal one has to use high energy probes such as X-rays or high energy electrons. As in the phonon problem where acoustic phonons describe the the low temperature properties of ion vibration, the particle-hole excitations dominate the behavior of Fermi liquids at low temperatures. Observe that all the elementary excitations we are discussing here have a counterpart in the physics of elementary particles. In order to create a particle from the vacuum one has to pay an energy price which is twice the rest energy of the particle. Thus optical phonons and plasmons behave like actual particles although they are collective modes which are born out of the interaction between ions and electrons. The result for the plasmon frequency is only valid for three dimensional solids. In lower dimensions the situation is more complicated because the electric eld is not the one created by a capacitor plate but by a line of charge. In this case we can show that the plasmon dispersion and the plasmon mode becomes gapless.
102
(a)
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + d d + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
(b)
E = 4
Figure 6.7: Formation of a plasmon in a non-interacting electron gas: (a) uniform positive background plus electron gas; (b) displaced electron gas relative to background.
spin. This is the Zeeman effect that you probably have seen in the study of the Hydrogen atom. The second effect is due to the Lorentz force that the electron feels when is in motion. These are relativistic effects but they have different origins. While the Lorentz force can be applied to classical particles, the Zeeman effect is a pure quantum mechanical effect. We are going to study each effect separately.
where me is the electron magnetic moment. For the electron spin S the magnetic moment is written as me = e where e = gs B 1.16 108 (6.25) S h (6.24)
in units of eV/Gauss. Here gs 2 is the g-factor and B = e /(2mc) is the Bohr magneton. In order to h study the Zeeman effect we have to add the new term (6.23) to the free electron Hamiltonian. Without any loss of generality we choose the magnetic eld to be along the z axis. Thus, it is clear that the problem can be completely diagonalized in the basis |k, where =, where means that the spin is parallel to the eld and that the spin is anti-parallel to the eld (Sz | = ( /2)| and Sz | = ( /2)| ). The h h
103
Ek EF n B
2 B
B
k
k
F
k
k
F
(6.26)
Thus, the dispersion of each mode is shifted relative to each other by a quantity 2B B. The ground state of the system is obtained by lling up the states starting from the lowest energy state up to the Fermi energy, EF , as shown in Fig.6.8. Since the up and down spins are in equilibrium with each other it means that the chemical potential is the same, that is, the eld does not change the Fermi energy of the system but only the number of species of electron as depicted in Fig.6.8. Thus, in the presence of a magnetic eld the number of spins per unit of volume, , is different from the number of spins, . This implies the system has been magnetized with a magnetization M per unit of volume given by M = B ( ) . (6.27)
Observe that the change in the number of the two species of electrons is equivalent to a change in the Fermi momentum of them. Thus we can dene two new Fermi momenta, kF, and kF, which are related to the number of fermions as in (6.4): = Sd kd . d(2)d F, (6.28)
Since the total number of electrons is conserved in the presence of the magnetic eld we have a constraint that + =
d d kF, + kF, =
d(2)d . Sd
(6.29)
The other constraint that we discussed before is that the chemical potential is the same for each species of
h 2 kF,
(6.30)
Equations (6.29) and (6.30) have to be solved at the same time. It is useful to parameterize the unknowns as kF, = k0 cosh kF, = k0 sinh so that (6.30) is solved at once: k0 = 4mB B h 2 (6.32) (6.31)
since cosh2 sinh2 = 1. Substituting (6.31) on (6.29) one gets coshd + sinhd = d(2)d d k0 Sd (6.33)
which is a complicated equation for . The quantity we are after is the magnetization of the system which is given in (6.27): M=
d Sd B k0 coshd sinhd . d(2)d
(6.34)
Instead of the general solution of this problem we are going to look in the limit of weak magnetic eld. In this limit (B 0) we see that from (6.32) that k0 0 and therefore the right hand side of (6.33) diverges. It implies that . In this case we can approximate cosh sinh e /2 and we nd e = In this approximation we can write (6.34) as M
d B 21d Sd k0 (d2) e (2)d 2 N (0)B B
2d1 d(2)d Sd
1/d
1 k0
(6.35)
(6.36)
where N (0) is the density of states at the Fermi energy which is given in (6.8) for E = EF . Notice that the magnetization is directly proportional to the density of states at the Fermi surface. The magnetic susceptibility of the system is given by P (T ) = lim M = 2 N (0) B B0 B (6.37)
is called the Pauli susceptibility and it is an exact result since only the linear term in the magnetic eld contributes. Thus, the measurement of the magnetic susceptibility of a free electron gas is a direct measurement of its density of states at the Fermi energy.
105
Let us consider a simple case of the general problem discussed here. Let us consider d = 2. In this case one can calculate the magnetization exactly from (6.34) and (6.32), M= m2 B B h 2 (6.38)
which is strictly linear with the magnetic eld. In this case (6.36) is not approximate but exact. Finally we observe that by increasing the eld one can fully polarize the electron gas (that is, = ). As we can see from Fig.6.8 this happens at a critical eld, Bc , given by Bc = EF . 2B (6.39)
For a metallic system this will happen at elds of the order of 108 G or 104 Tesla (1 Tesla= 104 Gauss). These are very large elds.
(6.40)
Classically the radius of the orbit depends on the velocity of the electron. This is not entirely correct in quantum mechanics. In quantum mechanics the angular momentum is always quantized in units of . For h the circular motion the angular momentum is just pR. Thus, from the quantization of the angular momentum we nd pR = n , h Rn = nl0 , where l0 = c h , eB (6.42)
(6.41)
is the so-called cyclotron radius that gives a measure of the size of the electron wavefunction in the presence of an external magnetic eld. Observe that apart from the magnetic eld the cyclotron radius only depends on universal quantities. If we choose the magnetic eld to be given in Tesla, the cyclotron radius can be written as 250 l0 = . B (6.43)
106
We see, therefore, that the cyclotron radius is usually much larger than the typical separation between atoms in a solid. In order for l0 to be of order of 1 one needs elds of the order of 6 104 T. These are incredibly large elds. As we saw previously at these elds the electron gas is completely spin polarized. As you already know the Hamiltonian for a charged particle in a magnetic eld is H= 1 e p A 2m c
2
(6.44)
where A(r) is the vector potential which is related to the magnetic eld via B = A. Assume for simplicity that the eld points in the z direction, that is B = Bz. As you know the problem of a particle in an electromagnetic eld has what is called gauge invariance, that is, we can replace the vector potential by A A + (where (r) is an arbitrary function) without changing the value of the actual magnetic eld since = 0 is a mathematical identity. This gauge invariance allows us to choose any form of A(r) such that B = Bz holds. Here we will choose the so-called Landau gauge: A = Byx . Substitution of (6.45) directly into (6.44) leads to H= 1 2m px + eB y c
2
(6.45)
p2 p2 y + z . 2m 2m
(6.46)
Observe that the Hamiltonian does not depend explicitly on x and z. Therefore, in the x and z directions one has free motion. In this case the wavefunction of the problem can be written as kx ,kz (r) = ei(kx x+kz z) (y) where (y) is unknown. In this case the Hamiltonian of the problem reduces to H= 1 2m h kx + eB y c
2
(6.47)
p2 h 2 2 kz y + . 2m 2m
(6.48)
A simple redenition of variables helps a lot y = y + y0 h kx y0 = mc and the Hamiltonian of the problem becomes H=
2 p2 mc h k2 y + (y)2 + z 2m 2 2m
(6.49)
(6.50)
which is the Hamiltonian for a harmonic oscillator plus a free particle part. We have therefore solved the problem entirely. The wavefunction is given by (6.47) with (y) Hn (y y0 ) is a Hermite polynomial of order n. Since we have mapped the problem into the harmonic oscillator the spectrum is giving by En (kz ) = hc n + 1 2 + h 2 2 kz 2m (6.51)
and is shown in Fig.6.9. The physics here is straightforward: the electron undergoes circular motion in
107
(a) hc kz
n=0
E (b) h c
n=0
kz
Figure 6.9: Landau levels of a free Fermi gas: in (b) the eld is larger than in (a).
the plane perpendicular to the eld (due to the Lorentz force) and free motion along the direction of the eld (since the Lorentz force vanishes). Circular motion is essentially harmonic motion in the x and y coordinates. We are interested in the physical quantities such as the magnetization of the system. Notice that when we change the magnetic eld by an innitesimal amount B the internal energy of the system changes by U = M B. Thus, the magnetization is simply M (B) = U . B (6.52)
Observe, however, that the energy does not depend on kx and therefore the problem is highly degenerate. The reason for this degeneracy can be understood by looking into (6.49). Observe that the wavefunction of the electron is centered at y0 which is called the center of orbit. The energy of the problem is invariant if we shift the center of the orbit in the plane perpendicular to the eld because the eld is homogeneous in the perpendicular plane. Physically one has to impose the condition that 0 y0 Ly . It implies from (6.49) that the momentum in the x direction is bounded by 0 kx mc Ly / . h (6.53)
But from the periodic boundary conditions we require that kx = 2nx /Lx and therefore the number of
mc Lx Ly = 2 h 0
(6.54)
where = BLx Ly is the total magnetic ux through the plane perpendicular to B and 0 = in Gauss-cm2 is the so-called ux quanta. As we have seen so far the interesting part of the electron motion happens at the plane perpendicular to the eld. It is possible to build articial semiconducting structures such that the electrons are conned to a very thin layer of thickness Lz . In this case kz = 2nz /Lz is quantized. From now on we focus on the two-dimensional electron gas and forget about the motion in the third direction. It is easy to generalize our discussion here to the three dimensional situation in ordinary solids. From now on we just forget the kz dependence in (6.51) (as you can easily convince yourself all we are going to do here is equivalent to work with xed kz ). The energy levels are shown in Fig.6.10. Each level can comport N electrons (we forget spin for the moment being). If there are a total of Ne electrons in the system the ground state is obtained by lling up the lowest energy states. Observe that the number of states per electron is simply N B 1 = = Ne 0 where = is the planar density of electrons. Consider initially the case where N > Ne , that is, only the rst Landau level has electrons (0 < < 1). Clearly it implies very strong magnetic elds or small electronic densities since we must have from (6.56) that B > 0 . In semiconductors we have ordinarily 1011 cm2 which implies B > 4 104 G (or B > 4 Tesla). In this case the total energy of the system is simply E0 = Ne E0 Ne From (6.52) we nd h e M (B) = e = Ne 2mc (6.59) = h c 2 h eB . 2mc Ne Lx Ly (6.57) (6.56) ch 4 107 e (6.55)
(6.58)
is the electronic magnetic moment (notice that 2e B = hc ). Observe that this magnetization is constant as far as N > Ne or < 1. Suppose we decrease the magnetic eld in the system with xed number of electrons. Then the fractional occupation increases and the distance between Landau levels decreases. When = 1 the rst Landau level is fully occupied if we reduce the magnetic eld even further N < Ne < 2N (or 1 < < 2) and some of the electrons, say, N = Ne N have to occupy the second Landau
109
E
(n1/2) h /2 (n1/2) h /2
N n N N
n n1
level. In this situation the energy of the system is E1 = N E1 Ne and the magnetization is 4 M1 = e 3 + Ne . (6.61) 3 c h h c + N 2 2 3 1 = h c 2
(6.60)
Observe that something drastic happened because at = 1 there is a jump in the magnetization from e to +e . This jump occurred because the electrons in the lowest levels move to higher Landau levels as the eld decreases. Observe that this effect occurs every time is an integer number, that is, each time a Landau level is fully occupied. The generalization of the above argument is simple. Assume that the last fully occupied Landau level is n 1 so that n is occupied by N nN electrons (see Fig.6.10). In this case n < < n + 1 and the total energy of the system is
n1
En = N
m=0
h c m +
1 2
+ (Ne nN ) c n + h
1 2 (6.62)
En Ne
= h c n +
1 n(n + 1) 2 2
110
M N 1
0.5
-0.5
-1
which is shown in Fig.6.11 as a function of . Notice that the magnetization of the system is a periodic function of for > 1. Thus the magnetization of the electron gas is a periodic function of 1/B as we see from (6.56). This effect is observed experimentally in the magnetization of solids and it is known as the de Haas-van Alphen effect.
outside of the solenoid since A = = 0. Moreover, the magnetic ux through the plane is given by =
A
B dS =
A dl =
dl =
(6.65)
111
B C
Figure 6.12: Geometry of the Bohm-Aharonov effect.
where A is the area of the solenoid, C is a closed curve outside of the solenoid and l is a vector that parameterizes this curve. Observe that the ux through the plane depends only on the change of the function around a closed curve. The Schrdinger equation for this problem is
2 e 1 i A(r) (r) = E(r) h 2m c 2 ie h 2 + A(r) (r) = E(r) . 2m c h
(6.66)
For the region outside the solenoid we can use (6.64) and rewrite the above equation as h 2 2m + ie (r) c h
2
(r) = E(r)
(6.67)
and a simple inspection of this equation leads to the conclusion that the wavefunction has the form
h (r) = e c (r) (r) ie
(6.68)
obeys the free particle problem. As expected by gauge invariance the spectrum of the problem does not depend on the magnetic eld (since there is none) but the wavefunction now has a new phase factor that depends on the magnetic eld. This result has amazing consequences and one of them is the so-called Bohm-Aharonov effect. Since there is no eld outside the solenoid we must require that the wavefunction is invariant under closed curves that circulate the solenoid. Suppose the electron starts at a point R with a wavefunction B (R) and returns to the same point after going around the solenoid as in Fig.6.12 From (6.68) the wavefunction of the electron after the closed curve is
h A (R) = e c B (R) ie
(6.70)
112
y
Figure 6.13: Quantization of the magnetic ux in the Landau gauge. but since A (R) = B (R) we must require e = 2n c h where n is an integer. Using (6.65) we obtain that = n0 (6.72) (6.71)
where 0 = hc/e is the ux quantum. Our argument shows therefore that for a free electron moving in a closed orbit around the solenoid the magnetic ux through the orbit must be quantized. The relevance of this result for the Landau level problem comes from the fact that the degeneracy of each Landau level is the total magnetic ux measured in units of the ux quantum, (6.54). Physically, this can be understood if we recall that in the Landau gauge the electron motion in the x direction is free while in the y direction we have harmonic motion as shown in Fig.6.13. Observe that the wavefunction in the y direction is peaked around y0 and because kx = 2nx /Lx is quantized the values of y0 from (6.49) are also quantized as y0 (nx ) = (0 /BLx )nx . Thus, the magnetic ux through a strip of size y0 is simply, By0 Lx = 0 . Thus, exactly one ux quantum goes through the strip.
(6.73)
where the last line is valid if N 1 which is the case under consideration here. At zero temperature, in the free Fermi gas, the energy required to put one extra electron in the system is essentially the Fermi energy,
113
EF since as we put one extra electron it has to go to an unoccupied state at the Fermi surface. In the phonon problem the chemical potential is zero since phonons can be produced with innitesimally small energies. We observe that in the Fermi gas the excitations occur close to the Fermi surface, or chemical potential, thus it is convenient to measure all the energies relative to the Fermi energy (the natural energy scale in metals). Observe that one usually calculates the total energy of the system assuming a xed number of particles. It turns out, however, that for most calculations at nite temperature it is more convenient to work with xed chemical potential. Moreover, one actually have the systems of interest in contact with reservoir of particles (such as batteries, for instance) which keep the chemical potential constant and allow the number of particles uctuate. In the thermodynamic limit, N , these uctuations are of order N and therefore irrelevant compared to N itself. Thus we dene the free energy, F , of the system as F = E N , (6.74)
which is known as the Legendre transformation of the thermodynamic potential. Thus, using (6.73), we nd dF = dE dN N d = N d, that is N = F (6.75)
which gives the average number of particles as a function of the chemical potential. Let us consider here the simple problem of the N non-interacting electrons described by a Hamiltonian H with single energies . By the Pauli principle we know that the occupation n of each state can be only 0 or 1. A state n of the system is described by the occupation of each single particle state, that is, each state is represented by a set of numbers n = (n1 , n2 , n3 , ....). The total number of electrons in the system is N=
n ,
(6.76)
n .
(6.77)
n|e(HN ) |n
=
n =0
e
1
n ( )
=
n =0
e( )n 1 + e( )
(6.78)
ln 1 + e( ) .
(6.79)
1 e( ) + 1
(6.80)
which by comparison with (6.76) allows us to dene the mean occupation per state as n = 1 e( ) +1 (6.81)
which is the so-called Fermi-Dirac occupation number. Naturally the average energy density of the system in simply 1 E= V n
(6.82)
and other physical quantities can be calculated as well. As in the phonon case we dene the density of states N (E) = 1 V (E ) (6.83)
and transform all the sums into integrals, in particular, the mean energy density is E= dE N (E) E n(E) . (6.84)
Observe that the Fermi-Dirac distribution has the properties one expects for electrons. Consider for instance the zero temperature limit, that is, . If E > the exponent in the exponential in (6.81) is positive and therefore when the Fermi-Dirac occupation vanishes. If E < the exponent is negative and at zero temperature it vanishes leading to n = 1. The conclusion is that at T = 0 we have a step function n(E) = ( E) (6.85)
as in Fig.6.14(a) and only states below the chemical potential are occupied, as expected. At low temperatures kB T only states in a region of width kB T with energy close to are thermally excited, as shown in Fig.6.14(b).
These properties of the Fermi-Dirac distribution are very useful if we use the fact that the derivative of a theta function is a Dirac delta function, that is, n = ( E) . E (6.86)
At nite temperatures the delta function is broadened by kB T . In order to explore the usefulness of the properties of the Fermi-Dirac distribution consider an integral of the distribution with an arbitrary function
115
(a)
n
K T
B
(b)
Figure 6.14: Fermi-Dirac distribution: (a) zero temperature; (b) nite temperature .
f (E)
+
I(, T ) =
+
dEf (E)n(E) dE
(6.87)
F (E) =
dE f (E ) .
(6.88)
Observe that the last line was obtained by integration by parts. We assume that the integrand vanishes in the E limit. Observe that the derivative of the Fermi-Dirac distribution at low temperatures is highly peaked around E = and assuming that F (E) is a smooth function close to this point we make a Taylor series expansion for F (E): F (E) = F () + F (n) () (E )n , n! n=1
(6.89)
where F (n) () is the nth derivative of the function at E = . We further observe that
+
dE
n E
= 1,
(6.90)
and that the Fermi-Dirac function is an even function of E . By direct substitution of (6.89) into (6.87)
116 we obtain
I(, T ) =
f (E)dE +
dE(E )2n
n E
(6.91)
I(, T ) =
f (E)dE +
n=1
fn ()(kB T )2n ,
(6.92)
where (n) is a Riemann zeta function. Equation (6.92) gives the nite temperature correction to the zero temperature result in powers of T . This is called the Sommerfeld expansion. We also notice that the corrections to the zero temperature result is given in even powers of the temperature. In particular, at low temperatures the rst correction will always be of order T 2 , thus, the rst term in (6.92) can be replaced by
EF
f (E)dE =
EF
f (E)dE +
EF
f (E)dE , (6.94)
f (E)dE + ( EF )f (EF ) .
EN (E)dE
0
(6.95)
N (E)dE + ( EF )N (EF ) +
2 N (E)(kB T )2 . 6
(6.96)
Notice that number of particles at nite T is the same as at zero temperature and therefore from (6.96) we get EF which when substituted in (6.95) gives E0 2 + N (0)(kB T )2 . E V 6 (6.98) 2 N (EF ) (kB T )2 , 6 N (0) (6.97)
117
(6.99)
which is independent of the dimensionality of the problem (that only modies the density of states which for the case of free electrons is given in (6.8)).
Eq. (6.99) could have been obtained without all this mathematical workout if we have used the physical meaning of the specic heat which is the number of excitations in the system. Obviously in the case of a system with a Fermi surface the number of excitations available is proportional to the kB T times the number of states per unit of energy which is simply N (0) which gives the correct order of magnitude of the specic heat. Since the average energy of the particle-holes is kB T the excitation energy will be N (0)(kB T )2 which misses the correct value by a factor 2 /6. Another way to calculate the specic heat is to use the analogy we made before between the particle-hole excitations and acoustic phonons. As we showed in (6.15) the particle-hole excitations have a dispersion relation of one-dimensional acoustic phonons since it depends only on the component of the momentum perpendicular to the Fermi surface. The specic heat of one dimensional phonons, according to (4.59), indeed behave like T . In order to show that this is more than a coincidence let us calculate the specic heat of these particle-hole excitations. First we observe that the density of states at the Fermi surface of a Fermi gas can be written as in (6.8) N (0) = 2 dd k (EF k ) . (2)d (6.100)
Observe that only states at the Fermi surface, that is, with k such EF = k contribute to the integral. Thus we make a simple change of variables k = kF + q and therefore k EF + q kF = EF + q|| vF (6.101)
(6.102)
where we have used (6.16). As expected only the component of q normal to the Fermi surface appears in the energy in agreement with (6.15). Therefore it is natural to rewrite q = q|| kF + q kF (6.103)
where q is tangent to the Fermi surface. Observe that the density of states at the Fermi energy can be written as N (0) = 2 = 2 dq|| dd1 q (q|| ) (2)d vF d1 q d 1 d v (2) F
(6.104)
involves only the component of q perpendicular to the Fermi surface. Now assume for the moment being that the particle-hole excitations have bosonic character. Then the partition function for these bosons is the
(6.105)
The specic heat is simply CV V = = 2 4 dq|| dd1 q (vF q|| )2 (2)d sinh2 (vF q|| /2)
N (0)vF 2 4 2 N (0)T , 3
dq||
= 2
where we have used (6.104). Observe that this is twice than the expected result (6.99). This happened because we have double counted the number of particle-hole excitations. The mapping of the particle-hole problem into bosons has an extra requirement that q|| 0, that is, we count only the excitations outside the Fermi surface. If we take this into account we get the correct result. The connection between the particle-hole excitations of an electron system goes beyond what has been discussed here. We can show that in dense Fermi systems there is an operator identity between fermions and bosons. This identity is called bosonization for obvious reasons.
1 V 1 V
(E k + B B) , (E k B B) , (6.107)
1 N (E + B B) , 2 1 N (E B B) . 2
(6.108)
dE n(E) (N (E + B B) N (E B B)) ,
(6.109)
119
(6.111)
N (E) = 2N
n=0
(E c (n + 1/2)) , h
(6.112)
where N gives the degeneracy of each level and the factor of 2 account for the two spin orientations. The free energy of the system is given in (6.79) F = 2N
h ln 1 + e( c (n+1/2)) , n=0
(6.113)
It is very hard to perform this sum. Since we are interested on the susceptibility of the electron gas we just have to evaluate the free energy at very small elds, that is, kB T c . In this case we rewrite the sum as h 2B F = S 0 where S = Lx Ly is the total area and xn = n = ( c )n . h kB T /( c ) h (6.115)
h ln 1 + e( c /2) exn n=0
(6.114)
Thus, when kB T c ( c 1) the separation between the allowed values of xn is very small and we h h can replace the sum by an integral. However, we are not interested in strictly zero magnetic eld and we have to keep the rst correction. This is done with the help of the so-called Euler-MacLaurin formula,
r
f (xn ) =
n=0
1 h
xr
dxf (x) +
x0
+
n=1
where xn = x0 + nh and Bn is a Bernoulli number. Application to this formula to (6.114) leads to F 2B = S 0 where z = e , (6.118)
0 h dx ln 1 + z e c x +
e2 B 2 z , 2z+1 24mc
(6.117)
120
is called the fugacity of the system. Observe that all the dependence of the free energy on the chemical potential can be obtained from the fugacity. Thus any derivative in respect to the chemical potential can be replaced by a derivative with respect to the fugacity, in particular, from (6.75) one has N = F F = z . z (6.119)
(6.120)
Li2 (x) =
x
dy
ln(1 y) xn = , y n2 n=1
(6.121)
is a polylogarithmic function of second degree. The total number of electrons is obtained from (6.75) = e2 B 2 m z N = 2 ln(1 + z) + 24mc2 (z + 1)2 , S h (6.122)
which is a transcendental equation which gives z as a function of . Let us rst consider the B = 0 case, then 1 + z0 = eEF , (6.123)
where we have used (6.4). Since we are interested in the low density limit it is clear that corrections to z0 are going to be very small. We therefore write z z0 1. Substituting this result in (6.120) we nd e2 B 2 F E0 + S 24mc2 (6.124)
where E0 is the ground state energy of the system in the absence of an external eld. Finally the magnetization density is given by (6.52), N (0)2 B e2 B M B = = S 12mc2 3 (6.125)
where N (0) = m/( 2 ) is the density of states of a two dimensional electron gas (see (6.8) with d = 2) h and B = he/(2mc). Therefore, the susceptibility is L = N (0)2 P B = 3 3 (6.126)
where we have used the result (6.37). Observe that the sign is negative implying that the magnetization is contrary to the direction of the eld in accord with Lenzs law of electromagnetism. This susceptibility was originally derived by Landau and carries his name. Observe therefore that the total susceptibility of an electron gas is T = P + L = 2 P . 3 (6.127)
6.4. PROBLEMS
121
6.4 Problems
1. Prove eq.(6.11) and calculate for Al (rs 2) and Cs (rs 5.5) their respective Fermi energies. Show that in 3 dimensions we have: EF = where the energy is measured in eV. 2. Prove (6.12). 3. Calculate the Fermi velocity in d dimensions and show that in 3 dimensions we can write: vF = 4.2 106 rs 50.1 2 rs
where the velocity is measured in m/s. What is the value of the Fermi velocity for Al and Cs? 4. Prove (6.22). 5. Prove eq.(6.36). 6. Calculate the exact expression for the magnetization of a electron gas for d = 1 and d = 2. 7. What happens with the magnetization of a free electron gas for B > Bc ? 8. Prove (6.41). 9. Show that the mean square deviation (y)2 is proportional to the cyclotron length squared, that is, 2 l0 . 10. Solve the problem of an electron in a magnetic eld in the symmetric gauge: A = B(yx + xy)/2. 11. Prove eq. (6.62). 12. Calculate the jump in the magnetization and + 1 for a two dimensional electron gas. 13. Prove (6.111).
122
Chapter 7
Figure 7.1: Motion of an electron in the presence of scattering centers. When the number of imperfections is small the electrons will move ballistically between impurities. This is called the weak disorder limit. If however the number of impurities is large then new effects can arise since there can be a destructive interference of the electron wavefunction due to disorder. We are going to briey discuss these two extreme limits.
dvk eE 1 = vk dt m
(7.1)
that is known as the Langevin equation and describes the motion of a particle in a viscous environment. Observe that can depend on the momentum k but as explained before we are going to consider momenta close to the Fermi surface where the momentum dependence of is negligible when the scattering is weak. Observe that equation (7.1) leads to an exponential decay of the velocity with time in a time scale given by . If we wait long enough we obtain the steady state condition, that is, the electron reaches a nal velocity vk = which allows us to dene the electron mobility, , as vk = E where = e . m (7.4) (7.3) e E m (7.2)
Notice that vk has nothing to do with the actual velocity (that is, the Fermi velocity) of the electron. This velocity is the net effect of the collisions. In between impurities indeed the electron moves with the Fermi velocity but because the direction of motion of the electron is changing all the time its velocity along the electric eld is smaller than its velocity in between collisions. The steady state current in the system is just J = e vk where is the average electronic density. Using (7.2) and (7.5) we can write J = E where = e2 m (7.7) (7.6) (7.5)
is the electric conductivity of the system. The resistivity, R , is dened as its inverse, namely, R = m 1 = 2 . e (7.8)
Since we are talking about electrons at the Fermi surface with a Fermi velocity it is clear that in a time period the electrons will move a distance such that = vF (7.9)
which is called the mean free path of the electrons. Equation (7.7) is known as the relaxation time approximation. At this point the main element of the theory, namely, the relaxation time is a phenomenological parameter. We can estimate its value from a microscopic model if we assume that the number of impurities is
125
very small so that single impurity scattering dominates the relaxation mechanism. Consider the scattering of an electron from a state |k to a state |k in the presence of an impurity potential V . In this case we just use Fermi Goldens rule tells us that the scattering rate is giving by Wk,k = 2Ni (k k )| k|V |k |2 h (7.10)
where Ni is the number of impurities. The total scattering rate is obtained by integrating (7.10) over all nal states |k . We have to remember, however, that not all the states contribute. In Fig.7.2 we show the actual situation in which the electric eld is applies in some arbitrary direction. Observe that only the component of the momentum in the direction out of the eld contributes to the relaxation. Forward scattering events in the direction of the eld cannot contribute since the electron retains its momentum in this direction. Thus, the important fact is not that the electron is scattered but that the amount of momentum along the electric eld is changed in the scattering process. If we take this effect into account we can write the expression for the relaxation time as 1 = Wk,k 1 cos k,k (7.11)
Figure 7.2: Scattering of an electron wave by an impurity in the presence of the eld. Equation (7.10) is valid only in the Born approximation. A better approximation can be obtained if we observe that in the Born approximation the scattering amplitude fk (k,k ) is given by fk (k,k ) In general the differential cross-section is given by () = |fk ()|2 which allows us to rewrite (7.10) in a more general way, that is, Wk,k = (2 )3 Ni h (k k )(k,k ) . (V m )2 (7.14) (7.13) m V 2 k|V |k . 2 h (7.12)
126
d sin()[1 cos()]()
(7.15)
where c = Ni /V is the impurity concentration. From the knowledge of the differential cross section of the impurity we can calculate the relaxation time from (7.15). We can get further insight into (7.15) if use the partial wave expansion of the wavefunction in the presence of the impurity. As we know from elementary quantum mechanics the asymptotic value of the wavefunction of the electron due a central potential at the origin can be written as
k (r, )
(7.16)
where l (k) is the phase shift of the electron wavefunction in the presence of the potential (obviously, l (k) = 0 if V = 0). From (7.16) one can obtain the scattering amplitude in terms of the phase shifts, fk () = 1 k
(7.17)
and the differential cross-section is given in (7.13). By direct substitution of (7.17) into (7.15) one nds 1 4cvF = 2 kF
l=1
(7.18)
It is now interesting to investigate the effect of the impurities in the spectrum of the system. Assume that an impurity is at the center of an sphere of radius R which is of the order of the size of the system. We impose the condition that the wavefunction vanishes at the surface, that is, k (R, ) = 0. Then, from (7.16), we nd kn,l = n
l (k)
l 2
(7.19)
where n is an integer. In the absence of the impurity (l (k) = 0) we easily see that kn,l = (n + l/2)/R and therefore, for xed l, the distance between two allowed values of k is /R. In the presence of the impurity this energy levels are shifted by l /R. Consider rst the case without impurities. Suppose we ll up all the states up to the Fermi momentum kF = (Nmax + l/2)/R where Nmax is the label of the last occupied state. In the presence of the impurity we will have kF = (Nmax l (kF )/ + l/2)/R. Since the density of the system is kept constant the Fermi momentum cannot change. Thus we must require that Nmax = Nmax + l (kF )/ which implies that the number of available states has changed. Taking into account that the degeneracy of each state is 2(2l + 1) (the factor of 2 comes from the spin component) the total number of new electrons required to ll up the levels to the same wave-vector kF is
Z=
l=0
2(2l + 1)
l (kF ) .
