Chapter 8: Orbital angular momentum and molecular rotations

There are two types of angular momenta in quantum mechanics: orbital angular momentum and spin angular momentum. Orbital angular momentum corresponds to the quantum analogue of classical angular momentum. The spin angular momentum represents a purely quantum mechanical phenomenon that vanishes in the classical limit. In this chapter, we consider orbital angular momentum. The material covered in this chapter has many important applications in chemistry, including to rotational (microwave) spectroscopy, electronic structure of atoms and molecules, and scattering theory.

Angular momentum in classical mechanics

Consider a classical point particle of mass m , whose position and momentum vectors, in terms of a Cartesian coordinate system, are given by r = ( x, y, z ) and p = ( px , p y , pz ) , respectively. Then, the angular momentum of such a particle is a vector, denoted by L , and defined as the vector product of r and p :

L=rp

(1)

According to the definition of a vector product, the amplitude of L is given by L = rp sin( ) , where is the angle between r and p , and its direction is perpendicular to the plane defined by r and p . The right-hand rule is used in order to decide whether L is pointed above or below this plane (see Fig. 1). It should be noted that L = 0 if r and p are parallel to each other, which means that a particle moving in 1D has zero angular momentum. Put differently, in order to have an angular momentum, the particle must orbit or rotate around the origin of the coordinate system.

Figure 1: The right-hand rule for the vector product C = A B .

A more convenient, and completely equivalent, definition of L can be given in terms of its Cartesian components (stated without proof): Lx = ypz zp y

Ly = zpx xpz Lz = xp y ypx

Eq. (2) can be written in a more compact manner in the form of a determinant:
ux L= x px uy y py uz z pz

(2)

(3)

where,

ux , u y , uz

are

unit

vectors

along

the

x, y , z

axes,

respectively

( L = Lx u x + Ly u y + Lz u z ). The time evolution of L can be obtained from the classical equations of motion, and is given by (see proof below):
dL = rF dt

(4)

where F = V = ( Fx , Fy , Fz ) is the force acting on the particle. The quantity r F is called the torque, and plays a role with respect to angular momentum which is analogous to that played by the force with respect to momentum (remember that dp / dt = F ).
Proof of Eq. (4): dp p dLx d dy dp dz p = ( ypz zp y ) = pz + y z p y z y = y pz + yFz z p y zFy dt dt dt dt dt dt m m (5) = yFz zFy = r F x Similarly, it can be shown that dLy / dt = r F , dLz / dt = r F . y z

Angular momentum is particularly important in the case of a central potential. In this case, the potential energy is spherically symmetrical and only depends on the distance from the origin, r = x 2 + y 2 + z 2 (the gravitational and Coulomb potentials are examples of central potentials). As it turns out, angular momentum is conserved for a classical particle moving in a central potential (see proof below):
dL / dt = 0 (for a particle moving in a central potential).

(6)

This implies that angular momentum is a constant of the motion in the case of a central potential. Since L does not change its direction as well as its amplitude, the motion in this case must take place in a plane.

Proof of Eq. (6):

dLx V (r ) V (r ) dV (r ) r dV (r ) r = yFz zFy = y +z = y +z dt dr z dr y z y dV (r ) z dV (r ) y 1 dV (r ) = y +z = ( yz + zy ) = 0 dr r dr r r dr

(7)

Similarly dLy / dt = dLz / dt = 0 . It is important to note that the force vector in the case of a central potential is given by: dV r F = V = (8) dr r and lies along the radius (a radial force), which means that it is parallel to r , which is why the torque is zero [see Eq.(4)]. Intuitively, this makes sense, since a force must have a non-radial component in order to change the velocity component perpendicular to the position vector and thereby affect the rotational motion (remember that F = mdv / dt ).

Orbital angular momentum in quantum mechanics

Operators and commutators.

The operators that represent the Cartesian components of angular momentum can be obtained from the corresponding classical variables, via the quantization rules [see Eq. (2)]: Lx = ypz zp y i y z y z
Ly = zpx xpz i z x z x Lz = xp y ypx i x y x y

(9)

We will also define the operator that represents the square of the amplitude of the angular momentum vector: L2 = L2 + L2y + L2 . (10) x z The commutators of the four operators L , L , L , L2 play a central role in the theory of
x y z

quantum mechanical angular momentum, and are given by:

 Lx , Ly = i Lz Ly , Lz = i Lx Lz , Lx = i Ly Lx , L2 = Ly , L2 = Lz , L2 = 0

(11)

Proof of Eq. (11):

Lx , Ly = ypz zp y , zpx xpz = [ ypz , zpx ] [ ypz , xpz ] zp y , zpx + zp y , xpz = i ypx 0 0 + i xp y = i ( xp ypx ) = i Lz

, = ypx [ pz , z ] xy [ pz , pz ] p y px [ z , z ] + xp y [ z pz ]
y

(12)

Similarly, Ly , Lz = i Lx , Lz , Lx = i Ly . Also, Lx , L2 = Lx , L2 + L2y + L2 = Lx , L2y + Lx , L2 x z z = Lx , Ly Ly + Ly Lx , Ly + Lx , Lz Lz + Lz Lx , Lz = i L L +i L L i L L i L L = 0

z y y z y z z y

(13)

Similarly Ly , L2 = Lz , L2 = 0 . Eq. (11) implies that the Cartesian components of orbital angular momentum do not commute. This means that simultaneous knowledge of the three components of the angular momentum vector is not possible! At the same time, each of the Cartesian components commutes with the square of the angular momentum amplitude, which implies that simultaneous knowledge of the length of the angular momentum vector and one of its components is possible.
Spherical (polar) coordinates.

