Relativistic Quantum Fields 1

Mark Hindmarsh
University of Sussex
m.b.hindmarsh[NOSPAM]sussex.ac.uk
Autumn Term 2004
1
Contents
0 Course information 3
1 Preliminaries 7
1.1 Special Relativity . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2 Lorentz transformations and space-time interval . . . . . . . . 8
1.3 Relativistic notation . . . . . . . . . . . . . . . . . . . . . . . 8
1.4 Matrix representation of Lorentz transformations . . . . . . . 9
1.5 Space-time metric . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.6 General Lorentz transformations: the Lorentz group . . . . . . 11
1.7 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.8 4-vectors and the scalar product . . . . . . . . . . . . . . . . . 12
1.9 The dAlembertian . . . . . . . . . . . . . . . . . . . . . . . . 13
1.10 Lorentz covariance . . . . . . . . . . . . . . . . . . . . . . . . 13
2 Relativistic wave equations 14
2.1 Klein-Gordon equation . . . . . . . . . . . . . . . . . . . . . . 14
2.2 Maxwells equations in covariant form . . . . . . . . . . . . . . 15
2.2.1 Gauges . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.3 Lorentz gauge solutions . . . . . . . . . . . . . . . . . . . . . . 21
3 Lagrangian formulation of eld theory 23
3.1 Lagrangian and Hamiltonian mechanics . . . . . . . . . . . . . 23
3.1.1 Lagrangian Mechanics . . . . . . . . . . . . . . . . . . 24
3.1.2 Hamiltonian Mechanics . . . . . . . . . . . . . . . . . . 26
3.2 Lagrangian mechanics for the real scalar eld . . . . . . . . . 27
3.3 Lagrangian mechanics for the electromagnetic eld . . . . . . 31
3.4 Noethers theorem and conservation laws . . . . . . . . . . . . 32
3.4.1 Conservation of energy-momentum . . . . . . . . . . . 36
4 Canonical quantisation of the scalar eld 36
4.1 Quantisation of nite systems . . . . . . . . . . . . . . . . . . 37
4.2 The real scalar eld . . . . . . . . . . . . . . . . . . . . . . . . 38
4.3 States of the scalar eld; zero point energy . . . . . . . . . . . 42
4.4 Particle interpretation of states . . . . . . . . . . . . . . . . . 46
5 Canonical quantisation of the electromagnetic eld 47
5.1 Coulomb gauge quantisation . . . . . . . . . . . . . . . . . . . 48
5.2 Canonical quantisation in the Lorentz gauge . . . . . . . . . . 51
2
6 Fermions 52
6.1 Plane wave solutions . . . . . . . . . . . . . . . . . . . . . . . 54
6.2 The interpretation of negative energy states . . . . . . . . . . 56
6.3 Quantising the spinor eld . . . . . . . . . . . . . . . . . . . . 57
6.4 Conserved charge . . . . . . . . . . . . . . . . . . . . . . . . . 60
7 Scalar elds in an expanding Universe 61
7.1 Scalar eld in curved spacetime . . . . . . . . . . . . . . . . . 61
7.2 Classical scalar eld uctuations . . . . . . . . . . . . . . . . . 63
7.3 Quantum scalar eld uctuations . . . . . . . . . . . . . . . . 65
8 Interacting elds 1 67
8.1 The interaction picture . . . . . . . . . . . . . . . . . . . . . . 69
8.2 The S-matrix and transition amplitudes . . . . . . . . . . . . . 72
8.3 Example:
4
. . . . . . . . . . . . . . . . . . . . . . . . . . . 74
8.4 Feynman propagator for scalar eld . . . . . . . . . . . . . . . 75
8.5 Wicks theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 76
8.6 Using Wicks Theorem . . . . . . . . . . . . . . . . . . . . . . 77
8.6.1 S
(0)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
8.6.2 S
(1)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
A Dirac matrices 81
A.1 Standard representation of Dirac matrices . . . . . . . . . . . 81
A.2 Pauli matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
B Problem Sheets 83
B.1 Problem Sheet 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 83
B.2 Problem Sheet 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 85
B.3 Problem Sheet 3 . . . . . . . . . . . . . . . . . . . . . . . . . . 87
B.4 Problem Sheet 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 89
3
0 Course information
Course outline
Aim: to introduce 1st year physics graduate students to the theory of
quantum elds.
Relativistic wave equations (Ryder Ch 2, G & R Ch 4.1, 6.2). Rela-
tivistic notation; Klein-Gordon equation.
Lagrangian formulation (Ryder Ch 3, G & R Ch 2). Lagragian particle
mechanics; Euler-Lagrange for real scalar eld; Noethers theorem for
scalar elds.
Canonical quantisation (Ryder Ch 4, G & R Ch 4/7). Canonical com-
mutation relations for real scalar eld; Fock space; Number, energy and
momentum operators; Propagators.
Fermions (Ryder 4.3, 6.7). Dirac equation; Lagrangian formulation;
Canonical quantisation;
Interacting elds 1 Interaction picture; S-Matrix and transition ampli-
tudes; Wicks theorem; Matrix elements in perturbation theory; Scat-
tering cross section 2 2 particles.
Scalar Quantum Electrodynamics Maxwell equations; Lagrangian for-
mulation of Electromagnetic eld equations; Canonical quantisation of
EM eld in Lorentz gauge; Gauge symmetry; Feynman rules for SQED;
photon polarisation sums in scattering probabilities.
Scalar eld theory in an expanding Universe Friedmann-Robertson-
Walker metric; Lagrangian;
Learning outcomes
By the end of the course the student should be able to: perform four-
vector algebra; derive the Euler-Lagrange equations for any eld theory;
understand and apply Noethers Theorem; give an account of the canonical
quantisation procedure for elds; explain how transition amplitudes are cal-
culated in eld theory; write down and calculate simple Feynman graphs;
give an account of quantum uctuations of the eld in the vacuum.
4
Teaching methods
There will be 3 lectures a week throughout the term. Problem sheets will
be given every two weeks.
5
Assessment
MSc students Problem sheets will count for 40% of the total mark for the
course, the remaining 60% will come from the 1.5 hour exam at the
end of the year, which will have a choice of two questions from three.
DPhil students Problem sheets will count for 100% of the total mark.
Reading list
Field Quantization, W. Greiner and J. Reinhardt (Springer, Berlin,
1996),
Quantum Field Theory, F. Mandl and G. Shaw (Wiley, Chichester,
1984).
Quantum Field Theory, L. Ryder (C.U.P., Cambridge, 1984).
An introduction to quantum eld theory, Michael E. Peskin, Daniel V.
Schroeder (Addison-Wesley, Reading, Mass; 1995).
Introduction to Gauge Field Theory, D. Bailin & A. Love (Adam Hilger,
1993)
Quantum eld theory: a modern introduction, Michio Kaku (O.U.P.,
Oxford, 1993)
Fields, W. Siegel (http://insti.physics.sunysb.edu/ siegel/plan.html).
The books are complementary in some way. Greiner & Reinhardt is
extremely thorough. Ryder has a good balance of topics and starts fairly
simply, but used path integral quantisation which is not the most pedagogical
way to introduce eld theory. Mandl and Shaw is pedagogical but (like
Greiner & Reinhardt) adopts dierent Fourier transform and normalisation
conventions which can confuse the unwary. Peskin & Schroeder covers the
most material and is probably the book of choice for serious students of eld
theory. There are quite a few errors: a list can be found at
http://www.slac.stanford.edu/~mpeskin/QFT.html
Bailin & Love starts at a higher level than this course, but is concise
and has a unique discussion of the S-matrix generating functional. Kaku is
very complete but rather rushed. Siegels PDF lecture notes are the most
complete exposition I have seen. They are very advanced and will cost you
a lot to print at over 800 pages.
6
Prerequisites
The course assumes that you have already encountered
Quantum Mechanics (Schrodinger equations, free particle solution, har-
monic oscillator).
Special Relativity (Lorentz transformations, space-time interval, Lorentz
invariance).
Electromagnetism (Maxwells equations in free space).
Cauchys theorem (contour integration).
Fourier transforms.
The Dirac -function.
Scattering cross-sections.
It will help to have come across the following subjects, mostly covered in the
Sussex undergraduate course Theoretical Physics II, as only a brief review is
given here.
Lagrangian formulation of classical mechanics (e.g. Classical mechan-
ics, T.W.B. Kibble and F.H. Berkshire, London: Longman, 1996).
4-vector notation in Special Relativity (e.g. Chapter 2 in General rel-
ativity and cosmology, T.L. Chow, Winnipeg: Wuerz, 1994).
Natural Units (Mandl and Shaw, Section 6.1).
Course Lecturer
Mark Hindmarsh, Scitech, Arundel 213.Telephone: 8934
E-mail: m.b.hindmarsh[NOSPAM]sussex.ac.uk
Course web: http://www.pact.cpes.sussex.ac.uk/~markh/RQF1/
Oce hour: Tuesday 23pm, or by arrangement (email is best).
7
1 Preliminaries
The reason for studying quantum eld theory is that it is the theory which
best describes elementary particles and their interactions. Field theory rec-
onciles quantum mechanics and special relativity and is automatically a
quantum-mechanical theory of many particles. Where we can calculate with
great precision, such as in Quantum Electrodynamics, spectacularly accurate
agreement with experiment is found. However, eld theory is not entirely
satisfactory. Compared to the elegant geometrical framework of General
Relativity, it is a mess, and when one rst tries to use it to calculate simple
physical quantities, we immediately encounter innite results. Making sense
of these innities occupies a large proportion of any quantum eld theory
course, but fortunately most of them can be ignored if we neglect interac-
tions between particles. There are still non-trivial physical predictions to
be made in non-interacting (free) eld theory, associated with non-trivial
boundary conditions, such as in the Casimir eect, or with strong gravita-
tional backgrounds, such as we expect to nd near black holes or in the early
Universe.
This course is aimed towards computing the amplitude of uctuations in
a real scalar eld in an exponentially expanding space-time, and does just
enough (but no more) to get there. We will prove the famous result that in
this spacetime, which is a limiting case in the class of inationary universes,
the power spectrum of the uctuations in a massless scalar eld is (H/2)
2
,
where H is the expansion rate. The scalar eld which causes ination, the
inaton, is eectively massless during ination, and these vacuum uctua-
tions form the basis of the explanation of the perturbations we observe in
the Cosmic Microwave Background today.
1.1 Special Relativity
We start by quoting a formulation of the Principle of Special Relativity:
Fundamental physical laws are the same for all observers moving
with constant velocity relative to one another.
Special Relativity is a rather peculiar theory as it appears to be a theory
about other theories. However, when applied to dynamics it does have
testable consequences, which have of course corroborated the theory to great
accuracy. It is in fact true in only a restricted domain, where gravitational
elds are weak and can be neglected. Although it may not seem like it,
the gravitational eld at the surface of the earth it suciently weak to be
neglected for most purposes in relativity.
8
1.2 Lorentz transformations and space-time interval
Suppose a frame of reference F

is moving at velocity v in the x direction

relative to another F. Special Relativity tells us that coordinates measured
in F

are related to coordinates measured in F by a Lorentz transformation

t t

= (t vx/c
2
) ,
x x

= (x vt) ,
y y

= y,
z z

= z,
(1.1)
where
= 1/
_
1 v
2
/c
2
. (1.2)
Although neither the distance nor the time interval between two events is
observer-independent in special relativity, there is a concept of space-time
interval, which contains a little of both, and is something that all observers
can agree upon. In its innitesimal form, the interval ds between two events
at (t, x, y, z) and (t + dt, x + dx, y + dy, z + dz) is given by
ds
2
= c
2
dt
2
dx
2
dy
2
dz
2
. (1.3)
One can easily check that the value of ds does not change under this trans-
formation. We say that ds is Lorentz invariant.
1.3 Relativistic notation
If a theory which is usually formulated in terms of a set of equations
is consistent with the Principle of Special Relativity we often say that
it is Lorentz covariant. There is a notation which, if followed correctly,
automatically ensures that equations are Lorentz covariant. Quantities are
assembled into 4-vectors (and later on we will encounter 4-tensors) which
transform in a simple linear way when one compares their values between
observers in dierent states of uniform motion. Just as one can form a 3-
component vector x
i
from three spatial coordinates
x
i
= (x
1
, x
2
, x
3
) (x, y, z), (1.4)
one can dene the 4-vector space-time coordinate for an event by
x

= (x
0
, x
1
, x
2
, x
3
) (ct, x, y, z). (1.5)
We shall adopt the following conventions when labelling vectors:
Greek letters from mid-alphabet for space-time indices (, , , , . . .);
Roman letters from mid-alphabet for spatial indices (i, j, k, . . .);
bold face will also be used for 3-vectors, e.g. x.
9
1.4 Matrix representation of Lorentz transformations
Recall the form of a Lorentz transformation between two frames of reference
moving with velocity v in the x direction relative to one another (1.1). This
linear transformation can be represented by matrices

acting linearly on
the coordinates:
x

=
3

=0

, (1.6)
or in dierential form
dx

dx

=
3

=0

dx

, (1.7)
The matrix has entries

=
_
_
_
_
_
v/c
v/c
1
1
_
_
_
_
_
, (1.8)
where labels the rows and the columns.
In order to save writing a large number of summation signs, we often use
Einsteins convention that repeated indices in an expression are summed over
automatically without the need for a summation sign. In this convention,
Equation (1.7 becomes
dx

dx

dx

, (1.9)
1.5 Space-time metric
It is convenient in special relativity (and fundamental in general relativity!)
to dene a matrix g

, which is used in the expression for the space-time

interval:
ds
2
=
3

,=0
g

dx

dx

. (1.10)
We call g

the metric. In Special Relativity, the metric always takes a

particular constant value,
g

_
_
_
_
_
1
1
1
1
_
_
_
_
_
. (1.11)
10
We often call

the Minkowski metric. In Einsteins repeated index con-

vemtion, (1.10) can be written as
ds
2
=

dx

dx

. (1.12)
Note that the position of the indices is important. There is a dierence
between vectors with superscript indices and those with subscript indices,
which shows up in their properties under Lorentz transformations. We may
use the metric tensor to dene an innitesimal coordinate with a lower index:
dx

dx

(1.13)
Vectors with subscript indices are termed covariant, to distinguish them from
their contravariant partners with superscript indices. They transform oppo-
sitely to contravariant 4-vectors, with the inverse of the Lorentz transforma-
tion matrix :
dx

dx

= (
1
)

dx

. (1.14)
One can check that this is true by noting that the space-time interval can be
written ds
2
= dx

dx

, and then substituting the transformation laws (1.7)

and (1.13). One nds
ds
2
ds

2
= dx

dx

= (
1
)

dx

dx

. (1.15)
Note that we have not just blindly substituted (1.7): we have the repeated
index from a to a . The meaning is still the same: that index is to be
summed over the values 0,1,2,3. However, if we have left the index as
equation (1.15) would be ambiguous, as we would not know how to pair o
indices in the summations. This is an important rule with index notation:
never use repeated indices twice on the same side of an equation.
Continuing with equation (1.15), we note that the indices are paired
and can be summed over. We can see that we are multiplying a matrix
by its inverse
1
and therefore must obtain the identity, which expressed in
index notation is
(
1
)

. (1.16)
Here we introduce the Kronecker delta, dened by

=
_
1, = ,
0, ,= .
(1.17)
Note that the spatial components of a 4-vector with its index lowered (or
a covariant vector) have the opposite sign to its counterpart with a raised
index (a contravariant vector):
x
0
= x
0
, x
i
= x
i
. (1.18)
11
Lastly, one can also dene the metric tensor with raised indices as the matrix
inverse of the covariant metric tensor

, (1.19)
1.6 General Lorentz transformations: the Lorentz group
One can explicitly verify that the transformation law (1.1) leaves the space-
time interval ds invariant. By choosing space coordinates so that the relative
velocity of two inertial frames is along the x direction, it follows that all
Lorentz transformations leave the interval invariant. Let us see what we can
infer about the matrices from this condition. The interval transforms as
ds
2
ds
2
=

dx

dx

= ds
2
. (1.20)
But ds
2
may be written ds
2
=

dx

dx

: thus we may infer that

. (1.21)
Hence, any matrix which leaves the metric

invariant under the trans-

formation (1.21) represents a Lorentz transformation. These matrices form a
group of transformations known as the Lorentz group. When combined with
translation symmetry, x

= x

+ a

, with a

a constant 4-vector, it
forms a larger group known as the Poincare group.
1.7 Derivatives
It is through studying derivatives that we discover the utility of the idea of
covariant 4-vectors. Coordinate vectors are naturally contravariant, and it
seems slightly perverse to introduce a covariant version with lowered indices.
However, it turns out that there are 4-vectors which are naturally covariant,
and the principal among these is the partial derivative 4-vector.
The standard partial derivatives with respect to spatial coordinates can
be collected together into a 3-vector:

i
=

x
i
=
_

x
1
,

x
2
,

x
3
_
. (1.22)
The
i
are the components of the gradient operator in an orthonomal
basis e
i
, or
= e
i

i
. (1.23)
12
The relativistic generalisation is a covariant 4-vector

, which is dened as

=

x

=
_
1
c

t
,

x
i
_
. (1.24)
The transformation law for

may be found from the chain rule:

=
x

(1.25)
However, by dierentiating
x

= (
1
)

, (1.26)
which follows from equation (1.6), we nd that

= (
1
)

. (1.27)
Hence the 4-vector dierential really is a covariant 4-vector.
1.8 4-vectors and the scalar product
To recap, a contravariant 4-vector can be dened as any collection of four
quantities which transform the same way as dx

under a Lorentz transfor-

mation. Similarly, a covariant 4-vector can be dened as a collection of
four quantities which transform the same way as

. One can map any con-

travariant vector into a covariant one with the metric

, an operation called
lowering the indices, or one can go in the opposite direction with the inverse
metric

(raising the indices).