(7.20)
Observe that in the presence of the impurity we have an extra amount of charge Z in the system. So, in order for the system to be neutral we have to require that Z equals the amount of electrons given by the impurity to the system, that is, the valence difference between the impurity and the host metal. Of course an impurity
127
with Z = 0 cannot be distinguished from the metal itself. This is the so-called Friedel sum rule and reects the charge neutrality of the solid: the electric charge of the impurity must be neutralized by an excess of electrons in its vicinity. Of course, far away from the impurity the the system relaxes to the case without impurities where kF is given by the value of the average density. Observe that Friedel sum rule is very useful since it relates the phase shift to the valence of the impurity. Suppose for simplicity that only s-wave scattering is important (l = 0) and Z = 1 (notice that if Z > 1 then s-wave scattering alone cannot account for the phase shift). Then from (7.20) one has 0 = 2 (7.21)
which is the value of the phase shift at a resonance. That is, the impurity is resonant with the Fermi system. From (7.18) we have 1 4cvF = . 2 kF Thus under these conditions the resistivity is given by R = 4c e2 kF 1/3 c = 4 3 e2 (7.22)
(7.23)
where in the last line we used the free electron result (6.5). For typical electronic densities one has /c 4 cm.
which implies that the wavelength of the electron F = 2/kF has to be larger than the mean free path. This is nothing but the condition of ballistic motion. Condition (7.25) can also be written in a different form if we realize that the Fermi wave-vector, given in (6.5) is approximately given by /a where a is the lattice spacing. Therefore, we have the condition that a which implies that the mean distance between impurities has to be much larger than the lattice spacing. The question that comes to mind is: what happens when kF 1 or a? It is clear that in this case we have, in average, one impurity per unit cell, that is, c 1/a3 . Let us assume that the impurity potential
V (r) = V0
er/a . r/a
(7.26)
Using the simplest Born approximation (7.10) we nd that the differential cross section can be written as () 4(m )2 a6 V02 h 4 (7.27)
in the limit where a 0 (that is ka << 1). Substituting (7.27) into (7.18) and using (5.82), (7.9) and c = 1/a3 one nds 16a3 V02 (m )2 1 4 = = 4 a h V0 t
2
(7.28)
and observe that the condition of < a implies V0 t, that is, the strength of the disordered potential is of order of the kinetic energy. It is clear that in this limit it is wrong to do perturbation theory around the plane wave state! So, if the plane wave state is incorrect what is the nature of the ground state? In order to solve the problem let us consider the one-dimensional potential of Fig.7.3(a). As we have seen the solution for this problem is a Bloch wavefunction which in the atomic limit can be well described by (5.69) and is depicted on Fig.7.3(b). Suppose we now have a potential like the in Fig.7.4(a) where the heights of the potential wells uctuate around some mean value V and with variance V0 . What we are going to argue is that the wavefunction looks like Fig.7.4(b), that is, the wavefunction is localized around the site in the lattice with lowest energy. That is, the wavefunction, instead of be given by (5.69), looks like er/ (r) = N eikT A (r T) . (7.29)
where is the so-called localization length and gives the envelope of the function in Fig.7.4(b).
V(x) (a) E
(b) || 2
Figure 7.3: (a) Ordered potential; (b) Module square of the Bloch wavefunction. In order to understand how localization arises in this problem consider our familiar problem of the molecule with two atoms. We now have only two sites 1 and 2 and the state of the electron localized on site
129
Figure 7.4: (a) Disordered potential; (b) Module square of the wave function.
1 (2) is |1 (|2 ). We now generalize the Hamiltonian (1.43) in order to allow the two states to have different energies and the tunneling is still described by (1.45), that is, the complete Hamiltonian reads H = V |1 1| + (V + V0 )|2 2| + t (|1 2| + |2 1|) . (7.30)
which is our toy model for the disordered problem. Since V is just a shift in the energy of the problem we just set it to zero. This problem can be solved exactly as before by writing linear combinations of the states |1 and |2 |B = |1 + |2 |A = |2 |1 (7.31)
which are the analogue of the bonding and anti-bonding states. In this new basis the Hamiltonian is diagonalized and it looks like H = EA |A A| + EB |B B| . (7.32)
The coefcients , , EA and EB depend on the parameters t and V0 and are left for you to calculate. Let us consider the extreme cases where our physical intuition can work for us. The perfect lattice happens when the two potential wells are the same, that is, V0 = 0. If V0 << t we have 1/ 2 and EB EA 2t that is the case we discussed previously. In this case the two wavefunctions for the bonding and anti-bonding states have equal weight in the two wells. This is the case equivalent to the Bloch state where the electron, by tunneling between atoms, is extended over the crystal. Consider however the opposite case in which V0 t. In this case you can show that V0 1 2t (7.33)
which means that the probability of nding the particle in the site 1 is much larger than the probability of nding the particle in site 2, that is, the particle is trapped inside of the well. If we now turn to the innite crystal this argument suggests that there should be not only uctuations in the phase of the wavefunction
130
as we go from site to site but also uctuations in the amplitude! These uctuations become larger as the ratio V0 /t increases. If V0 /t is very large we expect the wavefunction of the electron on that particular site to be very little affected by the presence of the other sites in its neighborhood and therefore should decay exponentially away from the site as the wavefunction (7.29) implies. Thus, our expectation is that for a given electron energy there must be a minimum value of V0 /t for which the electron becomes localized This is the so-called Anderson localization transition. The critical value of V0 /t is a non-universal number which depends on the type of lattice, dimensionality and type of disorder in the material. In particular, it has been shown that in one dimension all the electronic states are localized. The fact that the localization depends on the energy of the electron is clear from Fig.7.4 since electrons with large energies do not care for the disordered potential but electrons with low energies will feel the bottom of the wells. Thus, for a given value of V0 /t above the critical value where localization happens, there is also a minimal energy EC , below which all the states are localized and above which the states are extended. This is the so-called mobility edge. If we make a plot of the density of states of a such material it will look like the one on Fig.7.5. Below the mobility edge the states are localized and above it they are extended.
N(E)
Delocalize Localized
Ec
It was originally proposed by Mott that the transition between metallic to insulating behavior due to disorder would occur when kF 1. The reason for this is that on a lattice the smallest distance between impurities is one lattice spacing a. Therefore, = a is the smallest value possible for the mean free path. Since kF 1/a we must conclude that kF 1 at the transition. This argument would imply that the smallest conductivity in a system, according to (7.34), would be given by min 2Sd (2)d d e2 d2 . k h F (7.35)
131
Thus, according to Mott the conductivity of a disordered system, as a function of the disorder strength of kF would look like as in Fig.7.6 with a discontinuous drop of the resistivity at the transition between metal and insulator.
min kF a kF l
Figure 7.6: Behavior of the conductivity at the metal insulator transition according to the Mott minimum. The prediction of the Mott minimum of conductivity was not fully conrmed experimentally and was also questioned theoretically because it does not take into account the statistical nature of the disordered state. It turns out that the metal insulator transition is a continuous transition in terms of the disorder strength and not discontinuous as Motts argument would make it. In order to understand this transition consider the difference between a localized and delocalized state. A localized state, having a characteristic size should be insensitive to boundary conditions in a system of size L > . On the other hand, in a metallic system, the wave-function is extended over the entire volume of the system and therefore it should be sensitive to the boundary conditions. Now imagine taking a system of volume V and breaking it into pieces of size Ld so that L . It does not make a lot of sense to talk about the conductivity of a piece of a material of size Ld < V since the conductivity is an extensive quantity. Nevertheless, we can talk about the dimensionless conductance of this piece that is dened as: g(L) = Ld2 . (7.36)
Notice that, according to (7.34) this quantity is proportional to (kF L)d2 . In a metallic system the conductivity in one piece of a material does not change from piece to piece and therefore the dependence of g(L) with L is essentially g(L) Ld2 1 in the limit of L . However, in the insulating case one expects g(L) eL/ 1 since the wavefunction is localized in when L , . These two limits are the asymptotic limits of the conductance. However, we are interested in the limit of the conductance close to the metal-insulator transition where g(L) 1. In order to understand the behavior of the conductance close to the metal-insulator transition one has to understand the behavior of the conductance for all values of g. This seems to be an impossible task since we need to know details about the disorder strength, distribution of disordered sites, etc. Nevertheless, we can make some progress if we make a few assumptions about the behavior of g. We rst notice that the logarithmic derivative of g with L must be weakly dependent on L since d ln(g)/d ln(L) (d 2) when g 1 and d ln(g)/d ln(L) ln(g) when g 1. The scaling theory of localization makes the assumption that the function (g) = d ln(g) d ln(L) (7.37)
is only dependent on g but not L. This assumption has strong consequences since the asymptotic behavior of g(L) imposes constraints in the behavior of (g). In fact, if one plots (g) as a function of ln(g) one gets Fig.7.7.
132
(g)
1 0 1 gc
ln(g)
Figure 7.7: Behavior of (g) according to scaling theory of localization. Notice that while (g) vanishes at g = gc in d = 3 it is always negative for d = 2 and d = 1. The meaning of this result becomes obvious when one looks carefully at what happens close to g = gc . Let us expand (g) to rst order close to gc : (g) 1 g gc gc (7.38)
where we have assumed that (g gc )/gc 1 and = (gc d(gc )/d ln(g))1 . Using (7.37) we can calculate g(L) explicitly: g(L) gc + (g0 gc )(L/L0 )1/ (7.39)
where g0 is the value of the conductance at the scale L = L0 . Let us consider rst the case of g0 > gc . In this case equation (7.39) predicts that g(L) should grow as L grows indicating a metallic state. However, g(L) cannot grow indenitely as indicated in (7.39) because we have assumed that g g0 gc and one has to stop (7.39) at some scale L = M where (g gc )/gc 1. It is easy to see that M (g0 ) A L0 [(g0 gc )/gc ] (7.40)
where A is a number of order of unit and indicating that (g) diverges at g = gc with a critical exponent . The physical meaning of is clear: for regions of size L < M the system behaves as an insulator while for L > M it looks like a conductor. At g = gc the system becomes an insulator since M becomes of the size of the system. For g0 < gc we rewrite (7.39) as: g(L) gc (1 (gc g0 )/gc (L/L0 )1/ ) (7.41)
indicating that g(L) decreases as L grows indicating insulating behavior. Notice that this expression would predict that g(L) would vanish at L = loc dened by: loc (g0 ) L0 . [(gc g0 )/gc ] (7.42)
However, by denition g(L) > 0 and the vanishing of g(loc ) is just a consequence of the approximation of expanding closing to g = gc . In fact one realizes that (7.41) can be rewritten as: g(L) gc e(L/loc )
1/
(7.43)
7.4. PROBLEMS
133
that has the same asymptotic behavior of (7.41). Notice that for L loc the conductance vanishes indicating that loc is the localization length of the system: for L > loc the system looks like an insulator and for L < loc the system looks like a metal. From the denition (7.36) one can nd the actual conductivity of the material: for g < gc the system is an insulator and = 0 while for g > gc the system is a metal and for L > M in d = 3 the conductivity can be written as: (g) g(M ) gc = B [(g gc )/gc ] M L0 (7.44)
where B is a number of order unit, indicating that the conductivity vanishes continuously at the metal insulator transition with an exponent as shown in Fig.7.8.
INSULATOR
METAL
kF l
Figure 7.8: Actual behavior of the conductivity at the metal insulator transition. Notice that there are various ways to cross the metal-insulator transition. One of them is to change the strength of the disorder and therefore to change . However, we can also cross the metal insulator transition by changing the electronic density, that is, changing kF . In the later case one has to change the Fermi energy across the mobility edge, namely, we can write, to rst approximation close to the transition that g(EF ) gc + dg(Ec ) (EF Ec ) dEF (7.45)
so that for EF > Ec the system is metallic (g > gc ) and for EF < Ec the system is an insulator (g < gc ). Finally we comment about the problem in dimensions smaller than 3. In those cases, as shown in Fig.7.7, we have (g) < 0 for all values of g indicating that (L) 0 as L grows. Thus, the scaling theory of localization predicts that for d < 3 the electronic system should be localized even in the presence of an innitesimal amount of disorder. This result has been challenged by recent experiments in two-dimensional electron systems conned in heterostructures.
7.4 Problems
1. Using (7.11) and (7.14) and assuming k = h2 k2 /(2m ) prove (7.15). 2. Prove that in the Born approximation the scattering amplitude for the potential (7.26) is given by: fk () = 2m V0 a3 / 2 h . 2 sin2 (/2) + 1 4(ka)
134
3. Diagonalize the Hamiltonian (7.30) exactly and calculate , , EA and EB . 4. Consider the case the problem of localization in two dimensions. In this case we nd that (g) = 0 in the limit of g 1. Corrections to this extreme limit can be written as power series in 1/g as: (g) a/g + O(1/g2 ) . (7.46)
(i) What is the sign of a? Why? (ii) Calculate the localization length loc in this case and show that it is an exponential function of the conductance.
Chapter 8
Semiconductors
As we have seen in the previous chapters we can classify clean non-interacting electron systems either as metals or insulators depending whether they have a partially lled or completely lled band, respectively. Metals do not have a gap in the charge spectrum and can conduct electricity very well, insulators, however, cannot conduct charge if the temperature is much smaller than the gap energy, Eg . When the thermal energy, KB T , at room temperature is of order of Eg then one can promote electrons from the valence band to the conduction band, making electrical conduction possible. Systems where Eg KB T at room temperature are called semiconductors. It is quite amazing that the electronic revolution of the 20th century was not driven by the development of metals but by the progress in the understanding of semiconductors. These are the basic elements for the production of transistors and other electronic elements in modern computers. It is therefore of interest to understand their physics. Here we are going only to consider the case where the conduction band and the valence band electronic energies can be written as:
c Ek = Ec + v Ek
h 2 k2 2mc h 2 k2 = Ev 2mv
(8.1)
where mc and mv represent the different curvatures of the conduction and valence band, respectively. Notice that in terms of our denition (5.88) of the effective mass, the valence band has negative mass while the conduction band has positive mass. The bands described in (8.1) are not generic because many times the top of the valence band is not at the same point as the bottom of the conduction band (these are called indirect gap semiconductors). Many times the bands are shifted by a nite momentum Q. However, there are some systems where the two edges are at the same point in momentum space (these are called direct gap semiconductors). Another simplication in (8.1) is that we are using innite parabolic bands. As we know this approximation is only good if we are considering states close to the edge of the bands. If the temperature becomes too high one starts to excite particles at higher energy states and the deviation from parabolicity becomes an important factor. Here, however, we are going to consider cases where kB T Eg = Ec Ev in which case we constrain ourselves to states close to the edges of the bands. For simplicity, from now on we are going to set Ev = 0 so that Eg = Ec . It is a very simple exercise to calculate the density of states of a semiconductor as a function of energy. Using the results of the previous sections we nd that for E Eg : Nc (E) = 1 2 2 2mc h 2 135
3/2
E Eg
(8.2)
136
CHAPTER 8. SEMICONDUCTORS
is the density of states for the conduction band and for E 0 we have Nv (E) = 1 2 2 2mv h 2
3/2
(8.3)
is the density of states for the valence band. Observe that the density of states is zero everywhere in the gap, 0 < E < Eg . When an electron with momentum ke is excited from the valence band to the conduction band it leaves a hole behind in the valence band. Since the total number of electrons do not change we can either describe the problem in terms of electrons or holes. The total momentum of the system before and after the transition is the same which implies that the momentum of the hole, kh , has to be ke . Moreover, in describing the system in terms of holes the energy of the holes is obtained by inverting the valence band so that Eh (kh ) = E(ke ) which implies that the mass of the hole is given by mh = mv . Because the properties of semiconductors depend strongly on the number of electrons and holes we will be always talking about the number of such carriers in the system. Let us calculate the number of electrons in the conduction band, nc , and holes in the valence band, pv , at a xed temperature T . It is clear that
nc =
Eg
(8.4)
and
0
pv =
(8.5)
is the Fermi-Dirac distribution function ( = 1/(kB T )). In what follows we are going to assume that the temperature is such that Eg kB T
kB T
(8.7)
so that the Fermi-Dirac distribution function (8.6) can be replaced by f (E) e(E) . (8.8)
Because the chemical potential depends on the amount of charge in the system these conditions can only be checked a posteriori. We are going to see that this is indeed the case of interest. Using (8.8) we can write (8.4) as nc 1 2 2 2mc h 2
3/2
dE
Eg
E Eg e(E) (8.9)
= Nc (T )e(Eg )
137
(8.10)
that should not be confused with the density of states (8.2). By the same token we can show that pv = Pv (T )e where Pv (T ) = 1 2 2 2mv kB T h 2
3/2
(8.11)
(8.12)
Equations (8.9) and (8.11) give the density of electrons at the conduction band and holes in the valence band. Observe, however, that these expressions still depend on the chemical potential which is undened. However, if we multiply (8.9) and (8.11) we nd that nc pv = Nc Pv eEg (8.13)
which is independent on the chemical potential. This expression is very important in the theory of semiconductors.
This result implies that at nite temperatures the number of electrons in the conduction band or holes in the valence band is exponentially small with temperature. Using (8.14) and (8.15) in (8.9) we can immediately obtain the value of the chemical potential; = 3kB T Eg + ln 2 4 mv mc (8.16)
which shows that at low temperatures the chemical potential is essentially in the center of the energy gap. Observe that in this case the conditions (8.7) are obeyed as long as kB T Eg /2 which is obeyed in most semiconductors where the energy gap is of order of a few electron volts. (8.17)
138
CHAPTER 8. SEMICONDUCTORS
In semiconductors we can nd 20 100. The consequence of (8.18) is that we can replace e2 by e2 /. Putting together the effects of the electron-ion interaction and the electron-electron interactions we see that the binding energy of an electron by an impurity is of order: Eb = mc 1 me4 m 2 2 h (8.19)
which can be 104 times smaller than the biding energy of an electron in the Hydrogen atom! In fact for a host of Ga atoms (valence +4) with As impurities (valence +5) it is found that the binding energy is of order of 0.013 eV. For this reason, the radius of motion of the electrons is also increased from a few Bohr radius, a0 , to r0 = m a0 . m (8.20)
Thus, the conclusion is that in a semiconductor the impurity levels are weakly bound to the impurities and can be easily ionized. In other words, at T = 0 the impurity level is occupied by a single electron which can be promoted to the conduction band at nite temperatures. Because these are bound states of electrons they are localized just below the bottom of the conduction band by an amount given by Eb . We call Eg = Ed +Eb
139
Analogously to the electron doped case, we can also dope holes in a semiconductor. We just have to choose an impurity atom that has a valence 1 relative to the host. In this case we are taking one proton and one electron from the host. The situation here is one of a localized negative charge with a hole bounded to it. For the same reasons described above this will lead to a bound state of holes above the top of the valence band. These are called acceptor states and the semiconductor is said to be of p-type. Observe that the ionization of a acceptor state is equivalent to a hole going to the valence band or an electron moving from the valence band to the impurity level. At T = 0 the acceptor state has one hole but at nite temperatures such that kB T = Ea = Eb (where mc is replaced by mv in (8.19)) an electron can be promoted to the acceptor state while a hole goes to the valence band. Because the donor and acceptor states are such that |Eb |, Ea Eg one can easily dope the conduction or the valence bands with external charges. In this case the nc and pv as dened previously will change by an amount n, that, is: n = nc pv (8.21)
which is equivalent to a change in the chemical potential of the system. Independent of this change the condition (8.13) is still valid. Using the denition of ni given in (8.15) we can rewrite (8.13) as n c pv = n 2 . i Solving (8.21) and (8.22) for nc and pv we nd nc = pv = n 2 n 2
2
(8.22)
+ n2 + i
2
n 2 n 2 (8.23)
+ n2 i
which gives the number of electrons and holes as functions of the intrinsic density and the extrinsic density of impurities. When n > 0 there are more donors than acceptors and the system is n-type. If n < 0 we have the opposite situation and the system is p-type. In order to calculate the value of n we need to know how many electrons (holes) are ionized and go to the conduction (valence) band and how many remain bounded to the impurity atoms. Let us call the density of donors Nd , the density of acceptors Na , density of electrons bounded in the donor states nd and the density of holes bounded in acceptor states pa . From the fact that the total number of electrons and holes has to be conserved it is easy to see that: n = nc pv = Nd Na (nd pa ) . Since Nd and Na are xed for a given semiconductor we are left with the calculation of nd and na . Let us rstly consider the problem of the donors. The occupation Nj of the donor level can be N0 = 0 (no electrons), N1 = 1 (spin up or down) or N2 = 2 (one electron with spin up and another with spin down). The doubly occupied state requires the donor state to be charged relative to the host and therefore is energetically not stable. We therefore disregard double occupancy. The donor state with zero electrons makes no contribution to the energy and therefore E0 = 0. The donor state with one electron has energy (8.24)
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CHAPTER 8. SEMICONDUCTORS
E1 = Ed . Thus, in thermal equilibrium the average number of electrons in donor sites will be n d = Nd = Nd =
j
Nj e(Ej Nj )
(Ej Nj ) je 2e(Ed )
1 + 2e(Ed ) Nd . (Ed ) e +1 2
(8.25)
Observe that because Ed Eg we expect that for kB T Eg Ed we must have nd Nd , that is, almost all the donor levels are ionized the electrons are in the conduction band.
In the case of acceptors the situation changes. The occupation Nj can be 0 holes (or two electrons Nj = 2), 1 hole (or one electron with either spin up or down, Nj = 1), or 2 holes (that is, zero electrons Nj = 0). The state with two holes is again unstable because it is positively charged relative to the host. The energy of the state with one hole is zero, E1 = 0, while the state with no holes has energy Ea (this is the energy cost to take one electron from the valence band and put in the acceptor level, in this case, the hole that was bound in the acceptor level goes into the valence band). In this case, the number of electrons in the acceptor level can be calculated as before: <n> = = 2e(Ea 2) + 2e e(Ea 2) + 2e e(Ea ) + 1 . e(Ea ) +1 2
(8.26)
Since the maximum occupation of the acceptor level is 2, the average number of holes in an acceptor level is: pa = Na (2 < n >) = Na e(Ea ) 2 +1 . (8.27)
Since we expect Ea kB T Eg the acceptor levels to be fully ionized at room temperature. In this case the holes will be all essentially in the valence band.
at room temperature. Substitution of (8.28) in (8.23) allows us to give a better denition to intrinsic and extrinsic semiconductor. The intrinsic semiconductor is the one in which ni |Nd Na | and therefore nc ni + pv Nd N a 2 Nd N a ni 2
(8.29)
and the number of electrons in the conduction band and holes in the valence band only change by a small
141
(8.30)
(8.31)
for Nd < Na . Thus, in an extrinsic semiconductor there is large difference between the number of charge carriers. It is exactly the exibility in controlling the number of charge carriers that makes semiconductors so attractive for making devices.
Ea (x) = Ea e(x)
(8.32)
by the same token the edges of the conduction band, Ec , and the valence band, Ev , also change with (x) so that the chemical potential is the same in both sides of the junction: Ec (x) = Eg e(x)
Ev (x) = e(x) .
(8.33)
Because of this change in the energies of electrons and holes, their densities will also change. In fact, we can write: nc (x) = Nc e(Eg e(x)) pv (x) = Pv e(+e(x)) . (8.34)
CHAPTER 8. SEMICONDUCTORS
nc (x +) = Nd = Nc e(Eg e(+)) pv (x ) = Na = Pv e(+e()) and if we multiply the two equations we get: Na Nd = Nc Pv e(Eg e) = (+) () which can also be written as: e = Eg + kB T ln Nd Na Nc Pv
(8.35)
(8.36)
(8.37)
which gives the total drop of the potential across the junction. Observe that because usually Eg kB T the drop is essentially given by the gap in the semiconductor. Notice also that the actual values of () do not have especial physical signicance because we can x one of them to be zero. Only the difference of potential has physical meaning. Using (8.35) we can rewrite (8.34) as nc (x) = Nd ee((+)(x)) pv (x) = Na ee((x)()) . (8.38)
The last equation gives the total drop in the potential across the junction but does not tell us how does the potential changes as a function of x. In order to know that we have to remember that the potential obeys the Poisson equation: 2 = 4 d2 = (x) dx2 (8.39)
where is the dielectric constant of the semiconductor and (x) is the local charge density in the system that can be written as = en where n is given in (8.24): (x) = e(Nd (x) Na (x) nc (x) + pv (x)) (8.40)
where Nd (x) = Nd (x) and Na (x) = Na (x) with (x) = 1 (1) if x > 0 (x < 0). If we use (8.40) and (8.38) into (8.39) it is clear that we will nd a highly non-linear problem to solve. Instead we are going to simplify the problem and assume the that (x) varies only considerably in a region of size dp < x < dn across the junction (dp and dn have to be calculated), that is, for x < dp we have (x) = () , and for x > dn we have (x) = (+) . (8.42) (8.41)
In this case, using (8.38) it is clear that for x > dn we have nc (x) = Nd and for x < dp we have pv (x) = Na . In this case we can break the (8.39) into different parts: for x > dn or x < dp we have d2 =0 dx2 (8.43)
143
(8.44)
(8.45)
We have to solve these equations using the boundary conditions (8.41) and (8.42) and impose that the potential and its derivative are continuous across the interfaces at x = dp , x = 0 and x = dn . A simple calculation shows that for dp < x < 0 we have (x) = () + and for 0 < x < dn we have (x) = (+) where dn = dp = Na /Nd Na + Nd 2e Nd /Na . Na + Nd 2e (8.48) 2eNd (x dn )2 (8.47) 2eNa (x + dp )2 (8.46)
These equations provide the full prole of the potential generated by the junction between two semiconductors. Suppose that an external potential V is applied to the junction. It is clear that this case the difference of potential can be either increased or decreased and that in general we must have = 0 V (8.49)
where 0 is the potential difference in the absence of the applied potential and is given in (8.37). Because of this new potential we see that the regions of variation of potential given by dp and dn change to: dn,p (V ) = dn,p (0) 1 V 0 (8.50)
where n,p is given in (8.48). Thus, by applying an external potential one can change the region where the electronic density varies because of the junction. This effect can be used to make a device called a rectier. In the absence of an external voltage V the number of electrons (or holes) crossing the junction in one direction is the same as the number of electrons crossing the junction in the opposite directions since the system is in equilibrium. Thus, if a hole current owing from the p-type to the n-type is called Ih,0 , there is an equal but opposite current from the n-type to the p-type. The same is true for the equilibrium electron current Ie,0 . When a voltage V is applied to the junction we see that the total potential across the junction is modied according to (8.49). Since e0 is the barrier height between the two sides of the junction when V = 0, it becomes clear that when a potential V < 0 is applied to the n-type side of the barrier, the barrier
144
CHAPTER 8. SEMICONDUCTORS
height for holes is decreased relative to the other side. In this case, holes coming from p-type side of the junction see a smaller potential barrier while the holes on n-type side of the junction see essentially the same potential as before. In a tunneling junction the current increases exponentially with the barrier height since it depends on the tail of the wave-function below the barrier. Thus, while the current for holes going from the n-type to the p-type is I0 , the current for holes going from p-type to n-type is I0 eeV and thus, the net hole current in the system is: If = Ih,0 (eeV 1) . (8.51)
This is called the forward current. For electrons the situation is the same. When a potential V > 0 is applied to the n-type side of the junction the situation is reversed: holes moving from the p-type to the n-type side of the junction see a larger potential and its current is decrease do Ih,0 eeV while holes moving from the n-type toward the p-type side of the junction have essentially the same current Ih,0 . In this case the net current is simply Ir = Ih,0 (1 eeV ) . (8.52)
This is called the reverse current. Thus, at zero temperature ( ) one sees If while Ir = Ih,0 . This implies, that for positive voltages the current through the junction is very high while for negative voltages the current is essentially the equilibrium current. Because of this highly non-linear behavior of the current in regards to the sign of the voltage this device is called a rectier.