The expressions for the angular momentum operators in terms of Cartesian coordinates are not separable ( x, y, z , px , p y , pz are coupled). As it turns out, the angular momentum operators can be made separable when cast in terms of another set of coordinates the spherical coordinates.1 The position of the particle in terms of spherical coordinates is specified by the following three independent variables (see Fig. 2): 1. Its distance from the origin, 0 r . 2. The (small) angle between the position vector and the z axis, 0 .

The origin for this is the inherent spherical symmetry of rotational motion.

3. The (counter-clockwise as viewed from the positive z axis) angle between the projection of the position vector onto the xy plane and the positive x axis, 0 2 . The transformation between Cartesian and spherical coordinates, which is based on elementary geometry, is given below:
x = r sin ( ) cos ( ) y = r sin ( ) sin ( ) z = r cos ( ) r = x2 + y 2 + z 2 1 2 2 2 = cos z / x + y + z = tan 1 ( y / x )

(14)

The transformation of the infinitesimal volume element from Cartesian to spherical coordinates is given by (stated without proof):
dxdydz = r 2 sin drd d

(15)

For example, we can calculate the volume of a sphere of radius R by performing the following integral:
r3 2 drr d sin d = drr d sin d = 3 [ cos ]0 [ ]0 0 0 0 0 0 0 0
R R 2 2

(16) 4 R 3 R3 = [ (1) (1) ] 2 = 3 3 The volume of an infinitesimally thin spherical shell of radius r and thickness dr is given by
drr 2 d sin
0

d = drr
0

= drr 2 [ cos ]0 [ ]0 = 4 r 2 dr

(17)

which is simply equal to the area of the surface of the shell, 4 r 2 , times its thickness, dr .

Figure 2: The spherical coordinate system.

A wave function which is given in terms of spherical coordinates, (r , , , t ) , will be normalized when
2 drr d sin 0 0

d ( r , , , t )
0

=1 .

(18)

A few examples of how to evaluate probabilities in terms of spherical coordinates follow: 1. The probability density for finding the particle at a distance r from the origin, regardless of the orientation, i.e. the values of and , is given by:

P(r ) = r 2 d sin
0 2

d ( r , , , t )
0

(19)

The r factor is a geometrical factor that originates from the fact that the surface of a spherical shell of radius r grows like 4 r 2 . This factor increases the 2 probability when r is increased. At the same time, one expects that (r , , , t )
will decay to zero at large r , such that
r2 r2

drP(r ) = 1 . It should also be noted that

2

the probability of finding the particle at a distance between r1 and r2 is given by P (r1 r r2 ) = drP(r ) = drr
r1 r1

d sin d (r , , , t )
0 0

(20).

2. The probability density of finding the particle at a given direction, specified by and , regardless of its distance from the origin, r , is given by P ( , ) = drr 2 sin (r , , , t ) .
2 0

(21)

Thus, P(1 2 , 1 2 ) = d d P( , )
1 1 2 2

(22)

The angular momentum operators in terms of spherical coordinates. In order to see why it is advantageous to use spherical coordinates when dealing with angular momentum, we need to rewrite the angular momentum operators in terms of spherical coordinates. This leads to the following expressions: + cot cos Lx = i sin

cot sin Ly = i cos Lz = i L2 =

2

(23)

2 2 1 2 + cot + 2 2 sin

Proof of Eq. (23): The derivation is based on Eq. (14) and the chain rule, and is rather tedious. We will outline it below in the case of Lz :

r Lz = i x y = i r sin cos + + x y y r y y . (24) r r sin sin + + x r x x By using Eq. (14) it is easy to show that r cos cos sin = sin cos ; = ; = r r sin x x x (25) r cos sin cos = sin sin ; = ; = r y y y r sin Substituting the identities in Eq. (25) into Eq. (24) and some algebra leads to the final result: Lz = i / . The major simplification that results from the transformation to spherical coordinates is that the angular momentum operators in Eq. (23) only depend on and , and do not depend on r . Thus, what appeared to be a 3D problem in terms of Cartesian coordinates (i.e., the operators explicitly involve x, y, z ), turns out to be an effectively 2D problem in terms of spherical coordinates (i.e., the operators only involve , ).