There are many important quantities than can be assembled into 4-
vectors: for example, energy E and momentum p of a particle belong to-
gether in a single momentum 4-vector p

, dened by
p

= (E/c, p). (1.28)

We are used to the interval ds
2
= dx

dx

being a Lorentz invariant quantity,

but it is straightforward to see that one can form a Lorentz invariant quantity
from any pair of 4-vectors a

, b

. This is the scalar product, written

a b p
2
= a

= a

= a
0
b
0
a
1
b
1
a
2
b
2
a
3
b
3
. (1.29)
We can take the scalar product of the momentum 4-vector with itself and
recover the relativistic relation between energy, momentum and mass:
p
2
= E
2
/c
2
p
2
c
2
= m
2
c
2
, (1.30)
13
Another invariant can be constructed by contracting the momentum 4-vector
with the position 4-vector
p x = Et px. (1.31)
Another important object is the velocity 4-vector. In Newtonian mechanics,
the velocity is dx
i
/dt, so it might be thought that dx

/dt is the analogous

quantity. However, dt is not Lorentz invariant, so the object as a whole does
not transform like dx

as a 4-vector should. Instead, we should dierentiate

with respect to a Lorentz invariant quantity. There is one ready-made for
us in the form of the space-time interval ds, from which we can dene the
proper time d = ds/c. The 4-velocity of a particle can then be written as
v

=
dx

d
. (1.32)
1.9 The dAlembertian
There is a second order dierential operator which students of electromag-
netism will already have come across, called the wave operator or the dAlembertian.
It is sometimes given its own symbol 2:
2 =
1
c
2

2
t
2

2
. (1.33)
This is already Lorentz invariant, for it may also be written as

. You
may recall that it was partly consideration of the theory of electromagnetic
waves that led Einstein to formulate the special theory of relativity: elec-
tromagnetism (in free space) is automatically a theory which is consistent
with the special theory of relativity, a fact which becomes obvious when it is
written down in terms of 4-vectors and tensors. This is not at all obvious if
it is written down as Maxwell originally did, component by component.
1.10 Lorentz covariance
The power of 4-vector formalism is that when one uses it consistently to
write down equations, the equation will keep the same form under a Lorentz
transformation: that is, it will be Lorentz covariant. For example, we might
want a relativistic generalisation of Newtons Second Law, F = dp/dt. We
have already seen that dierentiations with respect to time should be changed
to dierentiations with respect to proper time, so the analogous equation in
relativistic dynamics should be
F

=
dp

d
. (1.34)
14
The zeroth component of the force 4-vector, F
0
, is the (proper) rate of change
of the energy of the particle.
A potential source of confusion is whether physical quantities are rep-
resented by covariant or contravariant vectors. As a rule, coordinates and
momenta naturally have their indices up and derivatives naturally have their
indices down.
2 Relativistic wave equations
2.1 Klein-Gordon equation
The Klein-Gordon equation was originally thought up by Schrodinger, who
wanted a relativistic wave equation describing the electron, even before he
settled on what we now call the Schrodinger equation,
i h

t
=
h
2
2m

2
. (2.1)
Klein and Gordon published rst, so they got the credit. We will see later
why Schrodinger dropped this equation. One of the ways of understanding
the Schrodinger equation is to recall that in quantum mechanics, physical
quantities are represented by operators:
E i h

t
, p i h, (2.2)
(where the double-headed arrow symbol means is represented by). Hence
the Schrodinger equation (2.1) represents the non-relativistic equation for
the kinetic energy E = p
2
/2m.
As we know, the relativistic relation between energy and momentum is
E
2
= p
2
+ m
2
, which seems to suggest that a relativistic version of the
Schrodinger equation ought to be

2
t
2
=
2
+ m
2
, (2.3)
or, in a manifestly Lorentz covariant form,
(
2
+ m
2
) = 0. (2.4)
(It is traditional to use rather than in this context).
Solutions to this equation with denite energy and momentum are
(x) = Ae
iEt+px
(2.5)
15
with E = =

(p
2
+ m
2
), and A an arbitrary complex constant.
In fact if is a solution to the Klein-Gordon equation it cannot be
interpreted as a wave function as Schrodinger discovered. A wave func-
tion is a probability amplitude, whose modulus squared is the probability
of nding the particle at a particular position (or with a particular mo-
mentum). For example, in non-relativistic quantum mechanics, the prob-
ability density = [[
2
, which is associated with a probability current
j = i(

)/2m, in the sense that together they make up a

probability conservation equation
+ j = 0, (2.6)
an equation which one can check by dierentiating with respect to time
and using the Schrodinger equation.
The KG equation also has a conserved density and a 3-vector current,
which are
=
i
2
(

t

t

), j =
i
2
(

). (2.7)
However, this density cannot be interpreted as a probability density, as it is
not positive denite. This was the reason that Schrodinger chose the non-
relativistic form for his equation. This was reasonable for his purposes, but
the Klein-Gordon equation makes a comeback later, when we shall see that
can be interpreted as a charge density, which is allowed to take both positive
and negative values.
2.2 Maxwells equations in covariant form
This section recaps some important results, and introduces the formulation
of the theory in an explicitly Lorentz covariant manner.
Firstly, we recall Maxwells equations in free space, writing them down
in natural units, in which the permittivity and permeability of free space
0
and
0
are both unity:
Homogeneous Inhomogeneous
B = 0 E =

t
B+ E = 0

t
E + B = j
(2.8)
The homogeneous and inhomogeneous (i.e. having a source term on the right
hand side) equations have diering status. The homogeneous equations imply
the existence of potentials and A from which the physically measurable
16
quantities E and B can be calculated. These potentials are specied only
up to a gauge transformation,

and A A+ , where is an
arbitrary function of space and time. The inhomogeneous equations imply
that the source terms must obey a current conservation equation, +j = 0.
To summarise:
Homogeneous Inhomogeneous
potentials , A current conservation
B = A
E =

A A+
+ j = 0
(2.9)
All these quantities can be assembled into explicitly Lorentz covariant ob-
jects. The gauge potentials belong together in a 4-vector potential A

=
(, A), while the charge density and the current density j can be put to-
gether into a 4-vector current density j

= (, j). Recall that putting quanti-

ties together into 4-vectors is not just a matter of notation: it means that the
quantities transform just like the space-time coordinates x

under a Lorentz
transformation.
The electric and magnetic elds E and Balso belong together in a Lorentz
covariant object, as they are mixed up by Lorentz transformations (an ob-
server moving through a magnetic eld also sees an electric eld). However,
this object cannot be a 4-vector as there are a total of 6 components of the
electric and magnetic elds, when they are taken together. In fact, the object
is an antisymmetric tensor, the eld strength tensor F

, which is dened in
terms of the gauge potential 4-vector A

:
F

. (2.10)
The Lorentz transformation law for F

follows from those of

and A

:
F

, (2.11)
from which we explicitly see that F

is a covariant tensor of rank 2. Let us

examine the components where = 0 and ranges over spatial indices i:
F
i0
=
i
A
0

0
A
i
=
i
A
0

A
i
=
i


A
i
= E
i
, (2.12)
where we have used the expression for the electric eld in terms of the gauge
potentials in equation (2.9). The magnetic eld is contained in the entries
where both and take spatial values:
F
ij
=
i
A
j

j
A
i
=
i
A
j
+
j
A
i
=
ijk
B
k
, (2.13)
17
where we have introduced the Levi-Civita symbol
ijk
. The Levi-Civita sym-
bol is dened by

ijk
=
_

_
+1 if i ,= j ,= k cyclic,
1 if i ,= j ,= k anticyclic,
0 otherwise.
(2.14)
Written in matrix form, the eld strength tensor is:
F

=
_
_
_
_
_
0 E
1
E
2
E
3
E
1
0 B
3
B
2
E
2
B
3
0 B
1
E
3
B
2
B
1
0
_
_
_
_
_
. (2.15)
The reverse relations may be written
E
i
= F
i0
, B
i
=
1
2

ijk
F
jk
, (2.16)
which can be veried with the help of the identity

ijk

klm
=
il

jm

im

jl
. (2.17)
When expressed in terms of 4-vectors and tensors, electromagnetism looks
very simple and beautiful. For example, the covariant expression of the
current conservation equation is simply

= 0. (2.18)
Maxwells equations become:
Homogeneous Inhomogeneous

= 0,

= j

.
(2.19)
The rst of these expressions looks as if it contains many more equations
than the 4 of the original homogeneous Maxwell equations. However, the
antisymmetry of the eld strength tensor (F

= F

) means that the

expression is trivial if any of the two indices are equal. Thus all the indices
, and must take dierent values, and the number of ways of choosing
three dierent numbers from a set of four is
4
C
3
, which is equal to 4, precisely
the number of equations we started with.
We can use the four-dimensional Levi-Civita tensor to re-express the ho-
mogeneous equations more compactly. This tensor has four indices, and is
dened by

=
_

_
+1 if ,= ,= ,= , symmetric,
1 if ,= ,= ,= , antisymmetric,
0 otherwise.
(2.20)
18
A symmetric permutation of the indices in one in which an even number of
indices are exchanged, while an antisymmetric permutation is one for which
an odd number of indices are exchanged. Thus, for example,
0123
=
1032
=
+1,
1023
=
0132
= 1, but
0012
= 0.
There is also a version with the indices raised:

, (2.21)
whose symmetric permutations take the value 1 and antisymmetric ones
+1.
Using this totally antisymmetric (i.e. antisymmetric on the exchange of
any two indices) tensor, the homogeneous Maxwell equations become

= 0. (2.22)
One of the problems uses this expression to give you practice with the sum-
mation convention as applied to these tensors.
2.2.1 Gauges
Recall that the gauge potential A

is specied only up to an arbitrary func-

tion of space and time (see equation 2.9). The 4-vector version of a gauge
transformation is written
A

, (2.23)
under which one can check that F

is unchanged. Indeed, using (2.10) one

nds
F

(A

(A

) = F

) = F

. (2.24)
The last step follows because we can take the partial derivatives in any order.
In order to use the gauge potential to solve the eld equations, one should
eliminate this freedom to make gauge transformations, otherwise one can get
misled into thinking that the solutions one obtains are all physically distinct.
There are many ways to do this, but some of the most common are as follows.
Coulomb gauge This is also called radiation gauge, and is dened by
A = 0, (2.25)
which implies that the rst of the inhomogeneous equations in (2.8) becomes

2
= . (2.26)
19
This gauge is therefore very useful in solving electrostatics problems, and we
shall use it later on when studying the Casimir eect. However, it is not so
often used in relativistic applications, as the gauge condition does not respect
Lorentz invariance.
The second of the inhomogeneous Maxwell equations becomes, on using
the gauge condition,

t
(

A)
2
A = j, (2.27)
which we may write as
(

2
t
2

2
)A = j
T
, (2.28)
where j
T
= j+

. One can show, using the equation of current conservation
and Eq. (2.26), that
j
T
= 0. (2.29)
This of course is necessary for the consistency of Eq. (2.28), because if one
takes the divergence of the left hand side one gets zero as well.
In free space, that is, in the absence of charges and currents, we may take
= 0, and the equation for the vector potential becomes
(

2
t
2

2
)A = 0. (2.30)
One of the important discoveries of the last century was that this equation has
plane wave solutions, which carry energy and momentum: electromagnetic
waves. Such a solution may be written
A(t, x) = ae
it+ikx
(2.31)
where a is a constant 3-vector, and = [k[.
The gauge condition (2.25) imposes a condition on the vector a, for
A = ika e
it+ikx
= 0, (2.32)
which implies
ka = 0, (2.33)
Hence the gauge potential A is orthogonal to the wave vector k, which re-
moves one of the three apparent degrees of freedom of the eld. Thus the
eld oscillates in directions orthogonal to the direction of propagation: we
say that the waves are transverse.
20
Lorentz gauge This is very often the gauge to choose if one is interested
in wave propagation in electromagnetism. It is dened by

= 0, (2.34)
and is manifestly Lorentz invariant. In this gauge the equation of motion for
the gauge eld is simplied, for

=
2
A

= j

. (2.35)
However, an irritating feature of this gauge is that it does not quite specify
A

fully. One can still make a gauge transformation A

which
satises the Lorentz gauge condition (2.34), as long as the function satises

2
= 0. Such functions are called harmonic. In classical eld theory this is
not too much of a problem, but in setting up the quantum theory of gauge
elds care must be taken.
In free space, the eld equation is (2.35)
2A

(

2
t
2

2
)A

= 0, (2.36)
which looks very much like four copies of the Klein-Gordon equation for elds
with zero mass. Plane wave solutions may be written
A

(t, x) = a

e
ikx
(2.37)
where we recall that k x = k
0
t kx, and k
0
= [k[, Again, a

is a constant
4-vector.
The Lorentz gauge condition (2.34) are

= ik

e
ikx

= 0, (2.38)
which implies that
k a = 0. (2.39)
Once again the eld is orthogonal to the wave vector k

, but this time in the

4-vector sense.
Once the gauge condition is taken into account there are apparently three
degrees of freedom, one more than in the Coulomb gauge. However, one of
those is unphysical as a result of the remaining freedom to make harmonic
gauge transformations.
21
2.3 Lorentz gauge solutions
The general solution may be constructed from a superposition of plane wave
solutions
A

(x) =
_
d
3
k
2
k
_
a

(k)e
ikx
+ a

(k)e
ikx
_
, (2.40)
where the amplitudes a

(k) must all satisfy k a(k) = 0.

It would be more convenient to be able to choose the amplitudes of the
components of the gauge eld independently, and to this end we introduce a
set of four basis 4-vectors, or polarisation vectors,
A

(k), with A = 0, 1, 2, 3,
for each wavevector k. They must satisfy a completeness condition in order
for them to be considered as basis vectors, and it is also convenient to make
them orthonormal, in a 4-vector sense.

(k)
B

(k)

=
AB
(Orthonormality), (2.41)

(k)
B

(k)
AB
=

(Completeness). (2.42)
A good choice is

0
= (1; 0),

1
= (0, n k)/[ n k[,

2
= (0, k ( n k))/[k ( n k)[,

3
= (0;

k),
(2.43)
where n = (0, 0, 1) is a unit 3-vector. We refer to these polarisation vec-
tors as timelike (A = 0), transverse (A = 1, 2), and longitudinal (A = 3)
respectively.
Using this basis we can write a

= a
A

, and nd that the gauge condi-

tion implies k a = k
0
a
0
[k[a
3
= 0, or
a
0
(k) = a
3
(k). (2.44)
Thus the general solution is now more easily written down as
A

(x) =
_
d
3
k
2
k
_
a
A
(k)
A

(k)e
ikx
+ a
A
(k)

(k)e
ikx
_
. (2.45)
There now appears to be, for each k, three independent degrees of freedom,
down from the original four. However, one of the amplitudes, a
3
, does not
aect the physical value of the electric and magnetic elds. This is a result of
the remaining freedom in the Lorentz gauge to make gauge transformations
22
with a harmonic function (x), which may be decomposed into plane waves,
according to
(x) =
_
d
3
k
2
k
_

(k)e
ikx
+

(k)e
ikx
_
. (2.46)
Hence under such a gauge transformation,
a

(k) a

(k) = a

(k) ik

(k). (2.47)
By multiplying with
3
we see that this is equivalent to shifting a
3
(k),
a
3
(k) a

3
(k) = a
3
(k) i[k[(k). (2.48)
Thus although the Lorentz gauge does not entirely remove the freedom to
make gauge transformations, it does conne the ambiguity in the solution to
one and only one of the sets of constants in the solution. There are only two
physical amplitudes to choose, a
1
and a
2
.
Let us end this section by calculating the Hamiltonian in the Lorentz
gauge.
H
Lg
=
_
d
3
x
_

L
Lg
_
=
1
2
_
d
3
x(

A

+ A

). (2.49)
We already notice from this expression that the Hamiltonian is not positive
denite, as the timelike component of the gauge eld A
0
appears with a
negative sign. It turns out that we have already cured this problem in the
classical theory. To see this, let us substitute the plane wave expansion of
the eld operator, obtaining
H
Lg
=
_
d
3
k
2
k

AB
a
A
(k)a
B
(k). (2.50)
We saw earlier that the Lorentz gauge condition Eq. 2.34 implies a
0
= a
3
,
and so we nd that
H
Lg
=