8.4 Problems
1. Calculate Nc (T ) in (8.9) explicitly. 2. Show (8.29), (8.30) and (8.31). 3. Solve (8.39) explicitly and show that (8.46), (8.47) and (8.48) are indeed correct. 4. Consider a potential barrier between two metals 1 and 2. The potential prole of the barrier can be written as V (x) = (x)(a x) V1 + (V2 V1 ) x a (8.53)
where a is the thickness of the barrier and V1 > V2 . In the absence of any applied external potential the electric current owing from left to right, I12 = I0 , is equal but opposite to the current owing from right to left (I21 = I0 ) because the two sides of the junction are in equilibrium, that is, I12 + I21 = 0. (i) Make a plot of the potential as a function of x. (ii) If a potential V > 0 is applied to the left side of the junction, what is the net current owing in the system, that is, what is the value of I = I12 + I21 ? (iii) If a negative potential V < 0 is applied to the left side of the junction what is the net current in the system? (iv) Make a schematic plot of the current versus the voltage V at a nite temperature T . Can you use this device as a rectier?
Chapter 9
Electron-phonon interactions
So far we have studied the problem of non-interacting systems such as free phonons and free electrons. In this chapter we are going to start the study of interacting systems. It is very useful, however, to introduce a new language for the problem which is usually called "second quantization and allows us to work directly with operators that act on states with well dened occupation number. In the next two sections we are going to discuss this new representation for bosons and fermions, respectively. We are going to see that this new representation makes our life much easier when calculating physical quantities of experimental relevance.
[bi , bj ] = 0 , and when applied to a state with a well dened occupation, |ni , give b |ni i that implies b bi |ni i = ni |ni . bi |ni ni |ni 1 , = ni + 1|ni + 1 , =
(9.2)
(9.3)
For obvious reasons Ni = b bi is called the number operator. A many-body state of bosons can be described i in terms of the occupation of each state |i with i = 0, ..., as (b )n0 (b )n1 (b )n2 1 2 0 ...|0 |n0 , n1 , n2 , ... = n0 ! n1 ! n2 ! (9.4)
where |0 is the vacuum state with no bosons. The Hilbert space of all states which can be represented in this way is called the Fock space. A simple example is a problem of free bosons inside of a box of volume Ld (you can convince yourself 145
146
that this is actually the case of phonons). In this case the eigenstates of the problem are plane waves p (r) = 1 ipr e , Ld/2 (9.5)
with = x, y, z. Thus the many-body state can be constructed from operators bp in the way described before providing the occupation np to each state |p . We can invert this picture to real space by dening the operator (r) = 1 Ld/2
p
eipr bp ,
(9.7)
which annihilates bosons in a superposition of momentum states with amplitude eipr / Ld that is just the amplitude of having a particle annihilated at position r. Since a superposition of plane waves is an envelope function in real space (r) actually destroys a particle at position r. Similarly we can dene the creation operator of a particle at position r as (r). Another way to understand the physical meaning of (r) comes from usual space representation of a state in Dirac notation, namely, |r . In this representation we dene the operation |r = (r)|0 , (9.8)
that is, the operator (r) creates the state |r . This allows us to change representations using closure relations. By denition the state |i is created by acting with b on the vacuum i |i = b |0 , i where |i forms a complete set, that is,
i
(9.9)
|i i| = 1 .
(9.10)
=
i
i|r b . i
(9.12)
Observe now that i|r is nothing but the complex conjugate of the projection of the state |i into the space
147
(r)b . i i
(9.13)
This result is the generalization of (9.7) to any quantum state |i . Thus, we can always expand an operator in a complete basis of single-particle states. Observe that the wavefunction is a coefcient of the expansion. For this reason this process is called second quantization. The commutation rules of these new operators dened in real space can be obtained immediately from (9.1), (r), (r ) (r), (r ) = 1 Ld eipr+ip r bp , b = (r r ) , p (9.14)
p,p
= 0.
Thus, we can represent the many-body state of N bosons in real space as 1 |r1 , r2 , ..., rN = (rN )... (r2 ) (r1 )|0 . N! Observe that because of the commutation rule (9.14) we have the following identity |r1 , r2 , ..., rN = |r2 , r1 , ..., rN , which implies that if we exchange two bosons the wavefunction does not change. Furthermore, (r)|r1 , r2 , ..., rN = N + 1|r1 , r2 , ..., rN , r , (9.16) (9.15)
(9.17)
so that adding a particle produces the state with the correct symmetry. We further notice that for a single particle one can recover the usual one-particle picture quite easily. Since (r) creates a particle at position r the wavefunction of the particle can be obtained by (look at (9.8)): | (r)|0 0|(r)| where (r) is the usual wavefunction of the particle. = (r) , = (r) , (9.18)
148 way,
c|0 c|1 c |0
= 0, = |0 , = 0. = |1 ,
c |1
(9.19)
The condition c |1 = 0 is the Pauli principle, that is, we cannot put two electrons in the same quantum state. Moreover, from (9.19) we can prove that (cc + c c)|0 (cc + c c)|1 (c ) |0 = c |1
2
= |0 , = 0,
c |1 = c|0
= |1 , = 0. (9.20)
Since the states |0 and |1 generate the whole Hilbert space of the problem we can immediately write cc + c c = {c, c } = 1 , {c, c} = {c , c } = 0 , (9.21)
which are valid for all states of the problem. Observe that instead of commuting with each other we say that the fermions anti-commute. As for the case of the bosons we can dene a number operator N = c c , which has the required properties N |0 = c c|0 = 0 , = c c|1 = c |0 = |1 . (9.23) (9.22)
N |1
Thus we can dene the state by its occupation, |N , as in the case of bosons. Let us now consider the situation where we have two states with occupation N0 and N1 and the states can be labeled by their occupation, that is, |N0 , N1 . In this case we have four different states in the Hilbert space, namely, |0, 0 , |1, 0 , |0, 1 and |1, 1 . We now dene the creation and annihilation operators exactly as before, say c0 and c1 . In analogy with the single level case we have: c0 |0, 0 = 0, = |0, 0 , = 0, = |1, 0 , = 0, = |0, 0 , = 0. = |0, 1 , (9.24)
c0 |1, 0
c |0, 0 0 c |1, 0 0
c1 |0, 0
c1 |0, 1
c |0, 0 1 c |0, 1 1
Observe that we have not dened how the operators behave when we apply them to a state which is already
149
occupied, say, c0 |1, 1 . The reason is that we need to build into the operator language the correct symmetry of the problem, that is, the Pauli principle: the wavefunction has to be anti-symmetric when we exchange two electrons. Suppose we want to exchange two particles in the state |1, 1 . We have to perform the following operations: (i) we destroy, say, the particle in the state 1, c1 |1, 1 = |1, 0 , (9.25)
which denes the way c1 acts on this state; (ii) we destroy a particle in the state 0 following (9.24): c0 |1, 0 = |0, 0 ; (iii) we create a particle in state 1 again using (9.24): c |0, 0 = |0, 1 ; (iv) and nally put a particle 1 on state 0, that is, c |0, 1 . This series of operations can be written as 0 c c c0 c1 |1, 1 = c |0, 1 , 0 1 0 (9.26)
which by Paulis principle must have the opposite sign of the state |1, 1 . Observe that this is possible if we impose the condition c c0 = c0 c , 1 1 and use the number operators N0 = c c0 and N1 = c c1 as 0 1 N0 N1 |1, 1 = |1, 1 . Thus we have, c |0, 1 = |1, 1 . 0 By the same token we can show that c |1, 1 0 = c0 |0, 1 = 0 , = |1, 1 , = c1 |1, 0 = 0 , = |0, 1 . (9.29) (9.28) (9.27)
c |1, 0 1 c |1, 1 1
c0 |1, 1
(9.30)
From these considerations we see that the correct way to express the Pauli principle with creation and annihilation operators for different states is by imposing anti-commutation relations {ci , c } = i,j , j
{ci , cj } = {c , c } = 0 . i j
(9.32)
It is clear from the argument given above that the procedure can be extended to as many states as we wish. In this case the many-body state represented by electron occupation n0 , n1 , n2 , ... with ni = 0, 1 can be written as |n0 , n1 , n2 , ... = (c )n0 (c )n1 (c )n2 ...|0 , 0 1 2 (9.33)
150
with the anti-commutation rules given in (9.32). Observe that the total number of particles in this state is well dened and it is an eigenstate of the operator
N=
i=0
c ci . i
(9.34)
1 Ld/2
p
eipr cp, .
(9.35)
= (r r ), , = 0, (9.36)
(9.37)
in complete analogy with the boson many-body state. Moreover, from the anti-commutation rules we obtain immediately |r1 , 1 ; r2 , 2 ; ...; rN , N = |r2 , 2 ; r1 , 1 , ..., rN , N , (9.38)
in accordance with Paulis principle. We can represent any operator in terms of creation and annihilation operators. Consider, for instance, number operator given in (9.34). Using (9.35) we can write cp, = and from (9.34) one gets, N =
p,
1 Ld/2
dd reipr (r) ,
(9.39)
c cp, , p,
dd r (r) (r) ,
(9.40)
(r) =
i=1
(r ri ) =
(r) (r) ,
(9.41)
since N =
dd r(r). The kinetic energy, K, of the electrons can also be obtained in the new operator
151
language if we recall for each momentum p the kinetic energy is simply 2 p2 /(2m) and therefore one just h has to count how many electrons with momentum p there are, that is, K=
p,
h 2p c cp, . 2m p,
(9.42)
Using the identity (9.39) one can rewrite the kinetic energy as K=
dd r (r)
h 2 2 2m
(r) .
(9.43)
Other operators like the current operator, J(r) = and the spin operator at position r S(r) =
, (r)S, (r) ,
(9.44)
(9.45)
where S = h /2 with = x, y, z are written in terms of Pauli matrices , can be easily written in this new language.
Observe that the tunneling term makes the Hamiltonian non-diagonal. In order to diagonalize this Hamiltonian we dene, in analogy with (1.55), the anti-bonding and bonding operators, cA = cB = 1 (c1 c2 ) , 2 1 (c1 + c2 ) , 2
(9.47)
152
which obey the anti-commutation rules. In terms of this new operators the Hamiltonian (9.46) becomes H = (E0 + t)c cA + (E0 t)c cB A B (9.48)
as expected. Observe that the state of the system is now given by |nA , nB and the ground state is simply |0, 1 in this new basis. The molecule problem just described can be easily generalized to the case of the solid. Here we assume that the electrons to be in the tight binding approximation and only one atomic orbital A is involved. In complete analogy to the physics discussed at the end of Chapter 5 we only allow the electrons to hop between its nearest neighbor atoms. Let Ri be the atom site in a static lattice and ci, the electron operator that destroys an electron in a certain orbital at site i with spin . The Hamiltonian of the system as H = EA
i,
c ci, tA i,
c cj, + c ci, i, j,
<i,j>,
(9.49)
where < i, j > means that sites i and j must be nearest neighbor sites. The problem imposed by (9.49) can be easily solved by Fourier transforming the electron operator 1 cp, = N eipRj cj,
j
(9.50)
observe that by the translation symmetry the momentum p is dened only in the rst Brillouin zone. For simplicity we will assume a cubic lattice. You can easily show that in this case the Hamiltonian (9.49) can be written in terms of the operators in (9.50) as H=
p,
Ek c ck, k,
(9.51)
(9.52)
where re,n is the position of the nth electron and ri,m the position of the mth ion. As previously we are going to assume that the departures of the ions from their equilibrium positions are very small so, as in Chapter 4, we write ri,m = Rm + um , and expand the interaction (9.52) to leading order in the deviations u. Observe that to zeroth order in u we have only n,m Vei (re,n Ri,m ) which is the potential of the static periodic lattice which can be readily incorporated into the electron part He which in this case is given by (9.51). The uctuation part of interest is, Hep um Vei (re,n Ri,m ) . (9.53)
n,m
153
1 N
(9.54)
i N
n,m,p
(9.55)
(9.56)
where
N
(p) =
dd r
n=1
(r re,n )eipr =
dd r(r)eipr ,
(9.57)
is the Fourier transform of the electronic density operator. Moreover, from (4.43), one has eipRm um =
m k
ek
m
ei(kp)Rm
h ak + a k , 2M N k (9.58)
=
G
ep+G
N h ap+G + a pG . 2M p+G
Using (9.56) and shifting the sum over the momentum we rewrite the electron-phonon interaction as Hep where V (p) = i h Vei (p)p ep 2s M (p) (9.60) 1 Ld/2
p,G
V (p + G)(p + G) ap + a p
(9.59)
and s = N/Ld is the density of the lattice. Using (9.41) and (9.35) we have (r) =
(r) (r) =
1 Ld
ei(kp)r c cp, , k,
p,k
1 Ld
eiqr
q p
c p+q, cp, ,
(9.61)
where in the last line we have change variables to q = k p. Comparing (9.61) with (9.57) one has (q) =
p,
c p+q, cp, .
(9.62)
154
Thus the Hamiltonian, written in terms of creation and annihilation operators for electrons and phonons, becomes Hep 1 Ld/2
p,G,k, V (p + G)c k+p+G, ck, ap + ap .
(9.63)
Let us consider rst the case when Vei = 0 which is well-known: the ground state of the problem has the electrons in a Fermi sea with ground state energy, E0 and Fermi momentum kF which depends on the density as discussed in Chapter 5, and there are no phonons (since we are talking about zero temperature). Let us label this state by |F S . Observe that the perturbation has actually two different types of operators (consider only the case where G = 0), namely, c k+p, ck, ap and ck+p, ck, ap . In the rst case a phonon with momentum p is destroyed, an electron state with momentum k is destroyed and an electron state with momentum k + p is created. There is a simple way to represent this process graphically as in Fig.9.1 (a). Let us represent the momentum of the electron by a continuous line and the momentum of the phonon by a dashed line. Physically the process is one in which an electron absorbs a phonon and it is scattered to a new momentum state. In the second process an electron of momentum k emits a phonon of momentum p and is scattered to a new momentum state k + p as shown in Fig.9.1 (b). The diagrams which are shown in Fig.9.1 are known as Feynman diagrams and they are very useful since they allow a clear representation of the physical process of scattering. Furthermore, they allow us to rewrite perturbation theory in very simple terms.
p k+p (a) k
k+p k (b) p
Figure 9.1: Scattering process: (a) phonon absorption; (b) phonon creation.
155
(9.64)
(9.65)
(9.66)
At long wavelengths (G = 0) and for longitudinal phonons both the electric eld and the polarization eld are parallel to p and we must have from (9.66) that Ep = 4Pp . (9.67)
Observe now that the electric eld will create a potential due to the electric dipole term which is simply given by E(r) = (r) = i peipr p .
p
(9.68)
(r) is the potential felt by the electrons due to the phonons. Comparing (9.67) with (9.68) we obtain ipp = 4Pp , that allows us to compute the potential once the polarization is known. Since the displacement of the ions create a dipole eld one must also have Pp = eup = eep qp . (9.70) (9.69)
where is a real constant and qp is the phonon displacement given in (4.24). Using ep = ip/|p| (since e = ep ), together with (9.69) and (9.70) we nd: p p = 4e qp , |p| (9.71)
4e ipr e |p|
h ap + a p . 2s Ld 0
(9.72)
e(re,n ) = Ld/2
dd r(r)e(r) , (9.73)
1 (p) ap + a p , |p|
156 where
= 4e2
h 2M s 0
(9.74)
is the electron-phonon coupling constant. Observe that this result could be obtained from (9.60) with Vep (p) 1/p2 which is the case of the Coulomb potential in three dimensions. Substituting (9.62) into (9.73) and rewriting the whole Hamiltonian of the electron phonon problem we have H =
k,
Ek c ck, + k,
p
h 0 a ap p (9.75)
Ld/2
p,k,
which has the form described in (9.63). Let us consider the effect of the phonons on the electron ground state in perturbation theory, that is, we write H = H0 + Hep where H0 is the free electron-phonon problem and treat Hep as a perturbation. The ground state of H0 , as we said before, is a lled Fermi sea and no phonons are present. In rst order we have F S|Hep |F S which vanishes since the perturbation does not conserve the number of phonons and therefore cannot have a nite expectation value. We have to rely on second order perturbation theory. For each value of k the shift in the energy is given by E =
n=0
| n|Hep |0 |2 , 0 0 E0 En
(9.76)
0 where H0 |n = En |n . Since Hep has one creation and one annihilation operator the excited states that participate in the sum (9.76) must have at least one excited phonon. Moreover, since ap |0 = 0 because the ground state has no phonons we are left with a |0 = | p . Moreover, by applying the operator p ck+p, ck, to the ground state of the electron system we create a particle-hole pair exactly as in Fig.6.3. Thus we have to ensure that |k| < kF and |p + k| > kF in order to get a nite result. Observe that the 0 energy of such excited state is En = E0 Ek + Ek+p + h0 . Thus, for each value of k one can write,
Ek = 2
2 Ld
(9.77)
157
(9.78)
2m0 / . Since at low energies the momentum of the electron is small we perform an h where P0 = expansion in terms of k/P0 . We rewrite Ek
1 + 4m2 1 x2 8 du dx + (k/P0 )2 u2 x2 + 1 3 (1 + x2 )3 (2)2 2 P0 0 h 1 4m2 + (k/P0 )2 . = 2 2P 6 (2) h 0
(9.79)
Observe, therefore, that the total electron energy to second order in perturbation theory can be written as Ek E0 + where E0 = m2 2 2 P0 h 2mE0 3 2 P0 h 2 . (9.81) h 2 k2 2m (9.80)
m = m 1 +
So we see that up to second order in perturbation theory the electron-phonon interaction only changes the mass of the electron and replaces it by an effective mass m . This is a result of the dragging of the lattice due to the electron motion.
158
Another elegant way to re-express this part of the Hamiltonian is to use (9.41), (4.85) and (4.43) and write D Hep ns where K(r) = i Ld |p|eipr . (9.84) dd r
dd r K(r r ) (r) (r)(r ) ,
(9.83)
K(x) = i
dp ipx d pe = 2 dx
dp ipx e 2 (9.85)
d(x) . dx
Substituting the kernel into (9.83) and doing integration by parts one nds, Hep D
dx (x) (x)
(x) . x
(9.86)
Observe that in this new formulation the electron-phonon problem is described by the Hamiltonian H =
dx (x)
h 2 2 2m x2
(x) +
dx
c2 s 2 + s 2s 2
+ D
dx (x) (x)
(x) . x
(9.87)
Let us study the equations of motion for this Hamiltonian in the Heisenberg representation. You can easily show that the equations of motion are i h h 2 2 (x, t) (x, t) + t 2m x2 2 (x, t) 2 (x, t) c2 s 2 t x2 = D (x) = D s (x) x (x, t) (x, t) . x
(9.88)
Now observe that although we are dealing with equations of motion for operators we can project these equations into single particle states (since we are dealing with the problem of a single electron) as in (9.18). In this case instead of the adjoint of operators we are going to have the complex conjugate of the functions. Observe that the equation of motion for the phonon eld has the form of a relativistic wave equation with a light velocity cs . As we know from classical mechanics this equation has a traveling wave solution, that is, can be written as (x, t) = (x cs t) . (9.89)
159
In order to explore this type of solution we change variables to = x cs t. Moreover, we are interested in the stationary states of the electron and therefore we look for a solution of the type:
(x, t) = 0 (x cs t)e h (mvxE0 t) , i
(9.90)
where 0 is assumed to be real and v and E0 are the velocity and energy of the electron. In terms of the new variables (9.88) becomes mv 2 d h 2 d2 0 + D 0 () = E0 0 () , 2 2m d d 2 D d = 2 () , 2 v2 ) 0 d s (cs
(9.91)
where we have integrated the second equation in (9.88) once. Observe that the rst of these equations is very similar to the Schrdinger equation for a particle moving under the inuence of a potential given by V =D d , d (9.92)
but the second equation allows to rewrite this potential as V = D2 2 , s (c2 v 2 ) 0 s (9.93)
which shows that the potential itself depends on the wave-function. We say that the potential is determined in a self-consistent way. Observe that this potential is attractive if v < cs and repulsive if v > cs . Substituting the second equation of (9.91) into the rst one nds D2 h 2 d2 0 3 () = 2m d2 s (c2 v 2 ) 0 s E0 mv 2 2 0 () (9.94)
which is the so-called non-linear Schrdinger equation which appears in many different elds of physics. You can show that the solution for this equation reads g g (9.95) sech 0 () = 2 2 where g= and E0 = mv 2 2 g2 h . 2 8m (9.97) m 2 v2 ) s (cs D h cs /a
2
(9.96)
Observe that this solution exists only for v < cs and is unstable otherwise. In this case the potential felt by the electron is given by (9.93) and (9.95) as V = g h 2 g2 sech2 4m 2 . (9.98)
160
Physically what is happening is that the electron is deforming the lattice around it creating a phonon cloud. This phonon cloud responds to the electron by creating a trapping potential which is given by (9.98). In this way, electron and phonon-distortion move together as a single object called a soliton. This situation is depicted on Fig.9.2.
0.8
0.6
0.4
-0.4
Figure 9.2: Polaron formation: electron is trapped in the self-consistent potential created by the phonons.
Ek c ck, + k,
p
h p a ap p (9.99)
1 Ld/2
V (p) ap + a p (p) .
p
In the Heisenberg representation the equation of motion for the phonon operators can be easily obtained i h dap dt = [ap , H] = hp ap + Ld/2 1 V (p) (p) , (9.100)
161
c ck = kp
k,
c ck+p = (p) . k
(9.101)
A similar equation can be obtained for a Equation (9.100) has the form of the equation for a forced harp monic oscillator and the solution is straightforward: ap (t) = ap (t0 )eip (tt0 ) i V (p) h Ld/2
t t0
(9.102)
where t0 is some arbitrary time. We are going to assume that the interaction between the electrons and phonons is switched on very slowly (that is, adiabaticly) so that in the innitely distant past the interaction is off and one has the non-interaction problem. In this case one can pick t0 = . Observe that in this case the rst term in left hand side of (9.102) is strongly oscillating and can be neglected. The second term, however, has the electronic density on it and can produce some new effect.
k+p k
k+p
Figure 9.3: Scattering process mediated by phonons. If we substitute (9.102) into (9.99) we see that the Hamiltonian of the problem gets an extra piece which reads Hextra = 1 h Ld |V (p)|2
t
(9.103)
Observe that this extra term only carries electronic degrees of freedom and moreover, it is retarded. The physical meaning should be clear by now: the phonons generate a retarded density-density interaction between the electrons. This is exactly what photons do indeed. The only difference is that for photons we can choose the Coulomb gauge where the interaction is instantaneous. But if instead one decides to use the Lorentz gauge one would generate an interaction which very much looks like (9.103). Moreover, the Hamiltonian is a conserved quantity and therefore can be dened at any time t from now on we pick t = 0. It is convenient to rewrite the problem in terms of Fourier components, that is, we dene
+
(p, t) =
d it e (p, ) , 2
(9.104)
dt eit = i
eit , i
(9.105)
162
where 0. In this case (9.103) can be written as Hextra = d 2 d 2 |V (p)|2 p /( Ld ) h (p, )(p, ) . i)2 2 ( p (9.106)
The problem is not solved yet because we do not know what the electrons are doing. It could be very well that in (9.103) averages to zero. In order to understand what is going on one needs to look at the electronic problem as well. In order to do that we are going to assume that the Fermi energy EF of the electrons is much larger than any phonon energy (like the Debye energy, for instance) in this case the electron operator obeys the equation i h dck, = [ck, , H] Ek ck, , dt (9.107)
since the corrections to this equation are of order V /EF . Thus we write ck, (t) ck, (0)eiEk t , and therefore (p, t) Thus we get from (9.104), (p, ) and nally in (9.106) Hextra =
p,k,k ,, i(Ek+p Ek )t c . k+p, (0)ck, (0)e k,
(9.108)
(9.109)
k,
(9.110)
(9.111)
which is the nal result. Observe that the interaction appears in terms of four fermion operators. The reason for that is simple to understand in terms of the Feynman diagrams we used before. The process described in (9.111) is one in which two electrons, one with momentum k and another with momentum k + p, are destroyed and two other electrons with momentum k + p and k are created. This is a scattering process where a momentum p is transferred between electrons. It is depicted on Fig.9.3. The important point about the process described by Hamiltonian (9.111) is the fact that if |Ek+p Ek | < p it has a negative sign which implies attraction. Consider the case of optical phonons, for instance, where p = 0 . In this case electrons states sufciently close to the Fermi surface can feel an attraction between them. The constraint is that EF 0 < Ek+p , Ek < EF + h0 which implies that there is a region of h thickness of size 2 0 around the Fermi surface where the electrons attract each other due to the interaction h with phonons (see Fig.9.4). For acoustic phonons we just replace 0 by the Debye frequency D but the result is essentially the same. The physical reason for which the attraction appear can be traced back to the basic physics of the problem: as the electron moves it attracts the ions in the lattice. The lattice is an elastic medium which can propagate this information and therefore other electrons, which will also attracted by the lattice, will feel this information as if the electrons were attracting each other as depicted on Fig.9.5. The net effect is the attraction between electrons.
163
Figure 9.4: Region of attraction around the Fermi surface due to the electron-phonon coupling.
where p is the relative momentum of the pair and V (r) is the attractive interaction given in (9.111). In momentum space the Schrdinger equation becomes k2 1 E0 (k) + d m L V (k k )(k ) = 0 . (9.113)
We will assume that the interaction is attractive and exists only when the electron is inside of the shell in Fig.9.4. Since k and k are very close to kF we will assume that the matrix element is essentially constant close to the Fermi surface, that is, V (k k ) = V0 (0 |Ek |) . In this case the solution of (9.113) reads (k) = where = V0 Ld (0 |Ek |)(k ) . (9.116)
k2 m
(9.114)
E0
(9.115)
(0 |Ek |)
k (k )2 m
E0
1 = V0
EF
N () . 2 E0
We will also assume that the density of states is constant over the interval of integration we have 1= which can be easily solved E0 = 2EF + 2 0 h 1 e N(0)V0 that is the binding energy of the two electrons. Observe that when the attraction is turned off V0 0 one obtains the expected result, that is 2EF , which is energy of two electrons at the Fermi surface. Observe however that when V0 = 0 we have E0 < EF , that is, the energy of the system is lowered by the formation of a bound state. What this implies is that the Fermi surface is unstable with attractive interactions since the electrons will lower the energy of the system by forming pairs. The binding energy of these pairs is simply = 2EF E0 = 2 0 h e
2 N(0)V0 2
V0 N (0) ln 2
2(EF + h0 ) E0 2EF E0
(9.119)
(9.120)
2 0 e h
2 N(0)V
(9.121)
which is a non-analytic expression in terms of the interaction and therefore cannot be obtained in perturbation theory (which always gives a power series expansion). These pairs are called Cooper pairs. Cooper pairs are the heart of a phenomena called superconductivity where pairs condense to form a macroscopic quantum state in clear violation of Paulis principle. The reason for that is that Cooper pairs, being formed of two fermions, behave very much like free bosons and therefore can Bose-Einstein condense. This is a completely new state of matter. Observe, moreover, that the formation of Cooper pairs only requires an net attractive interaction between electrons and therefore does not care very much about the mechanism for the attraction. The electron-phonon interaction is only one of many mechanisms that can make the electron-electron interaction attractive. More on this in the future...
p + + + + + + + +
9.4. PROBLEMS
165
9.4 Problems
1. Using the orthogonality and closure of the states |i and the creation operator in real space dened in (9.12) show that the operators (r) and (r) obey bosonic commutation relations. 2. Show that for fermion operators we have cs |n0 , n1 , n2 , ... = (1)Ss ns |n0 , n1 , ...ns 1, ... = (1)Ss ns + 1|n0 , n1 , ...ns + 1, ...