The simultaneous eigenfunctions of L2 and Lz (the spherical harmonics) We have seen that simultaneous knowledge of the three Cartesian components of the angular momentum vector is not possible. At the same time, each of the Cartesian components commutes with the square of the angular momentum amplitude, which implies that simultaneous knowledge of the length of the angular momentum vector and one of its components is possible. We will choose Lz as the component whose value is known with certainty (the choice of the z axis is completely arbitrary, and corresponds to the standard convention). In the following, we will seek a basis of simultaneous eigenfunctions of L2 and Lz (we know that such a basis exists since those two operators commute). Since the angular momentum operators only depend on and , the simultaneous eigenfunctions of L2 and L are effectively functions of only those two variables. We
will denote those eigenfunctions by Y ( , ) . Note that the overall wave function of the system will also include a radial part, such that (r , , ) = R(r )Y ( , ) , where R(r ) is any function of the variable r (as long as the boundary conditions are satisfied and the overall wave function is well behaved). Thus, if Y ( , ) is an eigenfunction of Lz and
z

 L2 , then so is R (r ) Y ( , ) This is because L2 and Lz dont operate on R (r ) , such that Lz [ R(r ) Y ( , ) ] = R(r ) LzY ( , ) and L2 [ R(r ) Y ( , )] = R(r ) L2Y ( , ) . We next consider the possibility that Y ( , ) is separable such that: Y ( , ) = ( ) ( ) . (26) Normalization of (r , , ) = R(r )Y ( , ) than implies that
2 2 drr d sin d (r , , ) = drr R(r ) 2 0 0 0 0

d sin ( ) d ( ) = 1
2 2 0 0 2

drr 2 R (r ) = 1 ;
2 0

(27)

d sin ( )
0

=1 ;

d ( )
0

=1

We first require that Y ( , ) be an eigenfunction of Lz : LzY ( , ) = Y ( , ) i [( )( )] = [( )( )] ( ) i = ( ) The solution of this elementary differential equation is given by (check this!): ( ) = Nei / . The normalization function can be found via Eq. (27):
2

(28)

(29)

1 (30) 2 0 0 0 Note that at this point in the development, the eigenvalue can be any real number (it must be real since Lz is hermitian). However, we still need to impose the condition that the wave function is well behaved, which implies that it is single valued, such that ( ) = (2 + ) (a complete rotation around the z axis will return us to the same point and therefore the same value of the function). Thus, ei / = ei (2 + ) / ei 2 / = 1 / = 0, 1, 2,... = 0, , 2 ,... (31) Thus, the z component of the angular momentum vector has discrete eigenvalues! It should be noted that the choice of the z axis is completely arbitrary. This result therefore applies to the component of the angular momentum along any axis of ones choice. For example, the eigenvalues of Lx and Ly are also given by 0, , 2 ,... It should be noted that the corresponding eigenfunctions of L and L are not given by Eq. (29), since

d ( ) = N
2

i / i / d e e = N

d = 2

N =1 N =
2

the wave function is given in terms of spherical coordinates, which differentiate between the x,y and z axes. The results so far imply that Y ( , ) = ( )eim / 2 , where m = 0, 1, 2,... is the quantum number that corresponds to the eigenvalue m of L (it should be noted that
z

has angular momentum units, such that the quantum number m is unitless). We also need Y ( , ) = ( )eim to be an eigenfunction of L2 . Using the expression for this operator in Eq. (23), then leads to the following differential equation:

 2 1 1 im 1 im 2 + cot + 2 ( ) e = ( ) e 2 2 sin 2 2 (32) 2 m2 2 + cot ( ) = 2 ( ) sin 2 The resulting differential equation for ( ) is far from trivial. Nevertheless, it can be solved analytically (although we are not going to go through the derivation here). Selecting only those solutions which are well behaved then leave us with wave functions of the following from: 2 2l + 1 (l | m |)! |m| ( ) = Pl [ cos ] , (33) 2 (l + | m |)! where the eigenvalues of L2 are given by
2

l (l + 1)

, l = 0,1, 2,3,....

(34)

Here, l is the corresponding quantum number. Furthermore, for a given value of l , only the following 2l + 1 values of the quantum number m are allowed: m = l , l + 1,..., l 1, l . (35) Finally, { Pl|m| [ x ] l = 0,1, 2,...; m = 0, 1,..., l} correspond to a family of functions, which are known as the associated Legendre functions. Those functions are defined for 1 x 1 [note that x = cos in Eq. (33) automatically satisfies this condition since 1 cos 1 ]. They can be obtained from the following general formula: l +|m| m /2 d l 1 Pl |m| [ x ] = l (1 x 2 ) x 2 1) (36) l +|m| ( 2 l! dx The first few associated Legendre polynomials are given by: P00 = 1
P 0 = cos , P1 = sin 1 1 1 ( 3cos2 1) , P21 = 1 sin cos , P22 = sin 2 2 2 To summarize, the simultaneous eigenfunctions of L2 and Lz are given by: P20 =

(37)

Yl m ( , ) =

2l + 1 (l | m |)! |m| Pl [ cos ] eim with l = 0,1, 2,... and m = 0, 1,..., l (38) 4 (l + | m |)!
2

and the corresponding eigenvalues are l (l + 1)

for L2 and m for Lz : (39)

L2Yl m ( , ) = l (l + 1) 2Yl m ( , ) ; l = 0,1, 2,... LzYl m ( , ) = m Yl m ( , ) ; m = 0, 1,..., l The functions

{Y ( , ) | l = 0,1, 2,... ;
m l

m = 0, 1,..., l} are

called

the

spherical

harmonics. The first few spherical harmonics are given below:

2

For the derivation, see for example, Chapter 5 of Quantum Chemistry by Ira N. Levine.