A=1,2
_
d
3
k
2
k

k
a
A
(k)a
A
(k), (2.51)
which is clearly positive denite, and depends only on the physical tranversely
polarised components.
23
3 Lagrangian formulation of eld theory
3.1 Lagrangian and Hamiltonian mechanics
The starting point for classical mechanics is Newtons Second Law of Motion,
which states that the rate of change of momentum of a particle is proportional
to the applied force. For a system of P particles, we may write
p
A
= F
A
, (3.1)
where the index A runs from 1 to P. The momentum p
A
= m
A
v
A
, where
v
A
= x
A
. If the mass is constant in time, we obtain the formulation m
A
x
A
=
F
A
.
If the force is conservative, which means that the work done by the force
around a closed path is zero, then the force can be written as the gradient
of a potential energy V :
F
A
=
A
V (x
1
, . . . , x
P
). (3.2)
There may also be constraints on the motion: for example, the particles
may be forced to move on a sphere. This reduces the number of degrees
of freedom of the system. A very common form of constraint is one which
relates some or all of the coordinates through a function (which may also
depend explicitly on time):
f(x
1
, . . . , x
P
, t) = 0. (3.3)
These are known as holonomic constraints. For example, the rst particle
may be constrained to move on a sphere of radius a, or x
2
1
= a
2
. Thus we
lose one degree of freedon, the radial coordinate of the particle. If there are k
such constraints, then the eective number of degrees of freedom is reduced
by k: P particles moving in 3 dimensions under k constraints have 3P k
degrees of freedom. Thus not all of the coordinates are needed to describe
the motion of the system, and it is convenient to introduce the notion of
generalised coordinates q
a
which result from the solution of the constraint
equations. In the system of particles there will be N = 3P k of them, and
one can express the original coordinates as functions
x
1
= x
1
(q
1
, . . . , q
N
, t)
.
.
.
.
.
. (3.4)
x
P
= x
P
(q
1
, . . . , q
N
, t)
Generalised coordinates are also useful in systems without constraints: for
example, orbits around central potentials, where it is convenient to use spher-
ical polar coordinates.
24
3.1.1 Lagrangian Mechanics
There is a later and more sophisticated formulation of classical mechanics
due to Lagrange. We dene the kinetic and potential energies T and U in
terms of the generalised coordinates of a system as
T =
1
2

a
m
a
q
2
a
, U = U(q
a
, q
a
) (3.5)
(we have allowed for the possibility of a velocity-dependent potential). We
then introduce a new function called the Lagrangian, dened as
L = T U. (3.6)
The equations of motion of the system are then
L
q
a

t
_
L
q
a
_
= 0. (3.7)
These equations are known as the Euler-Lagrange equations.
For example, suppose there is no velocity dependence in the potential.
Then
L
q
a
=
U
q
a
,
L
q
a
= m
a
q
a
, (3.8)
so that

t
(m
a
q
a
) =
U
q
a
. (3.9)
Thus we return to Newtons Second Law. The quantities L/ q
a
can be iden-
tied as a momentum associated with the coordinate q
a
if the q
a
are simply
Cartesian coordinates, then they are the ordinary mechanical momenta, but
in general we refer to them as conjugate or canonical momenta, which are
dened as the derivative of the Lagrangian with respect to the rate of change
of the generalised coordinates:
p
a
=
L
q
a
. (3.10)
The Euler-Lagrange equations can be derived from a variational principle,
known as Hamiltons Principle or the Principle of Least Action (which is a
misnomer). The action for a system between times t
1
and t
2
is dened as
I =
_
t
2
t
1
dt L(q
a
, q
a
, t). (3.11)
25
The Principle of Least Action states that the action is an extremum for the
path of the motion.
To see this, let us consider a small variation in the path, vanishing at t
1
and t
2
, sends q
a
(t) q

a
(t) = q
a
(t) +q
a
(t) (see Figure 3.1). The action also
changes, I I

= I + I, where to rst order in q

a
(t),
I =
_
t
2
t
1
dt L(q
a
, q
a
, t) =
_
t
2
t
1
dt
_
L
q
a
q
a
+
L
q
a
q
a
_
. (3.12)
Integrating by parts we nd
I =
_
t
2
t
1
dt
_
L
q
a

t
_
L
q
a
__
q
a
+
_
L
q
a
q
a
_
t
2
t
1
. (3.13)
The second term vanishes, as we are keeping the end points of the path xed,
and so if the variation of the action I vanishes we nd that
L
q
a

t
_
L
q
a
_
= 0, (3.14)
which are precisely the Euler-Lagrange equations (3.7).
q(t)
q
q(t)
q(t )
q(t )
1
2
t
Figure 3.1: A path q(t) for a system with one coordinate, and a small
variation q(t) to that path. Classical paths are ones for which the
action (see Equation 3.11) is stationary under such small variations.
26
3.1.2 Hamiltonian Mechanics
There is yet another formulation of classical mechanics due to Hamilton.
It may seem excessive to have so many ways of doing the same thing, but
each formulation has its own advantages: for example, the Lagrangian and
Hamiltonian formulations are very powerful in bringing out conservation laws
in dynamical systems, and relating them to symmetries.
Recall that associated with each coordinate q
a
there is also a conjugate
momentum p
a
, dened by Eq. (3.10). We can therefore express the Euler-
Lagrange equations as p
a
= L/q
a
. We now dene an important function
called the Hamiltonian from the Lagrangian:
H(p
a
, q
a
, t) =

a
p
a
q
a
L(q
a
, q
a
, t). (3.15)
The idea is to replace q
a
by p
a
, having solved equation (3.10). The rst of
Hamiltons equations follows from the Euler-Lagrange equations (3.7), after
noting that L/q
a
= H/q
a
. It is
p
a
=
H
q
a
. (3.16)
The second follows straightforwardly from the denition of the Hamiltonian
(3.15) upon partial dierentiation with respect to p
a
:
q
a
=
H
p
a
. (3.17)
If the Lagrangian does not depend explicitly on time, the Hamiltonian is a
constant of the motion, as the following calculation shows. Recall that H is
a function of q
a
, p
a
, and possibly t only. Hence
dH
dt
=
H
t
+

a
_
H
p
a
p
a
+
H
q
a
q
a
_
. (3.18)
Using Hamiltons equations (3.16,3.17) we see that
dH
dt
=
H
t
=
L
t
. (3.19)
Thus if L/t = 0, the Hamiltonian is constant. One can also show that
H = T + U, (3.20)
where T and U are the kinetic and potential energies introduced in (3.5).
Thus the Hamiltonian can be identied with the total energy of the system.
27
As a trivial example, consider a set of N free particles with mass m. The
Lagrangian is
L =
1
2

a
m q
2
a
, (3.21)
from which we can easily derive the Euler-Lagrange equations
0

t
(m q
a
) = 0, (3.22)
or that the acceleration of each particle is zero. The conjugate momentum
is p
a
= m q
q
, and hence the Hamiltonian becomes
H =
1
2

a
p
2
a
m
. (3.23)
This is of course the familiar non-relativistic expression for the kinetic energy.
3.2 Lagrangian mechanics for the real scalar eld
A eld is an object which takes a value (e.g. a real number) at every point
in space-time. We are probably most familiar with the electric and magnetic
vector elds E(t, x) and B(t, x), which we have seen can be unied into the
electromagnetic eld strength tensor F

(x). The Klein-Gordon eld (t, x),

which takes complex values, is another example.
A very powerful way of classifying elds is to organise them according to
their properties under Lorentz transformations. The Klein-Gordon eld is
an example of a scalar eld, which transforms as a scalar (a pure number):
(x)

(x

) = (x), (3.24)
where x

.
In eld theory, the dynamical coordinates, the analogues of the q
a
of
classical Lagrangian mechanics, are the values of the elds at every point.
A eld therefore has an uncountably innite number of degrees of freedom,
which is the source of many of the diculties in eld theory. Sums over the
label a are replaced by integrals over space: for example, the Lagrangian can
be expressed as the integral over space of an object called the Lagrangian
density L. For a scalar eld, the Lagrangian density can be expressed as a
function of the eld, its time derivative, and also its spatial derivatives:
L =
_
d
3
x L(,

, , t). (3.25)
The appearance of spatial derivatives may at rst sight be puzzling, but
arises naturally from terms coupling neighbouring space points, in the limit
28
that the separation goes to zero. The puzzle is perhaps more that higher
derivatives dont appear: in practice, they seem not be relevant for most
applications in particle physics.
t
1
2
t
x
y
R
t
B
Figure 3.2: A space-time diagram (with the z coordinate sup-
pressed) of the four-dimensional region in which the variation of
the eld (x) is non-zero.
From a Lagrangian we can derive Euler-Lagrange equations, by the ap-
plication of Hamiltons Principle to the action for a scalar eld,
I =
_
t
2
t
1
Ldt =
_
t
2
t
1
dtd
3
x L(,

, , t). (3.26)
Consider an arbitrary variation in the eld, (x) (x) + (x), where
(x) vanishes at t
1
and t
2
, and outside an arbitrary spatial region R with
boundary B (see Figure 3.2). This variation causes a small change in the
action:
I I

= I + I. (3.27)
Taylor expanding the Lagrangian density, we nd
I

_
t
2
t
1
dtd
3
x
_
L +
L

+
L

+
L

_
, (3.28)
where
L

=
_
L
(
x
)
,
L
(
y
)
,
L
(
z
)
_
. (3.29)
29
We can separate out the variation in the action I, and integrate the last
term by parts, to obtain
I =
_
t
2
t
1
dtd
3
x
_
L

+
L

t
_
L

_
+
_
L

_
t
2
t
1
. (3.30)
In this last expression the last term vanishes because of the conditions placed
on at t
1
and t
2
. Hence, upon integrating by parts again, we get
I =
_
t
2
t
1
dtd
3
x
_
L

t
_
L

_
+
_
t
2
t
1
dt
_
B
dS
_
L

_
.
(3.31)
Again, the last term vanishes, as we supposed that the variation vanished
on the boundary B. Hence we are left with
I =
_
t
2
t
1
dtd
3
x
_
L

t
_
L

__
. (3.32)
The argument proceeds by noting that not only is arbitrary, but so is the
region R. Hence the condition that the action be stationary reduces to
L

t
L

= 0. (3.33)
This is the Euler-Lagrange equation for a scalar eld.
We can put it into a more obviously relativistic form by noting that

= (
t
,
i
), so that the Euler-Lagrange equation becomes
L

i
L
(
i
)

0
L
(
0
)
= 0, (3.34)
or
L

L
(

)
= 0. (3.35)
If there are many elds
a
we can derive an Euler-Lagrange equation for each
of them,
L

L
(

a
)
= 0. (3.36)
We are now in a position to ask the question: from what Lagrangian
density does the Klein-Gordon equation follow? The Klein-Gordon eld may
be split into its real and imaginary parts, (x) =
R
(x) + i
I
(x), each of
which obey the same equation
(

+ m
2
)
R
(x) = 0, (

+ m
2
)
I
(x) = 0. (3.37)
30
Consider rst the following Lagrangian density for the real part,
L
R
=
1
2

1
2
m
2

2
R
. (3.38)
We can immediately derive
L

R
= m
2

R
,
L
(
i

R
)
=
i

R
,
L
(
0

R
)
=
0
. (3.39)
Hence,
L
(

R
)
= (
0

R
,
i

R
), (3.40)
and so
L

L
(

R
)
= m
2

R
= 0. (3.41)
Thus the Lagrangian (3.38) implies the correct equation for the real part
of the Klein-Gordon eld: exactly the same argument may be made for the
imaginary part of the eld, so we can write for the total lagrangian
L = L
R
+L
L
=
1
2

1
2
m
2

2
R
+
1
2

1
2
m
2

2
I
=
1
2

1
2
m
2

. (3.42)
Lastly in this section, we will write down the Hamiltonian for a single
real scalar eld. Drawing on our experience with nite numbers of degrees
of freedom, it seems natural to dene a conjugate momentum
(x) =
L

(x)
. (3.43)
Thus for our example (3.38), the conjugate momnetum (x) =

(x). The
Hamiltonian for the real scalar eld is dened by
H =
_
d
3
x (x)

(x)
_
d
3
xL(,
i
,

). (3.44)
If the eld obeys the Klein-Gordon equation, its Hamiltonian is
H =
_
d
3
x
_
1
2

2
+
1
2
()
2
+
1
2
m
2

2
_
. (3.45)
We talk of the three terms in the Hamiltonian as kinetic, gradient and po-
tential terms respectively. One can allow more general functions of in the
potential term than just a quadratic:
H =
_
d
3
x
_
1
2

2
+
1
2
()
2
+ V ()
_
. (3.46)
31
In classical eld theory, V can be any function of , as long as it is bounded
below, otherwise the energy of the eld can become arbitrarily negative.
Quantum eld theory is much more restrictive about the terms it allows: it
turns out that the theory does not make sense in four space-time dimensions
unless V () is restricted to be a polynomial of degree no more than four.
3.3 Lagrangian mechanics for the electromagnetic eld
The electromagnetic gauge potential A

(x) is an example of a 4-vector eld,

which is a eld which transforms like a covariant 4-vector. Its behaviour
under Lorentz transformations is
A

(x) A

(x

) =

(x), (3.47)
where x

. In this section we will write down a Lagrangian density,

and show that the resulting Euler-Lagrange equations are just the inhomo-
geneous Maxwell equations. The homogeneous ones will be shown to be an
identity, rather than arising from any variational principle. The Lagrangian
is
L =
1
4
F

, (3.48)
where F

. The generalised coordinates of this system are

the four components the gauge eld A

. We can therefore derive Maxwells

equations by looking at the Euler-Lagrange equations:

L
(

)

L
A

= 0. (3.49)
Let us rst consider the variation with respect to A

, which gives
L
A

= j

= j

= j

. (3.50)
The variation with respect to

is a little more complex, and it will help

to dene X

, so that F

= X

. Hence

L
(

)
=
1
4
F

1
4
F

,
=
1
4
_

_
F

1
4
F

) ,
=
1
4
(F

)
1
4
(F

)
= F

. (3.51)
32
Thus we can substitute (3.50) and (3.51) back into the Euler-Lagrange equa-
tion (3.49) to obtain

+ j

= 0, (3.52)
which we recognise as the covariant form of the inhomogeneous Maxwell
equations.
We now turn to the homogeneous Maxwell equations, and demonstrate
that they are an identity, which is sometimes known as the Bianchi identity.
This can be seen by examining the 4-vector k

=
1
2

= 0. Sub-
stituting the expression for the eld strength tensor in terms of the gauge
potential, we obtain
k

=
1
2

). (3.53)
By exchanging the labels and in the second term we nd
k

=
1
2

1
2

, (3.54)
=

, (3.55)
with the last step following by the antisymmetry of the Levi-Civita tensor.
Now we exchange the labels and , from which we can deduce

. (3.56)
The antisymmetry of the Levi-Civita tensor then shows that

. (3.57)
The only number that is equal to its negative is zero, hence
1
2

= 0. (3.58)
Thus we have demonstrated that the Principle of Least Action applied to
the Lagrangian (3.48), combined with the Bianchi identity (3.58), results in
Maxwells equations.
3.4 Noethers theorem and conservation laws
It is a profound feature of both classical and quantum dynamics that con-
servation laws, such as conservation of energy or momentum, follow from
symmetries of the dynamical equations. In this section we shall show how
this feature is realised in classical eld theory.
33
A conservation law can be expressed in two equivalent ways. Firstly, and
most obviously, one can say that there is a quantity Q which is constant in
time, i.e.
dQ
dt
= 0. (3.59)
Entirely equivalently, one can also say that there is a 4-vector current j

(x)
which satises a continuity equation,

(x) = 0. (3.60)
If we integrate (3.60) over all space, we nd
_
d
3
x

t
j
0
(x) +
_
d
3
x j = 0. (3.61)
Dening Q =
_
d
3
xj
0
(x), we nd using Gausss Law that
dQ
dt
=
_

dS j, (3.62)
where the surface integral is taken over a sphere at spatial innity. With the
assumption that all currents die o at innity, we have
dQ
dt
= 0. (3.63)
Hence the existence of a 4-vector current satisfying a continuity equation

(x) = 0 automatically implies that there is a conserved quantity, which

is the integral over all space of the time component of the current.
Now consider a theory with N real elds
a
(x), with a = 1, . . . , N, and
an action
S =
_
d
4
xL(
a
,
a
). (3.64)
Noethers theorem for this theory can be stated as follows.
For every transformation
a
(x) =

a
(x

) which leaves the action S in-

variant, there is a conserved current.
The proof of this theorem is quite involved, and we rst need some de-
nitions.
Let us rst dene the total variation in the eld,

a
(x) =

a
(x

)
a
(x). (3.65)
Note that this variation can be split into two pieces, the rst due to the fact
that the functions
a
are changing:

a
(x) =

a
(x)
a
(x), (3.66)
34
which denes what we mean by the operator

. The remaining piece can
be interpreted as the change in the eld functions due to the change in
coordinates,

a
(x) =
a
(x

)
a
(x)

a
(x)x

, (3.67)
where x

= x

. Hence

a
(x)

a
(x) +

a
(x)

a
(x) +
a
(x), (3.68)
where we have dropped at term of second order in the variation, as

a
(x) =

a
(x) + O(x).
One can show straightforwardly from the denition of

that

a
(x)) =

a
(x)), (3.69)
but that
(

a
(x)) =

(
a
(x))

a
(x)

. (3.70)
Hence the partial derivative operator commutes with

, but not with .
Now consider the eect of the variation in Eq. 3.65 on the action:
S =
_

d
4
x

(x

)
_

d
4
xL(x), (3.71)
where L

(x

) = L(

a
(x

),

a
(x

)), and is an arbitrary region in spacetime.

