1Classical field theory
III Quantum Field Theory
1.1 Classical fields
Definition
(Field)
.
A field
φ
is a physical quantity defined at every point of
spacetime (x, t). We write the value of the field at (x, t) as φ(x, t).
The “physical quantities” can be real or complex scalars, but later we will
see that we might have to consider even more complicated stuff.
In classical point mechanics, we have a finite number of generalized coordi-
nates
q
a
(
t
). In field theory, we are interested in the dynamics of
φ
a
(
x, t
), where
a
and
x
are both labels. Here the
a
labels the different fields we can have, and
(
x, t
) labels a spacetime coordinate. Note that here position has been relegated
from a dynamical variable (i.e. one of the q
a
) to a mere label.
Example.
The electric field
E
i
(
x, t
) and magnetic field
B
i
(
x, t
), for
i
= 1
,
2
,
3,
are examples of fields. These six fields can in fact be derived from 4 fields
A
µ
(x, t), for µ = 0, 1, 2, 3, where
E
i
=
∂A
i
∂t
−
∂A
0
∂x
i
, B
i
=
1
2
ε
ijk
∂A
k
∂x
j
.
Often, we write
A
µ
= (φ, A).
Just like in classical dynamics, we would specify the dynamics of a field
through a Lagrangian. For particles, the Lagrangian specifies a number at each
time
t
. Since a field is a quantity at each point in space, we now have Lagrangian
densities, which gives a number at each point in spacetime (t, x).
Definition
(Lagrangian density)
.
Given a field
φ
(
x, t
), a Lagrangian density is
a function L(φ, ∂
µ
φ) of φ and its derivative.
Note that the Lagrangian density treats space and time symmetrically. How-
ever, if we already have a favorite time axis, we can look at the “total” Lagrangian
at each time, and obtain what is known as the Lagrangian.
Definition
(Lagrangian)
.
Given a Lagrangian density, the Lagrangian is defined
by
L =
Z
d
3
x L(φ, ∂
µ
φ).
For most of the course, we will only care about the Lagrangian density, and
call it the Lagrangian.
Definition
(Action)
.
Given a Lagrangian and a time interval [
t
1
, t
2
], the action
is defined by
S =
Z
t
2
t
1
dt L(t) =
Z
d
4
x L.
In general, we would want the units to satisfy [
S
] = 0. Since we have
[d
4
x] = −4, we must have [L] = 4.
The equations of motion are, as usual, given by the principle of least action.
Definition
(Principle of least action)
.
The equation of motion of a Lagrangian
system is given by the principle of least action — we vary the field slightly,
keeping values at the boundary fixed, and require the first-order change
δS
= 0.
For a small perturbation φ
a
7→ φ
a
+ δφ
a
, we can compute
δS =
Z
d
4
x
∂L
∂φ
a
δφ
a
+
∂L
∂(∂
µ
φ
a
)
δ(∂
µ
φ
a
)
=
Z
d
4
x
∂L
∂φ
a
− ∂
µ
∂L
∂(∂
µ
φ
a
)
δφ
a
+ ∂
µ
∂L
∂(∂
µ
φ
a
)
δφ
a
.
We see that the last term vanishes for any term that decays at spatial infinity,
and obeys δφ
a
(x, t
1
) = δφ
a
(x, t
2
) = 0.
Requiring δS = 0 means that we need
Proposition
(Euler-Lagrange equations)
.
The equations of motion for a field
are given by the Euler-Lagrange equations:
∂
µ
∂L
∂(∂
µ
φ
a
)
−
∂L
∂φ
a
= 0.
We can begin by considering the simplest field one can imagine of — the
Klein–Gordon field. We will later see that this is a “free field”, and “particles”
don’t interact with each other.
Example.
The Klein–Gordon equation for a real scalar field
φ
(
x, t
) is given by
the Lagrangian
L =
1
2
∂
µ
φ∂
µ
φ −
1
2
m
2
φ
2
=
1
2
˙
φ
2
−
1
2
(∇φ)
2
−
1
2
m
2
φ
2
.
We can view this Lagrangian as saying L = T − V , where
T =
1
2
˙
φ
2
is the kinetic energy, and
V =
1
2
(∇φ)
2
+
1
2
m
2
φ
2
is the potential energy.
To find the Euler-Lagrange equation, we can compute
∂L
∂(∂
µ
φ)
= ∂
µ
φ = (
˙
φ, −∇φ)
and
∂L
∂φ
= −m
2
φ.
So the Euler-Lagrange equation says
∂
µ
∂
µ
φ + m
2
φ = 0.
More explicitly, this says
¨
φ − ∇
2
φ + m
2
φ = 0.
We could generalize this and add more terms to the Lagrangian. An obvious
generalization would be
L =
1
2
∂
µ
φ∂
µ
φ − V (φ),
where
V
(
φ
) is an arbitrary potential function. Then we similarly obtain the
equation
∂
µ
∂
µ
φ +
∂V
∂φ
= 0.
Example. Maxwell’s equations in vacuum are given by the Lagrangian
L = −
1
2
(∂
µ
A
ν
)(∂
µ
A
ν
) +
1
2
(∂
µ
A
µ
)
2
.
To find out the Euler-Lagrange equations, we need to figure out the derivatives
with respect to each component of the A field. We obtain
∂L
∂(∂
µ
A
ν
)
= −∂
µ
A
ν
+ (∂
ρ
A
ρ
)η
µν
.
So we obtain
∂
µ
∂L
∂(∂
µ
A
ν
)
= −∂
µ
∂
µ
A
ν
+ ∂
ν
(∂
ρ
A
ρ
) = −∂
µ
(∂
µ
A
ν
− ∂
ν
A
µ
).
We write
F
µν
= ∂
µ
A
ν
− ∂
ν
A
µ
.
So we are left with
∂
µ
∂L
∂(∂
µ
A
ν
)
= −∂
µ
F
µν
.
It is an exercise to check that these Euler-Lagrange equations reproduce
∂
i
E
i
= 0,
˙
E
i
= ε
ijk
∂
j
B
k
.
Using our F
µν
, we can rewrite the Lagrangian as
L = −
1
4
F
µν
F
µν
.
How did we come up with these Lagrangians? In general, it is guided by two
principles — one is symmetry, and the other is renormalizability. We will discuss
symmetries shortly, and renormalizaility would be done in the III Advanced
Quantum Field Theory course.
In all of these examples, the Lagrangian is local. In other words, the terms
don’t couple φ(x, t) to φ(y, t) if x 6= y.
Example. An example of a non-local Lagrangian would be
Z
d
3
x
Z
d
3
y φ(x)φ(x − y)
A priori, there is no reason for this, but it happens that nature seems to be
local. So we shall only consider local Lagrangians.
Note that locality does not mean that the Lagrangian at a point only depends
on the value at the point. Indeed, it also depends on the derivatives at
x
. So we
can view this as saying the value of
L
at
x
only depends on the value of
φ
at an
infinitesimal neighbourhood of x (formally, the jet at x).