Y00 ( , ) = Y10 ( , ) =

1 4 3 3 cos , Y11 ( , ) = sin e i 4 8

5 15 Y20 ( , ) = ( 3cos2 1) , Y21 ( , ) = 8 sin cos ei , 16 Y22 ( , ) = 15 sin 2 e 2i 32

(40)

Properties of the eigenvalues of L2 and Lz :

1. Angular momentum is characterized by two quantum numbers, l and m , which reflect the fact that it can be described in terms of two degrees of freedom, and . 2. The eigenvalue l (l + 1) 2 of L2 is (2l + 1) -fold degenerate. In other words, there are (2l + 1) independent and orthogonal eigenfunctions that have this eigenvalue. Simultaneously, those are also non-degenerate eigenfunctions of L , corresponding to
z

the eigenvalues: l , (l 1) ,..., (l 1) , l . In fact, we know that they are orthogonal because they are the non-degenerate eigenfunctions of a hermitian operator ( L ).
z

3. The amplitude of the angular momentum vector is quantized. Individual measurements of the amplitude can only yield the eigenvalues l (l + 1) ,with l = 0,1, 2,... Thus, the amplitude of the angular momentum vector can be given by 0, 2 , 6 , 12 ,... , and no intermediate values are allowed! This should be contrasted with classical mechanics, where the amplitude of the angular momentum vector is a continuous quantity. Indeed, since is an extremely small number on the macroscopic scale, the angular momentum of a macroscopic object would be characterized by an astronomically large value of l , such that the amplitude of the angular momentum vector is quasi-continuous. However, the quantization must be taken into account in the case of microscopic objects, such as electrons and atoms, where the value of l is small. 4. For a given value of l , or equivalently the amplitude of the angular momentum vector, l (l + 1) , only certain orientations of the angular momentum vector relative to the z axis are allowed, namely those that correspond to one of the following components along the z axis: 0, ,..., l . This phenomenon, which is sometimes referred to as space quantization, should be contrasted with classical mechanics where the orientation of the angular momentum vector relative to the z axis is continuous. As for the angular momentum amplitude, the orientation becomes quasicontinuous for macroscopic objects, since is an extremely small number on the macroscopic scale. However, it must be taken into account when dealing with microscopic objects.

10

5. It should be noted that except for the case l = m = 0 , m is always smaller than
l (l + 1) . This is a reflection of the Heisenberg uncertainty principle. More

specifically, m = l (l + 1) would have implied that the angular momentum amplitude is equal to the size of its component along the z axis. This would have implied that the angular momentum vector lies along the z axis, and that therefore its components along the x and y axes are equal to zero. This would have meant that the x, y and z components of the angular momentum can be known simultaneously. However, the uncertainty principle dictates that 2 1 1 Lx , Ly = Lx Ly i Lz = | m | (41) 2 2 2 Thus, unless l = m = 0 , Lx Ly > 0 and Lx , Ly cannot be known with certainty if Lz

is.

Figure 3: In quantum mechanics, only the amplitude and z component of the angular momentum can be known simultaneously. An infinite number of vectors which form the cone shown in the picture can have the same amplitude and z component. This cone represents our uncertainty regarding the actual angular momentum vector.

Figure 4:The allowed orientations of the angular momentum vector for l = 2 .

11

Shapes and properties of the spherical harmonics

The spherical harmonics are functions of the angles and . Specifying the values of those two angles defines an orientation. A convenient and informative way of visualizing the spherical harmonics is by associating with each set of values of and , a vector that starts at the origin, has an orientation which is set by the specific values of and , and whose amplitude is equal to the corresponding value of Yl m ( , ) . The tips of all those vectors then form a surface whose shape provides a visualization of Yl m ( , ) . Areas where the function is positive or negative can be designated by + and , respectively. We will now demonstrate this in the case of the first few spherical harmonics. The first spherical harmonic is Y00 ( , ) = 1/ 4 . (42) This function is constant and independent of and . Thus, the above mentioned vectors will have the same amplitude in all directions, and the surface formed by their tips will correspond to a sphere (of radius 1/ 4 ). Physically speaking, this spherical symmetry implies that the probability of finding the particle at different orientations is the same, i.e. uniform (there is no preferred direction). Next, consider the spherical harmonic Y10 ( , ) = 3 / 4 cos . (43)
Y10 ( , ) is independent of , and is therefore invariant with respect to rotation around

the z axis. At the same time, Y10 ( , ) is explicitly dependent on , and the dependence is rather simple and goes like cos . It should be remembered that 0 correspond to the angle relative to the positive z axis. Thus, Y10 ( , ) is positive above the ( x, y ) plane ( z 0, 0 < < / 2 ) and maximal in the direction of the positive z axis, where = 0 and cos = 1 . Y10 ( , ) is negative below the ( x, y ) plane ( z 0, / 2 < < ) and reaches its most negative value along the negative z axis, where = and cos = 1 . When the orientation lies in the ( x, y ) plane and does not have a z component, = / 2 , cos = 0 and Y10 ( , ) = 0 . The shape of the resulting surface resembles that which would be obtained by rotating the number 8 around the z axis. We next consider the two other spherical harmonics that correspond to l = 1 : Y11 ( , ) = 3 / 8 sin e i (44) In order to proceed, we need to address the problem posed by complex wave functions. One approach is to look at the probability density, which happens to be the same for those two wave functions: Y11 ( , ) = (3 / 8 ) sin 2 . Similarly to Y10 ( , ) , this probability density is independent of and therefore has a rotational symmetry around the z axis. However, it goes to zero along the z axis ( = 0, ). Thus, the surface that
2