Ss = n1 + n2 + ... + ns1 . 3. Show that for fermion operators Sz (r) = Sx (r) = h (r) (r) (r) (r) 2 h (r) (r) + (r) (r) 2
and using the anti-commutation rules between electrons show that Sx (r), Sy (r ) = i (r r )Sz (r) . h (9.122)
4. Show that (9.47) obeys the anti-commutation relations and prove that (9.48) is indeed correct. Write down all the states of the system from the new basis in terms of the old basis. 5. Using (9.50) prove that (9.51) is indeed correct. 6. Prove (9.79). 7. Expand (9.78) to all orders in k/P0 and calculate the exactly value of the integral. What happens when k becomes large? What is the physical meaning of your result? 8. Prove (9.88). 9. Show that (9.95) is a solution of (9.94) by direct substitution. 10. (i) Find the displacement eld prole, (x), for the static polaron (v = 0) by solving (9.91). What do you conclude about the lattice distortion around the electron? (ii) Show that due to translational invariance of the problem the polaron mass is given by
+
M0 = s
dx
(9.123)
and calculate the ratio between the bare electron mass m and M0 . 11. (i) In the polaron problem the potential felt by the electron is given in eq. (9.98). Consider the case where the polaron is at rest (v = 0) and solve the Schrdinger equation in this case (9.91). Show that one of the eigenstates is given by (9.95) with eigenenergy given in (9.97). (Hint: The solution can be
166
12. Obtain the time evolution of a for the electron-phonon problem described by Hamiltonian (9.99). p 13. Prove equation (9.102). 14. Prove that the Heavyside Theta function can be written as
+
(t) =
d eit 2i i
Chapter 10
Magnetism
One of the most important problems in condensed matter physics is magnetism. Magnetism is a quantum mechanical phenomenon since it has its origins on the spin of the electron, its orbital motion and the Coulomb interaction between electrons. Naturally, magnetism is associated with the response of the electrons to a magnetic eld. We have seen that for a free non-interacting electron gas we have two main types of response, paramagnetic when the effect of the magnetic eld is only the magnetic polarization of the medium, and diamagnetic when the response of the medium is against the magnetic eld. In isolated atoms and molecules a similar type of response appears, as we have seen in Chapter 1. It is however the interaction between different atoms in a solid that leads to the phenomenon of magnetic ordering that is observed in natural magnets like iron. As we have seen in Chapter 1 the magnetic response of isolated atoms can be summarized in the three Hunds rules that are phenomenological rules based on simple energetic arguments. In short, the Hunds rules provide statements about the magnetic ground state of many electron atoms when the Coulomb energy is minimized. In a solid, however, the atoms are localized in well dened positions that are dictated by the symmetry of the lattice. Thus, the atoms are subject to electromagnetic elds that are called crystal elds. We are going to discuss the effect of crystal elds and the Coulomb interaction among electrons in different atoms lead to the phenomena of magnetism in solids.
168
the expectation value of the angular momentum has to vanish, that is, L = 0, even though we still have L2 = L(L + 1). Consider, for instance the case of p-orbitals in a free atom. These orbitals can be written as xf (r), yf (r) and zf (r) where x, y, z refers to the rst Legendre polynomial (l = 1) and f (r) is the solution of the radial equation. In a free atom these orbitals are degenerate. It is easy to show that these orbitals do not carry well dened angular momentum projection which are obtained as linear combinations of them: (x + iy)f (r) with mz = +1, zf (r) with mz = 0 and (x iy)f (r) with mz = 1. But since the orbitals are degenerate it really does not matter how we represent them. In the presence of an external electric eld the energy of the orbitals is split as in Fig. 10.1. If the electric eld is very asymmetric one can split the energy of all the states. Consider the case where the orbital yf (r) has the lower energy. In this case the angular momentum of this state can be computed immediately Lz = but observe that ( Lz ) = Lz (Lz ) Lz = Lz = Lz (10.2) dryf (r) 1 i x y y x yf (r) (10.1)
and therefore Lz = 0. Where in the rst line above we have integrated by parts and in the second line we used the fact that Lz is a hermitian. From the classical point of view this can be understood as the precession of the angular momentum in the crystal eld such that its magnitude is xed but it has zero average. In this case we have J = S which is known as quenching of the orbital angular momentum.
z
+ p p x z
y x
p y
y x z
+
C r y s t a l
f i e l d
p p p x y z
Figure 10.1: Crystal eld effects for a p-orbital. Thus, in 3d electron systems the spin-orbit interaction is a perturbation on the crystal eld effects. Let us consider this problem more closely. The spin-orbit effect is described by a Hamiltonian (1.25). Let us
169
(10.3)
that we consider as a perturbation of the problem in the presence of a crystal eld. Moreover, we are only interested in the spin degrees of freedom since J = S . Therefore we perform perturbation theory in the orbital degrees of freedom alone. Let |0 be the ground state of the problem in a crystal eld. Observe that rst order perturbation theory gives E1 = 2B B S (10.4)
due to the quenching of the orbital angular momentum. Let us consider now the second order perturbation theory in the absence of the eld. If |n is an orbital eigenstate of the system in the presence of crystal eld effects then second order perturbation theory leads to a change in energy given by E2 = 2
n=0
| n|
E0 En
a Sa La |0
|2
= 2
n=0 a,b
= where a,b = 2
a,b Sa Sb
a,b
n=0
n|La |0 n|Lb |0 . En E0
(10.6)
Observe that in this case the spin-orbit effect induces an interaction between different components of the spin. This is called single ion anisotropy. If one takes the principal axis of the crystal as x, y and z and express the components of a,b as x , y and z , one can show that (10.5) can be written as Eso = + (x + y 2z ) 2 S(S + 1) (x + y ) + Sz 3 2 (y x ) 2 2 Sx Sy . 2
(10.7)
The rst term in (10.7) is just an overall shift of the energy of the problem. The second term, because it appears as a sum of two is usually more important than the last term which appears with the difference. 2 For integer S the term with Sz splits the energy levels into doubly degenerate levels with Sz = S, (S 1), ..., 1 and a non-degenerate level with Sz = 0. For half-odd integer S there are doubly degenerate 2 2 2 2 levels with Sz = S, (S 1), ..., 1/2. The term proportional to Sx Sy S+ + S induces transitions between states that have Sz = 2. Thus, for integer S this term lifts the degeneracy of the degenerate states which can be linked by 2. For instance, for S = 1 the Sz = 0 state is non-degenerate and the Sz = 1 states have their degeneracy lifted by the spin-orbit. For a half-odd integer S the degeneracy persists because Sz is always an odd integer, thus, the degenerate ground state has always Sz = 1 and therefore are not linked by the perturbation. Thus the degeneracy persists even with spin-orbit coupling. The case with integer S corresponds to an even number of electrons in the atom while a half-odd integer S corresponds to an odd number of electrons. So for atoms with an odd number of electrons even in the presence of crystal elds and spin-orbit interactions the ground state is a doublet. This degeneracy has its
Observe that although spin-orbit and crystal elds reduce considerably the symmetry of the atom, there is one symmetry that persists in the absence of external magnetic elds: time reversal symmetry. Time reversal symmetry changes t t and therefore changes p p but keeps the position invariant, that is, r r. It implies that by inversion of the time direction the angular momentum is reversed, that is, L L. Classically it just means that we reverse the direction of rotation. Time reversal also inverts the spin, S S, because spin and angular momentum are indistinguishable in quantum mechanics. In the absence of magnetic elds the Hamiltonian is invariant under time reversal. Indeed, consider the general problem of an electron moving under the inuence of an external potential and the spin-orbit effect, H= h 2 2 + V (r) + S L 2m (10.8)
and it is clear that under time reversal the Hamiltonian is invariant. Notice that the addition of a magnetic eld leads to a new term of the form B (L + 2S) which is not invariant under time reversal since this term changes sign. We can obviously dene a quantum mechanical operator K such that K|r K|p K|S and it is clear that [H, K] = 0 (10.10) = |r
= |p
= |S
(10.9)
which means that if | is an eigenstate of the Hamiltonian with energy E (H| = E| ) then K| is also an eigenstate of the system with the same energy. This simple exercise in group theory proves that even if all the symmetries in a quantum mechanical systems are broken and no magnetic elds are present the ground state of the system has to be doubly degenerate, that is, it must be a doublet. This is the so-called Kramers theorem. In the presence of an external magnetic eld the energy shift of the ground state is given by E = Eso +
a,b
B ga,b Ba Sb
Ba a,b Bb
(10.11)
is the so-called g-tensor. Observe that (10.11) has a deep physical meaning: when a magnetic eld is applied to an atom the spin does not necessarily responds in the direction of the eld because of the spin-orbit coupling. The magnetization of the atom is Ma = = E Ba
b
B ga,b Sb 2
a,b Bb
(10.13)
171
a,b
(10.14)
is the induced orbital momentum which arises from the spin-orbit coupling. Observe that in this case an applied eld in one direction can have a response in another different directions. This is a direct consequence of the crystal elds. Observe that in this case the magnetic moment of the atom can be written as ma = B
b
ga,b Sb
(10.15)
as a consequence of spatial anisotropy. Thus, in this case the application of the magnetic eld in one direction can lead to a magnetization in another direction.
(10.16)
where ci, (c ) destroys (creates) an electron in a localized state at atom i (i = 1, 2) with spin ( =, ). i, As before we can diagonalize the problem by a simple linear combination of electron operators cA, = cB, = 1 (c1, c2, ) 2 1 (c1, + c2, ) . 2
(10.17)
c cA, c cB, A, B,
(10.18)
The ground state for the case where two electrons are present is |B = c c |0 B, B, 1 c c + c c + c c + c c |0 = 1, 2, 2, 1, 2, 2, 2 1, 1, 1 = c1, c1, + c c c c + c c |0 1, 2, 1, 2, 2, 2, 2
(10.19)
where |0 is the empty state. Notice that in the last line of (10.19) we used the anti-commutation rules between the electrons. Observe that the state represented in (10.19) is anti-symmetric as required by the Pauli principle. Another way to represent this state is to use the occupation of each atom (say |n1, , n2, )
172 where |B =
1 (| , 0 + | , | , + |0, ) . 2
(10.20)
In this new notation one has to be very careful about the order of the spins because of Fermi statistics. Observe, however, that the Hamiltonian (10.16) cannot be a good approximation for the real problem of the molecule since it completely neglects the Coulomb repulsion between the electrons. In particular it neglects the large Coulomb repulsion for electrons in the same atom. If this Coulomb energy is very large then the double occupancy of the atom costs a lot of energy. But we see from (10.19) that for the non-interacting problem this states are present. In the Heitler-London theory it is assumed that the Coulomb energy is very large and that double occupancy is avoided. In this case we see that a good approximation for the wave-function for the problem can be obtained from (10.19) by suppressing the operators that create two electrons in the same atom. In this case we see that the approximate ground state can be written as 1 |s = c c c c |0 1, 2, 2 1, 2, which in the other notation is simply 1 |s = (| , | , ) 2 (10.22) (10.21)
which is the so-called singlet state we discussed before. Thus the singlet state has the best of two worlds: it minimizes the Coulomb repulsion and maximizes the kinetic energy and therefore is a good starting point for the calculation of the true ground state of the problem.
where (r) is the electronic density. If we write the density in terms of the electron operators (9.41), we have HC = dd r dd r
,
(10.24)
Since the electron operators create or destroy electrons at the position of the atoms we can rewrite them as
2
(r) =
i=1
i (r)ci,
(10.25)
where i (r) is the wavefunction of the electron localized at the atom i. We consider the case where the number of electrons in each atom is constant. If the number of electrons do not change it means that electronic motion is completely disregarded. This approximation is only possible when we are dealing with a very large Coulomb repulsion in the atom. Actually this Coulomb repulsion
173
has to be much larger than the kinetic energy gain for an electron to jump from one atom to another so that in rst approximation one can neglect the kinetic energy. This is the case of insulating systems and therefore can only describe what is called localized magnetism which should be contrasted with the case of itinerant magnetism, as we shall see later. For the H2 molecule the condition of single occupancy implies ni, + ni, = 1. Direct substitution of (10.25) directly into (10.24) lead to 16 different terms. We consider only the ones that do not lead to hopping of the electron from one atom to another. There are, therefore, three different types of terms that interest us. The rst one is HU = U 2
2
ni, ni,
, i=1 2
= U
i=1
ni, ni, +
U 2
ni,
i=1
(10.26)
where in the last line we used that n2 = ni, for a fermion and i, U = e2 dd r dd r |i (r)|2 |i (r )|2 |r r | (10.27)
represents the Coulomb interaction in the same atom and it is the type of interaction we talked about when discussing the Hunds rules. Observe that the last term in the right hand side of (10.26) only shifts the local energy of the electrons and can be included in the non-interacting part of the Hamiltonian.
The second type of term is the one that involves interaction between electrons in different atoms. Although this term is not important if we constraint the atoms to single occupation we shall see later that this term is very important for virtual processes. It is written as: HV = V
,
n1, n2,
(10.28)
And nally the third term is called the exchange term and is given by HJ = JC where JC = e2 dd r dd r (r)1 (r ) (r )2 (r) 2 1 . |r r | (10.31) c c1, c c2, 1, 2,
,
(10.30)
The exchange term can be written in a more illuminating way if we use the representation of the spin operators in terms of electron operators (9.45). We dene Si = hsi /2 where sa = i
, a c , ci, i,
(10.32)
174 where a is a Pauli matrix (a = x, y, z). We need to calculate s1 s2 = and for that we can show that
a a , , = 2, , , , . a a , , c c1, c c2, 1, 2, a ,,,
(10.33)
(10.34)
And therefore we have from (10.30) we have HJ = JC 2 n1, n2, 2JC s1 s2 (10.36)
where the rst term can be absorbed into (10.28). We are more interested in the last term which represents a spin-spin interaction between spins in the two atoms. Moreover, since JC > 0 the energy is minimized if the spins are such that they are aligned pointing in the same direction. This is called a ferromagnetic coupling. It is very tempting now to conclude that if the electrons are localized in the atoms, that is, the tunneling is very small then the electrons should have their spins aligned. This of course contradicts our previous conclusion that in a bonding state the electrons should have their spins pointing in the opposite direction in a singlet state. What is wrong here? What is wrong is that even if the electrons cannot actually hop from one atom to another, they can still virtually hop! This is a pure quantum mechanical effect since second order perturbation allows a quantum systems to undergo transitions over excited states. Indeed let us consider the case where only interaction in the same ion are taking into account. In the limit where the ions are far apart these are the terms that dominate the physics and they are giving by (10.26). Let us treat now the tunneling Hamiltonian (10.16) as a perturbation. The ground state of (10.26) is given by one electron in each atom since the Coulomb repulsion is minimized, that is, in the ground state has ni, + ni, = 1. The energy of this state is zero. First order perturbation theory in the hopping gives a null result because the occupations of the atoms change. In second order perturbation theory one electron can hop back and forth from one atom to another. In this process it has to go through an excited state with double occupancy and energy U . By the Pauli principle this can only occur if the spins in different atoms are pointing in opposite direction. Indeed, in second order perturbation theory this process can be written as HA = 4t2 U c c2, c c1, 1, 2,
,
(10.37)
and using again (10.35) we can rewrite the above interaction as HA = 2t2 s1 s2 U (10.38)
which has the same form of (10.36) but with the opposite sign implying that the spins tend to become anti-aligned. This is called a kinetic exchange interaction.
175
If we now put put together (10.36) and (10.38) together we get the total exchange interaction between the electrons in the atoms is given by HE = Js1 s2 where J= 2t2 2JC U (10.40) (10.39)
which can be ferromagnetic or antiferromagnetic depending on the relative values of t2 /U and JC . For the H2 molecule it turns out that the antiferromagnetic coupling is larger and therefore the ground state is a singlet.
Consider now the case of a 3d electron atom which has a quenched orbital momentum. The magnetic moment of the atom is only given by the spin degrees of freedom and the magnetic moment, due to crystal eld effects, is given in (10.15). Direct substitution of (10.15) into (10.41) leads to Hdip =
a,b
(10.42)
(10.43)
Thus the total interaction between two atoms can be written as the sum of (10.39) and (10.42) H=
a,b
(10.44)
We can diagonalize Ja,b and nd the principal magnetic axis of the system. Since Ja,b is a 3 3 matrix we can have 3 different eigenvalues. It the problem has symmetry around an axis then two of this eigenvalues must be degenerate and so on. These anisotropies lead to a Hamiltonian that is more general than (10.39)
176 and reads H = Jz S1,z S2,z + Jy S1,y S2,y + Jx S1,x S2,x , where Jz , Jy and Jx can be different from each other.
(10.46)
(10.47)
The extreme limits of this Hamiltonian can be obtained by varying the parameters in (10.47). For instance, when Jz Jy , Jx we have HI = Jz
i,j
Si,z Sj,z
(10.48)
which is the so-called Ising model. When Jy = Jx = Jxy >> Jz we have Hxy = Jxy
i,j
(10.49)
Si Sj
(10.50)
is the Heisenberg model. Each one of these models have different symmetries. While the Heisenberg model has full rotation symmetry in spin space, the XY model only has rotation on a plane (we say that it has an easy plane) and nally the Ising model is the one with lowest symmetry since the spins are constrained to be on a xed axis (we say that the system has an easy axis). The interesting thing about these systems is that because they have different symmetries they order magnetically in different ways leading to what is called universality classes. In nature, however, it is difcult to nd a magnetic system that can be classied as a pure Ising, XY or Heisenberg.
10.5.1 Problems
1. Prove (10.7). 2. Prove (10.11). 3. Consider the H2 molecule and obtain the expression for the anti-bonding state in terms of the localized states of the electrons.
177
H0 | ,
H0 |0, H0 |,
H0 | ,
= t (|0, + |0, ) = t (| , | , ) = 0.
= +t (| , 0 + |0, ) (10.51)
5. Use (10.25) and write down all the 16 terms that appear in (10.24). 6. Using the Fourier transform of the Coulomb potential show that JC > 0. 7. Prove (10.34). 8. Prove (10.38). 9. Prove (10.42).
178
Using the identities = () 2 and 2 (1/r) = 4(r) we nd BS = = mS + 4mS (R) |R| 2 mN 8 + mN (R) . 3 |R| 3
(10.55)
It is easy to show the rst term in (10.55) is (10.41) and the second term is a contact interaction at the atom. Thus, the Zeeman energy for the electron interacting with the magnetic moment of the atom is HZ = that is known as the Fermi contact interaction. In a lattice with N magnetic atoms we can generalize (10.56) to HZ =
j,n,a,b
8 me mS (R) 3
(10.56)
(10.57)
where Rn is the position of the nth atom, rj the position of the j th electron, and mS = S S where S is the atom magnetic moment. From (10.15) we have, Ja,b (rj Rn ) = 8 B S ga,b (rj Rn ) 3 (10.58)
is the exchange between the spin of the electron and the magnetic moment of the nucleus in the presence of crystal elds and spin-orbit effect. For simplicity we assume that the metal that surrounds the magnetic nucleus is described by plane waves (although the same can be done with Bloch states) and is described by He =
k,
Ek c ck, k,
(10.59)
where Ek = h2 k2 /(2m). We want to represent the new term in (10.57) in terms of plane waves as well. For that we need the representation of the electron spin in terms of eld operators as in (9.45). It is easy to show that the Hamiltonian can be written as HJ =
k,k ,n
where we have used S = Sx iSy . Observe that the effect of the interaction between the conduction electrons and the nucleus is a scattering process where a plane wave with wavevector k scatters to k . In this process the electron can ip its spin. In order to simplify our problem let us consider rst the highly anisotropic limit of (10.60) in which J = 0. Moreover, we are going to assume that there is just one impurity at the origin (Rn = 0). In this
179
=
j
In this limits it is easier to write the Hamiltonian of the problem in real space in terms of the eld operators dened in (9.45): H= where V (r) = Jz Sz (r) and z is a Pauli matrix and Sz is the eigenstate of the operator Sz (that is, Sz |Sz Sz = S, .., S). (10.63) = Sz |Sz where dd r
,
(r)
h 2 2 z , + V (r), (r) 2m
(10.62)
Observe that we can diagonalize the problem entirely if we expand the electron eld operator as (r) =
E,
E, (r)aE,
(10.64)
where aE, is a fermion operator ({a , aE , } = E,E , ) and E, h 2 2 E, (r) + V (r)E, (r) = EE, (r) 2m (10.65)
is a Schrdinger equation for the magnetic scattering of a particle by an impurity since it depends on the z electron spin (here we have used that , = , ). In this case the Hamiltonian of the problem in the new basis reads H=
n,
En a n , aEn , E
(10.66)
Let us consider, for simplicity, the case where the electrons interact only weakly with the magnetic impurity and we can use perturbation theory. In (10.65) dene E = h2 k2 /(2m) and U = 2mV / 2 and h rewrite (10.65) as 2 k, (r) + k2 U (r) k, (r) = 0 and transform it to Fourier space 1 k, (r) = N eiqr k, (q)
q
(10.67)
(10.68)
(10.69)
where U (p) is the Fourier transform of U (r). In order to solve (10.69) we have to impose the condition that when U = 0 the solution of the problem is a plane wave, that is, k, (q) = q,k . Thus, the solution of (10.69) is k, (q) = q,k +
p
U (p q) k, (p) k2 q 2
(10.70)
which is an integral equation for k, (q). We rst observe that because of (10.63) we have U (p) = U = 2mJz Sz / 2 is independent of momentum. Moreover, in rst order the solution of the problem is given by h the substitution of the delta function in the rst term on the l.h.s. of (10.70) into the second term on the l.h.s. which gives, 1 k, (q) N which when substituted into (10.68) leads to 1 k, (r) eikr + U k2 q 2 N q=k eiqr (10.72) q,k + k2 U q2 (10.71)
which is correct to rst order in U and it is known as Born approximation. In order to evaluate (10.72) we have to avoid the point where p = k where the sum diverges. Moreover, in the thermodynamic limit we replace the sum by an integral over q. We have the problem of avoiding the points where the integral diverge. The best way to do it is to add an innitesimal imaginary part to the integral and write eiqr = k2 q 2 dd q eiqr (2)2 k2 q 2 i (10.73)
q=k
where 0 at the end of the calculation. This is the principal value of an integral. So let us evaluate this integral eiqr dd q (2)2 k2 q 2 i = = = 1 (2)2 1 (2)2 1 (2)2 r eiqr cos k2 q 2 i 0 0 +1 eiqru dqq 2 2 du k q 2 i 0 1 + q sin(qr) dq 2 k q 2 i d sin dqq 2
(10.74)
where in the last line we used that the integrand is even in q. The integrals can be performed by contour integration. For instance,
+
dq
qeiqr = 2i q 2 k2 + i
eikr 2
= ieikr
(10.75)
181
since we have to close the contour in the upper half-plane in order for the integral to converge. The nal result is eiqr cos(kr) = 2 2 2 q2 k r (10.76)
q=k
Observe that we can now calculate various properties using (10.64) and (10.77). In the approximation used, the electrons are only scattered by the magnetic impurity but their energy remains the same. Thus the ground state is obtained by lling up the momentum states as usual. Observe, however, that the presence of the impurity modies the charge density for different spins. Indeed, n (r) = =
k,k
=
k
where kF is the Fermi momentum which is related to the electron density by kF = (3 2 n)1/3 . Using (10.77) to leading order in U we nd n (r) = = = where F (x) = sin x x cos x . x4 (10.80) n + 4 2 U 2 d3 k cos(kr) cos(k r)(kF k) (2)3 r
(10.79)
Notice that the electron density far away from the impurity reaches is bulk value n/2 but close to the impurity it oscillates strongly. These are called Friedel oscillations. Moreover, the oscillations depend on the spin of the electron. This happens because the interaction between the electron and the impurity atom is magnetic. Consider now the problem of two magnetic impurities at positions Rn and Rm . The spin Sn at position Rn polarizes the electron gas in its vicinity in the way given by (10.79). Another spin Sm at Rm feels this polarization that allows the two moments to interact. In order to get the interaction Hamiltonian between these magnetic moments we use (10.61) and (10.79) to nd HRKKY = Jz Sz (Rn ) (n (Rm ) n (Rm )) F (2kF |Rn Rm |)Sz (Rn )Sz (Rm ) (10.81)
n,m 2 Jz
EF
n,m
182
where EF is the Fermi energy. This interaction between magnetic moments is called RKKY interaction due to Ruderman, Kittel, Kasuya and Yosida. Observe that the interaction is oscillatory and decays like 1/r 3 as shown in Fig.10.2.
0.004 0.002 0 -0.002 -0.004 0 5 10 x 15 20
Figure 10.2: Plot of F (x) Notice that at short distances (r 1/(2kF )) the interaction is ferromagnetic but at large distances it changes sign and can be antiferromagnetic. Thus, depending on the position of the spins in the lattice this interaction can be effectively ferromagnetic or antiferromagnetic. In the case of disordered alloys both ferromagnetic and antiferromagnetic couplings are possible. At zero temperature this lead to a frustrated magnetic state called spin glass. In the generic case where J is not null the form of the RKKY is given by: HRKKY = a,b F (2kF |Rn Rm |)Sa (Rn )Sb (Rm ) (10.82)
n,m,a,b
<i,j>,
where h.c. means hermitian conjugate and J > 0 is the Hunds coupling between localized and itinerant electrons. This is a problem of great complexity since, as we have seen before, the electrons mediate the interaction between spins via the RKKY interaction. Let us consider the case where we have a large localized spin S (S 1/2). In this case the spin can be considered as a classical variable that can be parameterized by angles i and i relative to some xed axis (for instance, S z = S cos(), S x = sin() cos(), S y = S sin() sin()). Note that the Hunds coupling forces the spin of the electron to orient along the direction of S. If the local spins are all aligned with each other (ferromagnetic situation) then the second term in (10.83) gives a negative energy contribution, reducing the energy of the system. If the spins are oriented degrees from each other (antiferromagnetic situation) the electrons cannot hop from site to site because the hopping term in (10.83) does not ip the spin
183
and therefore it costs an energy J to frustrate the Hunds coupling. Consider the situation on Fig.10.3(a): the electron spin (empty circle) is oriented with all the localized spins (lled circles) and can move freely over the lattice. In Fig.10.3(b) the antiferromagnetic orientation of the spins makes the hopping from one atom to the other difcult since it costs an energy J for the electron with spin up to move to an atom with spin down. When J t the hopping is completely suppressed.
a) J b)
Figure 10.3: (a) Hopping in a ferromagnetic situation; (b) Hopping in a anti-ferromagnetic situation.
This argument shows that in order to gain the electron kinetic energy the spins tend to form a ferromagnetic state where all the spins are aligned to allow for the electron motion. Notice that because of the symmetry of the interaction in (10.83) the change in energy can only depend on the relative angle, i,j , between the localized neighboring spins. When i,j = the effective hopping between spins, ti,j , is zero, when i,j = 0 the effective hopping is simply ti,j = t. In order to study this formally Let us consider the case of two magnetic atoms with spins S1 and S2 . The Hamiltonian of the problem (in accord with (10.83) is: H = t (c c2, + h.c) J(S1 s1 + S2 s2 ) 1, (10.84)
In the reference frame of the local spins the energies of the states are +JS or JS depending if the electron spin is anti-parallel or parallel to the local spin, respectively. Let us call these states |i, and |i, + , respectively (i = 1, 2). Thus, the whole problem reduces to four states, namely, |1, + , |1, , |2, + , and |2, . The spins S1 and S2 form an angle between them, that is, S1 S2 = S 2 cos(). If we choose a quantization axis along S1 (we write S1 = Sz), for instance, we have to rotate to project the states of S2 into this axis. This can be done by rotating the states of in atom 2 via a spin rotation operator around the Y axis: U
h = eiSy / = eiy /2
= cos(/2) iy sin(/2)
(10.85)
where y is the Y -Pauli matrix. The states of atom 2 relative to the quantization axis of atom 1 can be written as: |2, = U, |2 , (10.86)
=1
where |2 , refer to the states in the rotated frame. The rotation can be written more explicitly as: |2, + |2, = cos(/2)|2 , + + sin(/2)|2 , = sin(/2)|2 , + + cos(/2)|2 , . (10.87)
184 Moreover, from (10.84) and (10.87) we see that: i, |H|i, 1, |H|2, Thus, the Hamiltonian can be written as: JS 0 0 JS [H] = t cos(/2) t sin(/2) t sin(/2) t cos(/2) E= = JS,
= tU, .
(10.88)
t cos(/2) t sin(/2) JS 0
which can be diagonalized in the usual way. The eigenenergies are: (JS)2 + t2 2JSt cos(/2) .
where the rst term corresponds to the Zeeman energy of the electron in the eld of the localized moment and the second term corresponds to the effective hopping energy for the spins between the different atoms. Notice that for = 0 it gives t and for = it gives zero, in agreement with the previous discussion. Notice further that while in the RKKY mechanism the rst contribution to the energy is of order J 2 /EF (see (10.81)) the current mechanism produces an effect of order J and therefore should be dominant at small J. The above argument shows that, when generalized to a lattice, the effective Hamiltonian of the problem can be written, in the limit of J t, as: Hef f = t cos(i,j /2) c cj, + h.c. i,
i,j, ,
(10.91)
where i,j is the angle between neighboring spins. For a given lattice one has to nd out the spin conguration that minimizes the energy in (10.91). The model described here was originally proposed by Zener in order to describe the physics of certain oxides with Mn and the term double exchange comes from the fact that the exchange between Mn atoms occurs via the lled p-orbitals of the O atoms.
10.7.1 Problems
1. Show that in two dimensions the RKKY interaction decays like 1/r 2 at long distances. 2. Consider the problem of an electron jumping between two atoms with large spins in the limit of J t. (i) What is the ground state energy in this case? (ii) What is the physical interpretation of the solution in this limit? (iii) Will the electron travel over the system for any spin conguration? 3. Consider the problem of electrons moving in a d-dimensional hyper-cubic lattice with a density n of electrons in the limit of J t. Assume that the angle between adjacent spins is always the same, that is, i,j = independent of i, j. (i) What is the spin conguration in this case? (ii) What is the ground state energy as a function of angle? (iii) Assume that the angle between adjacent spins is very small and that the distance between atoms is a. Dene k = /a and show that the energy of the system behaves like k2 . How does the electron spin behave in this case ?