By adding and subtracting L(x) to the rst integrand, we see
S =
_

d
4
x

(x) +
_

d
4
x

L(x)
_

d
4
xL(x). (3.72)
We now need to express the integration measure d
4
x

in terms of d
4
x,
using the Jacobian of the coordinate transformation:
d
4
x

= d
4
x

(x

0
, . . . , x

3
)
(x
0
, . . . , x
3
)

= d
4
x[ det M[, (3.73)
where the components of the matrix M are given by
M

=
x

. (3.74)
In matrix notation, we may write M = 1 + M, and using the standard
matrix identity
ln det M = tr ln M, (3.75)
we see that
det M 1 + tr M = 1 +

. (3.76)
35
Substituting into the varation of the action Eq. (3.71), we nd
S =
_

d
4
x(1+

)L

(x)+
_

d
4
x(1+

)L(x)
_

d
4
xL(x). (3.77)
Dropping terms which are second order in the variation we arrive at
S =
_

d
4
xL(x) +
_

d
4
x

L(x) (3.78)
Now we can again split the variation in the Lagrangian function into a piece
arising from the change in the function itself, and that due to the change of
coordinates:
L(x) =

L(x) +

L(x)x

. (3.79)
The change in the Lagrangian function arises because of the change in the
functional form of the elds, so

L(x) =
L

a
(x)

a
(x) +
L
(

a
)
(x)

a
(x)). (3.80)
At this point, the commutativity of

and

stated earlier means that

S =
_

d
4
x
_
L

a
(x)

a
(x) +
L
(

a
)
(x)

a
(x))
+
_

d
4
x (

L(x)x

+L(x)

)
_
. (3.81)
Some simple algebra shows that
S =
_

d
4
x
_
L

L
(

a
)
_

a
(x)+
_

d
4
x

_
L
(

a
)

a
+L(x)x

_
.
(3.82)
Note that the rst term is precisely the variation in the action we found
when deriving the Euler-Lagrange equation, where we considered variations
in the elds alone, vanishing on the boundary of . Suppose we consider a
particular eld conguration (x) satisfying the Euler-Lagrange equation. If
we then make an innitesimal transformation on it which leaves the action
invariant, i.e. which results in no rst-order change in the action, we must
have that

= 0, (3.83)
where
f

=
L
(

a
)

a
(x) +Lx

, (3.84)
36
or substituting back for

(x),
f

=
L
(

a
)

a
(x)
_
L
(

a
)

a
(x) L

_
x

. (3.85)
Thus we see the appearance of an innitesimal 4-vector, f

, which satises
a continuity equation.
3.4.1 Conservation of energy-momentum
Consider a translation on the coordinates, with the elds unchanged:
x

,
a
= 0, (3.86)
where

is a constant 4-vector. This clearly includes both space and time

translations. Hence we can write
f

, (3.87)
where we have introduced the canonical energy-momentum tensor

=
L
(

a
)

a
(x) L

. (3.88)
Because the translations

are arbitrary, translation invariance of the La-

grangian implies that there must be four continuity relations,

= 0. (3.89)
The conserved charge associated with these four continuity relations is the
total 4-momentum,
P

=
_
d
3
x
0

(x). (3.90)
4 Canonical quantisation of the scalar eld
We have seen that the Klein-Gordon eld (x) does not have the interpre-
tation as a single-particle wave-function, as it does not have a conserved
probability current, and it possesses negative energy states which cannot
form a Dirac sea. However, as we shall see in this section, the theory
of the quantum eld operator

(x) is in fact a relativisitic many-particle
quantum theory. The process of developing the theory of a quantum eld
is sometimes misleadingly called second quantisation, perhaps because of
some idea that one is quantising a wavefunction, itself a quantum object. It
37
should be emphasised at the outset that the scalar eld, real or complex, is
not a wavefunction.
In this section we go through the quantisation procedure for free elds.
By free we mean non-interacting: once a state has been set up there are no
transitions to any other states. In practice this means that the Lagrangian
can be written as a quadratic function of the elds. Free elds may sound
somewhat irrelevant, but nearly all quantum eld theories have to be treated
as small perturbations away from free eld theories, so it is an appropriate
place to start. There are non-trivial eects in free eld theory which come
from boundary conditions: one of these eects is the Casimir eect in the
free electromagnetic eld.
4.1 Quantisation of nite systems
In this section we shall proceed with quantising the real scalar eld. Firstly,
we will go through the quantisation procedure for a system with a nite
number of degrees of freedom, to show that quantising a eld is not unlike
quantising a more familiar system. Suppose the system has N coordinates
q
a
with associated velocities q
a
, and Lagrangian
L =
1
2
N

a=1
m q
2
a
U(q
a
) (4.1)
(we are supposing for simplicitys sake that all the masses are the same).
Each coordinate is also associated with a conjugate momentum p
a
, dened
as the dierential of the Lagrangian with respect to the velocity q
a
, which in
this case has the value m q
a
. The Hamiltonian, dened as
1
2

a
q
a
p
a
L, is
then
H =
N

a=1
p
2
a
2m
+ U(q
a
). (4.2)
The procedure of canonical quantisation replaces the classical conjugate vari-
ables q
a
, p
a
by operators obeying the following commutation relations
[ q
a
, p
b
] = i
ab
, [ q
a
, q
b
] = 0 = [ p
a
, p
b
]. (4.3)
The Hamiltonian is also replaced by an operator

H, which generates the
time evolution of the system. We are probably used to thinking about time
evolution as the change in time of the state of the system, governed by the
Schrodinger equation
i

t
[(t)) =

H[(t)). (4.4)
38
However, there is another way of thinking about the time evolution of a
quantum mechanical system known as the Heisenberg picture (to distinguish
it from the Schrodinger picture above).
In the Heisenberg picture, the states [)
H
that are time-independent, and
the time evolution of the system is transferred to the operators O
H
(t), which
satisfy the Heisenberg equations
i
d
dt
O
H
(t) = [O
H
(t), H]. (4.5)
The states and operators in these two pictures can be related by a unitary
transformation e
iHt
(a unitary operator is one whose inverse is its Hermitian
conjugate). Thus, if we suppose that the two sets of operators and states
coincide at t = 0, they are related at all times by
O
H
(t) = e
iHt
O
S
e
iHt
[)
H
= e
iHt
[(t))
S
. (4.6)
This ensures that the matrix elements are the same. To see this, let us
consider the matrix elements between any two states [)
H
and [

)
H
in the
Heisenberg picture, and apply these transformations:
H
[O
H
(t)[

)
H
=
S
(t)[e
iHt
e
iHt
O
S
e
iHt
e
iHt
[

(t))
S
=
S
(t)[O
S
[

(t))
S
. (4.7)
This demonstrates that the Heisenberg picture matrix elements are the same
as those in the Schrodinger one, which means there is no dierence in the
physical predictions one makes.
4.2 The real scalar eld
We begin with the Lagrangian density
L =
1
2

1
2
m
2

2
, (4.8)
from which we can dene the conjugate momentum
(x) =
L

(x)
. (4.9)
The Hamiltonian density is then found from
H = (x)

(x) L =
1
2

2
+
1
2
()
2
+
1
2
m
2

2
. (4.10)
39
In free quantum eld theory we usually work in the Heisenberg picture,
where the elds carry the time dependence, although it is possible to use
the Schrodinger picture. The eld theory is quantised by constructing the
equal time canonical commutation relations
[(t, x), (t, x

)] = i h
3
(x x

), (4.11)
[(t, x), (t, x

)] = 0, (4.12)
[(t, x), (t, x

)] = 0, (4.13)
where
3
(xx

) = (x
1
x
1
)(x
2
x
2
)(x
3
x
3
). One can see the analogy
between the eld commutation relations and (4.3) if one recalls that the space
coordinate x is like the label a in the system with a nite number of degrees
of freedom. The -function acts in the same way for a continuous label x as
the Kronecker does for a discrete label.
So in quantum eld theory, the elds become operators, obeying the
Heisenberg equations

(x) =
1
i
_

(x),

H
_
,

(x) =
1
i
_
(x),

H
_
, (4.14)
where

H is the quantum Hamiltonian, constructed by replacing the classical
elds by operators in the original:

H =
_
d
3
x
_
1
2

2
+
1
2
(

)
2
+
1
2

2
_
. (4.15)
It is straightforward to show for a free eld that

(x) = (x),

(x) = (
2
m
2
)

(x). (4.16)
Hence the eld operator obeys the same equation of motion as the classical
eld. It is convenient to expand the eld operator in terms of eigenfunctions
of the operator (
2
+ m
2
),
(
2
+ m
2
)e
ikx
=
2
k
e
ikx
. (4.17)
Writing

(x) =
_
d
3
k
(2)
3
(f
k
(x) a(k) + f

k
(x) a

(k)) . (4.18)
with f
k
(x) = A
k
(t)e
ikx
, we see that f
k
(x) obeys the eld equation (
2
+
m
2
)f
k
(x) = 0 if
A
k
(t) = ^
k
e
i
k
t
(4.19)
40
The normalisation factor ^
k
is essentially arbitrary, although almost ev-
eryone uses one of two conventions outlined below. The properties of the
operators a(k) and a

(k) can be deduced from the commutation relations of

and . We therefore need expressions for a(k) and a

(k) in terms of

and
, where
(x) =

(x) =
_
d
3
k
(2)
3
_
i
k
^
k
a(k)e
ikx
+ i
k
^

k
a

(k)e
ikx
_
. (4.20)
Firstly, we take the Fourier transform of the eld operator and its conjugate
momentum:

(k

, t) =
_
d
3
x

(x)e
ik

x
, (k

, t) =
_
d
3
x (x)e
ik

x
, (4.21)
and use the relation
_
d
3
xe
i(kk

)x
= (2)
3

3
(k k

) (4.22)
to obtain

(k

, t) = ^
k
a(k

)e
i
k
t
+^

k
a

(k

)e
i
k
t
(4.23)
(k

, t) = i
k
^
k
a(k

)e
i
k
t
+ i
k
^

k
a(k

) a

(k

)e
i
k
t
. (4.24)
Hence
2
k
^

k
a(k

) =
_
d
3
x(i (x) +
k

(x))e
ik

x
. (4.25)
By complex conjugation one obtains
2
k
^

(k

) =
_
d
3
x(i (x) +
k

(x))e
ik

x
. (4.26)
One can then derive the commutation relation
[ a(k), a

(k

)] = (2)
3

3
(k k

)/2
k
[^
k
[
2
, (4.27)
This looks very similar to the commutation relations for simple harmonic
oscillators. We recall that the harmonic oscillator of unit mass is described
by the Hamiltonian
H =
1
2
p
2
+
1
2

2
x
2
, (4.28)
where is the angular frequency of oscillation. Upon quantisation, the oper-
ators a = (i p+ x)/

2 and a

= (i p+ x)/

2 satisfy the commutation

relations
[ a, a

] = 1. (4.29)
41
This correspondence suggests that one choice for the normalisation factor:
^
k
=
1

2
k
, (4.30)
in which case we nd for the complete set of commutation relations
[a(k), a

(k

)] = (2)
3

3
(k k

), (4.31)
[a(k), a(k

)] = 0, (4.32)
[a

(k), a

(k

)] = 0. (4.33)
The right hand side of the rst equation, with its -fucntion, is the closest we
can get to unity when we are dealing with functions of a continuous variable
like a(k). However, we shall adopt a convention which keeps the Fourier
expansions of the eld operators Lorentz invariant,
^
k
=
1
2
k
, (4.34)
which means that the rst commutation relation becomes
[ a(k), a

(k

)] = 2
k
(2)
3

3
(k k

). (4.35)
It is also quite common to work in a nite volume V , which means that
instead of integrals over k we have sums over allowed wavevectors k =
2(n
1
, n
2
, n
3
)/L, which go over to integrals in the innite volume limit:
1
V

_
d
3
k
(2)
3
. (4.36)
All wavefunctions are multiplied by a factor 1/

V : hence the plane wave

expansion of the eld operator is written
(x) =

k
1

2
k
V
_
a(k)e
ikx
+ a

(k)e
ikx
_
. (4.37)
The non-trivial equal time canonical commutation relation then becomes
[ a(k), a

(k

)] =
k,k
, (4.38)
where
k,k
=
n
1
n

n
2
n

n
3
n

3
.
To summarise, we have discovered that the eld theory behaves very
much like innitely many harmonic oscillators, labelled by their wavenumber
k, and with a dierent frequency
k
=

(k
2
+ m
2
). We will see that the
occupation number n of each oscillator actually corresponds to the number
of particles with that particular momentum in the state considered. The
raising and lowering operators a

(k) and a(k) are also called creation and

annihilation operators, because they change the number of particles with
3-momentum k in a state.
42
4.3 States of the scalar eld; zero point energy
In this section we will drop the hat notation for operators, unless there could
be ambiguity, as all elds will be assumed to be operators unless otherwise
specied. We saw in the previous section that the scalar eld could be
decomposed into a sum over operators obeying oscillator-like commutation
relations, one pair for each wavevector k. It will therefore be no surprise to
see that the states of the quantum eld theory are just towers of oscillator
states, one for each k. The ground state is dened as the state annihilated
by all the annihilation operators:
a(k)[0) = 0, k. (4.39)
We can use the creation operators a

(k) to construct excited states: for

example, the rst excited state in mode k may be written
[k) = a

(k)[0). (4.40)
The normalisation of these states follows from deciding that the vacuum state
has unit norm, that is 0[0) = 1. Thus
k

[k) = 0[a(k

)a

(k)[0) (4.41)
= 0[a

(k)a(k

)[0) +0[0)(2)
3
(k k

). (4.42)
The rst term on the right hand side vanishes, and so we have that the rst
excited states satisfy the normalisation condition
k

[k) = (2)
3
(k k

). (4.43)
The corresponding completeness relation is
_
d
3
k
(2)
3
1
2
k
[k)k[ = 1. (4.44)
The states [k) are interpreted in quantum eld theory as single-particle
states, an interpretation discussed further below. We could continue ap-
plying creation operators a

(k
1
), a

(k
2
), . . . to make states
[k, k
1
, k
2
, . . .) = . . . a

(k
2
)a

(k
1
)a

(k)[0). (4.45)
The general state could have multiple applications of creation operators with
the same momenta, in which case it is conventional to divide by a symmetry
factor, to obtain
[n
k
) =

k
(a

k
)
n
k

n
k
!
[0), (4.46)
43
where n
k
is the number of times the creation operator with label k is applied.
The Hilbert space spanned by the set of all states [n
k
) is called Fock space.
The basis states are interpreted as multiparticle states, which demonstrates
one of the special features of eld theory: it is a quantum mechanical theory
of many particles. Ordinary non-relativistic quantum mechanics describes
only one particle at a time.
There is an operator which allows us to count the number of particles in
a given state. Let us rst introduce the set of operators n(k) by
n(k) = a

(k)a(k). (4.47)
It is straightforward to see that they return zero when acting on the ground
state,
n(k)[0) = 0, (4.48)
from the denition of the vacuum state (4.39). When acting on 1-particle
states, we nd
n(k)[k

) = a

(k)a(k)a

(k

)[0) (4.49)
= a

(k)
_
a

(k

)a(k) + 2
k
(2)
3
(k k

)
_
[0), (4.50)
where we have used the commutation relations (4.33) to change the order of
a(k) and a

(k

). Thus,
n(k)[k

) = 2
k
(2)
3
(k k

)[k). (4.51)
The operators n(k) are used to construct the number operator
N =
_
d
3
k
(2)
3
1
2
k
n(k), (4.52)
which tells us the total number of particles in a particular state Thus, as
expected, N[0) = 0, meaning that there are no particles in the ground state.
In the 1-particle state [k

), we nd
N
tot
[k

) =
_
d
3
k
(2)
3
N(k)[k

) =
_
d
3
k
(2)
3
(2)
3
(k k

)[k) = [k

). (4.53)
The state [k

) is indeed an eigenstate of the number operator, with eigenvalue

1 there is 1 particle in state. When applied to the general state (4.46), it
can be shown that
N[n
k
) =
_
d
3
k
(2)
3
n
k
[n
k
). (4.54)
44
The energy of the states can be computed from the quantum energy operator

E =
_
d
3
x
0
0
(x) =
_
d
3
x
_
1
2

2
+
1
2
(

)
2
+
1
2

2
_
, (4.55)
which is immediately seen to be identical to the Hamiltonian. Since the sim-
ple states we have seen so far have been constructed from creation operators
a

(k) acting on the vacuum, it is more convenient to re-express the Hamilto-

nian operator H in terms of creation and annihilation operators. Recalling
the expansions of and in terms of a(k) and a

(k) in equations (4.18) and

(4.20), we nd
H =
_
d
3
k
(2)
3
1
2
k

k
2
(a

(k)a(k) + a(k)a

(k)) . (4.56)
If we try to change the order of the operators, we run into a problem, for

H =
_
d
3
k
(2)
3
1
2
k

k
2
_
2a

(k)a(k) + (2)
3
(0)
k
_
. (4.57)
The second term in the brackets has the formally innite term (2)
3
(0).
The source of this innity is the fact that we are working in innite spatial
volume. Let us revert to a nite cubic region with sides of length L, in which
case we can write
(2)
3
(0) = lim
L
lim
k

k
_
L/2
L/2
d
3
xe
ix(kk

)
= lim
L
_
L/2
L/2
d
3
x = V, (4.58)
where V is a formal volume factor, and the limit of innite volume should
strictly be taken at the end of the calculation of any physical quantity. Thus
we can interpret the quantity multiplied by V as an energy density. However,
even this energy density is innite, in general. Suppose we try to nd the
energy density of the ground state
0
, for which
H[0) =
_
d
3
x
0
[0) =
_
d
3
k
(2)
3

k
a

(k)a(k)[0)+
1
2
_
d
3
k
(2)
3
1
2
k

k
(2)
3
(0)[0).
(4.59)
Hence,

0
=
1
2
_
d
3
k
(2)
3

k
. (4.60)
This integral is divergent. If we integrate the momenta over the range 0 <
[k[ < , with the upper cut-o in the integration range , we nd that
the dominant behaviour in the integral is

0
=
1
(2)
2
_

0
dkk
2
(k
2
+
2
)
1
2
=
1
(4)
2

4
(1 + O(
2
/
2
)). (4.61)
45
As we take to innity, the ground state energy diverges as the fourth
power of the cut-o in the momentum we say that the integral is quartically
divergent. Thus the ground state energy density of a eld theory appears to
be innite. This is the rst of many innities in quantum eld theory, and
the way of dealing with this, and all others, is to renormalise, that is, to
subtract o unobservable quantities.
As the theory stands, we have no way of measuring the absolute value
of the the energy of the ground state: all we see is dierences between that
state and others. Thus we measure all energies relative to the ground state
by subtracting the term which gives the innite value for the ground state
energy,
1
2
_
d
3
k
k
(0), and we dene a new, renormalised Hamiltonian
H
ren
= H
1
2
_
d
3
k
k
(0) =
_
d
3
k
(2)
3
1
2
k

k
a

(k)a(k). (4.62)
This has the sensible property that

H
ren
[0) = 0 the ground state energy is
zero. You might like to check that the energy of the state [k

) is indeed
k
:
a simple way to start is to note that
H
ren
=
_
d
3
k
(2)
3

k
n(k), (4.63)
and use Eq. (4.51).
This conclusion is not sound one we start thinking about gravity. To see
this, consider the Friedmann equation in cosmology
H
2
=
8G
3
, (4.64)
where H = a/a is the Hubble expansion rate, a is the scale factor of the
Universe, is the total energy density, and G is Newtons constant. This
equation seems to tell is that we can measure the energy density of the vac-
uum by measuring the expansion rate of the Universe, and subtracting o the
contributions from known sources to which we have more direct experimental
access. In practice, measuring the vacuum energy density (or equivalently a
cosmological constant) dicult, but through a combination of supernova and
Cosmic Microwave Background observations, the value expressed in natural
units is

10
47
GeV
4
. (4.65)
Furthermore, it is now generally recognised that the quantum theory of grav-
ity (if it exists at all) is not a eld theory, and that eld theories in general
are only an acceptable description of fundamental particles and forces up
46
to a momentum of around the Planck mass M
P
= 1/