12

represents Y11 ( , ) has the shape of a donut that lies in the ( x, y ) plane and is centered at the origin. Another approach is to separately plot the real and imaginary parts of Y11 ( , ) :
Y1x ( , ) = 2 Re Y11 ( , ) =

1 3 Y11 ( , ) Y11 ( , ) = sin cos 4 2

(45)

1 3 1 1 Y1 y ( , ) = 2 Im Y11 ( , ) = i 2 Y1 ( , ) + Y1 ( , ) = 4 sin sin It should be noted that Y1+1 ( , ) = Y11 ( , ) , such that those two functions have the same real and imaginary parts (except for the sign of the latter). Since the real and imaginary parts can be written as linear combinations of Y11 ( , ) , they are still eigenfunctions of L2 , with the eigenvalue l (l + 1) 2 (but they are no longer
eigenfunctions of Lz !).
*

Y1x ( , ) is obviously maximal when sin and cos are, that is along the x axis.
Y1x ( , ) is positive on the side of the positive x axis, negative on the side of the negative
x axis, and uniformly zero in the ( y, z ) plane. Similarly, Y1 y ( , ) is obviously maximal

when sin and sin are, that is along the y axis. Furthermore, Y1 y ( , ) is positive on the side of the positive y axis, negative on the side of the negative y axis, and uniformly zero in the ( x, z ) plane. In fact, these two functions have the exact same shape as Y10 ( , ) , except for the fact that they are oriented along the x and y axes. In fact,
Y10 ( , ) is often denoted Y1z ( , ) for this reason.3

The shapes of higher order spherical harmonics can be obtained in a similar manner, although they become increasingly more complicated.
Raising and lowering angular momentum operators

As in the case of the harmonic oscillator, one can define raising and lowering operators in the case on angular momentum. Those operators are very useful in practice because of their ability to simplify calculations. The raising and lowering angular momentum operators are defined by: L = L + iL ; L = L iL (46)
+
x y

respectively. It should be noted that although Lx and Ly are hermitian, L+ and L are not hermitian, and are in fact hermitian conjugates of each other (for the same reason that the

The shapes of Y1 , Y1 , Y1 are essentially those of the s, px , p y , pz atomic orbitals which you should be

familiar with from general chemistry. This is not a coincidence. We will explore this relationship in more detail in the next chapter.

13

 harmonic oscillator raising and lowering operators, a and a , are hermitian conjugates of each other): * (47) L+ = L .

Figure 5: The modulus square of the spherical harmonics.

Figure 6: The real and imaginary parts of the spherical harmonics.

14

 It is also important to note that Lx and Ly can be written in terms of L+ and L :

1 1 (48) Lx = L+ + L ; Ly = L+ L . 2 2i Actually, in practice, working with L+ and L , rather than with Lx and Ly , can save a lot of time in carrying out actual calculations (see below). The commutators between L+ and L , Lz and L2 , can be obtained from Eqs. (46) and (11): L+ , L = 2 Lz
Lz , L+ = L+ (49) Lz , L = L 2 L+ , L = L , L2 = 0 For example, the first commutator in Eq. (49) is obtained as follows: L+ , L = Lx + iLy , Lx iLy = i Lx , Ly + i Ly , Lx = i (i ) Lz + i (i ) Lz = 2 Lz (50)
The second commutator in Eq. (49) will be used in order to show that L+ is indeed a raising operator: Lz , L+ = L+ Lz , L+ Yl m = L+Yl m Lz L+Yl m L+ LzYl m = Lz L+Yl m m L+Yl m = L+Yl m (51) L L Y m = (m + 1) L Y m L Y m Y m +1
z

+ l

+ l

+ l

The proportionality constant can be obtained as follows:

2 L+Yl m = Yl m +1 = d sin 0

m d L+Yl
0

L+Yl m = d sin
0

d ( Y )
l 0

m *

L L+Yl m (52)

Since,

L L+ = Lx iLy We get

)( L + iL ) = L + L
x y 2 x

2 y

+ i Lx , Ly = L2 + L2y Lz x

= L2 + L2y + L2 L2 Lz = L2 L2 Lz x z z z
* 2 = d sin d (Yl m ) L2 L2 Lz Yl m z 0 0

(53)

= = Thus,

l (l + 1) m 2 m d sin
0

m m d (Yl ) Yl * 0

(54)

[l (l + 1) m(m + 1)] =
L+Yl m =

l (l + 1) m(m + 1)

l (l + 1) m(m + 1)Yl m +1

(55)