185
In the last sections we discussed Fermi contact interaction between conduction electrons and localized magnetic moments. The contact interaction is important for s-wave states where the electron has large probability of being at the atomic site. For p- or d-like orbitals that vanish at the origin, the contact interaction also vanishes. In this case the interaction between a conduction electron and a localized moment has a different origin which is related with the hybridization between conduction electrons and localized electrons. Consider the problem of a localized impurity at a position r = 0 in a Fermi gas as shown in Fig.10.4.
A conduction electron can hop back and forth from the electron gas into the impurity site. This hopping can be seen as a tunneling term like in (10.16). However, in order to do so, the electron has to pay an energy U if the impurity site is doubly occupied (which was studied in (10.26)). The simplest Hamiltonian that describes this situation is the Anderson Hamiltonian: HA =
k,
Ek c ck, + Ef k,
f f + V
c (r = 0)f + f c (r = 0)
+ U nf, nf,
(10.92)
Ek c ck, + Ef k,
V f f + N
c f + f ck, k, k,
+ U nf, nf,
(10.93)
where ck, and f are electron operators for electrons on the conduction band and the localized impurity state, respectively (N is the number of sites). Ef is the atomic energy at the impurity, V is the tunneling energy between the conduction band and the impurity and U is the Coulomb energy to put two electrons on the impurity (nf, = f f ).
The Hamiltonian describes a true many-body problem. In trying to understand the physics of this Hamiltonian we have to make approximations. Let us consider rst the problem where U = 0, that is, the Coulomb interaction is absent. One expects from the beginning that no magnetism can be described in this case since a magnetic moment can only exist in a isolated energy level if double occupancy of the level is not allowed. We will see later how the Coulomb energy can be introduced into the problem. When U = 0 the problem is quadratic in the operators and therefore can be diagonalized by an unitary transformation: f = f +
k
k ck, k,k ck ,
k
ck, = k f +
(10.94)
where f and ck, are the new electron operators which diagonalize the problem, that is, after the unitary transformation the Hamiltonian (10.93) with U = 0 can be written as H=
Ef f f +
k
E k c ck, k,
(10.95)
with E f and E k the new eigenvalues (still to be calculated). Notice that the coefcients , k , k and k,k depend on the parameters in the Hamiltonian.
One of the ways to solve this problem is to look at the equations of motion for the operators. Using the anti-commutation relations between the fermion operators it is easy to show that V [f , H] = Ef f + N ck,
k
V [ck, , H] = Ek ck, + f . N On the other hand, using (10.94) and (10.95) one has [f , H] = E f f +
k
(10.96)
k E k ck, k,k E k ck ,
k
[ck, , H] = k E f f +
(10.97)
187
V (Ef k + N
k ,k )ck,
k
k +
k k
V Ek k ,k )ck, + (Ek k + )f . N
(10.98)
V (E f E k )k = . N
(10.99)
In solving the above equations we have to exercise some care. We are interested in the case where E f is within the electron band which implies that the second equation in (10.99) has a singularity at E f = E k . This singularity can be avoided, however, if we interpret the second equation as the real part of V 1 k = , N E f E k i (10.100)
where 0. Moreover, we are interested in the case of a single impurity in 1023 electrons. Thus, in this case E k Ek plus corrections of order 1012 . Substitution of (10.100) into the rst equation of (10.99) leads to the desired nal result E f = Ef + 1 N V2 E f Ek i (10.101)
which is an equation for E f . Equation (10.101) shows that the energy of the f-state is shifted by an amount proportional to V 2 . One has to remember, however, that the original f-state is not an eigenstate of the Hamiltonian since it is actually coupled to the conduction band. Thus, when an f-electron is put into the many-body system it has to decay into f and c states which are the true eigenstates of the problem. Therefore, the f-electron has a nite lifetime. In order to estimate the life-time we go back to (10.101) and interpret the imaginary part of the f-electron energy as its decay rate (this step can be formally proved with the use of Greens function method). Thus, we have: 1 h = f N where we have used that: limit0 1 =P x i 1 x + i(x) (10.103) V2 E f Ek i = V 2 N (E f Ek ) (10.102)
where P means the principal value of the function. Notice that (10.102) makes a lot of sense. When V 0 the f-state is decoupled from the conduction band and therefore has innite lifetime. But as the coupling to the conduction band increases the lifetime of the f-state becomes shorter. It is interesting to compare the behavior of the density of states in both cases. When V = 0 the density
188
of states of the problem reduces to the density of states of the f-state and the conduction band Nf (E) = (E Ef ) = limit0 Nc (E) = 1 N
k
1 = (E Ef )2 + 2
1 E Ef i (10.104)
(E Ek ) .
h In the presence of the coupling V we replace Ef in the rst equation in (10.104) by E f i /f in order to get Nf (E) = = 1 E E f i /f h h /f 1 2 (E E f )2 + h2 /f
(10.105)
which is not a Dirac delta function but a Lorentzian with width 1/f with is given by h = V 2 Nc (E f ) f (10.106)
as we can readily see from (10.102) and (10.104). Thus, as a result the f-level is not sharp. This reects the fact that when an f-level hybridizes with a conduction band it is not an exact eigenstate of the system (indeed, one sees from (10.94) that the eigenstate is a linear combination of f and conduction band states). This situation is depicted in Fig.10.5.
N(E)
(a)
Ef E
N(E)
(b)
1/f
Ef
Figure 10.5: Density of states for the U = 0 Anderson model: (a) V = 0, (b) V = 0. In order to understand what happens in the presence of U one has to treat the interaction term in (10.93). It is obvious that this term cannot be treated in the same way we treated the V term because it contain four fermion operators. The problem is highly non-linear. Instead of trying to solve the problem exactly we search for an approximate solution. The approximation is called mean eld or Hartree-Fock approximation.
189
(10.107)
which corresponds to replace the actual value of the occupation at the f-level state by its average. If one make the substitution (10.107) into (10.93) one sees that the only modication is in the energy of the f-level state, that is, Hf =
Ef nf, Hf,M F =
Ef, nf,
(10.108)
that is, the energy of the f-level with spin, say, is increased by U times the average occupation of the state with spin and vice-versa. After this approximation is done we can proceed as earlier (in the case of U = 0) but taking into account that the local energy of the f-level depends also on the spin projection. In particular the density of states for each spin projection is given by: Nf, (E) = where E f, = Ef + U nf, + 1 f, = V 2 Nc (E f, ) . 1 Re N V2 E f, Ek i (10.111) h /f, 1 2 (E E f, )2 + h2 /f, (10.110)
For simplicity we assume that the electron density of states is essentially constant and therefore E f, = E f + U nf, and f, = f . In this case, the average occupation of the f-level with spin is given by the usual expression
nf,
1 arccot
which is a transcendental equation for the occupation. Thus we have traded a non-linear problem in terms of operators for a non-linear problem in terms of ordinary equations. It is common to rewrite the equation above in terms of new variables dened as: y = f U/ h x = Ef U (10.113)
(10.114)
Let us consider two simple limits of these equations. When U = 0 it is obvious that nf, = nf, = 1 2 (10.115)
since nf, + nf, = 1. This corresponds to the non-magnetic solution of the problem. Thus, as one could have guessed the Coulomb interaction is fundamental for the appearance of magnetic moments. In the limit of U the arccot(x) can give or 0 depending on the sign of the argument. It is obvious that the solution close to corresponds to the full occupation of the level and the one close to 0 corresponds to an empty level. Thus, let us assume that nf, 1 and nf, 0 and calculate nf, nf, 1 1 y(x nf, ) 1 y( nf, ) x
(10.116)
which implies that the local magnetization in the magnetic impurity is mf nf, nf, 1 1 y(1 x) V 2 N (0) 1 U =
(10.117)
where we have used (10.106) and (10.113). Notice that the above approximation is valid for V EF , U . That is, in the limit in which the impurity is weakly coupled to the electron gas and the Coulomb energy is large. In this case the charge uctuations in the impurity site are very small and only the spin of the impurity has dynamics. The transition between the magnetic states at large U and the non-magnetic states at small U can be studied within the mean-eld approach by solving (10.114) in the case of n = n = nc . Taking the derivative of this equation with respect to n we nd another equation, namely, y= . sin (nc )
2
(10.118)
Using (10.118) in (10.114) we nd that, at the transition, x has to be such that: x = nc sin(2nc ) . 2 (10.119)
Further notice that because n + n 2 we must have nc 1. A plot of x as a function of nc as given by (10.119) and /y as a function nc is shown in Fig.10.6. From these plots we can draw the phase diagram of the problem as a function of x and /y as shown in Fig.10.7. This phase diagram shows that moment formation in an atom embedded in a metallic environment is only possible if x < 1, that is the impurity level is close to the Fermi energy, E f < U , and /y < 1, that is the Coulomb interaction is large, U > /f . h Therefore, the conclusion of this calculation is that for the existence of local moments the Coulomb
191
Figure 10.6: Plot of (10.118): continuous line; and (10.119): dashed line.
x 1
1/2
Magnetic
NonMagnetic
/y 1
Figure 10.7: Phase diagram of the Anderson impurity problem at the mean-eld level. energy is fundamental. The physics of this problem is relatively simple: in the absence of the Coulomb interaction, U = 0, the singly occupied states are degenerate with energy Ef and the doubly occupied state has energy 2Ef . In the presence of U the energy of doubly occupied state is U + Ef > and therefore out of the Fermi surface. Thus, this state is always empty. It is exactly because the doubly occupied state is higher in energy that moment formation is stabilized.
192
an interaction between the conduction band and the electron impurity of the form Hexch| = Js(r = 0) S where J NV 2 1 1 + U + Ef Ef (10.121) (10.120)
which is always positive (we have xed the energy so that EF = 0). The Hamiltonian (10.120) is called the Kondo Hamiltonian and has exactly the same form as the Fermi contact interaction (10.57). Thus all our results about the RKKY interaction are valid for this model if we take the impurity spin to be 1/2. However, there is new physics in this model that we have not discussed yet.
As before we consider the anisotropic Kondo problem where the exchanges in the Z and X-Y directions are different from each other (happen in the presence of crystal elds and spin-orbit effects). The complete Hamiltonian reads: H =
k, +
k c ck, + Jz S z k,
k,k , +
c ck , k, (10.122)
+ J S s (r = 0) + s (r = 0)S
where Jz is the longitudinal or Ising coupling and J is the transverse of XY coupling. Notice that because Jz , J > 0 there is an effective attraction between the impurity and the electrons when they have opposite spins. Attraction between particles implies the possibility of forming bound states. In fact, for J = 0 we can use the results of the previous section, in particular, we write (10.69) as: (E h 2q2 )E (q) = Jz 2m E (p)
p
(10.123)
where we reestablished the usual units. This equation can be solved as E (q) = where C(E) = Jz Substitution of (10.124) into (10.125) leads to 1 = Jz 1
p h 2 q2 2m
C(E) E
h 2 q2 2m
(10.124)
E (p) .
p
(10.125)
(10.126)
which gives the equation for E. If we now use the denition of the density of states N (E) =
p
h 2 p2 2m
(10.127)
193
(10.128)
that has to be solved for E(Jz ), the bound state energy. Observe that (10.128) does not have any information about the occupation of the states in the system, that is, it does not carry a Fermi occupation number. However, it predicts the possibility of forming bound states that can be written as: | , or | , , where , represent the impurity spin and , represent the electron spin. In the above problem the two states are degenerate in energy. In this localized picture we immediately see that the coupling J in (10.122) lifts the degeneracy between these two states so that the nal states of the problem are: |s |t = = 1 (| , | , ) 2 1 (| , + | , ) 2
(10.129)
which are the singlet and triplet states. Notice that due to the transverse term in the Hamiltonian the single state is lower in energy by J relative to the triplet state. For the reasons given below the approach given above has a big aw associated with the fact that we are treating the Jz and J in a very non-perturbative way and we will show that this is incorrect physically. However, we borrow the ideas of the naive calculation above in order to simplify the problem to a treatable form. Let us rst dene new states by: |+ = | ,
= | , .
(10.130)
These states are not pure spin states since they involve the impurity spin and the localized electron spin. However, we see that the operator associated with J has a very simple effect when acting on those states, namely: S + s (r = 0) + s+ (r = 0)S |+ S s (r = 0) + s (r = 0)S
+ +
= | = |+
(10.131)
that is, the J operator acts as a Pauli matrix x on these states. By the same token, the operator S z has a very simple action on those states: S z |+
z
S |
= |+
= |
(10.132)
and therefore S z acts on these states as the Pauli matrix z . Thus, by changing the basis we can rewrite the Kondo Hamiltonian as: H=
k,
k c ck, + Jz z k,
k,k ,
c ck , + J x . k,
(10.133)
This problem still has too many degrees of freedom. As we know from our study of metals perturbations in metals can only affect states close to the Fermi energy and unfortunately (10.133) still contains states with high energy close to the bottom of the Fermi sea. If we consider the momenta k and k in (10.133) close to the Fermi surface we see the effect of the impurity is to scatter and electron with momentum k from an
194
occupied state inside of the Fermi sea (leaving a hole behind) and putting this electron outside the Fermi sea in a unoccupied state with momentum k . That is, it creates particle-hole pairs. Thus, it is more convenient to work in the language of particle-hole pairs instead of the language of the original fermions. As we showed in Chapter 6 (see Eq.(6.106)) the low energy excitations of the Fermi gas, the particlehole pairs, have bosonic character and at low temperatures provide a natural description of the specic heat, for instance. The energy of a particle-hole pair close to the Fermi surface is q = vF q (10.134)
where vF is the Fermi velocity and q > 0 is the module of the momentum transfer for the production of the particle-hole pair (k = k + q). Thus, particle-hole pairs behave like harmonic oscillators in momentum space (like acoustic phonons). Thus, at low energies we can replace the free electron Hamiltonian by: H0 =
k,
(k )c ck, Hph = k,
q>0
2 Q2 Pq 2 q + M q 2M 2
(10.135)
where is the chemical potential. Notice that the mass of the oscillators M is undened. The reason is given below. Furthermore, in the scattering process the electron transfer momentum to the impurity (momentum is not conserved in the scattering problem) and this leads to the creation of particle-hole pairs with nite momentum q. Since the momentum operator associated with the pair is Pq we see that the Jz operator in (10.133) can be replaced by: Jz z
k,k ,
c ck , Jz z k,
Pq .
q>0
(10.136)
Dening now creation and annihilation operators for the harmonic oscillators in the usual way: aq = a = q M q 2 M q 2 Qq Qq + i Pq M q i Pq M q (10.137)
q a aq + z q
q>0
qi(aq a ) + J x . q
(10.138)
We rewrite the Hamiltonian by redening the creation and annihilation operators by a phase factor: bq = iaq = ei/2 aq so that: H=
q>0
(10.139)
q b bq + z q
q>0
q(bq + b ) + J x , q
(10.140)
where = Jz M vF . 2 (10.141)
195
The Hamiltonian in (10.140) is called the dissipative two-level system model since it describes a problem of a two-state system (the eigenstates of z ) coupled to a heat bath of harmonic oscillators. Notice that the parameter M remains undened. The reason for that is that the oscillators are produced by the coupling of the impurity to the particle-hole excitations and therefore their mass is also determined by this coupling. The procedure presented here cannot tell how M behaves as a function of Jz and thus we can think of M (or ) as parameters of the problem that have to be obtained by different means. What is guaranteed from our construction is that the number of degrees of freedom in the problem is correct and that the low energy physics of Hamiltonian (10.140) is the same as the original fermionic problem (10.122). One way to obtain is to solve the problem via Bethe ansatz or numerical methods and obtain as a function of Jz . From now on we think of as a parameter of the problem in its own right. The Hamiltonian in (10.140) has another free parameter that is not explicit in the Hamiltonian. Notice that the bosons are a result of the linearization of the electronic dispersion close to the Fermi surface (see Fig.6.4) and therefore our procedure is only valid if the momentum of the particle-hole pair is smaller than a cut-off, . The size of this cut-off can be estimated to be of the order of the Fermi momentum, kF , which, in most metals, is proportional to the inverse of the lattice spacing, a. That is, our approximation assumes that: q kF 1 a (10.142)
which implies that our theory only describes wavelengths much larger than a. Moreover, the sums in (10.140) are bounded from above by . We have in our hands the tools to discuss the reason why perturbation theory fails for this problem. Notice that the operator J ips the pseudo-spin |+ to | and vice-versa. In this case the sign of the interaction in the Jz term changes from + to , that is, there is a sudden switch on of the interaction! As long as the system is in one of the eigenstates of z not much happens (there is a simple shift of the boson momentum) but once the pseudo-spin ips a potential is switched on and the ground state changes! When we study time dependent perturbation theory we learn that the perturbation potential has to be switched on very slowly so that states of the system do not change drastically. This is not the case here and this is why perturbation theory fails. In what follows We describe a way to treat this problem with a renormalization group (RG) method that gives results beyond perturbation theory. The main idea is to use the fact that the harmonic oscillators with energy close to vF = W EF are very fast compared with the impurity spin. In fact, the ipping time of the spin can be estimated by the uncertainty principle to be: tf lip h h J W (10.143)
since we assume that J W EF . In this case, the fast (high energy) harmonic modes always see the impurity spin as static (in the same sense that in the Born-Oppenheimer approximation the electrons see a static lattice of ions). Thus, the fast oscillators oscillate many periods before the spin ips and their only effect is to dress the spin (in the same way that electrons glue the ions). In order to understand how these high energy degrees of freedom dress, or more elegantly, renormalize the spin ips, let us consider the effect of changing the cut-off to d (see Fig.10.8). Observe that this is equivalent to change the energy scale from W to W dW (W = vF ).
196
W W-dW
Figure 10.8: Dispersion of the particle-hole excitations and the RG procedure of reducing the cut-off .
where |, 0 is the state without bosons and |, q is the state with one boson with momentum q such that d < q < . The subscript R indicates that this is the renormalized state and N is the normalization factor that can be easily calculated from the condition that | R = 1, that is, N 1+ vF
2 q=d
The state of the pseudo-spin can be now calculated in rst order perturbation theory in as: q 1 |+, 0 |+ R = bq |+, 0 |+, 0 + z q N q=d 1 1 = |+, 0 + |+, q vF q N q=d 1 1 |, 0 | R = |, 0 |, q vF q N
q=d
(10.144)
1 . q
(10.145)
In order to calculate the renormalized value of the coupling constants, that is, the value of the coupling constant when we reduce the cut-off we realize that the bare value of the coupling is given by: J = +, 0|H|0, + (10.146)
R as you can easily prove from (10.140). The renormalized value of the coupling, J is then given by: R J =
1 +|H|+ N 1 1+
vF vF 2 2
12
vF
1 q
(10.147)
197
where in the last line we have used the fact that we are doing perturbation theory. The sum over q can be easily performed:
q=d
1 = ln q
d .
(10.148)
Furthermore, the value of the renormalized cut-off is (see Fig.10.8): W R = W dW = W (1 d/) and therefore we can write
R J WR R J WR
= =
J 1 d W 1 d J W 1 + (1 )
(10.149)
(10.150)
which, like M and can be viewed as a undetermined function of Jz . Notice that (10.149) tells us how the dimensionless coupling constant g = J W (10.151)
changes as we change the scale of the problem, that is, g = g (). Eq.(10.149) can be rewritten in the form: g ( d) = g () (1 (1 )d ln(1/)) dg = (1 )g () d where we have dened = ln(0 /) = ln(W0 /W ) (10.153)
(10.152)
where 0 is the bare value of the cut-off. Eq.(10.152) is the RG equation for the coupling J . This equation can be solved as: g () = g (0)e(1) (10.154)
where g (0) = J /W is the coupling constant of the problem at the original scale while g () is the coupling at an arbitrary scale. From this result we immediately can conclude the following: if > 1 we have g ( ) 0 indicating that at longer and longer length scales (smaller ) the coupling vanishes and this term is irrelevant for the long wavelength physics; if < 1 we have g( ) and the coupling constant becomes arbitrarily large as we reduce the cut-off indicating that the coupling is relevant and dominates the physics in the long wavelength limit. The case of = 1 is called marginal and the RG 2 equations only have a contribution to order g . Thus, the conclusion of this RG calculation is that perturbation theory is only reliable in the case of > 1 since the perturbation becomes weaker and weaker at low energies. In this case we can effectively
198
make J = 0 in the Hamiltonian and we see that the problem simply reduces to the problem discussed previously of the formation of the bound state (or scattering) out of a impurity. The state of the impurity is an eigenstate of the z operator. For < 1 the coupling becomes arbitrarily large and perturbation theory fails miserably and in this case we have to nd a better way to solve the problem (which usually means that we have to solve the problem exactly!). We can make an educated guess of what happens in this case. Observe that the fact that J becomes relevant in this limit indicates that this is the dominant term in the Hamiltonian. This implies that the state of the impurity is an eigenstate of the x operator. Thus, we expect the ground state of the problem to be: 1 |s = (|+ 2
R
R)
(10.155)
which is the singlet state. This is the so-called Kondo singlet. Notice that our perturbation theory is only well dened for J W , that is, g 1. For < 1 the coupling g grows under the RG until it reaches g ( ) = 1 where, from (10.154): 1 ln(1/g (0)) 1 1 ln(W0 /J ) 1 J 1/(1) . = W0 W0 (10.156)
Using (10.153) we see that this scale is associated with an energy scale kB TK such that: ln(W0 /kB TK ) = kB TK
(10.157)
For energies or temperatures above TK the coupling is small and the problem can be treated perturbatively but for temperatures smaller than TK the system ows to strong coupling and the Kondo singlet is formed. This temperature scale is called the Kondo temperature of the problem. Observe that for J = 0 we have TK = 0 and therefore no Kondo effect. Thus we have found that the behavior of the problem changes dramatically when we go from > 1 to < 1. We have argued that has to be a function of Jz and therefore there must be a critical value of Jz for which the physics of the problem changes completely. The question is: can we look at our original Hamiltonian (10.122) and understand how Jz changes the physics of the Kondo problem? The answer is quite simple, in fact. Notice that the formation of a singlet state requires Jz > 0, that is, an antiferromagnetic coupling between the impurity and the electron gas. If Jz < 0, that is, if we had ferromagnetic coupling the state of the system would be a triplet, that is, | , or | , (degenerate). In this case notice that the J term does not connect these two terms and therefore is not able to lift this degeneracy in rst order perturbation theory. Thus, J is irrelevant in this case. Thus, we can easily assign the case of > 1 with the case of Jz < 0 while < 1 for Jz > 0. Thus, the quantity 1 is an odd function of Jz , that is, 1 (Jz ) can be written as a power series expansion in Jz with odd powers only. For Jz W0 we expect, based on this argument that: (Jz ) 1 C Jz W0 (10.158)
where C is a constant independent of Jz . Thus, in this limit the Kondo temperature can be written, from
199
= W0 exp
(10.159)
Notice that for Jz = 0 we have = 1 and therefore from (10.157) we have TK = 0 indicating that there is no Kondo effect in this case. These results show that in the Kondo effect both the Ising, Jz , as well as the transverse, J , couplings are fundamental for the occurrence of the Kondo effect. In the case of the isotropic Kondo effect (J = Jz = J) the RG ow can be calculated by different techniques and one can show that: dg = g2 d (10.160)
where g = J/W . In this case the Kondo coupling is marginal (very similar to the previous case with = 1). We can solve this RG ow as: g() = g(0) 1 + g(0) (10.161)
which shows that when = 1/g(0) the coupling constant diverges under the RG, indicating, as previously the failure of perturbation theory. From this result we obtain the Kondo temperature of the problem as: kB TK = W e1/g(0) W eW/J and therefore the Kondo temperature is exponentially small with the Kondo coupling J. The experimental consequences of the formation of the bound states are immediate: the magnetic susceptibility of the impurity, instead of following the Curie law, 1/T has to saturate at T = TK so that imp (T ) 1 ; TK (10.163) (10.162)
the entropy of the system, instead of being S = kB N ln(2) as it would be for a spin 1/2 atom has to go to zero below TK because of the formation of the bound state, this implies that the specic heat, CV = T dS/dT , has to behave like CV,imp T TK (10.164)
Moreover, the Kondo effect leads to the so-called Kondo minimum in the electric resistance of these systems. On the one hand, formation of the virtual bound states or scattering resonances implies strong scattering (with the phase shifts close to /2 also known as the unitarity limit). The scattering increases below TK and therefore one expects the resistivity to increase with decreasing temperatures. On the other hand, phonon scattering decreases with decreasing temperature because as the system gets cooler there are less phonons in the system. The nal result is therefore a minimum in the resistivity. Another interesting property of the Kondo effect is its non-locality. Since EF kB TK for TK T EF /kB the only energy scale in the problem is the Kondo temperature TK . The only information of the electron gas that is important for this problem is the Fermi velocity, vF , since the states at the bottom of the Fermi sea are irrelevant for the physics discussed here. Using the Fermi velocity and the Kondo temperature
for T < TK . All these effects are observed at low temperatures (TK 1 5 K) in magnetic alloys.
200
we can construct a quantity with dimensions of length, K , that is called the Kondo screening length: K = h vF . kB TK (10.165)
The physical meaning of this quantity can be understood if we calculate the average value of the correlation function between the local spin S and the electron spin s(r), namely, S s(r) er/K (10.166)
which decays exponentially with distance from the impurity. Although this result has not been demonstrated here it seems almost obvious when we realize that the only length scale in this problem is given by (10.165). This result shows that for distances larger than K the electrons and the impurity are uncorrelated. Only electrons in a region of size 3 around the impurity site participate in the Kondo screening process. We can K estimate the number of electrons, NK , as: NK n3 (kF K )3 K EF kB TK
3
(10.167)
3 where we used that the electronic density n is essentially proportional to kF and used (10.165) with EF 4 eV we see that N 1012 electrons. That is, a huge number h vF kF . Since EF 1 eV and kB TK 10 K of electrons participate in the screening process.
10.9.1 Problems
1. Consider the Hamiltonian for the Kondo problem (10.120) in its isotropic form (Jz = J ) and assume that J EF and that the electrons have a spherical Fermi surface. (i) Using the rst Born approximation, calculate the scattering amplitude t(1) (k, k ) for the scattering of an electron from an state |k, to a state |k , as a function of J and S z . (ii) Show that in the second Born approximation the scattering amplitude, t(2) is given by: t(2) (k, k ) =
k ,
+ nk k , |HK |k , k, |HK |k , .
Here nk is the Fermi-Dirac occupation number. Notice that the order of the matrix elements matters! Explain why. (iii) Show that the above expression can be approximately written as: t(2) (k) 2 (J)2 S z nk . Ek Ek (10.169)
Assuming a constant density of states N (0) for the electrons calculate the sum above and show that it can be approximately written as: t(2) (k) 2 (J)2 S z N (0) ln(W/|EF Ek |) where W is the bandwidth. Notice that t(2) (kF ) is divergent! (iv) The total scattering by the impurity, t(2) can be obtained by integrating (10.170) around the (10.170)
201
Fermi surface in a region of energy of size kB T . Show that in this case t(2) is logarithmically divergent with the temperature. (v) The total resistance generated by the presence of the impurity is written as: (T ) = 0 (1 t(2) ) where 0 is the temperature independent part of the resistance (see Chapter 7). If the phonon contribution to the resistance is given by P (T ) = AT 5 show that this gives rise to a resistance minimum in the behavior of the total resistivity of the material.