G, and therefore it
is makes no sense to let the upper limit of integration go beyond M
P
in
Eq. (4.61). There is an analogy here with the theory of phonons in a solid
lattice: there are no modes with momenta above 1/a, where a is the spac-
ing between atoms, but at low momenta we have a very good description of
phonons as the quantised oscillation in a massless eld whose waves moves at
the sound speed. Hence, in the absence of other interactions besides gravity,
the vacuum energy of a real scalar eld seems to be

0

1
(4)
2
M
4
P
, (4.66)
which is a factor of about 10
121
larger the the currently accepted value.
We can recover a more realistic result by starting from a theory with a
potential
V () = V
0
+
1
2

2
, (4.67)
with V
0
=
0
+

. However, it is quite incredible that nature would have

arranged such a precisely set value for V
0
:
0
and

must dier by a precisely

dened number which is a factor of 10
121
smaller than the individual terms.
This is an example of a ne-tuning problem.
Returning to Minkowski space, and ignoring gravity again, we can sub-
tract o ground state energies automatically by a procedure known as normal
ordering. When we place an operator in normal order, we write it so that all
the creation operators are to the left, and all the annihilation operators are
to the right, so for example
:a(k)a

(k

): = a

(k

)a(k). (4.68)
Hence,
:H: = H
ren
. (4.69)
4.4 Particle interpretation of states
We shall nish this section by returning to the justication for interpreting
the states of the eld as multiparticle states. Let us rst compute the energy
of a one-particle state:
H
ren
[k) =
_
d
3
k

(2)
3
1
2
k

k
n(k

)[k) =
k
[k). (4.70)
Hence the energy of the state we have decided is a one-particle state is
k
,
which we recall is equal to

(k
2
+m
2
), the energy of a relativistic particle of
47
mass m and momentum k. Thus in order to strengthen our case we should
check that the momentum of the state [k) is indeed k.
The momentum operator in eld theory is
P =
_
d
3
x

(x)(x) =
_
d
3
k

(2)
3
1
2
k

k n(k). (4.71)
There is in fact no need to normal order the momentum operator, as the
momentum of the vacuum is zero as a result of its translational invariance.
Thus we may evaluate the momentum of the state [k) through
P[k) =
_
d
3
k

(2)
3
1
2
k

n(k

)[k) = k[k). (4.72)

Thus the state [k) is an eigenstate of the momentum operator, with eigen-
value k, properties we expect from a one-particle state.
The state is also rotationally invariant: a property following from the
rotational invariance of the scalar eld, but should be explicitly checked by
constructing the angular momentum operators. Hence the state is a spinless
particle. Lastly, multiparticle states automatically have the correct Bose
symmetry. Consider a state of two particles with momenta k
1
and k
2
. Then
[k
1
k
2
) = a

(k
1
)a

(k
2
)[0)a

(k
2
)a

(k
1
)[0) = [k
2
k
1
), (4.73)
which shows that the quantum state is symmetric under the exchange of
particles. This property stems from the canonical commutation relations of
the eld operators.
5 Canonical quantisation of the electromag-
netic eld
In order to quantise the electromagnetic eld we shall follow exactly the same
procedure as we did for the real scalar eld. We shall nd the dynamical
coordinates, identify the conjugate momentum, and impose canonical com-
mutation relations between them. However, we should be aware that there
are extra complications not present for the real scalar eld. Firstly, there is
gauge invariance: recall that the transformation A

does not
change any physical quantity associated with the eld, which means we are
in danger of encoutering unphysical degrees of freedom when we use A

as
the dynamical coordinates. Secondly, the conjugate momentum to A

looks
like it should be

=
L
(
0
A

)
= F
0
. (5.1)
48
But this means that A
0
does not have a canonical momentum, as
0
vanishes
identically.
Our approach to dealing with these problems is to choose a gauge from
the outset, trying to leave only physical degrees of freedom in the Lagrangian.
We shall carry out this procedure in the Coulomb gauge and in the Lorentz
gauge.
5.1 Coulomb gauge quantisation
We recall that the Coulomb gauge is dened by
i
A
i
= 0. In this gauge, A
0
is entirely determined at any time by the charge distribution, through the
equation

2
A
0
= , (5.2)
and it is therefore not a true dynamical coordinate. In free space, that is, in
the absence of charges or currents, we are entitled to set it to zero.
We recall the action of the electromagnetic eld, in the absence of an
external 4-current:
S =
_
d
4
xL =
_
d
4
x
1
4
F

,
where F

. In the Coulomb gauge, this action is equivalent

to
S
Cg
=
_
d
4
xL
Cg
=
_
d
4
x
_
1
2

A
i

A
i

1
2

j
A
i

j
A
i
_
. (5.3)
Thus the momentum conjugate to A
i
is

i
=
L
Cg

A
i
=

A
i
= E
i
, (5.4)
and the Hamiltonian density is
H
Cg
=
i

A
i
L
Cg
=
1
2

A
i

A
i
+
1
2

j
A
i

j
A
i
. (5.5)
At this point it would seem natural to impose a commutation relation
1
[A
i
(x, t),
j
(x

, t)]
?
= i
ij
(x x

), (5.6)
based on our experience with scalar elds. This is incorrect, as dierentiating
both sides with respect to x
i
quickly shows. The gauge condition
i
A
i
= 0
implies that the left hand side vanishes. However,
[
i
A
i
(x, t),
j
(x

, t)] = i
j
(x x

) ,= 0. (5.7)
1
Note that as this is not a covariant equation, there is no need for covariant and
contravariant indices to balance on both sides of the equation.
49
The derivative of a -function is not zero. So we need a more general function

ij
(x, x

) for the right hand side of the CCR, which satises

i

ij
(x, x

) = 0
(and also

ij
(x, x

) = 0, as one can easily check by substituting for

j
in
the commutation relations). We also know that the dependence on x and x

must be in the combination x x

through translational invariance.

It is easier to nd the function
ij
(x x

) in Fourier space: we write

ij
(x x

) =
_
d
3
k
(2)
3

ij
(k)e
ik(xx

)
. (5.8)
Then

ij
(x x

) = i
_
d
3
k
(2)
3
k
i

ij
(k)e
ik(xx

)
= 0, (5.9)
from which we infer that
k
i

ij
(k) = 0. (5.10)
The solution to this equation is

ij
(k) =
_

ij

k
i
k
j
k
2
_

F(k), (5.11)
where F(k) is some function of k. The correct function is just F = 1: this
keeps the function
ij
(x x

) as close as possible to the -function. Thus

the canonical commutation relations in the Coulomb gauge are
[A
i
(x, t), E
j
(x

, t)] = i
_
d
3
k
(2)
3
_

ij

k
i
k
j
k
2
_
e
ik(xx

)
. (5.12)
This we can write
[A
i
(x, t), E
j
(x

, t)] = i
_

ij

2
_
(x x

), (5.13)
where
2
is the inverse of
2
=
i

i
, and can be thought of as being
dened in terms of its Fourier transform. Given any function f(x) with
Fourier transform

f(k),
2
f(x) is that function whose Fourier transform is
k
2

f(k). One can now verify that the Coulomb gauge commutation relations
5.13 are now consistent with the gauge condition
i
A
i
= 0.
As we saw in Section 2.2.1, the free space Coulomb gauge equations of
motion for the electromagnetic eld are
(
2
0

2
j
)A
i
(t, x) = 0,
50
which have the general solution
A
i
(t, x) =
_
d
3
k
(2)
3
1

2
k
_
a
i
(k)e
ikx
+ a
i
(k)e
ikx
_
,
where k
i
a
i
= 0. Thus the three components of the vector amplitude a
i
(k) are
not independent: they satisfy a constraint arising from the Coulomb gauge
condition. Rather than worry about this constraint all the time, it is much
more convenient to have two freely chosen functions of k, so we instead write
the general solution
A
i
(t, x) =

=1,2
_
d
3
k
(2)
3
1

2
k
_
a
()

i
()
e
ikx
+ a

()

()
i
e
ikx
_
. (5.14)
The
i
()
are called polarisation vectors, and have the properties
k
i

i
()
= 0,
i
()

i
(

)
=

. (5.15)
These conditions mean that, along with

k
i
= k
i
/[k[, the two polarisation
vectors form an orthonormal basis for 3-vectors. The completeness property
therefore follows:

i
(1)

j
(1)
+
i
(2)

j
(2)
+

k
i

k
j
=
ij
, (5.16)
or

i
()

j
()
=
ij

k
i

k
j
.
Upon quantisation, the independent amplitudes a
()
(k) become opera-
tors, and by substituting the expansions for A
i
(t, k) and E
j
(t, x

) into the
canonical commutation relations, and using the properties of the polarisation
vectors, one discovers the commutation relations for a
()
(k) and a

()
(k):
[a
()
(k), a

)
(k

)] =

(2)
3
(k k

). (5.17)
Thus each polarisation behaves like an independent scalar eld.
As before, we can dene a vacuum state [0) satisfying
a
()
(k)[0) = 0, k, . (5.18)
Other states are derived from the vacuum by acting with raising operators.
For example, there is a set of rst excited states
[k, ) = a

()
(k)[0). (5.19)
Note that there is an extra label, , which indicates the polarisation of the
electromagnetic eld in the state. The quantised states of the electromagnetic
eld are of course the particles known as photons.
51
Last in this section, we write down the Hamiltonian for the electromag-
netic eld in terms of the ladder operators in its normal-ordered form:
:H: =
_
d
3
x:H: =
_
d
3
k
(2)
3

k
a

()
(k)a
()
(k). (5.20)
This is just two copies of the renormalised scalar eld Hamiltonian (4.62),
one for each polarisation.
5.2 Canonical quantisation in the Lorentz gauge
In order to quantise this theory in the canonical manner, we must rst iden-
tify the canonical momentum conjugate to the degrees of freedom, impose
canonical commuation relations, identify the ladder operators and nd their
commutation relations, and construct a Fock space of states. We should also
calculate the Hamiltonian, and verify that it is positive denite.
As one might imagine, gauge invariance poses problems for the quantum
theory as well. Firstly, we have the problem mentioned in Section 2.2.1, that
the canonical momentum vanishes. This turns out to be cured by xing the
gauge, as we can verify in the Lorentz gauge:

=
L
Lg

=

A

. (5.21)
Hence the equal time canonical commutation relations can be guessed to be
those for four independent real elds
[A

(t, x),

(t, x

)] = i

3
(x x

), (5.22)
[A

(t, x), A

(t, x

)] = 0, [

(t, x),

(t, x

)] = 0. (5.23)
By substituting the plane wave expansion for the eld operator A

(x), one
can nd (after some algebra) that
[a
A
(k), a
B
(k

)] =
AB
2
k

3
(k k

), (5.24)
[a
A
(k), a
B
(k

)] = 0, [a
A
(k), a
B
(k

)] = 0. (5.25)
Hence we have four pairs of ladder operators, one pair for each polarisation.
We dene the vacuum state [0) by
a
A
(k)[0) = 0 k, A. (5.26)
and we can construct excited states by acting with the raising operators
a
A
(k). For example,
[k, A) = a
A
(k)[0). (5.27)
52
How do we impose the gauge constraint in the quantum theory? Firstly,
it is clear that demanding A = 0 is too strong a condition on the operator
A

, as it is inconsistent with the equal time canonical commutation relations

Eq. 5.22. A weaker condition is to ask that matrix elements of the gauge
condition vanish. This we cannot do for every possible state, but we only
accept as physical states for which this is true: i.e.

[ A[) = 0 (5.28)
for any two physical states [) and [

). Any equivalent condition is to

demand that the positive frequency part of the gauge condition gives zero
when acting on a physical state:
A
(+)
[) = 0, (5.29)
where
A
(+)
=
_
d
3
k
2
k
ik ae
ikx
. (5.30)
Hence physical states must satisfy
(a
0
(k) a
3
(k))[) = 0, (5.31)
which is the equivalent of Eq. (2.44) in the quantum theory.
Finally, we shall check that the Hamiltonian is positive denite for phys-
ical states. We must deal with the innity in the vacuum energy by normal
ordering, and study the matrix elements of the operator
:H
Lg
: =
_
d
3
k
2
k

AB
a
A
(k)a
B
(k). (5.32)
The quantum version of the gauge condition Eq. (5.31) implies that for phys-
ical states [) and [

),

[:H
Lg
:[) =

A=1,2
_
d
3
k
2
k

[a
A
(k)a
A
(k)[). (5.33)
In particular, the expectation value of the energy is clearly positive denite
for physical states.
6 Fermions
Dirac knew of the problem with the probability interpretation of the Klein-
Gordon equation, and traced it to the fact that it was second order in time
53
derivatives. He therefore set out to nd a relativistic wave equation with
only one time derivative. The requirement that the form of the equation be
unchanged under Lorentz transformations, which mix up /t and , means
that the equation must be rst order in spatial derivatives as well. Hence
Dirac proposed a relativistic free particle wave equation
i

t
= i + m. (6.1)
There clearly has to be something rather special about the objects , and
in order that Lorentz covariance be preserved. In fact, and are 44
Hermitean matrices, and is an object with four components called a spinor.
Now, we want a wave equation with plane wave solutions which satisfy
the relativistic energy-momentum relation E
2
= p
2
+ m
2
. To see how this
property emerges from the Dirac equation, we act on both sides of (6.1) with
(i
t
i ), to obtain
_
i

t
i
__
i

t
+ i
_
= m
_
i

t
i
_
. (6.2)
At this point, it is more convenient to switch to index notation for the 3-
vectors and , noting that =
i

i
. We expand out the brackets on
the left hand side of (6.2), and add and subtract i
i

i
on the right hand
side. The result is
_

2
t
2
+
i

j
_
= m
__
i

t
i
i

_
+ i
i

i
+ i
i

_
. (6.3)
It is now useful to dene the anticommutator, represented by curly brackets:
A, B = AB + BA. (6.4)
(Sometimes you will also see the anticommutator written as [A, B]
+
). We
rewrite the left hand side as (
2
t
+
1
2

j
+
1
2

i
), which we are
entitled to do by renaming the indices on and , to obtain
_

2
t
2
+
1
2

i
,
j

j
_
= m
2

2
+ i,
i

i
. (6.5)
In order to reproduce the relativistic relation between energy and momentum,
we must end up with an equation like the Klein-Gordon equation. If
i
and
satisfy

i
,
j
=
ij
, ,
i
= 0,
2
= 1, (6.6)
54
where 1 is the 44 identity matrix, satises the equation
_

2
t
2
+
2
_
= m
2
. (6.7)
Hence, each of the four components of satisfy the Klein-Gordon equation.
We can make the Dirac equation more explicitly relativistic by dening
four new 44 matrices:

0
= ,
i
=
i
. (6.8)
If we multiply both sides of the Dirac equation (6.1) by , we obtain
(i

m1) = 0. (6.9)
The conditions (6.6) on
i
and are neatly unied into the matrix equation

= 2

1. (6.10)
6.1 Plane wave solutions
We know that plane wave solutions exist, but what are they? We therefore
look for solutions of the form
(x) = ue
ipx
, (6.11)
with u a constant 4-component spinor. Although it appears from (6.7) that
u is arbitrary, this is not in fact true, as we shall see.
Substituting our plane wave ansatz (6.11) into the Dirac equation (6.9),
we nd
(

m)ue
ipx
= 0. (6.12)
At this point we shall introduce another piece of notation: we shall use the
slash symbol , to denote contraction of a 4-vector with the Dirac gamma
matrices:
a

=,a. (6.13)
Thus u satises the equation
(,p m)u = 0. (6.14)
Let us premultiply this equation by (,p +m), and use an important identity,
,a ,b = a b 1, (6.15)
55
to show that
(,p + m)(,p m)u = (p
2
m
2
)u = 0. (6.16)
This brings us back to the relation E
2
= p
2
+m
2
, although by a rather more
rapid route. It is conventional to choose p
0
E always to be positive, in
which case we must also include solutions proportional to e
+ipx
separately.
We write them
(x) = ve
ipx
, (6.17)
which satisfy
(,p + m)v = 0. (6.18)
Solutions with time dependence e
iEt
(e
iEt
) are called positive (negative)
energy solutions.
We can use the rst equality in Eq. (6.16) to show that the following
spinors are solutions to the Dirac equations for u and v:
u

(p) = N(,p + m)
_

0
_
v

(p) = N(,p m)
_
0

_
where N is a normalisation factor, and we have dened
+
and

to span
the space of 2-component spinors (which means they are spinor spanners):

+
=
_
1
0
_
,

=
_
0
1
_
. (6.19)
Any solution to the Dirac equation can be written as a superposition of the
four four-component basis spinors u

and v

.
To proceed further we need an explicit representation for the gamma
matrices, and we will use the so-called standard representation given in
Appendix A.1. It will turn out to be convenient to choose the normalisation
factor to be
N = 1/

(E + m). (6.20)
The basis for the 4-component positive energy solutions can therefore be
written
u

=
1

E + m
_
(E + m)

_
, (6.21)
and the negative energy basis is
v

=
1

E + m
_
p

(E + m)

_
. (6.22)
There are a couple of remarks that are worth making about the solutions
(6.21) and (6.22). Firstly, when the 3-momentum is small, that is when
56
[p[ m, two of the components are much smaller than the other two. For the
positive energy solutions, it is the lower two components that are negligible,
while the opposite is true for the negative energy solutions. Secondly, the
appearance of two degrees of freedom for each set of solutions (that is,

)
demands some explanation. One can verify that

are eigenstates of the

Pauli matrix
3
, with eigenvalues 1:

. (6.23)
This degeneracy is actually a result of the fact that the particles described
by the Dirac equation (such as the electron) have an extra property, spin,
which is a type of angular momentum intrinsic to the particle itself, unlike
ordinary angular momentum which particles gain by virtue of rotation. Spin
is quantised in half-integer units, with the spin of the electron being s =
1
2
.
As with other types of quantised angular momentum, the value of the spin
projected along any axis (such as the z axis) takes the values s, s1, . . . , s.
Thus Dirac-type particles have only two spin states, which we conventionally
call up and down.
With the given normalisation convention, the spinors have the properties
u