15

 It should be noted that L+Yl l = 0 ( m = l is the maximal allowed value of m , and

m cannot be raised above it). Similarly, the third commutator in Eq. (49) can be used to show that L is the lowering operator: LYl m = l (l + 1) m(m 1)Yl m 1 (56)

It should be noted that LYl l = 0 ( m = l is the lowest allowed value of m , and

m cannot be lowered below it). Example: The expectation value of Lx in the state Yl m can be obtained without explicitly solving any integrals, as follows: 1 Yl m Lx Yl m = Yl m L+ Yl m + Yl m L Yl m = 2 1 m m +1 + l (l + 1) m(m 1) Yl m Yl m 1 (57) l (l + 1) m(m + 1) Yl Yl 2 1 = l (l + 1) m(m + 1) 0 + l (l + 1) m(m 1) 0 = 0 2 It can be shown in a similar manner that Yl m Ly Yl m = 0 (verify that!).

The expectation value of any function of the angular momentum operators can be evaluated in a similar manner.

Rotations of diatomic molecules within the rigid rotor approximation

The rigid rotor approximation
In our previous discussion of vibrations in diatomic molecules, we have restricted ourselves to a non-rotating diatomic molecule (the molecular axis was assumed to coincide with the x axis, such that angular momentum is irrelevant). We are now ready to lift this restriction and see how the quantum theory of angular momentum can be used for describing the rotations of a diatomic molecule, as well as its interplay with the vibrational and translational molecular degrees of freedom. The first part of the analysis involves manipulations on the Hamiltonian, which can be performed in terms of classical mechanics. To this end, consider a heteronuclear diatomic molecule in 3D, with the following classical Hamiltonian: 2 p12 p2 (58) H= + + V ( r2 r1 ) 2m1 2m2

16

Here, r1 , r2 are the 3D position vectors of nuclei 1 and 2, respectively, p1 , p2 are the corresponding momentum vectors, and m1 , m2 are the masses.

V ( r2 r1 ) is the

internuclear potential that generally has the shape of an asymmetrical well. Viewed in terms of the coordinates of the individual nuclei, this appears to be a nonseparable 6D problem (2 particles in 3D). However, one can make the Hamiltonian separable by casting it in terms of the center of mass and relative degrees of freedom. The corresponding position and momentum variables are given by: m r + m2 r2 , pcm = p1 + p2 rcm = 1 1 m1 + m2 (59) m1 p2 m2 p1 r = r2 r1 , p = m1 + m2 The Hamiltonian in Eq. (58) can then be rewritten in terms of the center of mass and relative positions and momenta: p2 p2 +V (r ) (60) H = cm + 2M 2 where M and are the familiar molecular and reduced masses [ M = m1 + m2 , = m1m2 / ( m1 + m2 ) ]. The important thing is that the original 6D

Hamiltonian is now separable into two 3D Hamiltonians. The center of mass part, 2 pcm / 2M , is equivalent to that of a free particle of mass M in 3D. The relatively large mass implies that the center of mass motion can be treated classically (under most circumstances). The other part of the Hamiltonian is associated with the relative motion and given by p 2 / 2 + V (r ) . It is equivalent to the Hamiltonian of a single particle of mass , in 3D, which is subject to the central potential V (r ) . The following discussion will focus on this part. The first step in the analysis would be to switch from classical mechanics to quantum mechanics. The quantum Hamiltonian operator that corresponds to the relative motion is given by (61) 2 (note that the original Hamiltonian included the center of mass part, which is now left out). Further simplification and separability can be achieved by rewriting this Hamiltonian in terms of spherical coordinates. The potential part, V (r ) , is already given in terms of the spherical coordinate r . However, at this point we only know how to represent 2 in terms of Cartesian coordinates: 2 = ( 2 / x 2 + 2 / y 2 + 2 / z 2 ) . Nevertheless, obtaining 2 in terms of spherical coordinates by using Eq. (14) and the chain rule is straightforward (although rather tedious), and the final result is given below: 2 1 2 1 2 1 2 1 L2 2 = 2 + 2 2 + cot + 2 = 2 2 2 . (62) r r 2 sin r r r r r r r r Substituting this back into the Hamiltonian in Eq. (61) then enables us to rewrite it in the following form:
H =
2

2 +V (r )

17

 L2 1 2 r + V (r ) + . (63) 2 r 2 r r 2 r 2 The important thing is that the only dependence of this Hamiltonian on the angles and is through L2 , and that L2 is independent of r [see Eq. (23)]. Thus, if it wasnt for the r 2 in the denominator of the term L2 / 2 r 2 in Eq. (63), the Hamiltonian would have H =
2

been separable! Can we still somehow treat it as separable, at least approximately? As it turns out, we can! The argument is based on the fact that changes in r , which corresponds to the internuclear distance, are due to vibrations, and the amplitude of the latter is typically very small in comparison to the equilibrium internuclear distance, or bond length, req . This suggests the following approximation: L2 / 2 r 2 L2 / 2 r 2 (64)
eq