(k )c ck, + U k,
ni, ni, .
i
(10.171)
Here we have introduced the chemical potential and work in the canonical ensemble keeping xed and allowing the number of electrons to uctuate. The complete solution of (10.171) was only obtained in one dimension via the so-called Bethe ansatz. In higher dimensions one has to use approximate solutions. In this section we are going to treat the problem in the Hartree-Fock approximation. This is an uncontrolled approximation. Although this kind of approach has its limitations, it can provide a deep insight on the physics of the problem. The complexity of (10.171) is associated with the treatment the interaction term. Because it involves four electron operators it is a highly non-linear problem. In mean eld theory we replace the interaction term by ni, ni, ni, ni, + ni, ni, ni, ni, (10.172)
where the averages are evaluated in the ground state. Notice that the factorization is done so that ni, ni, = ni, ni, , that is, the uctuations in the up and down spin channels are independent. It is convenient to rewrite the problem in terms of the total number of electrons per site and the magnetization as Ni = Mi = or equivalently, ni, ni, = = Ni + Mi 2 Ni Mi 2 ni, + ni, ni, ni, (10.173)
(10.174)
U 2
U 4
In what follows we are going to assume that state is always homogeneous and therefore Ni = n = Ne N (10.176)
where Ne is the total number of electrons and N the total number of sites. In this case we can rewrite (10.175) as HM F =
k,
(k )c ck, + k, Mi2 U 2
U N n2 U n + 4 2
(ni, + ni, )
i
U 4
Mi (ni, ni, )
(10.177)
where the second term in the rst line on the r.h.s. of (10.177) is a total energy shift that can be absorbed into the denition of the chemical potential, the third term is just a shift of the electron energy, the rst term on the second line on the r.h.s. is the energy required to magnetize the system and the last term is the interaction between the magnetization and the electron themselves. This term has the form of a local applied magnetic eld. Observe that in this problem Mi is an unknown and has to be calculated self-consistently. As usual in mean eld approaches one has to guess the nature of the ground state, that is, the form of the magnetization. In a ferromagnetic system all spins point in the same direction and therefore Mi = M0 for all sites. In an antiferromagnetic system the situation is more complicated because one can have different sub-lattices in which all the spins are up or all the spins are down. For instance, for a square lattice the magnetization can be given by Mi,j = M0 (1)i+j (10.178)
where each site is located at (ai, aj) where a is the lattice spacing. This classical state of the electron system is called the Nel state and it is shown on Fig. 10.9. Observe that although the magnetization changes from point to point in space the Fourier transform of (10.178) is quite simple M (q) (q Q) where Q = (/a, /a) is the so-called ordering vector. For a ferromagnet we have Q = 0. Here we are going to consider only the case of the ferromagnet and leave the antiferromagnet as a problem for the reader. In the ferromagnetic case we can rewrite (10.177) as HM F =
k,
(10.179)
(k )nk, +
2 U M0 N U M0 + 4 2
2 N U M0 + 4
U M0 2
nk, + k +
203
where we have used i ni, = k nk, . Observe that (10.180) is already in diagonal form since it only depends on the occupation in momentum space. The total energy only depends on the occupation of each spin state, that is, in the ground state nk, = (EF, k ) (10.181)
where EF, is the Fermi energy of each spin state and at this point it is an unknown. Moreover, M0 in (10.180) and it is also unknown. In order to calculate this parameters one has to minimize the total energy of the system as a function of them. The energy of the system is written as: E = +
2 N U M0 + 4
U M0 2
(EF, k ) (10.182)
U M0 k + 2
(EF, k )
As usual in this kind of problems it is useful to introduce the electronic density of states N () = 1 V ( k ) (10.183)
d U M0 2
U M0 2
N () (10.184)
d +
N ()
204
where s = N/V is the solid density. In order to calculate the unknown parameters in the energy density we have to minimize it so that E E EF, E M0 = = 0 = 0 (10.185)
where = Ne /V is the electron density. Furthermore, in order to make sure that the solution is really a minimum and not a maximum one has calculate the second derivatives of the energy with respect to the variational parameters. The reader can do it as an exercise. From (10.185) one nds
EF, EF,
dN () +
dN () =
EF,
EF,
U M0 N (EF, ) = 0 2
EF,
s M0
dN () +
dN () = 0 .
(10.186)
We can rst solve for and M0 using the two equations for the spins: = M0 = EF, + EF, 2 EF, EF, U
(10.187)
dN () +
EF,
dN () = dN () = s
EF,
EF, EF, U
(10.188)
which is the set of equation with dene EF, and EF, . After these are calculated one can obtain the chemical potential and magnetization from (10.187). Using (10.187) we can rewrite (10.188) as
EF,
= 2 s EF, EF, U
EF,
dN () + s dN () .
EF,
Observe that the last equation in (10.189) can be rewritten as s EF, EF, = U
EF, EF,
dN ( + EF, )
0
(10.190)
which for xed EF, is a single equation for EF, EF, . Since the density of states has a nite bandwidth
205
the integral in (10.190) is a monotonic function of x = EF, EF, and saturates at large values x as shown in Fig. 10.10. The solution of (10.190) is the intercept of a straight line of slope s /U and the r.h.s. of (10.190). Observe that there is a critical value of U , say, UC , above which one nds a solution with x = x = 0 and therefore from (10.187) corresponds to a system with nite magnetization, that is, a ferromagnetic state. For U smaller than UC the only solution is x = 0 which is the paramagnetic state. In order to calculate UC one has to linearize the r.h.s. of (10.190) in order to get UC = s N (EF, ) (10.191)
=2
dN ()
(10.192)
which completes the solution of the problem at U = UC . The condition for the existence of a ferromagnetic state for U larger than UC as given in (10.191) is called Stoner criterion. This criterion, which is based on a mean eld approach, is only a estimate of how large U has to be (or how large the density of states has to be) in order for the system to have a spontaneous ferromagnetic magnetization. As we are going to see in the next chapter the mean eld solution is an extreme approximation that has its limitations. It provides us, however, with the information that the Coulomb interaction is fundamental for the creation of an ordered magnetic state.
U<UC
U=UC
U>UC
x*
Figure 10.10: Diagrammatic solution of (10.190): solid line is the r.h.s. of (10.190).
10.10.1 Problems
1. Consider the problem of itinerant magnetism in an antiferromagnet in a square lattice in two dimensions in which the magnetization is given in (10.178). This can be rewritten as M (r) = M0 cos(Q r) where Q = (/a, /a).
206
and show that the Hamiltonian of the problem can be written as H= where H, = k
U Ms 2 z U Ms 2 z k 2 U Ms + 4
k,,
[ (k)] H, [ (k)]
where Ms is the staggered magnetization and z is a Pauli matrix. (iv) Diagonalize the above Hamiltonian and nd the new electronic spectrum. Show that a gap opens in the spectrum at the Fermi surface. Interpret your result in terms of the symmetry of the problem. (v) Calculate the total energy of the problem by lling up all the energy states below the gap (since the system is at half lling) and by minimizing it nd the equation that determines Ms . (vi) In this item we are going to use the Debye theory for phonons in the context of the antiferromagnet. Expand the spectrum around the (/a, /a) point to second order in the momentum and perform the integral of item (v). Note that you have to introduce a cut-off in the upper limit of the integral so it converges. This cut-off, say , can be calculated by using Debyes theory for phonons. What is the functional form of Ms as a function of U and t.
Chapter 11
(11.1)
where Jij > 0. Observe that the Hamiltonian (11.1) has a symmetry for the reversal of the spins, that is, we can reverse all the spins and the Hamiltonian remains the same. Mathematically the Hamiltonian is invariant z z under Sj Sj . The question here is: is the ground state of the (11.1) invariant under this symmetry? Firstly let us assume that the ground state is invariant under the overturn of spins. We can specify any state of the system in the terms of the eigenstates of S z , that is,
z z z |0 = |S1 , S2 , ..., SN .
(11.2)
(11.3)
207
(11.4)
(11.5)
Thus, if the ground state has the symmetry of the Hamiltonian then the average magnetic moment is zero. It is possible, however, that the ground state does not have the symmetry of the Hamiltonian, that is, it is not z z z invariant under Sj Sj . In this case we cannot conclude (11.5) and we can have that 0|Sj |0 = 0. If this happens the symmetry of the Hamiltonian is broken and the system acquires nite magnetic moment. In a system without disorder, that is, a system with the translation symmetry of the lattice, all the spins are equivalent and therefore it does not make sense to speak of a particular spin. In this case we dene an average quantity over all spins 1 M= N
N j=1 z 0|Sj |0
(11.6)
where M = 0 in the phase with broken symmetry. M is the magnetization of the system and it can be nite because the spins interact ferromagnetically so they can align along a given direction. If the ground state of the system is such that M = 0 then we can say that the symmetry is spontaneously broken since the ground state does not have the symmetry of the Hamiltonian. Observe, however, that we have two possibilities: either all the spins point in the up direction or they point in the down direction. For these two states the symmetry is broken and they are degenerate since they have the same energy. This degeneracy can be lifted by applying a magnetic eld to the system. In the presence of an innitesimal eld the degeneracy of these two states is lifted and the system picks the state with lowest energy. We say that the external magnetic eld is the symmetry breaking eld. At high temperatures we expect the magnetic moments to act independently, that is, paramagnetically. At these temperatures we would have M = 0 since the spins point randomly. If the system at zero temperature has M = 0 then there is a critical temperature Tc such that above Tc the system is disordered, that, is M (T > Tc ) = 0, and above which the system is ordered, that is, M (T < Tc ) = 0. At T = Tc we have a phase transition. Since the magnetization changes from zero to a nite value at the transition we say that M (T ) is the order parameter of the magnetic problem. Moreover we have two possibilities at T = Tc : i) the magnetization can jump discontinuously from M (T = Tc + ) = 0 to M (T = Tc ) = 0 at T = Tc ( 0) in which case we say that the transition is of rst order as shown in Fig.11.1(a); ii) the magnetization can go continuously from M (T = Tc + ) = 0 to M (T = Tc ) = 0 at T = Tc in which case we say that the transition is of second order as shown Fig.11.1(b). Observe that the fact that the magnetization changes at T = Tc implies that the magnetic susceptibility, that is, (T ) = M B (11.7)
B0
diverges at T = Tc . In order to understand why this is so, consider the plot of the magnetization as a function of the magnetic eld at xed temperature as shown in Fig.11.2. At T > Tc the magnetization is just paramagnetic and is zero at B = 0. If T < Tc one has M (T ) = 0 for B = 0. Thus, at T = Tc , the magnetization has innite curvature implying a divergent susceptibility.
209
M M
(a)
Disordered Phase Ordered Phase
T M M
0
(b)
Ordered Phase
Disordered Phase
Figure 11.1: Magnetization of a system during a phase transition: a) First order; b) Second order.
Although we have talked about ferromagnets the concept of broken symmetry, order parameter and symmetry breaking eld can be generalized to any system with long range order. As an example consider the case of an antiferromagnet such as the one shown on Fig. 10.9. Observe that in this case the system z indeed has Sj = 0 at each site but the total magnetization as dened in (11.6) vanishes even in the ordered phase. This is because the homogeneous magnetization is not the order parameter of the antiferromagnet. Let us look at the magnetization of an antiferromagnet as the one in Fig. 10.9. It is easy to see that the local magnetization at position r = (n, m)a is M (r) = Ms (1)n+m (11.8)
where Ms is a constant. Observe that the magnetization oscillates from site to site and therefore vanishes if we sum over all sites. However, it is clear from the equation above that Ms has to be nite in the ordered phase. We can obtain Ms from (11.8) if we multiply both sides by (1)n+m and sum over all sites: Ms = 1 N (1)n+m M (r) =
r
1 N
N z (1)j Sj j=1
(11.9)
which is called the staggered magnetization of the system. Furthermore, observe that the ferromagnetic state retains the symmetry of the lattice while in the antiferromagnetic case the unit cell doubles (one needs at least one spin up and one spin down in the same unit cell) and then the translational invariance of the lattice is also broken. Consider Fourier transforming the local magnetization in each case. For a ferromagnet the local magnetization is uniform, M (r) = M0 , and therefore its Fourier transform has Dirac delta peaks at k = 0 and all other reciprocal lattice vectors G. Thus the ferromagnet has the symmetry of the lattice. In the antiferromagnetic case as given in (11.8) the Fourier transform has Dirac delta peak at Q = (/a, /a) as you can easily show. Finally, the symmetry breaking eld in this case is not an homogeneous magnetic eld
210
M T<Tc T>Tc
0 M0 T<Tc
since it cannot differentiate between the two degenerate states of the antiferromagnet (which are obtained by ipping all the spins in the system). In the antiferromagnetic case a staggered eld is the symmetry breaking eld.
for T very close (and smaller) than Tc . is known as a critical exponent. The magnetic susceptibility is given by (T ) (T Tc ) for T > Tc and (T ) (Tc T )
(11.11)
(11.12)
for T < Tc since the susceptibility has to diverge on both sides of transition but not necessarily with the same exponents. Another exponents are dened for other physical quantities. For instance, the specic heat at xed eld also diverges at T = Tc and one has CH (T ) (T Tc ) for T > Tc and CH (T ) (Tc T )
(11.13)
(11.14)
211
for T < Tc . Another interesting property is that the magnetization does not have to be linear with the magnetic eld at T = Tc and we dene another exponent by M |B|1/ sgn(B) . (11.15)
Besides the critical exponents that dene the thermodynamic functions there are exponents that dene dynamical correlations. Correlation functions are important for experiments that measure spatial and temporal correlations among spins. In the phonon problem we showed that the dynamical form factor S(k, ) is given in terms of a density-density correlation function. Remember that the neutrons interact with the system of interest via the spin-spin interaction. Thus, neutron scattering is very sensitive to magnetic order. Like in the case of scattering of neutrons by phonons we would have Bragg scattering of the neutrons below Tc when the system has long range order. Observe that as in the case of atoms in a solid the Bragg peaks are directly related with the magnetic ordering in the system. In the case of the ferromagnet the magnetic ordering does not break the symmetry of the lattice and in the case where the ordering is commensurate with the lattice the Bragg peaks are at the reciprocal lattice vectors G. Thus, in a ferromagnet magnetic peaks superimpose to lattice peaks. This is not the case of an antiferromagnet where a Bragg peak appears at Q which is not a reciprocal lattice vector. As in the case of phonons, perfect order implies the presence of innitely sharp Bragg peaks. In the magnetically disordered phase we do not expect these peaks to disappear immediately but to become broader due to lack of long range order. In momentum space this broadening is given by 1/ where is called the magnetic correlation length. In order to understand the meaning of the correlation length let us assume for simplicity that the shape of the peak (or intensity) in the magnetically disordered phase is given by a Lorentzian shape I(k) = I0 / (k Q)2 + 1/ 2 (11.16)
where Q is the ordering vector (Q = 0 for the ferromagnet and Q = 0 for the antiferromagnet) and I0 is a constant. Observe that when the Lorentzian in (11.16) goes to a Dirac delta function at k = Q as expected in the case of complete order. Thus is a function of temperature and diverges at T = Tc . Moreover, it is easy to show that the intensity is proportional to the spin-spin correlation function for spins z z at different sites, Si Sj . In order to see how the correlation function behaves one has to Fourier transform (11.16) to real space. It is a simple exercise to show that
z z I(ri rj ) Si Sj exp
|ri rj | (T )
(11.17)
Now the meaning of the term correlation length is clear: it tells what is the characteristic distance above which the spins are uncorrelated. At T = Tc one expects the spins to be strongly correlated and (T ) diverges. One then denes the critical exponents for (T ) as (T ) (T Tc ) when T > Tc and B = 0 and (T ) (Tc T )
(11.18)
(11.19)
when T < Tc and B = 0. In some cases the correlation function decays algebraically with the distance (in
212
1 |ri rj |d2+
(11.20)
with a new exponent for a system in d dimensions. It turns out that these critical exponents are not independent of each other because of the thermodynamic relations between the various quantities. This is related with the so-called scaling hypothesis. Consider the free energy F (T, B) which we parameterize in terms of the reduced temperature t= T Tc . Tc (11.21)
The scaling hypothesis asserts that, close to the phase transition, it is always possible to nd two parameters at and aB such that F (at t, aB B) = F (t, B) for any value of the real number . In order to understand how the exponents are relate let us differentiate (11.22) with respect to B in order to get the magnetization: aB F (t, B) F (at t, aB B) = (aB B) B aB at aB M ( t, B) = M (t, B) . (11.22)
(11.23)
Consider the case where B = 0 and t 0. From (11.23) we have: M (t, 0) = aB 1 M (at t, 0) that is valid for any . In particular for = we nd M (t, 0) = (t) But from (11.10) we have in the limit of t 0 = Let us now take t = 0 and let B 0 in (11.23) M (0, B) = aB 1 M (0, aB B) and choose = B 1/aB in order to get M (0, B) = B
1aB aB 1aB at
(11.24)
1 t
1/at
(11.25)
M (1, 0) .
(11.26)
1 aB . at
(11.27)
(11.28)
M (0, 1)
(11.29)
213
(11.30)
Similarly we can derive twice with respect to B in order to obtain 2aB (at t, aB B) = (t, B) . Taking the limit of B = 0 rst and choosing = (t)1/at we can prove = 2aB 1 at (11.32) (11.31)
implying that the susceptibility diverges with the same exponents on both sides of the transition. Moreover, combining (11.32) with (11.27), (11.30) and (11.33) one nds = ( 1) (11.34)
which relates three different exponents. Using exactly the same arguments for derivatives with respect to the temperature we can show that + ( + 1) = 2 and = and combining (11.35) with (11.34) one nds + 2 + = 2 . (11.37) (11.36) (11.35)
Many other relations between the exponents can be obtained in this way. Although the scaling hypothesis is an assumption and cannot be proved, it makes many predictions for critical exponents that can be checked experimentally or analytically for specic models. This hypothesis has been very successful in explain critical behavior in systems with second order phase transitions.
i,j
z z Si Sj gB
z Si B . i
(11.38)
(11.39)
z Moreover, we assume that the magnetization is uniform so that Si = M in which case Hamiltonian (11.38) is replaced by
HM F = gB where
z Bef f Si i
(11.40)
Bef f = B + M (T, B)
(11.41)
is the effective magnetic eld, Z is the number of nearest neighbors and M (T, B) is the magnetization given in (11.6) and = 2ZJ (gB )2 (11.42)
is the so-called molecular eld. Observe that (11.40) is the Hamiltonian of a set of independent spins S in a magnetic eld. This problem was studied for paramagnets in Chapter 8. In particular the magnetization of this problem is given in (1.37), M (T, B) = M0 BS [gSB Bef f ] = M0 BS [gSB (B + M (T, B))] (11.43)
where M0 = gB S. Notice that (11.43) has to be solved self-consistently to give the magnetization. For B = 0 this equation reduces to M = M0 BS [gSB M ] (11.44)
which is shown graphically on Fig.11.3. Observe that this equation has the trivial solution M = 0 for all T . This equation has also a non-trivial solution M = 0 if the slope of the straight line is larger than the slope of Brillouin function. Thus let us consider the Brillouin function close to M = 0. The expansion of BS (x) for small x is BS (x) (S + 1)(1 + 2S + 2S 2 ) 3 S +1 x x 3S 90S 3 (11.45)
and therefore there is a non-trivial solution for temperatures T < Tc . In order to calculate Tc we use (11.44) to nd where the slopes in both sides of (11.44) equal each other. That is, 1 = gB S S + 1 gB S 3S kB Tc (11.46)
215
(11.47)
(11.48)
B 1.5
0.5
M -4 -2 2 4
-0.5
-1
-1.5
For simplicity let us consider the case of a S = 1/2 spin in which the Brillouin function reduces to B1/2 (x) = tanh(x/2) and (11.43) reduces to M = M0 tanh (gB (B + M )/2) . In order to simplify our calculations we dene m = = M M0 T Tc (11.49)
(11.50)
gB B m + 2
(11.51)
which can be rewritten with the help of trigonometric identities as b= where b = tanh gB B 2 . (11.53) m tanh(m/ ) 1 m tanh(m/ ) (11.52)
Let us consider the case that T Tc where m 0 and expand (11.52) as bm 1 In zero eld, b = 0, (11.54) gives m2 3 2 1 1 (Tc T ) (11.55) 1 + m3 1 1 + 3 3 1 1 . (11.54)
At the critical temperature, that is, = 1, and small elds one nds m3 gB B 2kB Tc 3 which, by comparison with (11.15), gives M F = 3 . (11.58) (11.57)
In order to calculate the susceptibility we observe that (T ) = = If we differentiate (11.54) with respect b we get m b 1 1 + m2 3
1
M m b M = B m b B C m . T b
(11.59)
(11.60)
217
(11.61)
In the disordered phase (T > Tc ) we have m = 0 and therefore (T ) which from (11.11) gives M F = 1 while in the ordered phase (T < Tc ) we use (11.55) and nd (T ) which from (11.12) we nd M F = 1 .
C T Tc
(11.62)
(11.63)
1 C 2T Tc T
(11.64)
(11.65)
Observe that the scaling hypothesis predicts from (11.35) and (11.36) that M F = M F = 0 which implies that the specic heat does not diverge at the transition.
Ji,j i j
i B .
i
(11.66)
where we assume that Ji,j only depends on the relative distance |ri rj | between the two spins. The partition function of the problem is given by Z[B] =
{i }
eE({i },B)
(11.67)
where the sum is over all the spin congurations. Observe that (11.67) is quite useful since the magnetization at zero eld of the system is M = =
{i } j
E({i },B=0) {i } e
1 Z[B = 0]
.
B=0
(11.68)
218
dxex
2 /(4a2 )+sx
2a ea
2 s2
(11.69)
Using (11.66) in zero eld (B = 0) and the above identity we can write Z =
{i } N
i,j
Ki,j i j
dj e 4
1
i,j
1 i Ji,j j +
i i
{i } j=1 N j=1
dj e 4
i,j
1 i Ji,j j
e
{i }
i i
(11.70)
i i
=
i
2 cosh(i ) = 2e
ln(cosh(i ))
(11.71)
Moreover, we dene, i =
j
1 1 J j 2 i,j
(11.72)
dj e
i,j
i Ki,j j +
ln(cosh(2
Ki,j j ))
(11.73)
j=1
Observe that what we have done was to trade a sum over discrete variables for an integral over a continuous variable. But in doing so we have generated a non-linear term in the partition function.
(11.74)
Observe that in this case the partition function in (11.73) simplies considerably since we can change variables to a center of mass" variable = 1 N i
i
(11.75)
219
d eN {K
2 ln(cosh(2K))
(11.76)
which is a simple integral but with a very complicated integrand. Notice that the exponent in (11.76) scales with N the number of spins in the system. But we are interested in the limit of N in which case the integral is dominated by the saddle point value of the integral that is, the minimum of the function S() = K 2 ln (cosh(2K)) which is plotted of Fig.11.4. The minimum of this function is given by S = 0 = tanh(2K) (11.77)
(11.78)
which gives the value of . Observe that (11.78) is identical to (11.49) with M . In particular, if K < 1/2 (T > Tc = 2J/kB ) the only solution is = 0 and for K > 1/2 (T < Tc = 2J/kB ) a solution with = 0 exists. This simple calculation shows that mean eld theory is equivalent to solve the problem with interactions with innite range! Thus, we should suspect that in the real case with short range interactions uctuations around the mean eld solution are going to be very important. In order to understand the how the phase transition occurs let us consider the free energy for the order parameter as given by (11.77). Let us expand S() to fourth order in (ln cosh x x2 /2 x4 /12): S() K(1 2K) 2 + 4K 4 4 . 3 (11.79)
Observe that the quadratic term changes from negative to positive when K > 1/2 to K < 1/2. Therefore the curvature of the free energy at = 0 changes at K = 1/2. In this case the ground state of the system for K < 1/2 which is given by = 0 becomes metastable at K = 1/2. For K > 1/2 the state with = 0 becomes unstable and the ground state is given by the two solutions of (11.78) which are degenerate reecting the symmetry of the problem. It is clear that the full form of (11.77) is not needed. The physics of the problem is already present in (11.79). The term proportional to 4 is only need to keep the free energy bounded from below in the ordered phase. Without this term the system would be thermodynamic unstable since the ground state would be given by with innite negative energy when K > 1/2 which is clearly unphysical. In the disordered phase K < 1/2 the free energy is naturally bounded and the term in 4 is irrelevant for the physics.
(11.80)
On the one hand, if K(r) has innite range then K(k) is a Dirac delta function at k = 0. On the other hand, if K(r) is localized then K(k) is a smooth function of k. Therefore, if we work in momentum space we can
220
Sy 3.5
2.5
1.5
0.5 y -4 -2 2 4
Figure 11.4: Plot of eq. (11.77): dashed line: K < 1/2; continuous line: K > 1/2.
always work with functions that can be expand in power series. So let us introduce the Fourier transform of i , 1 (ri ) = N eikri (k) .
k
(11.81)
Moreover, because K(r) and (ri ) are real functions we must have K(k) = K (k) and (k) = (k). As we explained in the case of the interactions with innite range interactions we do not need to work with the full free energy (11.73). Instead we are going to use a truncated form of the free energy 2 4 3 4
i,j
i Ki,j j 2
For the moment being we are going to focus on the quadratic terms. It is very easy to show with the use of these Fourier transforms the following relations i Ki,j j
i,j
Ki,j j +
Ki,j j .
(11.82)
=
k
K(k)(k)(k)
Ki,j j
=
k
K(k)K(k)(k)(k) .
(11.83)
(11.84)
221
Observe that if the interaction was a Dirac delta function in real space (local interaction) then Kk) should be a constant. Let us assume that K(k) is a real function of its argument. Then (11.84) as F0 =
k
(11.85)
This equation is very important because it tells us that if there is a wave-vector Q for which K(Q) > 1 2 (11.86)
then the system is unstable at that particular wave-vector. Observe that this condition implies that the phase transition happens to the component (Q). Since is essentially the magnetization of the system is implies that the order parameter of the system is given by M (r) = M0 eiQr . This is in complete agreement with our previous discussion. (11.87)
Let us now go back to the ferromagnetic problem and study what happens at the unstable wave-vector, that is, k = 0. Close to k = 0 we can expand K(k) as K(k) K0 1 (k)2 + O(k4 ) (11.88)
where is a constant with dimensions of length. From now on we are going to disregard all the terms of order k4 and higher. Observe that K0 is the Fourier component of K(k) with zero momentum, that is, K0 =
r
K(r) = ZJ
(11.89)
where we have assumed that the interaction only couples the Z nearest neighbor atoms. Direct substitution of (11.88) into (11.84) one gets F0 =
k
(11.90)
Observe that, exactly like in the case of the mean eld theory when k = 0 this part of the free energy becomes negative when K0 > 1/2 which denes the critical temperature Tc = 2ZJ . kB (11.91)
At this level of approximation we are recovering the mean eld approximation. If we are interested only in temperatures close to Tc then we can write 1 + O(T Tc ) 2 T Tc + O(T Tc )2 1 2K0 Tc 4K0 1 1 + O(T Tc ) K0
(11.92)
T Tc + (k)2 |(k)|2 . Tc
(11.93)
It is usual to rewrite (11.93) in a slightly different way by dening (k) = (k) T Tc 2 = 2 Tc in which case (11.93) becomes F0 1 2 2 + k2 |(k)|2 . (11.95)
(11.94)
If one Fourier transform (11.95) back to real space we nd F0 = 1 2 dr ((r))2 + 2 ((r))2 (11.96)
which actually has a very interesting form. Let us calculate, for instance, the correlation function for for the elds when they are separated by a distance r, (r)(0) =
F0 r d(r)(r)(0)e F0 i d(ri )e
(11.97)
Notice however that the integration can be done easier if we Fourier transform since (r)(0) =
k,k
eikr (k)(k )
(11.98)
and (k)(k ) =
2 p d(p)d(p)(k)(k )e 2 p d(p)d(p)e
1 1
2 2 2 q ( +q )|(q)|
and therefore, after a simple Gaussian integral, (k)(k ) = and therefore from (11.98) we have (r)(0) =
k
k,k k 2 + 2
(11.100)
eikr . k 2 + 2
(11.101)
Observe that the integral above has poles at k = i which indicates that the correlation function will decay exponentially at large distances, that is, (r)(0) er/ (11.102)
223
(11.103)
is the magnetic correlation length. Observe that the magnetic correlation length diverges at the critical temperature when we lower the temperature from the disordered phase. Comparing with (11.18) we nd a new exponent M F = 1/2. Up to this point we have not talked about the non-linear term in (11.82). Observe that close to the phase transition the value of the magnetization, or , is very small and therefore we can keep only the k = 0 part of K(k). In this case the whole free energy simplies to a very simple form which is F = dr 1 2 ((r))2 + ((r))2 + ((r))4 2 2 4! (11.104)
4 where /4! = K0 /4 is a positive constant. (11.104) is known as the Ginzburg-Landau free energy which was proposed by Ginzburg and Landau only based on the symmetry and the existence of a second order phase transition. This kind of free energy is the heart of many discussions of the problem of phase transitions in magnetic and non-magnetic systems.
2 (r) 2
(r) 4!
B (r)
(11.105)
where B is the symmetry breaking eld. Observe that by construction the free energy (11.105) is invariant under rotations of the order parameter . This is the so-called O(N ) symmetry. The XY model which involves only the X and Y components of the spin is a particular case of N = 2 while the Heisenberg model has N = 3 since it involves all the components, namely, X,Y and Z. The main difference between (11.105) and (11.104) for N > 1 is the fact that one can rotate continuously the vector in the N dimensional space. Let us consider rst the case where N = 2 and the components of the vector are constant in space (in which case we disregard the derivatives in the action). The homogeneous
U (1 , 2 ) 2 2 = (1 + 2 ) + (2 + 2 )2 2 2 V 2 4! 1
(11.106)
where V is the volume of the system. The invariance of the system under rotations becomes obvious if we plot this free energy in the space of (1 , 2 ). If 2 > 0, that is, in the disordered phase, the minimum of the energy is at 1 = 2 = 0 but if 2 < 0 the potential energy looks as in Fig. 11.5. In this case the ground state is innitely degenerate, that is, we can rotate continuously the vector around the origin without changing the energy of the system.
20 U 0 2
-20 0 -2 f2
0 f1 2 -2
Figure 11.5: Energy potential for a O(2) model. The heart of Goldstone theorem is to state that it requires no energy to move the order parameter continuously along the minima of the potential. Thus, if an innitesimal eld is applied in this particular direction we can rotate the order parameter by a nite amount. The susceptibility of the system in this particular direction is therefore innity! That is, we get a nite response with an innitesimal cost. It implies that there must be excitations of the system which cost no energy in the direction transverse to the ordering direction. This is Goldstone theorem: if a system has a spontaneously broken continuous symmetry there is always an excitation of the system in the ordered phase that costs no energy to excite. We have seen already a good example of an application of this theorem: when a set of atoms forms a crystal the translational invariance (which is a continuous symmetry) is lost. This has to be contrasted with the liquid or gas phase where the translational symmetry persists! In the case of the crystal we know very well what is the order parameter: it is the Fourier transform of the density, (q) which now have Dirac delta peaks at the reciprocal lattice vectors G. The peaks only exist if the system is crystalline and they disappear in the gas phase. Thus, Goldstones theorem predicts the existence of modes with zero energy in this phase. But what are these modes? If you go back a few chapters you will nd that there are indeed modes which are gapless in this phase: acoustic phonons! Because acoustic phonons have zero energy at zero wavevector an innitesimal energy is required to excite them! There are many other examples of Goldstone modes which we will discuss in the next chapter and these are very important to understand the
11.5. PROBLEMS
behavior of systems in ordered phases with spontaneously broken continuous symmetries.