A
u
B
= 2E
AB
, v

A
v
B
= 2E
AB
(6.24)
We dene the adjoint spinors u and v by
u
A
= u

0
, v
A
= v

0
. (6.25)
The adjoint spinors have the further properties
u
A
u
B
= 2m
AB
, v
A
v
B
= 2m
AB
. (6.26)
Note that this is a dierent normalisation convention than that used by
Greiner and Reinhardt.
6.2 The interpretation of negative energy states
In the last section we called half of the solutions negative energy solutions
without proper explanation. Of course, the explanation is that they appear
to represent quantum states with negative energy, for if one acts with the
quantum mechanical energy operator, one obtains a negative number:
i

t
(ve
ipx
) = E(ve
ipx
), (6.27)
with E = [(p
2
+m
2
)
1
2
[. This raises an ugly possibility: that an electron with
positive energy E
1
could decay into one with negative energy E
2
, radiating
57
a photon with energy E
1
+E
2
. There indeed seems to be nothing to stop the
energy of the electron from tumbling down towards minus innity, releasing
an innite amount of energy in the process. This is clearly nonsense. Diracs
way of circumventing this problem was to suppose that all the negative en-
ergy states are lled already, and the Pauli exclusion principle prevents a
positive energy electron from radiating a photon and occupying a lled nega-
tive energy state. One can still imagine exciting one of these negative energy
electrons into a postive energy state, whereupon it becomes a real electron,
leaving behind a hole. A hole is an absence of a negative electric charge,
which has positive charge. Thus Dirac realised that his theory predicted
the existence of a positively charged spin
1
2
particle with exactly the same
mass as the electron. Initially, Dirac identied this particle with the proton,
hoping that somehow Coulomb interactions would provide a mass dierence,
but after others (including Weyl, Oppenheimer, and Tamm) showed that this
couldnt be right, he eventually had to abandon this conservative position
and predict a new particle now know as the positron. Its discovery in 1932
by Carl Anderson in cosmic rays settled the issue.
However, this picture of a negative energy sea of electrons leaves much
to be desired. Bosons, particles with integer spin such as the photon (spin
1) and the pion (spin 0) do not obey the Pauli Exclusion Principle and
yet they still have negative energy states. Thus there can be no Dirac sea
picture and we are left where we were before. It is only if we abandon simple
wave mechanics and turn to quantum eld theory that we nd a satisfactory
resolution of this problem.
6.3 Quantising the spinor eld
The Lagrangian density for a massive spinor eld is
L = i

, m

, (6.28)
where we recall that in the standard representation of the Dirac gamma
matrices the adjoint spinor

=

0
. Rather like the complex scalar eld,
one can obtain the eld equation (which is the Dirac equation) by treating

and as independent quantities, and demand that the action be stationary

with respect to arbitrary variations in either. The variation of the action
with respect to

gives
S =
_
d
4
x

= 0, (6.29)
(the Lagrangian is not a function of

in this formulation). Hence
_
d
4
x

(i , m) = 0, (6.30)
58
from which we derive the Dirac equation i , m = 0. Similarly, we can
vary with respect to and obtain
S =
_
d
4
x(i

, m

). (6.31)
Integrating by parts we nd
S =
_
d
4
x(i

) +
_
dS

, (6.32)
where the last term is an integral over the space-time surface at spatial
innity ([x[ ) with end-caps at [t[ . As usual, we suppose that the
variations die away at innity so that we can drop the surface term, so we
recover the equation for the adjoint spinor i

+ m

= 0.
The momentum conjugate to is given by
=
L

= i

0
, (6.33)
which shows that in the standard reresentation and i

are canonically
conjugate variables. Thus we can nd the Hamiltonian density:
H =

L = i

k
+ m

. (6.34)
The quantisation of the spinor eld follows a familiar pattern. We rst of
all suppose that (x) is an operator satisfying some commutation relations,
acting on some space of quantum states. We specify the equal time com-
mutation relations, and then try to nd the possible states. The problem
with the spinor eld is that it is not obvious from the outset what commu-
tation relations to impose on the eld operator, and the obvious relation
[(t, x), i

(t, x

)] = i(x x

) is not in fact correct.

To see why this is so, let us compute the zero-point energy of the eld.
Firstly, we expand the eld in terms of its operator-valued Fourier coecients:
(x) =

A=
_
d
3
p
(2)
3
1
2E
_
c
A
(p)u
A
(p)e
ipx
+ d

A
(p)v
A
(p)e
ipx
_
, (6.35)
where E
2
= p
2
+ m
2
, and u
A
(p) and v
a
(p) are 4-component spinors which
satisfy
(,p m)u
A
(p) = 0, (,p + m)v
A
(p) = 0. (6.36)
Similarly, the conjugate spinor has the expansion

(x) =

A=
_
d
3
p
(2)
3
1
2E
_
c

A
(p) u
A
(p)e
+ipx
+ d
A
(p) v
A
(p)e
ipx
_
. (6.37)
59
We can now compute the Hamiltonian, which is the spatial integral of the
Hamiltonian density H, or
H =
_
d
3
x(i

i
+ m

) (6.38)
The end result is
H =

A
_
d
3
p
(2)
3
1
2E
E
_
c

A
(p)c
A
(p) d
A
(p)d

A
(p)
_
. (6.39)
The vacuum state is traditionally dened as the state annihilated by all
annihilation operators c
A
(p) and d
A
(p):
c
A
(p)[0) = 0, d
A
(p)[0) = 0. (6.40)
We expect an innite zero point energy again, removed by normal ordering.
However, H is not positive denite, and nor is :H: unless we dene the
commutators of the operators such that
:d
A
(p)d

A
(p): = d

A
(p)d
A
(p). (6.41)
This means that the fermionic creation and annihilation operators must an-
ticommute, and we quantise the spinor eld by anticommutation relations
d
A
(p), d

B
(p

) = 2E
AB
(p p

), (6.42)
c
A
(p), c

B
(p

) = 2E
AB
(p p

). (6.43)
All other anticommutator brackets vanish:
c
A
(p), c
B
(p

) = 0 = d
A
(p), d
B
(p

) (6.44)
c

A
(p), c

B
(p

) = 0 = d

A
(p), d

B
(p

) (6.45)
c
A
(p), d

B
(p

) = 0 = c

A
(p), d
B
(p

). (6.46)
There are four kinds of 1-particle states in this theory, created by the four
operators c

A
(p) and d

B
(p). We write them
[p, A; 0) = c

A
(p)[0), [0; p, A) = d

A
(p)[0). (6.47)
As you can probably guess from the way the states are built up for the
complex scalar eld, c

A
(p) creates a particle with momentum p and spin
state A, while d

A
(p) creates an antiparticle.
Let us examine the 2-particle state
[p
1
, A
1
, p
2
, A
2
; 0) = c

A
1
(p
1
)c

A
2
(p
2
)[0). (6.48)
60
If we interchange the particles, we must interchange the creation operators,
which gives the state a relative minus sign:
[p
2
, A
2
, p
1
, A
1
; 0) = c

A
2
(p
2
)c

A
1
(p
1
)[0), (6.49)
= c

A
1
(p
1
)c

A
2
(p
2
)[0), (6.50)
= [p
1
, A
1
, p
2
, A
2
; 0). (6.51)
Thus we see something that has to be taken for granted in ordinary quan-
tum mechanics: the states of the spinor or Dirac eld are antisymmetric
under particle interchange. We call these states fermionic, and the particles
fermions. Furthermore, suppose we take both momenta and both spin states
equal to p and A respectively: we then nd
[p, A, p, A; 0) = [p, A, p, A; 0) = 0. (6.52)
Hence two fermions can never be in the same state. This is precisely the
Pauli exclusion principle, which is fundamental to the understanding of the
physics of the atom. On the other hands, bosons, like the scalar particle we
considered previously, have commuting eld operators, and so any number
of them can be in the same state.
Armed with the anticommutation relations for the creation and annihila-
tion operators, and the spinor completeness relations

A
u
A
a
u
A
b
= ( p)
ab
+ m
ab
,

A
v
A
a
v
A
b
= ( p)
ab
m
ab
, (6.53)
the elds can be shown to satisfy the equal time anticommutation relations

a
(t, x),

b
(t, x

) = i h
ab
(x x

), (6.54)

a
(t, x),
b
(t, x

) = 0 =

a
(t, x),

b
(t, x

) (6.55)
where the factor of h has been revived, and labels a and b have been put in
to keep track of the rows and columns of the spinors and

.
6.4 Conserved charge
One can show straightforwardly from the Dirac equation (and its adjoint)
that the 4-vector quantity
j

(x) =

(x)

(x) (6.56)
satises a continuity relation j = 0. Hence there is a conserved charge
Q =
_
d
3
x

0
=
_
d
3
x

. (6.57)
61
By substituting the plane wave expansion for the Dirac eld operator, one
can show that
Q =

A
_
d
3
p
(2)
3
1
2E
_
c

A
(p)c
A
(p) d

A
(p)d
A
(p)
_
. (6.58)
Applying this operator to to the single particle states, one can show that
Q[p, A; 0) = +[p, A; 0), Q[0, p, A) = [0; p, A). (6.59)
Hence the states created by c

A
(p) have equal and opposite charge to those
created by d

A
(p), which ts in with their interpretation as particles and
antiparticles.
One can also show that the existence of this conserved charge is a conse-
quence of a symmetry of the Lagrangian, as we should expect from Noethers
theorem. This symmetry is
(x)

(x) = e
i
(x),

(x)

(x) = e
i

(x). (6.60)
7 Scalar elds in an expanding Universe
In this section we shall look at a quantum eld theory in an expanding
space-time, as is appropriate for early Universe cosmology, and in particular
for the generation of perturbations in an inationary phase. We shall follow
our normal procedure of starting with an action, deriving the Euler-Lagrange
equations, nding the solutions, and promoting the eld and its canonical
conjugate to operators. We will nd the fascinating result that in a universe
expanding exponentially fast the vacuum is no longer quite empty: the eld
uctuates with an amplitude H, where H is the expansion rate. In the
inationary universe scenario is the traces of these uctuations which we see
in the Cosmic Microwave Background.
7.1 Scalar eld in curved spacetime
In order to dene a eld theory in a general spacetime background we must
go beyond Special Relativity and make it consistent with the postulates of
General Relativity. This means that we must be able to formulate it in such
as way that the equations of the theory take the same form for any observer,
not just those in a state of uniform motion. Thus the action must be invariant
under a general coordinate transformation
x

= x

(x). (7.1)
62
This is a much larger group of transformations than the Lorentz group of
Special Relativity. Under this transformation, a rank 2 tensor like the metric
transforms as
g

(x) g

(x

) = g

(x)
x

(7.2)
It is not too hard to show that the following expression for the action of a
real scalar eld is indeed invariant under general coordinate transformations,
and reduces to the Minkowski space action of the previous sections when
g

:
S =
_
d
4
x

g
_
1
2
g

V ()
_
, (7.3)
where g det g

. A at homogeneous and isotropic expanding Universe is

described by the metric g

= diag(1, a
2
(t), a
2
(t), a
2
(t)). By substitut-
ing this metric in the action (7.3) we quickly see that it is
S =
_
d
4
xa
3
_
1
2

1
2a
2
()
2
V ()
_
. (7.4)
Using the Euler-Lagrange equation we nd

+ 3
a
a

1
a
2

2
+ V

() = 0. (7.5)
The scale factor a(t) is determined from the Friedmann equation
_
a
a
_
2
=
8G
3
, (7.6)
where G is Newtons gravitational constant and is the energy density. Let
us suppose that the energy density is dominated by the energy density of a
scalar eld, which is
=
1
2

2
+
1
2
1
2a
2
()
2
+ V (). (7.7)
This must obey the covariant form of the energy-momentum conservation
equation

+ 3H( + p) = 0 (7.8)
where H is the Hubble parameter H = a/a, is the average energy density,
and p is the average of the pressure p, given by
p =
1
2

1
2
1
6a
2
()
2
V (). (7.9)
Let us look for solutions which consist of small uctuations around a homo-
geneous eld (t, x) =

(t) + (t, x), with [[ [

[. By substituting this
63
Ansatz into the eld equation (7.5), and neglecting the uctuations, we nd
for the homogeneous part:

+ 3
a
a

+ V

) = 0. (7.10)
Note that when the is large, the eld is strongly damped, so we look for
solutions in which we can neglect

in comparison to 3H

, so that
3H

V

(

). (7.11)
One can show that sucient conditions for this to be true are that the fol-
lowing two dimensionless quantities are small:
=
1
2m
2
P
_
V

)
V (

)
_
2
, = m
2
P
V

)
V (

)
, (7.12)
where m
2
P
= 1/8G is the square of the so-called reduced Planck mass.
These are the so-called slow roll parameters. In the limit 0, one nds
that

= 0, and that = p = V (

). Thus the energy density and pressure

are constant and dominated by the potential energy density V , and solving
the Friedmann equation one nds that the scale factor expands exponentially:
a exp(Ht), (7.13)
with H =
_
V/3m
2
P
. This rapid expansion is an example of ination. More
generally, we dene ination to be a situation where a > 0, and we nd that
is non-zero. In this case the expansion goes approximately as a power-law
a t
1/
. We can regard exponential expansion as the limit of power-law
expansion in the limit 0.
7.2 Classical scalar eld uctuations
We now examine the equation for the uctuations . By keeping only terms
rst order in in the full eld equation (7.5) we nd that the uctuations
obey
+ 3
a
a

1
a
2

2
+ V

) = 0. (7.14)
Using the slow-roll parameter ,
+ 3
a
a

1
a
2

2
+ 3H
2
0, (7.15)
64
where H = a/a is the Hubble parameter. One denition of an inating
universe is that a > 0 (in these coordinates). One can show that an inating
universe whose energy density is dominated by the potential of a scalar eld
V should generally have 1.
Let us now look for solutions to Eq. (7.15). First it is convenient to
introduce conformal time through the relation ad = dt, or
=
_
t
dt

/a(t

) (7.16)
If a t
p
, we have that t
1p
/(1 p), and hence a
p/(1p)
. By a
suitable choice of origin, one can also show that
=
p
1 p
1
aH
. (7.17)
Note that exponential expansion corresponds to the limit p , in which
case = (aH)
1
. Hence < 0, and the universe inates as 0

.
The general solution to Eq. 7.15 has the form
(t, x) =
_
d
3
k
2
k
_
a
k
f
k
(t)e
ikx
+ a

k
f

k
(t)e
ikx
_
(7.18)
where
k
=

(k
2
+ 3H
2
), and the functions f
k
satisfy

f
k
+ 3
a
a

f
k
+
k
2
a
2
f
k
+ 3H
2
f
k
= 0 (7.19)
Let us now suppose that the Universe is inating and we can neglect terms
of order . Then to simply the equation we can write f
k
(t) = a(t)u
k
(t), and
obtain the equation for u
k
(t)

2
u
k
+
_
k
2

2
_
u
k
= 0. (7.20)
One can easily show by direct substitution that the two independent solutions
are
f
k
(t) =
(k i)
ak
e
ik
= H
(k i)
k
e
ik
. (7.21)
and its complex conjugate f

k
(t).
Fluctuations in the scalar eld lead to uctuations in the energy density
. Because the energy density is dominated by the potential, the rst order
term in the uctuation is
(t, x) = (t, x) = V

(

)(t, x). (7.22)

65
The mean square uctuations in the energy density are therefore
_

2
(t, x)
_
= (V

))
2
_

2
(t, x)
_
=
H
2

2

2
_

2
(t, x)
_
. (7.23)
We will see that in an inationary universe, the uctuations are non-zero,
even in the vacuum state.
7.3 Quantum scalar eld uctuations
Let us now quantise the uctuation in the eld in the familiar canonical
manner, assuming exponential expansion, and zero mass. The conjugate
momentum is identied as
=
L

= a
3
, (7.24)
giving a Hamiltonian
H =

L = a
3
_
1
2

2
+
1
2a
2
()
2
+ V ()
_
, (7.25)
which we note is proportional to but not identical to the energy density. We
now promote and to operators and impose the canonical commutation
relations
[(t, x), (t, x

)] = i
3
(x x

) (7.26)
The wave functions f
k
and f

k
are already normalised so that the creation and
annihilation operators satisfy the standard relativistic commutation relations
[a
k
, a

k
] = 2
p
(k k

), [a
k
, a
k
] = [a

k
, a

k
] = 0. (7.27)
With these creation and annihilation operators we can dene the states in
the same way as in Minkowski space. The vacuum and 1-particles states [0)
and [k) are dened
a
k
[0) = 0 k, [k) = a

k
[0) (7.28)
respectively.
Let us now consider the mean square uctuations of the eld in the vac-
uum state,

2
= 0[
2
(t, x)[0), (7.29)
which we have seen determines the uctuations in the energy density. By
substituting the expansion of the eld in terms of its ladder operators, and
using their commutation relations, one nds

2
=
_
d
3
k
2k
[f
k
(t)[
2
=
_
dk
k
_
H
2
_
2
(1 + k
2

2
). (7.30)
66
Dene the power spectrum of the eld uctuations to be
T

(k) =
1
2
2
k
3
_
[
k
[
2
_
, (7.31)
we see that
T

(k) =
_
H
2
_
2
(1 + k
2

2
). (7.32)
As the universe inates, 0 from negative values, and we can neglect the
second term for an individual mode. Hence
T

(k)
_
H
2
_
2
. (7.33)
Thus we conclude that in an exponentially expanding Universe, a massless
scalar eld uctuates on all length scales with an amplitude H/2. Dening
the density uctuation on a comoving scale k by

k
=
1

_
d
3
x(t, x)e
ikx
, (7.34)
we recover the classic result
T

(k) =
H
2

2
_
H
2
_
2
. (7.35)
Strictly speaking, in pure de Sitter space, the energy density is constant and