This approximation treats the molecule as a rigid object, at least as far as the L2 / 2 r 2 term is concerned. It is therefore referred to as the rigid rotor approximation. It turns out to work rather well as long as the molecule is not too vibrationally excited, and is valid for most molecules in thermal equilibrium at room temperature. The main advantage of the rigid rotor approximation is that it makes the Hamiltonian in Eq. (63) separable into an r -dependent vibrational part and a ( , ) -dependent rotational part. The corresponding stationary Schrdinger equation is given by: 2 1 2 L2 (r , , ) = E (r , , ) + V (r ) + r (65) 2 2 2 req 2 r r r and we know that a solution of the form (r , , ) = R(r ) ( , ) will work, provided that R(r ) and ( , ) are solutions of the following simpler differential equations (note that we decomposed a 3D problem into simpler and independent 1D and 2D problems): 2 1 2 (66) + V (r ) R (r ) = E vib R(r ) r 2 2 r r r and L2 (67) ( , ) = E rot ( , ) . 2 2 req The overall energy also consists of vibrational and rotational contributions: E = E vib + E rot . Consider first the vibrational Schrdinger equation [Eq. (66)]. What is the meaning of 2 R(r ) ? According to Eq. (19), r 2 R(r ) correspond to the probability density of finding
an internuclear distance equal to r . So, if we want to think about Eq. (66) as a truly 1D Schrdinger equation, it makes sense to redefine the wave function as (r ) = rR (r ) , such that the probability density is given by (r ) . As it turns out, Eq. (66) indeed acquires
2

the usual form of a Schrdinger equation in 1D when written in terms of (r ) = rR (r ) (show that!):

18

 2 2 (68) + V (r ) (r ) = E vib (r ) 2 2 r We have considered this equation when we discussed vibrations in a diatomic molecule. We have seen that, within the harmonic approximation, V (r ) 2 (r req ) 2 / 2 , such that
vib E vib En = (n + 1/ 2) and (r ) n (r req ) are the energy levels and stationary

states of the harmonic oscillator ( n = 0,1, 2,.... ). Eq. (67) can also be easily solved since the rigid rotor rotational Hamiltonian is proportional to L2 , and we already know what the eigenvalues and eigenfunctions of this operator look like [see Eq. (34)]. More specifically: L2 l (l + 1) 2 m (69) Yl m ( , ) = Yl ( , ) ; l = 0,1, 2,... ; m = 0, 1,..., l 2 2 2 req 2 req It is important to note that the l -th rotational energy level is (2l + 1) -fold degenerate (corresponding to the 2l +1 possible values of the quantum number m ). For historical reasons, the letters J and M are used instead of l and m , in order to denote the angular momentum quantum numbers in the context of molecular rotations. It is also common to define the moment of inertia of the rotor as 2 (70) I = req such that :
L2 M J ( J + 1) YJ ( , ) = 2I 2I
2

YJM ( , ) ; J = 0,1, 2,... ; M = 0, 1,..., J

(71)

What we have accomplished until this point can be summarized as follows: 1. Within the harmonic and rigid rotor approximations, the rotational-vibrational energy levels of a diatomic molecule are given by: J ( J + 1) 2 En , J = (n + 1/ 2) + , n = 0,1, 2,... , J = 0,1, 2,... 2I with (2 J + 1) -fold degeneracy of the energy level En, J (72)

2. and the stationary wave functions that describe the state of the nuclei of a diatomic molecule are given by:

n (r req )YJM ( , )

(73)

where n (r ) are the stationary states of the harmonic oscillator, and YJM ( , ) are the spherical harmonics. The energy gap between the J -th and ( J + 1) -th energy levels of the rigid rotor is given by: 2 ( J + 1)( J + 2) 2 J ( J + 1) 2 (74) EJ +1 EJ = = 2( J + 1) 2I 2I 2I

Pure rotational spectroscopy and selection rules

19

Thus, the energy gap between neighboring levels grows linearly with J . The corresponding transition frequencies in Hz are given by E EJ h (75) = J +1 = 2( J + 1) 2 = 2( J + 1) B , 8 I h where h (76) B= 2 8 I

is the so called rotational constant (in Hz). The rotational constant is sometimes also defined in units of cm 1 : B h (77) B= = 2 c 8 cI such that the above transition frequencies in cm 1 are given by = 2( J + 1) B . (78) 25 26 The reduced mass of diatomic molecules is around 10 10 kg , and a typical bond length is around 1010 m , which gives a typical moment of inertia of 1045 1046 kg im 2 . This gives h 2 B = 2 = (1011 1012 ) Hz 2 B = ( 3 30 ) cm 1 (79) 4 I Those frequencies lie in the microwave and far IR regions of the electromagnetic spectrum and are much smaller than the frequencies of molecular vibrations, which fall in the IR region [ (102 103 ) cm 1 ].