225
11.5 Problems
1. Show that the Fourier transform of the local magnetization of a two dimensional antiferromagnet has Dirac delta peaks at Q = (/a, /a) and Q + G where G is a reciprocal lattice vector. 2. Consider the case of an one dimensional magnet with a Lorentzian neutron scattering line shape and show that the spin-spin correlation function decays exponentially with the correlation function . 3. What is the neutron line shape corresponding to the correlation function given in (11.20). 4. Prove (11.31), (11.32), (11.33) and (11.34). 5. Prove (11.35), (11.36) and (11.37). 6. Prove (11.52) and (11.54). 7. Prove that M F = M F = 0 by calculating the specic heat of a ferromagnet in the mean eld approximation. 8. Consider the O(N ) model where the magnetization in the direction is: M = = M n where M is the magnitude of the magnetization vector M and n is an unit vector. Dene susceptibility tensor = 2U , B B
where , = 1, ..., N are the components of the order parameter and B are components of magnetic eld in the N directions. Using the chain rule U B = = where B =
N 2 =1 B ,
B U B B B U B B
where 2U B 2 M 1 U = . (B) = B B B || (B) = Observe that || (B) gives the susceptibility of the system in the direction of the order parameter M while (B) gives the susceptibility in the direction transverse to M. While || (B) is nite in the
226
Chapter 12
Magnetic excitations
In the last chapter we have seen that systems with continuous broken symmetries exhibit gapless modes in the excitation spectrum. The clearest example of a Goldstone mode is the acoustic phonon that exists in crystals due to the breaking of the translation symmetry. However, crystals also support optical modes that have a gap. Optical modes are related to the breaking of a symmetry that is discrete: the doubling of unit cell. Therefore, when the system of interest has a particular symmetry that is broken a new mode appears in the spectrum. If the symmetry is continuous the mode is gapless and when the symmetry is discrete the mode has a gap. Since experimentally speaking we are always probing excited states of a given system it is important to understand the nature of these excitations. As we have seen, the nature of the ground state and the symmetries characterize it determine the nature of the excitation spectrum. In this chapter we are going to study the excitation spectrum of magnetic systems and distinguish between the various universality classes: the Ising model that has a discrete symmetry of spin inversion, the Heisenberg model that has the full symmetry of rotation in spin space, and the XY model that has the symmetry of rotation in a plane.
H=
z z z Ji i+1 + hi i=1
(12.1)
where J > 0 is the magnetic exchange and h = B gB is the magnetic eld energy. Since (12.1) is completely described in terms of the z operator we can use the eigenstates of this operator (namely, z i |i = i |i ) to write the energy of the system as:
N
E[{}] =
(Ji i+1 + hi )
i=1
(12.2)
which is a functional of the conguration {1 , 2 , ..., N } (we will assume periodic boundary conditions so that N +1 = 1 ). Observe that this problem is classical in the sense that the states of the system are generated by classical congurations of the spins. There are no quantum mechanical effects such as spin ips (that is, a tunneling of the spins between different congurations). The ground state of the system is obviously ferromagnetic and described by the conguration {+1, +1, ..., +1} if h > 0. Although the 227
228
ground state conguration is straightforward the excited states are not so obvious. Consider, for instance a spin ip in the system given by the conguration {+1, +1, 1, +1..., +1}. This spin ip costs 2J in energy (h = 0). Actually this state is highly degenerated because one can ip a spin anywhere along the chain. Now consider a two spin ip which is given by {+1, +1, 1, +1..., +1, 1, +1} which costs 4J in energy and compare with another two spin ip where the ipped spins are neighbors {+1, +1, 1, 1, ..., +1}. This last state costs only 2J of energy. Actually the state where the ipped spins are all together only cost 2J of energy, no matter how long the ipped sequence is. Thus, there is an enormous amount of degenerate states in the problem.
The question that comes out here is: at what temperatures does the system order magnetically? In order to answer this question one has to calculate the partition function for the problem, Z = tr(eH ) =
1 ,...,N =1
eE[{}] .
(12.3)
The whole problem in calculating this partition function is related with the problem of calculating the contribution of the interaction term. In order to evaluate this contribution we are going to use a trick which is valid for spin 1/2 particles. Consider the following matrix element: |e |
x
= cosh(), + sinh(),
where we used that ( x )2 = 1 and that x | = | . The above identity can be written in a more interesting way as: |e
x
= =
sinh() cosh()
cosh() sinh()
sinh(2) ln(coth()) e 2 2
(12.5)
as you can easily show. Thus, in (12.3) the interaction term in (12.1) can be written as eJi i+1 = sinh(2) 2
1/2
i |e |i+1
(12.6)
Observe that in this case the partition function can be written as:
N
Z=
1 ,...,N i=1
sinh(2) 2
1/2
i |eh e |i+1 .
i =1 |i
(12.8)
Now using the fact that the states are complete, that is,
229
| eh e tr eh e
z x
| (12.9)
Let us consider the case without the magnetic eld (h = 0). In this case (12.9) is straightforward to calculate: Z= sinh(2) 2
N/2
tr eN
(12.10)
which is the partition function of a completely different problem from the initial one, namely, one which is described by the Hamiltonian: H = x where = N N = ln(coth(J)) . 2 (12.12) (12.11)
Observe that we have mapped our original problem (12.1) of a chain of Ising spins (a classical onedimensional problem) into a problem of a single spin in a transverse eld of strength ( this is a zero dimensional problem). How this is possible? What we have shown here is an essential feature of relationship between quantum statistical mechanics in d dimensions and the classical statistical mechanics in d + 1 dimensions. The eigenstates of problem (12.11) are symmetric (|s = (| + 1 + | 1 )/ 2) and antisymmetric (|a = (| + 1 | 1 )/ 2) combinations of eigenstates of z with energy and +, respectively. Thus, if we prepare the state of the system at time t = 0, say, with spin up (that is, | + 1 ) quantum mechanics tells us that for t > 0 the state of the system will be: |(t) = 1 h h eit/ |s + eit/ |a 2 i sin (t/ ) | 1 h = cos (t/ ) | + 1 h (12.13)
that is, the spin will oscillate from up to down with time. Thus, if we imagine the time evolution as a line with N steps we see that along the line the spins are never ordered, except when = 0 in which case |(t) = | + 1 for all time t. In this case the chain is ferromagnetically ordered! Thus, with this argument one nds that the system is only ordered when = 0. Going back to (12.12) we see that this is possible in the limit of N when = , that is T = 0. For any nite , no matter how small, t as N and the system has to be disordered. Furthermore, from (12.10) we can compute the partition function and free energy of problem since we have a problem of a single 1/2 spin. We can diagonalize the operator that appears in the exponent of (12.10) which has the eigenvalues: E = N ln(coth(J)). 2 (12.14)
cosh
N ln(coth(J)) 2
(12.15)
1 F = ln(2 cosh(J)) N
(12.16)
which is the exact free energy of the 1D Ising model at any temperature. Observe that the free energy is a smooth function of the temperature: for kB T J we have F JN which is the energy of the fully aligned ferromagnetic state and for kB T J we nd F ln( 2)kB T .
In order to calculate the magnetization as a function of temperature we need to re-examine (12.9) again. Observe that because we are dealing with Pauli matrices we can write: eh e =
z x
cosh()eh sinh()eh
sinh()eh cosh()eh
(12.17)
and it is easy to show that the partition function in (12.9) becomes: Z = tr e(h+J) e(h+J) e(hJ) e(hJ)
N
(12.18)
Since the trace of a matrix is invariant to an unitary transformation (since tr(A) = tr(U 1 U A) = tr(U AU 1 )) we rewrite (12.18) as Z = tr(AN ) = tr U AU 1
N
= tr(N ) = N + N +
(12.19)
where is a diagonal matrix (U AU 1 = ) and its eigenvalues. Thus, in order to solve the problems we just have to diagonalize the matrix that appears in (12.18). It is a simple matter to show that the eigenvalues are: = cosh(h)eJ sinh2 (h)e2J + e2J (12.20)
which ends with the solution of the problem. Observe that + > and since we are interested in the limit of N we can rewrite (12.19) as: Z = eF = N + F 1 = ln cosh(h)eJ + N sinh2 (h)e2J + e2J
(12.21)
which is the exact expression for the free energy of the problem and reduces to (12.16) when h 0.
231
M 1
0.8
0.6
0.4
0.2
Figure 12.1: Magnetization of the one dimensional Ising model as a function of external eld for different temperatures (temperature increases from top to bottom).
Observe that the magnetization per spin in the problem is given by: M (h) = = F/N H sinh(h) sinh2 (h) + e4J
(12.22)
Notice that (12.22) is a quite interesting function. If we take the limit of h 0 with nite we nd M (h 0) = 0 that is, there is no magnetization in the system at any nite temperature. If however we take the limit of before we take the limit of h 0 we nd M (h 0) = 1 ! That is, the system is magnetized at zero temperature as we have argued before. Therefore, the magnetization is a singular function of the temperature since there is a discontinuous jump in the magnetization at T = 0. A most amazing effect! The approach to zero temperature is shown on Fig.12.1. The problem of phase transitions in one-dimensional systems can be understood very well by an argument due to Landau-Lifshitz. Consider a state with a nite magnetization M and free energy F . In the ordered state the free energy would be given by: F = E TS (12.23)
where E is the energy of the ordered state and S its entropy. Consider the congurations discussed at the
232
beginning of this section in which all the spins to the right of a point n are reversed creating a domain wall: (+1, +1, +1, ..., +1, 1, 1, 1). The internal energy of the system will be increased by an amount E 2M 2 J (M = 1 at T = 0). But we can choose the point n at any point of the lattice of N sites and therefore there is an entropy of the order S kB ln(N ) associated with the creation of the domain wall. Thus, the net change in free energy due to the creation of domain wall is F 2M 2 J kB T ln(N ) . Observe, therefore, that if we make N sufciently large, that is, N > N = e
2M 2 J kB T
(12.24)
(12.25)
the free energy of the problem becomes negative and the system becomes unstable. Thus, we have to conclude that M = 0 for any nite temperature. Observe that at T = 0 order is possible because N .
gB H
z Si i
(12.26)
which describes an anisotropic magnet in a external magnetic eld H. We rst point out that the total spin z z of the system, S2 = ( i Si )2 and the z component of the total spin, ST = i Si , commute with the T Hamiltonian. Thus, we can classify all the states with respect to their eigenstates, namely, ST (ST + 1) z z and ST . Observe that ST = SN, S(N 1), ..., 0 while ST = ST , ..., ST has 2ST + 1 possible values. We will consider two important case of this Hamiltonian: the ferromagnetic case with Jz > 0 and the antiferromagnetic case with Jz < 0. As we are going to see these two cases have completely different excitation spectrum and physical properties.
(12.27)
which the raising and lowering operators for spins. In this case Hamiltonian (12.26) is rewritten as H=
z z Jz Si Sj + i,j
J + + Si Sj + Si Sj 2
gB H
z Si i
(12.28)
which clearly shows that J ips the spins. Classically this term is equivalent to the precession of the magnetic moment around the z axis. Suppose the magnetic system is in the ordered phase. One expects the total magnetization of the system to be very close to N S. As in the case of phonons one expects that the uctuations can be described by
233
(12.29)
The problem is to nd out the relationship between these bosonic operators and the original spin operators. This was done by Holstein and Primakoff. They propose to write the spin operators as:
z S i = S b bi i + Si
2S 2Sb i
b bi 1 i 2S b bi 1 i 2S
1/2
bi
1/2
Si
(12.30)
which are known as the Holstein-Primakoff transformation. These transformations have nice properties: 1) using (12.29) it can be shown that the operators dened in (12.30) obey the spin algebra; 2) it is easy to show that S2 = S(S + 1) as it should be. i Observe that at zero temperature, because the excitations are bosons, we expect b bi = 0. Thus it is i reasonable to assume that b bi /S 1 and we can expand the transformation in (12.30) as i
z S i = S b bi i + 2Sbi Si Si 2Sb i
(12.31)
which, of course, is not a true operator relationship but can be shown to be a good approximation a posteriori. z Moreover, the operator Si is unchanged. The physical meaning of this transformation becomes clear now. z Si in (12.30) measures the number of overturned spins. This ipping of spins is generated by Si . If we imagine that the number of spins equals the number of bosons in the system then Si behave like creation and annihilation operators for bosons. If we substitute (12.31) into (12.28) and keep only the leading order in the operators we obtain: H = Jz N ZS 2 J N ZS gB HN S + H where H = (2Jz Z + gB H) b bi J S i b bi + b bj i j
i,j
(12.32)
(12.33)
Observe that H has exactly the same form of the tight binding Hamiltonian for electrons moving on a lattice with the difference is that now we have bosons. Thus, like electrons, the ground state is made out of Bloch waves. For the problem where there is just one atom per unit cell solve this problem exactly in the same way as before, that is, by Fourier transform: bk = bi = 1 N 1 N eikrj bj
k
eikrj bk
j
(12.34)
234
and since the system is periodic the wavevector is dened only in the rst Brillouin zone (also called magnetic Brillouin zone). The Fourier transform changes (12.33) into H= where Ek = 2Jz Z + gB H 2J SZk and k = 1 Z eik
Ek b bk k
k
(12.35)
(12.36)
(12.37)
where is the vector that links two nearest neighbor atoms. The excitations described in Hamiltonian (12.35) are called ferromagnetic magnons. Observe that if J = 0 the magnon spectrum is dispersionless and therefore there is no propagation! This is the case of the Ising model. If there is a nite J (or nite XY component) the magnons can propagate since there is a nite group velocity ck = Ek . Let us consider now the case of long wavelength magnons, that is, in the limit where |k | 1. In this case we can write k 1 (ka)2 /2 where a is the lattice spacing in a cubic system. Therefore from 12.36) we have Ek 2(Jz J )Z + gB H + J S(ka)2 which can be rewritten as Ek + where = 2(Jz J )Z + gB H 1 m = . 4J Sa2 k2 2m (12.39) (12.38)
(12.40)
Observe that (12.39) has the form of the dispersion of a free particle with mass m . This particle is the magnon. Observe that in the limit of J 0 the mass of the magnon diverges indicating that it localizes in the system. One could ask how these magnons are related to Goldstones theorem. Goldstones theorem requires the system to have a continuous symmetry which in the case of this Hamiltonian is obtained in the Heisenberg limit (J = Jz ) and in the absence of elds (H = 0) which leads to = 0, that is, there is no gap in the system at k = 0 implying that the gapless ferromagnetic magnon is the Goldstone mode. From (12.31) we can calculate the magnetization per atom of the system immediately M =S 1 N nk
k
(12.41)
where nk = b bk is the occupation number operator for magnons. This number is given by the Bosek Einstein occupation as for the case of phonons. We have M =S 1 dd k 1 eEk 1 (12.42)
235
where = N/V is the density. Observe that at low temperatures ( ) the integral is dominated by the long wavelength behavior and we can use (12.39) with d = 3: M = S
1 k2 dk k2 /(2m ) 2 2 0 e e 1 x1/2 1 . dx x = S 2 (2m kB T )3/2 2 e e 1 0
(12.43)
In the absence of external eld and in the fully isotropic system, J = Jz , we have = 0 and the last integral is a constant and therefore the magnetization decreases with a power T 3/2 of the temperature showing that indeed the magnons tend to disorder the classically ordered state. If = 0 the magnetization decreases exponentially with temperature. The energy per atom of the system is of course obtained directly from (12.35). Again, at low temperatures, we use (12.31): E = 1 N Ek nk
k
= +
(12.44)
which shows that in the absence of magnetic eld and anisotropy the energy of the system goes like T 5/2 and because of that the specic heat behaves like CV T 3/2 .
Let us now consider the case when Jz < 0 in (12.26). In this case the spins orient anti-parallel to each other in a structure similar to the one shown in Fig.10.9 which is called the Nel state. We are going to assume a similar structure where only two sublattices, A and B, are present each one with N/2 sites (that is, two different ferromagnetic lattices which are immersed into each other). This is due to the fact that an antiferromagnet breaks the translational symmetry of the original lattice and doubles the unit cell. This, of course, does not happen in a ferromagnetic system. In sublattice A the magnetization is essentially z z Si = +S while in sublattice B we have Si = S. Thus, in dening the spin operators in terms of bosons one has to be careful about each sublattice. The easiest way to proceed is to follow the recipe of the
z SB,i = S + b bi i + SA,i
2S
a ai 1 i 2S b bi 1 i 2S
1/2
ai
1/2
SB,i
2S 2Sa i 2Sa i
bi
1/2
SA,i
a ai 1 i 2S a ai 1 i 2S
1/2
+ SB,i
(12.45)
Notice that the role played by the lowering and raising spin operators are exchanged between the two sublattices because they have different orientation of the spins. Using the same approximations as in the ferromagnetic case and Fourier transforming the bosonic operators you can easily show that the Hamiltonian (12.28) z with J replaced by J and in the presence of a staggered magnetic eld gB Hs j (1)j Sj becomes: H = N ZJz S 2 N B Hs S + H where H = ZS J k a b + ak bk + Jz a ak + b bk k k k k
k
(12.46)
+ B H s
k
a ak + b bk k k
(12.47)
which has a structure which is very different from the ferromagnetic case since the operators from different sublattices mix with each other. The mixing of these operators becomes more obvious if one rewrites (12.47) as: H = N (Jz ZS + B H) + where k = and [H(k)] = Jz ZS + B Hs J ZSk J ZSk Jz ZS + B Hs . (12.50) ak b k (12.49) [H(k)]k k
k
(12.48)
237
(12.51)
(12.52)
with k and k are bosonic operators. The simplest transformation has the form U (k) = uk vk vk uk (12.53)
2 where uk and vk are real. Moreover, because the operators are bosons we must impose that u2 vk = 1. k Thus,we can write
uk = cosh(k ) vk = sinh(k ) . This implies that the inverse of U is U 1 (k) = uk vk vk uk (12.55) (12.54)
and therefore U is not unitary (U 1 = U ). Actually it is easy to see that U = U and therefore the equation for the energy in (12.51) is U HU = or HU = U 1 which is not our usual eigenfunction problem. Observe moreover that these matrix can be written in terms of Pauli matrices, H(k) = E0 I + E1 (k) x U (k) = uk I + vk x U 1 (k) = uk I vk x (12.56)
where E0 = Jz ZS + B Hs and E1 = J ZSk . Thus, the equation HU = U 1 leads to the following set of equations: ( E0 )uk E1 vk = 0
E1 uk + ( + E0 )vk = 0
(12.57)
k = = and moreover we nd
(12.58)
tanh(2k ) =
E0 . E1 (k)
(12.59)
238 Thus, the Hamiltonian becomes H = N E0 + which has the appropriate diagonal form.
k k + k k + 1 k k
(12.60)
Let us now look for the Goldstone mode. In the absence of a staggered eld and in the isotropic limit (Jz = J ) the dispersion (12.58) becomes k = 2JZS and in the long wavelength limit we nd (k) cs k (12.62)
2 1 k
(12.61)
where cs = 4 3JSa is the spin wave velocity in three dimensions. Observe that as indicated by Goldstone theorem the antiferromagnetic magnon is gapless with linear dispersion relation. This has to be contrasted with the ferromagnetic case where the dispersion goes like k2 . Like the case of acoustic phonons the antiferromagnetic magnons have a characteristic velocity. Ferromagnetic magnons, on the other hand, have Galilean invariance which is characterized by an effective mass which determines the curvature of the dispersion. Let us now consider the sublattice magnetization given by MA =
i z SA,i = N S
a ak k
k 2 + vk k k
= NS which at T = 0 becomes MA = N S
u2 k k k
k
(12.63)
2 vk = N S
sinh2 (k )
k
1 = NS 2
2 (1 k )1/2 1 .
(12.64)
Thus, the magnetization per unit of volume can be written as MA 1 = nS + n/2 V 2 which for a cubic lattice in three dimensions is given by MA = n(S 0.078) . V (12.66) dd k (2)d 1
2 1 k
(12.65)
This is the result of a numeric integration of the equation above. Observe that due to quantum uctuations (zero point motion of the spins) the staggered magnetization is reduced relative to its classical value.
12.3. PROBLEMS
239
12.3 Problems
1. Using the formalism of subsection (1.1) and the operator identity limM e M e M
A B
= eA+B
show that the one-dimensional Ising model in a transverse eld which is described by Hq =
n z z x J n n+1 n
where is the strength of the magnetic eld has the same partition function as the classical Ising model in a square lattice (with N 2 sites) which is given by Hc = J
z z z z n,m n+1,m + n,m n,m+1 . n,m
(12.67)
Find the relationship between J and to J, and M . 2. Prove equation (12.18). 3. Calculate (12.20) explicitly and show (12.22) to be correct. 4. Consider the one-dimensional Ising model with an interaction which has nite range, that is, H= where Ji,j = J |i j|n Ji,j i j
i,j
where n characterizes the power of decay of the interaction. Using the Landau-Lifshitz argument show that it is possible to have long range order at nite temperatures if n < 2. Show that for the system to have a well dened thermodynamic limit one has to require n > 1. Show that for n = 2 the same argument requires that the magnetization M cannot vanish continuously, in order words, either there is zero magnetization at all temperatures or the magnetization is non zero with a discontinuity at the transition. 5. Prove (12.28).
y x z 6. Using (12.29) show that [Si , Si ] = 2ii,j Si and prove that S2 = S(S + 1) . i
7. Prove equations (12.36). 8. Calculate the spin wave velocity for an isotropic spin 1/2 Heisenberg model in d space dimensions for a hypercubic lattice. 9. In the Debye model for the antiferromagnet one assumes the dispersion of the spin waves to be (k) = cs k but cuts-off the dispersion at large wavelengths, say, at . (i) Calculate by assuming that the number of states in the spherical zone is the same as in the cubic Brillouin zone. (ii) Calculate the sublattice magnetization for the Debye model as a function of temperature. (iii) Find the critical temperature TN above which the sublattice magnetization vanishes.
240
10. Consider the problem of antiferromagnetic magnons in a uniform magnetic eld H for the HeisenbergIsing Hamiltonian. Using the non-unitary transformation nd the magnon spectrum in this case. Show that the k and k modes are separated in energy. Calculate the uniform magnetization of the system at nite temperature assuming a Debye model for the magnons. 11. Using (12.42) and (12.65) nd the lower value of the dimensionality dc above which long range order is possible (at nite and zero temperature) for a ferromagnet and antiferromagnet, respectively.
Chapter 13
where e is the electric charge, ext is the external (applied) electron density. In many systems the polarization P is linearly related to the electric eld through P = e E where e is the electric susceptibility of the medium. Thus we immediately obtain that D = E (13.2)
where = 1 + 4e is the dielectric function of the medium. In systems like crystals the dielectric function is not a constant but a tensor that depends on the direction the eld is applied. Here we are going to consider the case of the isotropic electron gas where the dielectric constant is just a scalar. As an example consider a single charge Q at the origin in the presence of the electron gas (in this case ext (r) = Q(r)). If this charge is positive it attracts electrons close to it, if it is negative it repels. In principle one would think that this charge could attract an innity number of electrons. This is not true not only because the electrons repel each other and therefore it would lead to an enormous increase of the electrostatic energy of the system, but also because electrons are fermions and therefore cannot occupy the same position is space. Thus, basic thinking tells us that the positive charge attracts enough electrons so 241
242
that, as seen from far away, it looks like a neutral object. This is called screening. The same argument works for the case of a negative charge. Thus, in the presence of an external charge the electron gas can not be uniform. This non-uniformity can be described by (r), the local change in the density of the electron gas. Observe that in the jellium model of the electron gas there is always a background of positive ions that neutralize the average density 0 otherwise the whole system would be unstable. The main idea behind screening is that the electric led in the system is created by the total density ext + in such a way that E = 4e (ext + ) . (13.3)
In the general case the external charge can also change in time so that we can Fourier decompose it as ext (r, t) = dk dei(trk) ext (q, ) . (13.4)
One can now Fourier transform (13.1) and (13.3) and use (13.2) to nd (q, ) = = q D(q, ) ext (q, ) = q E(q, ) ext (q, ) + (q, ) 1 1+
(q,) ext (q,)
(13.5)
gives the dielectric function once the displaced charged can be measured (or calculated). Another way to obtain the dielectric function is to observe that in (13.5) only the longitudinal part of the displacement and electric eld vector enter in the denition of the dielectric function. This is because we assume the system to be isotropic. Associated with the longitudinal component of the eld there we can always dene a potential such that E = and D = ext which, accordingly to (13.1) and (13.3), obey the equations 2 ext (r, t) = 4eext (r, t) 2 (r, t) = 4e (ext (r, t) + (r, t)) . (13.6)
The potential energies associated with the potentials given above is V = e and Vext = eext . From (13.6) and (13.5) we immediately conclude that (q, ) = Vext (q, ) V (q, ) (13.7)
gives the dielectric function in terms of the external potential and the total potential generated by the external elds. Thus, in order to describe screening one has to calculate the potential function.
243
where EF,0 = h2 kF /(2m) is the Fermi energy of the system. Because of the electric eld the charge density 2 is not homogeneous and we have a displaced density (r) that is given by (r) = (r) 0 . Associated with this charge density there is a local shift in the Fermi energy of the system, EF (r) = EF,0 V (r) , (13.10) (13.9)
where V (r) is the total potential felt by the electrons. In the Thomas-Fermi approach we assume that (13.8) gives the change in the charge density due to the shift in the local chemical potential as given by (13.10), that is, (r) = (2mEF (r))3/2 3 3 2 h (2mEF,0 V (r))3/2 = 3 3 2 h V (r) 3/2 = 0 1 EF,0
(13.11)
gives the relationship between the potential energy and the displaced charge. From (13.6) we see that the potential energy obeys the Poisson equation 2 V (r) = 4e2 (ext (r) + (r)) = 4e2 ext (r) + 0 0 1
V (r) EF,0
3/2
(13.12)
where we have used (13.9) and (13.11). Notice that (13.12) is a non-linear equation for V (r). It turns out, however, that for practical purposes V EF,0 and therefore we can linearize (13.12) to give
2 2 qT F V (r) = 4e2 ext (r)
(13.13)
where
2 qT F
= =
(13.14)
where a0 = h2 /(me2 ) the Bohr radius. qT F is the so-called Thomas-Fermi screening length. To understand the physical meaning of this length scale let us solve (13.13) by Fourier transform V (k) = = 4e2 ext (k) 2 k 2 + qT F Vext (k) 1+
2 qT F k2
(13.15)
244 where
Vext (k) =
(13.16)
is the external potential created by the charge distribution ext . Comparing (13.15) with (13.7) one immediately concludes that T F (k) = 1 +
2 qT F k2
(13.17)
is the dielectric constant in the Thomas-Fermi approximation. Consider now the case of a single impurity of charge Q at the origin. In this case ext (q) = Q and therefore from (13.15) we nd that the total potential is given by V (r) = d3 k 4eQ 2 (2)3 k2 + qT F 2eQ k sin(kr) dk 2 = 2 r 0 k + qT F =
keikr eQ + dk 2 2 ir k + qT F eQ = eqT F r r
(13.18)
where the last integral is done by choosing the contour in the upper half complex plane. Observe that in the Thomas-Fermi approach the effective potential produced by the charge Q at the origin is not the bare potential Vext (r) = eQ/r but has a Yukawa form and decays exponentially with a characteristic length T F = 1 qT F = a0 . 4kF (13.19)
In this case we say that the electrons screen the bare potential to a distance of order T F . The Thomas-Fermi approximation gives a simple but powerful picture of the electrostatic behavior of metals. Screening comes from the fact that the system has a Fermi surface and therefore it has mobile electrons that can move around in order to screen the external charge. Thus, electrons far away from the impurity indeed cannot feel the potential created by the impurity and moves freely in the system.
where (r, t) = (r, t) (r, t) is the particle density. Observe that we have subtracted from the energy the energy of the positive background of charge that neutralizes the electron gas. If one Fourier
245
Vq (t)q (t)
(13.21)
where q =
k,
c k+q, ck, .
(13.22)
Notice that the term with q = 0 is excluded from the integral in (13.21) since q=0 = 0 is the average particle density. The physical meaning of (13.21) is quite straightforward: the potential created by the charges in the system induces charge uctuations that are particle-hole pairs described by (13.22) around the Fermi surface. The total Hamiltonian of the system can be written as H = H0 + HV where H0 =
k,
k c ck, k,
(13.23)
is the free electron Hamiltonian. In order to calculate the dielectric function one has to calculate the displaced charge. In order to do it we are going to look for the equation of motion for the density operator in the Heisenberg representation, that is, i h It is an easy exercise to show that H0 , c k+q, ck, HV , c k+q, ck, = (k+q k ) c k+q, ck, =
q Vq c k+q+q , ck, ck+q, ckq , .