= 0. This result should be understood as being the leading term in an

expansion in the slow roll parameters and . Further more, in order to
calculate the eect of these energy density perturbations today, one needs to
evolve them forward through the end of ination and past their eventually
re-entry through the horizon, when once again k = (aH). This requires cos-
mological perturbation theory which is discussed in Martin Kunzs lectures
in the current volume.
Based on our experience with the vacuum energy, we may be concerned
that the result for the mean square uctuations (7.30) contains divergent
quantities: the rst term on the right hand side is logarithmically divergent,
and the second term quadratically.
Let us examine the divergence in more detail. We again regularise the
integrals to make algebraic manipulations meaningful. We will again impose
an upper cut-o on the magnitude of the physical momentum variable k/a.
The cut-o is supposed to arise from a parent theory at higher energies, and
may even be the Planck mass m
P
. Because of the logarithmic nature of the
divergence in the rst term, we must also consider a lower cut-o on the
67
integral. This is actually supplied by the eective mass squared m
2
e
= 3H
in the more physical case ,= 0.
Hence the regularised version is

2
reg

_
H
2
_
2
_
log
_

m
e
_
+
1
2

2
a
2

2
_
(7.36)
Using the relation Eq. 7.17 in the limit p ,

_
H
2
_
2
log
_

m
e
_
+

2
4
2
. (7.37)
We note that the second term is independent of H, and therefore we should
suspect that is also present in Minkowski space. Let us denote the uc-
tuations in Minkowski space by
2
0
, where the wave-functions are f
k
(t) =
exp(i
k
t),

2
0,reg
=
_
d
3
k
2k
[f
k
(t)[
2
=

2
4
2
. (7.38)
Hence we see that the second term in Eq. (7.36) is identical to the Minkowski
space uctuations. The fact that
2
and
2
0
diverge should give us pause for
thought, but is not fatal to the theory. We should bear in mind that the
upper cut-o cannot be consistently taken higher than m
P
, and in any case
we do not observe the uctuations in the eld at a single point in space: we
always average over some region, giving a perfectly nite result.
To discuss the eect of these uctuations further would take us too far
aeld: we would end up having to consider eld theories with terms higher
than quadratic in the potential, for which we need to develop the machinery
of interacting eld theory.
8 Interacting elds 1
So far we have been dealing with free eld theories, which we have been able
to solve exactly. Thus we have found the equations of motion for the eld
operators, solved them in terms of a sum of operator-valued Fourier modes
(a(k), b(k) and their complex conjugates) and found all the possible quantum
states. The dynamics of a free eld theory are therefore rather trivial: one
decides which state (or superposition of states) the eld is in at some initial
time, and the eld then remains in this state for all subsequent times.
We would like to able to describe more realistic situations in which the
eld changes its state. Field theory was developed to describe the result
of scattering experiments in which a few particles head towards each other
68
t=-T t=+T
k
2
k
3
k
k
k
1
2
1
Figure 8.1: A scattering experiment. A set of particles with mo-
menta k
1
, k
2
come in from a large separation a long time in the
past, interact, and head out towards innity again in a dierent
state, where they have dierent momenta. There may also be a
dierent number of particles of various species in the nal state.
innity, interact,
from a long distance, interact, and the resulting particles (there may be more
than the incoming set) then head back to innity again (see Figure 8.1). The
goal is to calculate the quantum mechanical amplitude for the initial state
to change in to the nal state. From this we can calculate the transition
probability, which is usually expressed as a cross-section.
Unfortunately, it is almost always impossible to calculate these probabili-
ties exactly, and one must use a perturbation expansion in some small number
which parametrises the strength of the interaction a coupling constant. For
example, q, the charge of the complex scalar eld of the previous section,
is such an expansion parameter, which we must assume to be small. Most
of the technical complexity of quantum eld theory is due to the diculty
of writing this expansion in a manageable way, and in then in making sense
of the expressions once they have been written down. The breakthrough
that made eld theory calculations feasible was Feymans realisation that
the perturbation expansion could be written down in a graphical way with
Feynman diagrams. The procedure of making sense of the result, which is
formally innite, is renormalisation.
69
8.1 The interaction picture
In order to develop this expansion we must rst recall some results from ordi-
nary time-dependent perturbation theory in quantum mechanics. There are
three ways of looking at the time evolution of a state in quantum mechanics,
which are equivalent in the sense that all operators (which correspond to
observable quantities) have the same matrix elements. These ways are called
pictures.
Schrodinger picture This is probably the most familiar picture. The time
dependence of the system is carried in the states, while the operators
O
S
are time-independent. The time-dependent states [(t))
S
obey the
Schrodinger equation
i
d
dt
[(t))
S
= H[(t))
S
, (8.1)
where H is the Hamiltonian.
Heisenberg picture In this picture, it is the states [)
H
that are time-
independent, and the time evolution of the system is transferred to the
operators O
H
(t), which satisfy the Heisenberg equations
i
d
dt
O
H
(t) = [O
H
(t), H]. (8.2)
The states and operators in these two pictures can be related by a
unitary transformation e
iHt
(a unitary operator is one whose inverse
is its hermitean conjugate). Thus, if we suppose that the two sets of
operators and states coincide at t = 0, they are related at all times by
O
H
(t) = e
iHt
O
S
e
iHt
[)
H
= = e
iHt
[(t))
S
. (8.3)
This ensures that the matrix elements are the same. To see this, let us
consider the matrix elements between any two states [)
H
and [

)
H
in the Heisenberg picture, and apply these transformations:
H
[O
H
(t)[

)
H
=
S
(t)[e
iHt
e
iHt
O
S
e
iHt
e
iHt
[

(t))
S
=
S
(t)[O
S
[

(t))
S
. (8.4)
This demonstrates the point.
70
Interaction picture This picture is the one suited for situations in which
we cant solve the Schrodinger (or Heisenberg) equation exactly for our
Hamiltonian H, but we can split it into two parts, one solvable with
Hamiltonian H
0
, and the other H

which is small compared to H

0
:
H = H
0
+ H

. (8.5)
In the interaction picture, the time dependence is divided between the
operators and the states. The operators evolve according to the Heisen-
berg equations for the solvable Hamiltonian H
0
,
O
I
(t) = e
iH
0
t
O
S
e
iH
0
t
, (8.6)
which means that the states must be related to the Schrodinger picture
states by
i
d
dt
[(t))
I
= e
iH
0
t
[(t))
S
. (8.7)
Thus by dierentiating both sides with respect to time,
i
d
dt
[(t))
I
= e
iH
0
t
(H
0
+ H)[(t))
S
= (e
iH
0
t
H

e
iH
0
t
)e
iH
0
t
ket(t)
S
.
(8.8)
The piece is brackets o the right hand side is just the interaction picture
representation of the interaction Hamiltonian, which we write H

(t)
I
.
Thus
i
d
dt
[(t))
I
= H

(t)
I
[(t))
I
. (8.9)
Thus the interaction picture states evolve according to the non-trivial
part of the Hamiltonian only. In the limit H

0, the interaction
picture reduces to the Hesinberg picture.
The interaction picture state evolution equation has a formal solution
which is
[(t))
I
= U(t, t
0
)[(t
0
))
I
, (8.10)
where U(t, t
0
) is an operator which satises the equation
i

t
U(t, t
0
) = H

(t)
I
U(t, t
0
), (8.11)
with the boundary condition U(t, t
0
) = 1, the identity operator. We can
convert this into an integral equation
U(t, t
0
) = 1 i
_
t
t
0
dt
1
H

I
(t
1
)U(t
1
, t
0
), (8.12)
71
0
t
1
t
t
2
1
t
t
2
2
t > t
t < t
1
Figure 8.2: The simplication of the integration region in Equa-
tion (8.14). The orginal integration is over the darker shaded
area, but by using the symmetry of the integrand and careful
time ordering the region can be extended over the entire shaded
area.
which we solve by iteration. The starting point is to take U(t, t
0
) = 1. We
substitute this back into the integral equation, from which we obtain
U
(1)
(t, t
0
) = 1 i
_
t
t
0
dt
1
H

I
(t
1
). (8.13)
Substituting again, we get
U
(2)
(t, t
0
) = 1 i
_
t
t
0
dt
1
H

I
(t
1
)U
(1)
(t
1
, t
0
)
= 1 + (i)
2
_
t
t
0
dt
1
_
t
1
t
0
dt
2
H

I
(t
1
)H

I
(t
2
). (8.14)
We can simplify the integration region by exploiting the symmetry of the
integrand, and by using the time ordered product introduced in Section (8.4).
Note rst that we are integrating over the darker shaded area in Figure (8.2).
Let us now rewrite the second order term in Equation (8.14) as
(i)
2
2
_
t
t
0
dt
1
_
t
1
t
0
dt
2
H

I
(t
1
)H

I
(t
2
) +
(i)
2
2
_
t
t
0
dt
2
_
t
2
t
0
dt
1
H

I
(t
2
)H

I
(t
1
). (8.15)
72
All we have done here is to divide the integration into two and interchange
the integration variables in the second term. In doing so, we can see that
the second term is eectively an integration over the lighter shaded area. In
both terms the earlier time appears to the right. Thus we may write this as
(i)
2
2
_
t
t
0
dt
1
_
t
t
0
dt
2
T [H

I
(t
1
)H

I
(t
2
)] . (8.16)
It is not too hard to satsify oneself that the nth term in this iteration involves
the time ordered product of n copies of the Hamiltonian, with a symmetry
factor of 1/n!. Hence
U(t, t
0
) =

n
(i)
2
n!
_
t
t
0
dt
1
. . .
_
t
t
0
dt
n
T [H

I
(t
1
) . . . H

I
(t
n
)] . (8.17)
This can be written more compactly as
U(t, t
0
) = T
_
exp
_
i
_
t
t
0
dt

I
(t

)
__
, (8.18)
where the operation of time ordering on an exponential is dened by its
operation on the individual terms in the Taylor series.
8.2 The S-matrix and transition amplitudes
In principle, the matrix U(t, t
0
), whose formal solution we found in the pre-
vious section, completely determines the time evolution of the system. In
scattering processes, we are generally not interested in the intermediate times
when the particles are interacting, only in how the states in the distant past
change into other states in the future. We assume that interactions are lo-
calised in space and time, which amounts to assuming that H

I
(t) 0 as
t . Hence
[(t))
I
[())
I
, (8.19)
which are states of the free Hamiltonian H
0
. We dene the S-matrix to be
S = lim
t+t
0

U(t, t
0
), (8.20)
so that
[(+))
I
= S[())
I
. (8.21)
We call states [())
I
in states, and [(+))
I
out states. In order to
reduce the amount of typing, they will be denoted [i) and [f) respectively.
In a scattering experiment, we prepare the system in a state [(t))
I
at
t , and we would like to be able to compute the amplitude for a
73
quantum mechanical transition to any other state [(t)) as t +. The
probability amplitude for this transition is
A

= lim
t+
I
(t)[(t))
I
=
I
(+)[S[())
I
. (8.22)
Simplifying the notation, the probabilty amplitude for the transition from
an in state [i) to an out-state [f) is
A
fi
= f[S[i) S
fi
. (8.23)
That is, the matrix elements of the S-matrix give the transition probability
amplitudes between initial and nal states.
The S-matrix has a number of very important properties, the rst of which
is that probability must be conserved. That is, if we prepare an in-state with
unit probability, the sum of all the probabilities of the out-states must also
be unity. Thus we are assuming that i[i) = 1, and that this normalisation
is preserved throughout the evolution: that is,
I
(+)[(+))
I
= 1 also.
Hence
i[S

S[i) = 1. (8.24)
Let us insert a complete set of out-states:
1 =

f
[f)f[, (8.25)
so that

f
i[S

[f)f[S[i) = 1, (8.26)
or

f
S

if
S
fi
= 1. (8.27)
Thus the conservation of probability demands that the S-matrix be unitary
(S

S = 1).
Finally for this section it is important to point out a major problem with
this approach to dealing with interactin eld theories. The interaction picture
Hamiltonian H

I
(t) does not become negligible as t , because particles
are always interacting with themselves. The result of these self-interactions
is to shift the mass and charge of the particle by an innite amount, although
this only happens at second order in the perturbation expansion for S. These
innite shifts must be dealt with by the procedure of renormalisation.
74
8.3 Example:
4
We will consider a scalar Lagrangian densoty supplemented with a quartic
interaction term, with a strength parametrised by a constant :
L =
1
2

1
2
m
2

1
4!

4
. (8.28)
Given that the action S =
_
d
4
xL is dimensionless in Natural Units (it has
the same dimensions as h), and that the partial derivative has dimensions
of inverse length or mass M, we can immediately establish that the scalar
eld has dimension M and that the parameter is dimensionless. Parameters
such as multiplying non-quadratic monomials in eld theory lagrangians are
known as coupling constants, and they measure the strength of interactions
between particles.
From this Lagrangian density we can easily see that the Hamiltonian
density is
H =
1
2

2
+
1
2

2
+
1
2
m
2

2
+
1
4!

4
. (8.29)
A convenient decomposition into a solvable part and perturbation is to take
H
0
=
_
d
3
x
_
1
2

2
+
1
2

2
+
1
2
m
2

2
_
, H
1
=
_
d
3
x
1
4!

4
. (8.30)
Hence the S-operator in this theory is
S = T
_
exp
_
i
_
d
4
x
1
4!

4
I
(x)
__
, (8.31)
with the interaction eld operator satisfying
i

t

I
(x) = [
I
(x), H
0
]. (8.32)
From now on we will drop the subscript on the interaction picture elds, and
it will be assumed that all eld operators are in the interaction picture unless
otherwise specied.
Thus in order to nd S-matrix elements we will need to evaluate expres-
sions containing long products of eld operators in time order. We also have
the option to normal-order the interaction Hamiltonian - the advantage be-
ing that the vacuum state is left unaected by the S-operator. Finding an
algorithm to evaluate these expressions is the subject of the next few sec-
tions, which will lead towards a convenient graphical representations known
as Feynman diagrams (or graphs).
75
8.4 Feynman propagator for scalar eld
A special role is played by the vacuum expectation value of the time-ordered
product of two elds, which is termed the Feynman propagator:
i
F
(x
1
x
2
) = 0[T [(x
1
)(x
2
)] [0). (8.33)
Note that it is translational invariance which tells is that the function
F
(x
1
, x
2
)
can only depend on the dierence of its arguments. From the denition of
time ordering,
i
F
(x
1
x
2
) = 0[ ((x
1
)(x
2
)(t
1
t
2
) + (x
2
)(x
1
)(t
2
t
1
)) [0). (8.34)
Next we decompose the eld operator into positive and negative frequency
parts:
(x) =
(+)
(x) +
()
(x), (8.35)
with

(+)
(x) =
_
d
3
k
2
k
a
k
e
ikx
,
()
(x) =
_
d
3
k
2
k
a

k
e
+ikx
. (8.36)
Note that
()
(x) = (
(+)
(x))

. The positive frequency part annihilates the

vacuum, and so
i
F
(x
1
x
2
) = 0[
(+)
(x
1
)
()
(x
2
)[0)(t
1
t
2
) +0[
(+)
(x
2
)
()
(x
1
)[0)(t
2
t
1
),
= 0[[
(+)
(x
1
),
()
(x
2
)][0)(t
1
t
2
) +0[[
(+)
(x
2
)
()
(x
1
)][0)(t
2
t
1
).
(8.37)
We dene a new function

(+)
(x
1
x
2
) = i[
(+)
(x
1
),
()
(x
2
)] =
_
d
3
k
2
k
d
3
k

2
k

[a
k
, a

k
]e
ikx
1
+ik

x
2
= i
_
d
3
k
2
k
e
ik(x
1
x
2
)
, (8.38)
and its partner

()
(x
1
x
2
) = i[
()
(x
1
),
(+)
(x
2
)] = (
(+)
(x
1
x
2
))

. (8.39)
Hence
i
F
(x
1
x
2
) = i
(+)
(x
1
x
2
)(t
1
t
2
) + i
()
(x
1
x
2
)(t
2
t
1
). (8.40)
76
8.5 Wicks theorem
Wicks theorem re-expresses time-ordered products of elds in terms of normal-
ordered products (whose matrix elements are much simpler to evaluate) and
Feynman propagators.
In the following, we will use the notation
1
(x
1
) for economy. Hence

2
=
(+)
1

(+)
2
+
()
1

(+)
2
+
(+)
1

()
2
+
()
1

()
2
, (8.41)
which diers from the normal-ordered version only in the third term:
:
1

2
: =
(+)
1

(+)
2
+
()
1

(+)
2
+
()
2

(+)
1
+
()
1

()
2
. (8.42)
Hence

2
:
1

2
: = [
(+)
1
,
()
2
], (8.43)
or

2
= :
1

2
: + i
(+)
(x
1
x
2
). (8.44)
Armed with Eq. (8.44) we can rewrite the time-ordered product of two elds:
T[
1

2
] =
1

2
(t
1
t
2
) +
2

1
(t
2
t
1
),
=
_
:
1

2
: + i
(+)
(x
1
x
2
)
_
(t
1
t
2
) + (1 2). (8.45)
Given that :
1

2
: = :
2

1
:, we nd
T[
1

2
] = :
1

2
: + i
F
(x
1
x
2
) (8.46)
that is, the time ordered product of two elds is just the sum of the normal-
ordered product and the Feynman propagator, or the vacuum expectation
value of the time-ordered product of the two elds.
We will use a special notation to denote the vacuum expectation value of
the time-ordered product, the tie or contraction:
(x
1
)(x
2
) = 0[T[(x
1
)(x
2
)][0) (8.47)
Wicks theorem is the extension of Eq. (8.46) to many operators, and is
loosely stated by saying:
The time ordered product of any number of elds is the sum
of terms containing all possible contractions (including zero) with
the remaining operators in normal order.
Note that the contraction of a pair of operators already in normal order
is zero:
: (x
1
)(x
2
) : = 0. (8.48)
77
8.6 Using Wicks Theorem
As we know, transition amplitudes follow from S-matrix elements, which
can be evaluated in the Dyson-Wick expansion. If we want the transition
amplitude between states [) and [phi), the expresstion is
out
[S[)
in
=

n=0
S
(n)
(, ) (8.49)
with
S
(n)
(, ) =
(i)
n
n!
_
out
[T[:H
1
(x
1
): . . . :H
1
(x
n
):][)
in
d
4
x
1
. . . d
4
x
n
, (8.50)
where the interaction Hamiltonian for our real scalar eld theory is given by
H
1
(x) =
1
4!