The discussion above leads to the conclusion that the rotational-vibrational energy levels are organized in a hierarchical manner, such that each vibrational level is accompanied by a manifold of rotational levels. Microwave radiation can be used in order to induce transitions between rotational levels within the manifold that corresponds to the same vibrational energy level. This type of spectroscopy is called pure rotational spectroscopy. We next consider the selection rules that apply to pure rotational spectra. We first have to discuss the relevant dipole moment operator. Within the rigid rotor approximation, a diatomic molecule is made of two nuclei separated by a fixed distance (the bond length), which are surrounded by an electronic cloud. The details of the actual shape and density of this electronic cloud will be dealt with in a later chapter. However, one expects that the charge distribution will be symmetrical in the case of a homonuclear diatomic molecule, and asymmetrical in the case of a heteronuclear diatomic molecule. Thus, a homonuclear diatomic molecule will have a zero dipole moment, while a heteronuclear diatomic molecule will have a finite (permanent) dipole moment. The asymmetrical charge distribution in the latter case can be visualized by assigning a surplus positive charge, + q , to the more electropositive atom, and a negative charge, q , to the more electropositive atom. The dipole moment is then given by = qr (80)

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where, r is a vector of length req that points from the negative end of the molecule to its positive end. The x, y, z components of the dipole moment of a heteronuclear diatomic molecule in terms of spherical coordinates are therefore given by (81) x = qreq sin( ) cos( ) ; y = qreq sin( ) sin( ) ; z = qreq cos( ) Now, consider a transition between the state ( J i , M i ) and the state ( J f , M f ) . Assuming that the driving field is linearly polarized along the z axis, the selection rules are dictated by the following integral: Y
Mi Ji

z Y

Mf Jf

= qreq d sin ( ) d YJM i ( , ) cos ( ) YJ f f ( , ) i

*
M

(82)

As it turns out, this integral vanishes, unless J f = J i 1 and M f = M i (stated without proof). Similarly, if the field is polarized along the x axis, the selection rules are dictated by Y
Mi Ji

x Y

Mf Jf

= qreq d sin ( ) d YJM i ( , ) sin ( ) cos ( ) YJ f f ( , ) i

*
M

(83)

As it turns out, this integral vanishes unless J f = J i 1 and M f = M i 1 (stated without proof). The same result is also obtained if the field is polarized along the y axis. Thus the selection rules for pure rotational spectra can be summarized as follows: The molecule must have a permanent dipole moment (84) and the transition must satisfy: J = 1 ; M = 0, 1 It should be noted that the energy only depends on the quantum number J , such that the absorption spectrum will consist of transitions between the J -th and ( J + 1) -th energy levels. Thus, according to Eq. (75), the pure rotational absorption spectrum of a diatomic molecule with a permanent dipole moment (i.e. a heteronuclear diatomic molecule) would consist of a sequence of equally spaced lines at frequencies 2 B, 4 B, 6 B,... The relative intensity of each of those lines will depend on the fraction of molecules that populate the initial state at thermal equilibrium. The latter can be obtained from the corresponding Boltzmann distribution ( N is a constant which is needed to ensure that j =0 P( EJ ) = 1 ):
J ( J + 1)hB (85) P ( EJ ) = N (2 J + 1) exp k BT Thus, the most intense line will correspond to the value of J for which P ( EJ ) is maximal (which usually doesnt correspond to the lowest possible rotational energy level, because the degeneracy grows with J ).

Example. To a good approximation, the pure rotational spectrum of 1H 35Cl consists of a series of equally spaced lines, separated by 6.26 1011 Hz . This means that 2 B = 6.26 1011 Hz . We can use this information in order to calculate the bond length:

21

2B =

h 4 I
2

h 4 r
2 2 eq

req =

h 4 2 B
2

6.626 1034 Js = = 1.28 1010 m = 1.28 A 27 2 11 1 4 (1.63 10 kg ) ( 6.26 10 s )

(86)

What would the separation between the lines be if we substituted the Hydrogen by Deuterium? Since req remains the same (the only difference between deuterium and hydrogen is in the mass, and gravitational forces are negligible in the molecular domain): 2 B ( 2 H 35Cl ) ( 1H 35Cl ) 1 2 B ( 1H 35Cl ) 2 35 = 2 B ( H Cl ) = = 3.13 1011 s 1 . (87) 1 35 2 35 2 2 2 B ( H Cl ) ( H Cl )

Vibrational-Rotational Spectroscopy and selection rules

As we have seen in the previous chapter, shining a diatomic molecule with IR radiation can induce transitions between vibrational levels. However, we now see that each vibrational level actually comes with a manifold of rotational levels! Thus, the transitions are actually between rotational levels that correspond to different vibrational energy levels. This type of spectroscopy is called vibrational-rotational spectroscopy. The selection rules for vibrational-rotational spectroscopy are a combination of the selection rules for vibrational transition (within the harmonic approximation) and rotational transition (within the rigid rotor approximation): The molecule must have a permanent dipole moment (88) and the transition must satisfy: n = 1 ; J = 1 ; M = 0, 1

Thus, the absorption band in the IR that corresponds to a particular vibrational transition will actually consists of several lines that correspond to transitions between different rotational levels. More specifically, if the pure vibrational transition corresponds to the frequency (in the IR), then transitions of the form (n, J ) (n + 1, J + 1) will lead to lines at + 2B , + 4 B,... , which constitute what is known as the R branch, while transitions of the form (n, J ) (n + 1, J 1) will lead to lines at 2B , 4 B,... , which constitute what is known as the P branch (see Figs. 7 and 8).

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Figure 7: The allowed vibronic transitions in a heteronuclear diatomic molecule.

Figure 8: The R and P branches in the rotational-vibrational spectrum of a diatomic heteronuclear molecules.

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