(13.24)
(13.25)
Observe that because of the second term in (13.25) the equation for the density uctuation cannot be solved in full form. Thus one has to use some approximation. Consider now the rst quantized version of the density operator q that is given by q = 1 V eiqrn
n
(13.26)
where rn is the position of the nth electron (you can easily show that (13.26) is correct by the direct Fourier transform of (r) = n (r rn ))). Therefore, in the r.h.s. of the second equation in (13.25) we see that q+q appears which, in its rst quantized form is: q+q = 1 V ei(q+q )rn .
n
(13.27)
Observe that if the electrons are (statistically) randomly distributed in the system the above sum only gives a nite result if q = q because the sum involves random phases that add destructively. In RPA we
246
(13.28)
that leads to a solution for the equation of motion in (13.24) by Fourier transform in time: c k+q, ck, = Vq () nk, nk+q, h k+q + k nk, nk+q, . h k+q + k (13.29)
Eq. (13.30) can be rewritten in a slightly different way in terms of the charge density distribution q () = Vq ()(q, ) where (q, ) = nk, nk+q, h k+q + k + i (13.32) (13.31)
k,
is the so-called polarization function and depends on the electronic system alone. It is important to notice that the positive background of charge (associated with nq=0 ) is subtracted to (13.21) and therefore the sum in (13.32) has to be understood as the principal value. This can be accomplished by introducing a small imaginary part 0 in the denominator. Thus, the polarization function has actually a real and a imaginary part. Indeed from (13.32) we can write {(q, )} = P = P {(q, )} = =
k
k,
k,
k,
where where P means the principal value of the integral and in the last line we have changed k k q and used the fact that nkq, = nk+q, and kq = k+q .
We can now substitute the result (13.31) into the Poisson equation (13.12) Vq () = 4e2 (ext (q, ) + (q, )) q2 4e2 = 2 (ext (q, ) + eVq ()(q, )) q
(13.34)
247
(13.35)
(13.36)
is the RPA expression for the dielectric function. Observe that RPA gives a frequency dependence to the polarization function that is not present in the Thomas-Fermi result (13.17). This happens because the Thomas-Fermi result is purely static, that is, it is valid at = 0. In order to check if RPA can reproduce the Thomas-Fermi result let us investigate the static limit of (13.36). From the denition (13.32) at T = 0 and assuming a spherical Fermi surface one has {(q, 0)} = 2P d3 k (kF k) (kF |k + q|) . (2)3 k2 /(2m) (k + q)2 /(2m) (13.37)
This integral can be simplied if we take the limit of q 0 in that case we can write |k + q| k + q cos() where is the angle between k and q and therefore (kF |k + q|) (kF k) q cos()(kF k) (since (k) = d(k)/dk). Thus, to leading order in q we have {(q, 0)} 4mP which, when substituted in (13.36), leads to lim RP A (q, = 0) = 1 + 4me2 kF , q 2 (13.39) mkF 2 d3 k q cos()(kF k) (2)3 2qk cos() (13.38)
q0
that is identical to the Thomas-Fermi result (13.17). Thus, the Thomas-Fermi approach is a special limit of the RPA result when = 0 and q 0. This implies that the RPA result has more information about the screening process in a electron gas than the Thomas-Fermi approach. It should be clear that the real part of the dielectric function (this is simply related to the real part of the polarization function in the RPA approach) is related to the screening processes that go on in the electron gas. The screening process is a coherent process that involves the displacement of electrons in the system. By now you should be curious about the meaning of the imaginary part of the dielectric function and to what physical process it describes. In order to understand its meaning it is instructive to look at the frequencydependent electrical conductivity of the electron gas, (k, ). As we have seen in Chapter 7 in a disordered environment the electron gas has a nite conductivity or resistivity due to the elastic scattering with static impurities. But this is not the only source of scattering since the electrons interact with each other. By denition the conductivity of the system is dened by the equation (7.6), J = E, where J is the electric current density. This current is related to the particle current density j by J = ej and thus 1 j(k, ) = (k, )E(k, ) . e (13.40)
On the other hand the number of electrons in the system has to be conserved in the scattering and therefore
248
the current and the density are related by current conservation equation n(r, t) + j(r, t) = 0 . t (13.41)
We are only interested in the uctuations of the system around the equilibrium particle density 0 , that is, we write n(r, t) = 0 + (r, t) and thus k () = k j(k, ) . Using that E(r, t) = (r, t) = V (r, t)/e, (13.40), and (13.42) we nd that (k, ) = i(k, ) k2 V (k, ) . e (13.43) (13.42)
Using Vext (k, ) = 4e2 ext (k, )/k2 , (13.5) and (13.7) it is easy to show that (k, ) = 1 + 4i(k, ) (13.44)
that gives the relationship between the dielectric function and the conductivity. Direct comparison between (13.44) and (13.36) shows that in RPA we have RP A (k, ) = ie (k, ) . k2 (13.45)
Observe therefore that the real part of the conductivity that is related to the dissipative process in the system is related to the imaginary part of the polarization function. Or more generally, from (13.44) the dissipative processes are related to the imaginary part of the dielectric function. Indeed, from (13.44) one can write {(k, )} = 1 {(k, )} = 4{(k, )} 4{(k, )}
(13.46)
that relates these two functions. Observe that in experiments we always measure the real part of these functions and through the relations above one can get the imaginary parts as well. Thus the physical meaning of the real and imaginary part are clear now. Moreover, the static conductivity (the one that is measured when a static eld is applied to the system) can be obtained from the equations above by taking the limit of 0. However, a word of caution is in place at this point. In calculating that static conductivity one has to take the limit of q 0 before we take the limit of 0. The opposite limit, that is, 0 rst with q nite describes a static electric eld that is periodic in space with characteristic wavelength 1/q. In this case the charge distribution is not uniform since it tries to follow the periodicity in the eld. Thus, in calculating the static conductivity one has always to set q 0 rst.
249
dk
kF
1 h kF + kF
h kq m
2
m ln 2 h q
q 2 2 q 2 2
h 2 q2
2m 2 m h q 2 m h q
1 h +
h kq m
2
h 2 q2 2m
(13.47)
{(q, )} is depicted in Fig.13.1 for = 0. Observe that {(q, )} diverges when q = 2kF for = 0. This divergence comes from what is called the nesting of the Fermi surface.
Re@PD
Figure 13.1: Real part of the polarization function for the one dimensional electron gas as a function of q for = 0. Nesting it is related to the fact that in some systems it is possible to nd a vector k such that q+k = k for a xed q. In this case, from (13.32), there is a point in the sum that is divergent. If this is an isolated point then it is not important for the behavior of the polarization function. In one-dimension, however, the Fermi surface is made out of two points, +kF and kF and therefore the vector q = 2kF connects all the points in the Fermi surface and leads to the divergence in (13.47). In higher dimensions we need a nite density of points at the Fermi surface that are nested, otherwise their contribution to (13.32) is a set of measure zero. One of the classic cases where a nesting condition happens is the half-lled tight binding model in two space dimensions where the Fermi surface has the shape shown in Fig.13.2 and can be connected by a wave-vector (/a, /a). Because response functions, like the polarization function, have denominators that involve q+k k , nesting leads to instabilities in the electron gas.
250
kx /a /a
/a
Figure 13.2: Nesting of the half lled tight binding model in two dimensions.
{(q, )} = =
kF
dk h
h 2 kq 2 q 2 h m 2m
+ h
h h 2 kq 2 q 2 + m 2m
h kF q q 2 m h h h kF q q 2 + + + 2 m 2m m 2m h |q| 2 2 h h h kF q q h kF q q + + . + m 2m m 2m
(13.48)
Observe that {(q, )} has a dependence with q (that is, it behaves like 1/q) but it is not nite in the whole (, q) space. Actually it is easy to see that this function is only nite in the shaded area of Fig.13.3. This region is called the particle-hole continuum and it is source of dissipation in the system. Notice that the {(q, )} vanishes at = 0 for all 0 < q < 2kF . We can show that this behavior is very particular of one-dimensional systems and in higher dimensions {(q, )} is actually nite in this domain.
k c c,k + ,k
q
h q a aq + q
q
Uq q aq + a q
(13.49)
251
q kF q 2m m
2
0
2
2k F
q + kF q 2m m
where the electron-phonon coupling is the same as in (9.59). Observe the close resemblance of (13.49) and (13.21) if we dene Vq = Uq aq + a q . (13.50)
This is not just a coincidence since, as we said, the electrons are electrostatically coupled to the lattice. Thus, at least from a perturbative point of view we could see that the electrons respond to the phonons in the system and one would suspect that the response should be somewhat related to the polarization function of the electron gas. Using the polarization function (13.32) we add and subtract nk, nk+q, in the numerator, regroup the terms together, and change k k q in order to get: (q, ) = nk, (1 nk+q, ) 1 1 h k+q + k + i k + k+q + i h (13.51)
k,
that has the same structure as (9.77). Indeed, let us assume that the coupling between electrons and phonons is weak and use result (13.30) and substitute nq by nq in (13.49) using (13.50) in order to nd: H=
q
h q a aq q
|Uq |2 (q, q ) aq + a q
aq + a q
(13.52)
is a purely bosonic Hamiltonian. The electrons have been traced out of the problem and their effect only appears through the polarization function. Notice that in the phonon Hamiltonian we now have operators of the form aa and a a that do not conserve the number of bosons. This is happening because the electrons are now hidden" and from the point of view of the bosons every time an electron emits a boson it is as if bosons were being produced from vacuum. The full solution of the problem involves the diagonalization of the Hamiltonian (13.52) that is possible because the problem is quadratic but since we are assuming weak coupling we neglect the terms that do not conserve the number of bosons and, apart from a constant factor,
h q a aq q
q
(13.53)
where we have used that q = q because of the reection symmetry of the lattice and that in (13.32) (q, ) = (q, ). Eq. (13.54) shows that the electrons actually modify the bare frequency of the phonons. This should be expected since what is measured experimentally is not the frequency of the phonons isolated from their environment but the effective frequency that takes into account the interaction of the phonons with their surroundings. We observe two effects: the real part of (q, ) renormalizes the phonon frequency while the imaginary part of (q, ) produces damping or dissipation of the phonon modes, that is, when the bare phonon frequency enters a region where {(q, )} is nite the phonon mode becomes unstable and decay into the particle-hole continuum. Let us consider the application of (13.54) to the one dimensional case. Let us consider the renormalization of the phonon frequency using (13.47) for the case of the acoustic mode with q = vs q where vs is the sound velocity. In this case we nd 2m |Uq |2 ln q = vs q h 2 q kF + kF
q 2 2 q 2 2
mvs 2 h mvs 2 h
(13.55)
that is plotted in Fig.13.4. Observe that (13.55) has a singularity at q = 2kF (1 + vs /vF ) 2kF where vF = hkF /m is the Fermi velocity (notice that vs vF ). Because of the logarithmic singularity in (13.55) the phonon frequency vanishes close to 2kF in one dimension. In higher dimensions (13.54) predicts a nite frequency at 2kF with a divergent group velocity (dq /dq ) that leads to the so-called Kohn anomaly in metals. In one dimension the Kohn anomaly is more severe and leads to the vanishing of the phonon frequency and has strong consequences. Observe that in rst quantized language we can rewrite (13.53) as H=
q
Pq Pq M 2 q + Xq Xq 2M 2
(13.56)
and therefore the vanishing of q for some wave-vector q = Q leads to a Hamiltonian of a free particle for that particular mode. A free particle has an unbounded motion. This leads to the conclusion that the vanishing of the phonon frequency is associated to an instability of the lattice. What this result shows is that the lattice becomes soft" for this particular mode and that there is a gain in energy to produce a distortion with characteristic wavevector Q, that is, the lattice distorts with a modulation given by cos(Q r). In the case of the one dimensional system where Q = 2kF this distortion is only possible is the electronic density is commensurate with the lattice. To understand commensuration recall that Q = 2N /a where N is any integer is a reciprocal lattice vector. Thus, the condition Q = 2kF implies that kF = /(N a). In one dimension one has that kF = /a where n = N/Ns is the number of electrons per lattice site. Thus, for n the distortion to occur one needs n = 1 that is the condition for commensuration, that is, the number of lattice sites is a multiple of the number of electrons. The simplest case is n = 1 that is, the half-lled band. In this case Q = /a and therefore the distortion is of the form cos(x/a). In this case the unit cell doubles size as shown in Fig.13.5(a,b). Observe that in this case the unit cell doubles its size going from a to 2a. This is called the Peierls distortion and we say that the system is dimerized, that is, transformed to dimers
253
WHqL
q kF 1 2 3 4
Figure 13.4: Dashed line: bare phonon dispersion; Continuous line: result from (13.55).
of atoms. This implies that the Brillouin zone has to shrink from /a to /(2a) and a gap has to open at the zone boundary (exactly like in the case of phonons) as shown in Fig.13.5(c,d). It turns out however that the zone boundary is at the Fermi energy and therefore a gap has to open at the Fermi surface and the system should be insulating. The size of the distortion and the magnitude of the gap cannot be discussed in our picture because these are non-perturbative effects. This effect, however, is seen in one dimensional systems such as polymers which, by band structure arguments, should be conductors but by the effect of the Peierls mechanism are actually insulators.
(13.57)
254
u
(a) (b)
2a E
(c)
kF E kF k
kF kF
(d)
k
Figure 13.5: (a) Original Lattice; (b) Peierls distortion; (c) Original electron band; (d) Electron band after Peierls distortion. where Uq = 4e2 /q 2 (where the factor of 2 is to avoid double counting). In the mean-eld approximation one does the usual substitution: q q q q + q q that changes (13.57) to HM F = where Vq = Uq q . (13.59) 1 V Vq q
q
(13.58)
Observe that (13.58) has exactly the form of (13.21) for an external potential. In this picture each electron responds to the average charge density. Applying the arguments of the previous sections one would argue that the effective interaction felt by the electrons due to the other electrons would be given (in the static limit) by
ef Uq f =
Uq 4e2 = 2 2 (q, 0) q + qT F
(13.60)
that is the screened Coulomb interaction. Thus, from this point of view the electrons indeed interact via an screened (weaker) interaction than the original long range Coulomb interaction.
255
interacting electrons gas. The main problem here is how to incorporate the electron-electron interactions within the properties of the electron gas. Let us imagine that an electron with energy EF is introduced into the electron uid as in Fig.13.6(a). As the electron enters the electron gas the other electrons have to move around to screen its charge. In this case a correlation hole" with positive charge is formed around the electron in such way that, as seen from far away, the electron behaves like a neutral object as in Fig.13.6(b). One imagines that the same picture applies to all the electrons in the system (as we did in the mean eld theory described previously). In this case the electrons are dressed by the interactions with all other electrons in the system. Because of this dressing the elementary excitations in the system are not bare electrons but dressed electrons that we call quasiparticles (as shown in Fig.13.6(c)). Although this picture is quite appealing we would like to have a more quantitative description. For instance, one would like to know how interactions affect experimental quantities such as specic heat, magnetic susceptibility and so on.
Electron Gas
(a)
(b)
(c)
Figure 13.6: (a) Electron entering the electron gas; (b) Electron plus its correlation hole; (c) Quasiparticle gas. A major breakthrough in this eld was made by Landau in the 40s. Landaus theory of the Fermi liquid is based on two main intuitive" assumptions: (1)There is one-to-one correspondence between the quantum numbers of the non-interacting electron gas and the interacting electron gas. Thus, the quasiparticles carry one unit of electric charge, they are spin 1/2 excitations and they also carry the momentum quantum number k. Thus, in Landaus theory the Hilbert space for electrons and quasiparticles is the same and the only possible change in the problem is not dynamic but kinematic. This is also called the adiabatic principle because it assumes that the interactions can be turned on slowly so that nothing tragic happens in the system (in particular, no phase transition occurs). This principle tells us how the quasiparticles couple to external elds such as the electric and magnetic elds. Since the electric charge is the same as the electrons the electromagnetic eld should couple via minimal coupling" p eA/c plus the Zeeman energy S H.
(2)It is possible to write down an energy functional of the deviations of the quasiparticle occupation relative to the ground state. As we have seen before we can create excitations in the electron gas by creating particle-hole excitations around the Fermi surface. Accordingly to Landau the ground state of an interacting Fermi gas is also a Fermi sea of quasiparticles as dened by assumption (1). That is, we can dene an
1 e(Ek, ) +1
(13.61)
where Ek, is the quasiparticle energy and is the chemical potential of the system. In increasing the temperature or applying a eld to a system what one does is to modify the occupation by an amount nk, . In Landaus theory the total energy of the system can be written as an expansion in terms of these deviations: E[nk, ] = E0 + 1 V
0 Ek, nk, + k,
1 2V
fk,,k , nk, nk ,
k,,k ,
(13.62)
0 plus higher orders in nk, . E0 is just the ground state energy, Ek, is the bare dispersion of the quasiparticles and fk,,k , is so-called Landau parameter that describes the interaction between quasiparticles. Observe that the actual dispersion of the quasiparticles depend on their interaction. Since we can rewrite (13.62) as
E = E0 + Ek,
1 V
Ek, nk,
k,
1 0 = Ek, + V
fk,,k , nk ,
k,
(13.63)
where Ek, is the actual quasiparticle dispersion. In order to understand the physical meaning and consequences of (13.62) let us consider the problem of the specic heat of the electron gas in Landaus theory. The quasiparticle entropy is given by the usual Boltzmann expression S= kB V {nk, ln(nk, ) + (1 nk, ) ln(1 nk, )} . (13.64)
k,
A variation nk, in the distribution function leads to a variation S in the entropy that is given by S = = kB V nk, ln
k,
kB V
k,
(Ek, ) nk,
where we used (13.61). The specic heat is the variation of the entropy with temperature, S/T . Thus, one needs to calculate how the occupation nk, changes with temperature. This can be easily obtained from (13.61) as Ek, nk, = T T Thus, substituting (13.66) into (13.65) we nd 1 S = T V nk, Ek, Ek, T
2
nk, Ek,
(13.66)
(13.67)
k,
257
In order to evaluate the sum in (13.67) we transform it into an integral by using the usual denition of the quasiparticle density of states: N (E) = and rewrite (13.67) as S = T
+
1 V
k,
(E Ek, )
(13.68)
dEN (E)
E T
n(E) E
(13.69)
but observe that the derivative in the expression above is highly peaked around E = (Fermi statistics) while the rest of the argument is a smooth function around this point. Therefore, with great accuracy we can approximate the above integral as S T
2 = kB N (0) +
dxx2
2 3
2 kB N (0)
where we changed variables: x = (E ). The specic heat of the electron gas can be immediately obtained from (13.70) since CV = T S 2 2 = k N (0)T T 3 B (13.71)
looks identical to the free electron gas result (6.99). The temperature dependence of the specic heat is the same as in the free electron gas but the similarity is only apparent because in our denition of the density of states for the quasiparticles (13.68) we have used the actual quasiparticle dispersion Ek, which, accordingly to (13.63), also depends on the Landau parameters fk,,k , . Thus, in order to be complete the calculation one has to compute N (0) as a function of the interaction parameters. Firstly, we notice that due to Landaus rst assumption the number of electrons and the number of quasiparticles have to be the same. This implies that the Fermi momentum of the quasiparticles is the same as the Fermi momentum of the particles, kF = (3n)3/2 , since it depends on the density alone. Let us rst go back to the denition of the density of states (13.68) and rewrite it as an integral N (0) =
since by denition EkF , = . Since we are expanding the total energy to second order in nk, we just 0 keep the quasiparticle energy to rst order (take a look at (13.63)). In this case we can replace Ep, by Ep, in (13.72) that depends on the group velocity of the quasiparticles:
0 vp, = p Ep, .
(13.73)
In a non-interacting Fermi gas the group velocity is simply pF /m where pF is the Fermi momentum. By
258
analogy we dene the effective mass, m , of the quasiparticles as vp, = when substituted in (13.72) leads to N (0) = m p F 2 3 h (13.75) p m (13.74)
that is the quasiparticle density of states that depends only on the effective mass of the quasiparticles. Thus, what we have done is transferring the problem of calculating the density of states to the problem of calculating the effective mass. But life is simpler because the mass is a kinematical quantity that reects the inertia of the system. In the case of the interacting electron gas the effective mass depends on the interaction between the electrons since as one electron moves it has to push the other electrons around, that is, the inertia of the quasiparticle depends on its interaction with the environment.
The most direct way to compute m is use an argument due to Landau that assumes Galilean invariance of the system. Consider observing a Fermi gas from a moving frame with velocity V. If the interactions do not depend on the velocity of the particle (as in the case of the Coulomb interactions) then the only change in the Hamiltonian of the system is in kinetic energy that changes from
N
K=
i=1
p2 i 2m
to
N
K=
i=1
(pi mV)2 = 2m
N i=1
p2 i V 2m
pi +
i=1
Nm 2 V 2
(13.76)
since the velocities have to change from vi to vi V due to Galilean invariance. Thus, the total energy and momentum of the system in the moving frame is Emov = E P V + Pmov = P N mV Nm 2 V 2 (13.77)
where P = N pi is the total momentum in the rest frame of the electron gas. Consider now adding a i=1 quasiparticle with momentum p in the rest frame. This leads to an increase in the total mass of the system by m and the total momentum of the system by p. Thus, from (13.77) we see that momentum of the quasiparticle, as seen in the moving frame is, p mV while its energy is Emov,pmV, = Ep, p V + m V2 2 V2 . 2 (13.78)
Emov,p, = Ep+mV, p V m Now assume that V is very small and expand (13.78) to rst order in V: Emov,p, Ep, + V (mp Ep, p)
(13.79)
259
(13.80)
On the other hand, from the moving frame the distribution function of the quasiparticles has to change as nmov,p, = np+mV, np, + mV p np, np, = np, + mV p Ep, Ep, np, m . = np, + V p m Ep,
(13.81)
From the change of the occupation we can calculate the energy of the quasiparticle in the moving frame from (13.63) Emov,p, = Ep, + 1 V fp,,p ,
p ,
np , m V p m Ep ,
(13.82)
that has to be compared with (13.80). At this point the comparison is not straightforward because we have to evaluate the integral in (13.82). In dealing with an isotropic system (and therefore a spherical Fermi surface) we have an extra symmetry that is the rotations in momentum space. This implies that in the scattering between two quasiparticles with momentum p and p the interaction parameter fp,,p , only depends on the angle (p, p ) between p and p . In this case we can expand fp,,p , in Legendre polynomials, Pl (cos()) as
fp,,p , =
l=0
(13.83)
that can be inverted due to the orthogonality between the Legendre polynomials as fl,, = = 2l + 1 2 2l + 1 2
0 +1 1
d(p, p ) sin((p, p ))Pl (cos((p, p )))fp,,p , du(p, p )Pl (u(p, p ))fp,,p , (13.84)
where u = cos(). Direct substitution of (13.83) into (13.82) leads to Emov,p, = Ep, = Ep, = Ep, m m fl,,
,l
d3 p (2 )3 h fl,, Vp 6
np, Ep,
Pl (cos())V p
m N (0) m
s F1
m Vp m 3
(13.85)
(13.86)
the symmetric and anti-symmetric Landau parameters. Direct comparison of (13.85) with (13.80) we nd Fs m =1+ 1 m 3 (13.87)
that gives the renormalization of the mass in terms of the interactions. The density of states can now be easily obtained from (13.75). It is clear that many properties of the interacting electron gas can be directly calculated from Landaus assumptions. In this case the Landau parameters enter into the theory as phenomenological functions that describe the physics of the interacting Fermi gas. It is possible, therefore, to make many predictions about the physical quantities of interest and check experimentally. The simplicity and power of Landaus approach should be clear and its success in explaining the physical properties of many systems such as He3 and metals made it the standard approach to the study interacting electron systems.
13.7 Problems
1. Consider a system of electrons moving on a plane. Use a system of coordinates where the three dimensional vectors can be written as R = (r, z) where r is the vector in the plane. (i) Using the Thomas-Fermi approximation show that the Thomas-Fermi screening length is the Bohr radius a0 and does not depend on the density. (ii) Calculate the effective potential for a charge Q sitting at the origin of the plane (ext (r) = Q(z)(r)) and show that at long distances (r a0 ) the effective potential behaves like V (r) 4 2 a2 eQ 0 . r3
In this case the potential does not decay exponentially contrary to the problem in three dimensions. 2. Use (13.51) to calculate the polarization function for a three dimensional electron gas with a spherical Fermi surface. Hint: you can nd the full solution in Fetter and Walecka, Quantum Theory of ManyParticle systems" (McGraw-Hill, New York, 1971), pg.158. 3. Calculate the polarization function of a two dimensional electron gas in the RPA approximation. (i) Plot {(q, = 0) and show that it does not diverge at q = 2kF . (ii) In what region of the (, q) space {(q, )} is nite. (iii) How does the dielectric function compares with the result of question (2)? 4. Consider the Hamiltonian (13.52). (i) Calculate the commutator of aq with the Hamiltonian and show that the operators aq mixes with the operator a . (ii) Consider the unitary transformation q bq = uq aq + vq a q what is the condition on uq and vq in order to have canonical commutation relations? (iii) Show that the Hamiltonian can be completely diagonalized by the transformation above. What are the new phonon frequencies in this case?
13.7. PROBLEMS
261
5. Consider a one-dimensional electron gas at half lling with hopping matrix element t0 . Consider the case where the system undergoes a Peierls distortion of size u (that is, the size of the distortion in Fig.13.5). (i) Assume that wavefunction of the electron in each atom has an exponential form, (x) e|x|/a0 where a0 is the lattice spacing. Show that there is a change in hopping matrix element in the system from t0 to t0 2u depending on the site ( is a constant). (ii) From your result above show that in the tight binding approximation the Hamiltonian for the electrons of the problem can be written as He = (t0 + 2u(1)n ) c cn+1 + c cn . n n+1
n
(13.88)
(iii) Show based on general arguments that the elastic energy due to the Peierls distortion can be written as Hd = 2N Ku2 where N is the number of atoms and K is the spring constant. (iv) Fourier transform (13.88) and show that it can be written as He =
k
(13.89)
k b bk a ak + 2u sin(ka0 ) b ak + a bk k k k k
(13.90)
(v) Observe that (13.90) is quadratic and can be diagonalized. Dene the following unitary transformation k = uk ak + vk bk k = uk bk vk ak
where k = 2t0 cos(ka0 ), ak = ck and bk = ickQ (with Q = /a) and the sum over k goes from Q/2 to Q/2 (that is, the new Brillouin zone).
(13.91)
2 where uk and vk are constants. Show that u2 + vk = 1. Use (13.91) to diagonalize (13.90) and show k that the new Hamiltonian can be written as
He =
k
Ek k k k k
(13.92)
where Ek =
Make a plot of Ek and show that a gap opens at k = /(2a). Give a physical meaning to the operators k and k . (vi) Show that uk = vk = 1 2 1 2 1+ 1 k Ek k sgn(k) Ek
(v) Assume that the electron system is half lled and show that the total energy of the system (electron+distortion) is given by E(u) 4t0 = E 1 N where E[x] =
/2 dt 0
2u t0
+ 2Ku2
(13.94)
(vi) Show that for u t0 / the energy of the system can be written as 4t0 42 82 E(u) + 2K + ln N t0 t0 u2 . (13.95)
What happens when u ? From these two results prove that no matter how small there is always a value of u = u0 = 0 so that the energy is minimum. Thus, you just proved that, no matter how small the coupling constant , the one dimensional system is always dimerized (Peierls theorem). (vii) Using (13.95) show that u0
K 2t0 1 t02 4 e
for u t0 /. Observe that u0 is not an analytic function of . 6. In this exercise we are going to consider the compressibility of the electron gas in the presence of interactions. The physical quantity of interest is the compressibility, (or bulk modulus). The compressibility is dened as = 1 n 1 V = 2 V P n mu
where V is the volume, P is pressure, n = N/V is the particle density and is the chemical potential. Thus, in order to calculate one has to consider the change in the quasiparticle occupation, nk, in respect to the chemical potential. (i) Using (13.61) show that nk, = (Ek, ) (ii) Show that Ek, = and from that prove that Ek, = where n =
1 V k, s F0 n N (0)
nk, . Ek,
(13.96)
1 V
fk,,k , nk ,
k ,
(13.97)
13.7. PROBLEMS
(iii) Using the results of item (i) and (ii) show that = What is the physical meaning of this result? 1 N (0) s . n2 1 + F0
263
7. In the exercise you will calculate the magnetic susceptibility of the interacting electron gas in the Landau approach. Let us consider only the Zeeman energy that is given by
N
Hz = gB
i=1
Si H
where S is the spin of the quasiparticles which, according to Landaus assumption, is /2. h (i) What is the change in the quasiparticle energy in the presence of the eld? What is the difference from (13.97)? (ii) The change in the occupation is still given by (13.96). Notice however that the change in the chemical potential cannot depend on the direction of the eld H and therefore can only be of order H2 . Thus, neglecting the change in the chemical potential due to the eld and using the result of item (i) show that n = N (0)gB a Hz 2(1 + F0 )
1 where n = V k nk, , and Hz is the z-component of the magnetic eld. From the above result calculate the total magnetization of the electron gas and show that the magnetic susceptibility is given by
N (0)(gB )2 . a 1 + F0