4
(x). (8.51)
Hence
S
(n)
(, ) =
(i)
n
n!
_

4!
_
n _
out
[T[:
4
(x
1
): . . . :
4
(x
n
):][)
in
d
4
x
1
. . . d
4
x
n
.
(8.52)
Thus we see that the Dyson-Wick expansion can be viewed as an expansion
in powers of .
We will consider transition amplitudes between multiparticle momentum
eigenstates [k
1
, . . . , k
n
), which obey orthonormailty conditions
k

1
. . . k

m
[k
1
. . . k
n
) =
mn
2
k
1
. . . 2
kn

{p
i
}
(k
1
k

p
1
) . . . (k
n
k

pn
),
(8.53)
where p
i
is a permutation of 1 . . . n. That is, the states have zero inner
product unless the number of particles is the same and the momenta are the
same. This is easily demonstrated from the denition of the multiparticle
states in terms of creation operators a

k
i
acting on the vacuum state [0).
We will also need the following identities,
[
(+)
(x), a

k
] = e
ikx
[
(+)
(x), a
k
] = 0,
[
()
(x), a
k
] = e
ikx
[
()
(x), a

k
] = 0.
(8.54)
To be specic, we will consider the S-matrix elements for 2-particle in-states
and out-states:
[)
in
= [k
1
k
2
)
in
= a

k
1
a

k
2
[0)
[)
out
= [k

1
k

2
)
out
= a

1
a

2
[0). (8.55)
We now proceed to evaluate S
(n)
for n up to 2.
78
8.6.1 S
(0)
This is the trivial S-matrix element, as it has no powers of the coupling
constant and therefore is identical to the free eld theory:
S
(0)
= k

1
k

2
[k
1
k
2
) = 2
k
1
2
k
2
( (k
1
k

1
) (k
2
k

2
) + (k
1
k

2
) (k
2
k

1
)) .
(8.56)
The particles go straight through without interacting. Note that there are
two terms because the particles are indistinguishable.
8.6.2 S
(1)
Now we have the possibility of something interesting happening. Two parti-
cles may emerge with dierent momenta, which is the denition of scattering.
S
(1)
= i

4!
_
out
k

1
k

2
[:
4
(x
1
):[k
1
k
2
)
in
d
4
x
1
(8.57)
Let us expand the matrix element of the normal-ordered eld operator mono-
mial, dropping the in and out notation to simplify the expression. In
scattering amplitudes, this should be unambiguous as in-states are always on
the right and out-states always on the left.
k

1
k

2
[:
4
(x
1
):[k
1
k
2
) = k

1
k

2
[
_
(
(+)
1
)
4
+ 4(
()
1
)(
(+)
1
)
3
+ 6(
()
1
)
2
(
(+)
1
)
2
+ 4(
()
1
)
3
(
(+)
1
) + (
()
1
)
4
_
[k
1
k
2
).
(8.58)
Consider now the action of the positive frequency part of the eld operator
on the in-state:

(+)
(x
1
)[k
1
k
2
) = [
(+)
(x
1
), a

k
1
a

k
2
][0)
=
_
[
(+)
(x
1
), a

k
1
]a

k
2
+ a

k
2
[
(+)
(x
1
), a

k
1
]
_
[0)
= [k
2
)e
ik
1
x
1
+[k
1
)e
ik
2
x
1
. (8.59)
This can be wriiten in a tie notation:

(+)
(x
1
)[k
1
k
2
) =
(+)
(x
1
)[k
1
k
2
) +
(+)
(x
1
)[k
1
k
2
) (8.60)
Thus we learn that when a positive frequency part of a eld operator
(+)
(x)
is tied to a particle with momentum k in the in-state, both are removed and
79
replaced by a wave function factor e
ikx
. The operation is clearly repeatable,
so

(+)
(x
1
)[k
1
k
2
) =
(+)
(x
1
)
(+)
(x
2
)[k
1
k
2
) +
(+)
(x
2
)[k
1
k
2
)
=
(+)
(x
1
)
(+)
(x
2
)[k
1
k
2
) +
(+)
(x
1
)
(+)
(x
2
)[k
1
k
2
)
=
_
e
ik
1
x
2
ik
2
x
1
+ e
ik
1
x
1
ik
2
x
2
_
[0). (8.61)
By taking the Hermitian conjugate of these expressions, we learn that we
can tie negative frequency parts to out-states:
. . .k

. . . [
()
(x) = . . . [e
ikx
. (8.62)
Hence
k
1
k
2
[
()
(x
1
)
()
(x
2
) = 0[
_
e
ik

1
x
2
+ik

2
x
1
+ e
ik

1
x
1
+ik

2
x
2
_
. (8.63)
A very important point to notice is that once all particles in the in-state
have been tied to postive frequency operators, the vacuum state remains, so
the application of a further positive frequency operator annihilates the state.
Thus, in an expression where the eld operators have been expanded in terms
of their positive and negative frequency parts
the contribution to the matrix element vanishes if there are more pos-
itive frequency operators than in-state particles.
A similar statement obviously exists for the out-state:
the contribution to the matrix element vanishes if there are more neg-
ative frequency operators than out-state particles.
If there are fewer operators than particles, then once they have all been
tied o, we will be left with inner products between multiparticle states,
corresponding again to particles which travel straight through without in-
teracting. These terms will not generally be relevant for scattering.
Hence we conclude that the only term in Eq. (8.58) which does not vanish
is
k

1
k

2
[:
4
(x
1
):[k
1
k
2
) = k

1
k

2
[
_
6(
()
1
)
2
(
(+)
1
)
2
_
[k
1
k
2
),
= 60[2e
i(k

1
+k

2
)x
1
2e
i(k
1
+k
2
)x
1
[0),
= 4!e
i(k

1
+k

2
k
1
k
2
)x
1
. (8.64)
Hence
S
(1)
= i (k

1
+ k

2
k
1
k
2
). (8.65)
There are two important points to notice here:
80
1. Overall conservation of 4-momentum has appeared in the -function,
as a result of integrating the wave function factors over all space-time.
2. The 4! factor has cancelled. This happens because there are 4! ways
to tie the positive and negative frequency operators to external states
(which was of course the reason that the factor was included in the rst
place).
81
A Dirac matrices
Dirac matrices (in four space-time dimensions) are 44 matrices dened by
the anticommutation relations

= 2

1
4
, (A.1)
where

is the Minkowski metric, and 1

4
is the 44 identity matrix (which
is often dropped and left implicit in the equation).
Properties of the Dirac -matrices include:
(i). (

=
0

0
.
One can also dene another, linearly independent, -matrix

5
= i
0

3
, (A.2)
which has the easily derivable properties
(
5
)
2
= 1, (
5
)

=
0

0
,

,
5
= 0. (A.3)
The eigenvector of this matrix are known as chirality eigenvectors. Eigenvec-
tors with eigenvalue +1 are termed right-handed, with -1 left-handed. One
can dene projectors onto these eigenvectors:
P
R
=
1
2
(1 +
5
), P
L
=
1
2
(1
5
) (A.4)
A.1 Standard representation of Dirac matrices

0
=
_
1 0
0 1
_
,
i
=
_
0
i

i
0
_
,
5
=
_
0 1
1 0
_
. (A.5)
Here,
i
are the Pauli matrices, which are dened below, and 0 and 1 are the
22 zero and identity matrices respectively.
A.2 Pauli matrices
The Pauli matrices
i
(i = 1, 2, 3) are dened to be

1
=
_
0 1
1 0
_
,
2
=
_
0 i
i 0
_
,
3
=
_
1 0
0 1
_
. (A.6)
They are Hermitean, i.e.
(
i
)

=
i
, (A.7)
82
and they square to unity:
(
1
)
2
= (
2
)
2
= (
3
)
2
= 1. (A.8)
One can verify that

2
= i
3
, (and cyclic). (A.9)
Lastly, any two dierent Pauli matrices anticommute, or

2
+
2

1
= 0, (and cyclic). (A.10)
This set of relations can be neatly expressed using the 3 dimensional Levi-
Civita symbol
ijk
:
[
i
,
j
] = 2i
ijk

k
, (A.11)
and with the anticommutation relations

i
,
j
= 2
ij
. (A.12)
One may thus write the product of any two Pauli matrices as

j
=
1
2

i
,
j
+
1
2
[
i
,
j
] =
ij
+ i
ijk

k
. (A.13)
The Pauli matrices form a representation of the angular momentum al-
gebra with (spin) angular momentum
1
2
.
83
B Problem Sheets
B.1 Problem Sheet 1
1. In the following, x

is a 4-vector, and

.
(a) Show by explicit dierentation that

,
and hence that

.
Evaluate x
(b) Show that, if (x) = Ae
ip.x
, with p a constant 4-vector, then
(
2
+ m
2
)(x) = (p
2
+ m
2
)(x).
(c) Given the function f(x) = e

1
2
ax
2
, where a is a constant scalar,
and x is a 4vector coordinate, calculate

f(x) and show that

2
f(x) vanishes on the curve x
2
= 4/a.
2. (a)

, the tensor representing a Lorentz transformation on a four-

vector, is dened by the property

.
Show that if a

and b

are any two contravariant four-vectors,

then this property ensures that the scalar product a b is invariant
under a Lorentz transformation.
(b) Let (
1
)

be the inverse of

, such that (
1
)

.
Show from the transformation law x x

that

= (
1
)

,
where

.
(c) Show that
(
1
)

.
(d) Show that for a two-index tensor M

that M

and M

are not
necessarily equal (i.e. that the ordering of indices is important). In
the lectures the Kronecker delta was written

, not distinguishing
between

and

why do we not need to worry about index

ordering in this case?
84
3. (a) Let (x, t) be a wave function satisfying the Schrodinger equation
i h

t
=
h
2
2m

2
+ V (x).
Show that the probability density = [[
2
and the probability
current j = i h(

)/2m satisfy a continuity equation

+ j = 0.
(b) Let (x) be a complex-valued eld satisfying the Klein-Gordon
equation

2
+ m
2
= 0.
Show that the 4-vector current
J

=
i
2

+
i
2

.
satises a covariant continuity equation J = 0. Give a reason
why J

can not be a probability (4-)current.

4. A real scalar eld has Lagrangian density
L =
1
2

V (),
where the potential energy density V () is bounded from below.
(a) Calculate the canonical energy-momentum tensor

from this
Lagrangian density, using the formula derived in the lectures.
(b) Calculate the Hamiltonian density H, and verify that it is equal
to the energy density
00
(x).
(c) Verify, using the eld equation
2
+ V

() = 0, that the energy-
momentum tensor is conserved (satises

= 0).
85
B.2 Problem Sheet 2
1. (a) Write down the Lorentz transformation which takes the 4-vector
k

= (
k
, k
i
), where
k
= [(k
2
+ m
2
)
1
2
[, to a frame moving with
velocity v in the x direction relative to the original.
(b) By applying the Lorentz transformation of the previous part, or
otherwise, show that the momentum integration measure d
3
k/2
k
is Lorentz invariant,
2. Consider a real scalar eld whose Lagrangian density is L =
1
2

1
2
m
2

2
. Find the Hamiltonian operator

H, and show that the Heisen-
berg equations of motion

(x) =
1
i
_

(x),

H
_
,

(x) =
1
i
_
(x),

H
_
,
imply that (
2
+ m
2
)

(x) = 0.
3. The Hamiltonian for a free real scalar eld is H =
1
2
(
2
+
2
+m
2

2
).
(a) Show that
_
d
3
x m
2

2
(x) =
_
d
3
k
(2
k
)
2
m
2
_
a

k
a
k
+ a
k
a

k
+ a
k
a
k
e
2i
k
t
+ a

k
a

k
e
2i
k
t
_
.
(b) Calculate also
_
d
3
x ((x))
2
and
_
d
3
x

2
(x), and using your
results, show that
H =
_
d
3
k
(2
k
)

k
2
_
a

k
a
k
+ a
k
a

k
_
.
4. Let f
k
(x) = e
ikx
.
(a) Show that the functions f
k
(x) obey orthogonality conditions
i
_
d
3
x (f

k
(x)
0
f
k
(x)
0
f

k
(x) f
k
(x)) = 2
k

3
(k k

),
i
_
d
3
x (f
k
(x)
0
f
k
(x)
0
f
k
(x) f
k
(x)) = 0.
(b) Show that, for a real scalar eld,
a
k
= i
_
d
3
x
_
f

k
(x) (x)
0
f

k
(x)

(x)
_
,
where (x) =
0

(x) is the canonical momentum operator.

86
(c) Show that the non-trivial commutation relation for the ladder
operators is
[ a
k
, a

k
] = 2
k

3
(k k

).
5. The number operator for a real scalar eld is dened as

N =
_
d
3
k
2
k
a

k
a
k
,
where a

k
and a
k
are the ladder operators, which obey the commutation
relation of Problem 4.
(a) Show that [

N, ( a

k
)
n
] = n( a

k
)
n
.
(b) Hence, or otherwise, show that the state
[n
k
) =

k
( a

k
)
n
k

n
k
!
[0)
is an eigenstate of the number operator with eigenvalue
N = V
_
d
3
k
(2)
3
n
k
,
where V is a formal volume factor.
NB The product over the continuous variable k is dened as

k
= lim
L

m
,
where k = 2m/L, and m = (m
1
, m
2
, m
3
), with m
1
, m
2
, m
3
integers.
87
B.3 Problem Sheet 3
1. In the lectures we dened the functions

(+)
(x) = i
_
d
3
k
2
k
e
ikx
,
()
(x) = i
_
d
3
k
2
k
e
ikx
with k
0
=
k
=

(k
2
+ m
2
).
(a) Show that
(+)
(x) has the following properties:

(+)
(x

) =
(+)
(x), if x

(+)
(x) =
(+)
(x),

(+)
(x) = 0, if x
2
< 0.
Here,

is a general Lorentz transform.

(b) Find (x x

) = 0[[

(x),

(x

)][0) in terms of
(+)
and
()
.
2. Consider the one-parameter family of Lagrangians
L

=
1
4
F

2
( A)
2
.
(a) Find the eld equations for the gauge potential A

which result
from this family.
(b) Show that they all give the same eld equations in the Lorentz
gauge. What is special about the eld equations when = 1?
(c) Show that the Lagrangian
L =
1
2

gives the same eld equations as L

1
, and nd L L
1
. Why is it
that L and L
1
give the same eld equations?
88
3. (a) Show that the Lagrangian L =
1
2

gives the Lorentz

gauge eld equations for a free electromagnetic eld. Show that
the associated Hamiltonian density is
H =
1
2

1
2
A

.
(b) Given the expansion of the electromagnetic eld operator in a
plane wave basis,
A

(x) =
_
d
3
k
2
k
_
a
A
(k)
A

(k)e
ikx
+ a

A
(k)
A

(k)e
+ikx
_
,
with k
0
=
k
, show that
:H: =
_
d
3
x:H: =
_
d
3
k
2
k

AB
a

A
(k)a
B
(k).
(c) Show that if [) and [) are physical states (in the meaning of the
Gupta-Bleuler quantisation procedure) then
[:H:[) =
_
d
3
k
2
k

k
[(a

1
(k)a
1
(k) + a

2
(k)a
2
(k))[).
4. In the following, (x) is a real scalar eld operator.
(a) Show that :(x
1
)(x
2
): = :(x
2
)(x
1
): and : (x
1
)(x
2
) : = 0.
(b) Use Wicks theorem to expand T[(x
1
)(x
2
)(x
3
)(x
4
)] in terms
of contractions and normal ordered products.
(c) Expand also T[:(x
1
)(x
2
)::(x
3
)(x
4
):]
(d) How many ways are there of taking c contractions of n operators?
89
B.4 Problem Sheet 4
1. Consider the second order contribution to the S-matrix element for
2 2 scattering of real scalar particles
S
(2)
22
=
(i)
2
2
_

4!
_
2 _
d
4
x
1
d
4
x
2
k

1
k

2
[T[:
4
(x
1
)::
4
(x
2
):[k
1
k
2
)
(see e.g. Greiner & Reinhardt). Evaluate this expression and show that
there are three separate contributions iM
(2)
22
(where S
(2)
22
= iM
(2)
22

4
(k

1
+
k

2
k
1
k
2
) )
iM
(2,s)
22
=

2
2
_
d
4
p
1
(p
2
m
2
+ i)[(k
1
+ k
2
p)
2
m
2
+ i]
iM
(2,t)
22
=

2
2
_
d
4
p
1
(p
2
m
2
+ i)[(k

1
k
1
p)
2
m
2
+ i]
iM
(2,u)
22
=

2
2
_
d
4
p
1
(p
2
m
2
+ i)[(k

1
k
2
p)
2
m
2
+ i]
Write down the Feynman graphs representing these expressions.
2. Consider a real scalar eld theory with potential V () =
1
2
m
2

2
+
1
4!

4
.
A particular scattering process has two particles in the in-state, with
3-momenta k
1
, k
2
, and four particles in the out-state, with 3-momenta
k

1
and k

2
, k

3
, and k

4
.
The S-matrix element is written
S
42
= k

1
, k

2
, k

3
, k

4
[S[k
1
, k
2
).
(a) Write down the Dyson-Wick expansion for the S operator, and
show that the O(
0
) and O(
1
) terms in the corresponding ex-
pansion of the matrix element vanish.
(b) Apply Wicks theorem to T[:
4
(x
1
)::
4
(x
2
):] and identify the term
which contributes to the S-matrix element under consideration.
(c) Identify the ten Feynman graphs contributing to the S-matrix
element.
90
3. Calculate the symmetry factors for the following graphs:
(a)
(b)
(c)
(d)
(e)
91