2Free field theory
III Quantum Field Theory
2.3 Real scalar fields
We start with simple problems. We look at free theories, where the Lagrangian
is quadratic in the field, so that the equation of motion is linear. We will see that
the whole field then decomposes into many independent harmonic oscillators.
Before we jump into the quantum case, we first look at what happens in a
classical free field.
Example.
The simplest free theory is the classic Klein–Gordon theory for a
real scalar field φ(x, t). The equations of motion are
∂
µ
∂
µ
φ + m
2
φ = 0.
To see why this is free, we take the Fourier transform so that
φ(x, t) =
Z
d
3
p
(2π)
3
e
ip·x
˜
φ(p, t).
We substitute this into the Klein–Gordon equation to obtain
∂
2
∂t
2
+ (p
2
+ m
2
)
˜
φ(p, t) = 0.
This is just the usual equation for a simple harmonic oscillator for each
p
,
independently, with frequency
ω
p
=
p
p
2
+ m
2
. So the solution to the classical
Klein–Gordon equation is a superposition of simple harmonic oscillators, each
vibrating at a different frequency (and a different amplitude).
For completeness, we will note that the Hamiltonian density for this field is
given by
H =
1
2
(π
2
+ (∇φ)
2
+ m
2
φ
2
).
So to quantize the Klein–Gordon field, we just have to quantize this infinite
number of harmonic oscillators!
We are going to do this in two steps. First, we write our quantum fields
φ
(
x
)
and π(x) in terms of their Fourier transforms
φ(x) =
Z
d
3
p
(2π)
3
e
ip·x
˜
φ(p)
π(x) =
Z
d
3
p
(2π)
3
e
ip·x
˜π(p)
Confusingly,
π
represents both the conjugate momentum and the mathematical
constant, but it should be clear from the context.
If we believe in our classical analogy, then the operators
˜
φ
(
p
) and
˜π
(
p
) should
represent the position and momentum of quantum harmonic oscillators. So we
further write them as
φ(x) =
Z
d
3
p
(2π)
3
1
p
2ω
p
a
p
e
ip·x
+ a
†
p
e
−ip·x
π(x) =
Z
d
3
p
(2π)
3
(−i)
r
ω
p
2
a
p
e
ip·x
− a
†
p
e
−ip·x
,
where we have
ω
2
p
= p
2
+ m
2
.
Note that despite what we said above, we are multiplying
a
†
p
by
e
−ip·x
, and not
e
ip·x
. This is so that φ(x) will be manifestly a real quantity.
We are now going to find the commutation relations for the
a
p
and
a
†
p
.
Throughout the upcoming computations, we will frequently make use of the
following result:
Proposition. We have
Z
d
3
p
(2π)
3
e
−ip·x
= δ
3
(x).
Proposition. The canonical commutation relations of φ, π, namely
[φ(x), φ(y)] = 0
[π(x), π(y)] = 0
[φ(x), π(y)] = iδ
3
(x − y)
are equivalent to
[a
p
, a
q
] = 0
[a
†
p
, a
†
q
] = 0
[a
p
, a
†
q
] = (2π)
3
δ
3
(p − q).
Proof.
We will only prove one small part of the equivalence, as the others are
similar tedious and boring computations, and you are not going to read it anyway.
We will use the commutation relations for the
a
p
to obtain the commutation
relations for φ and π. We can compute
[φ(x), π(y)]
=
Z
d
3
p d
3
q
(2π)
6
(−i)
2
r
ω
q
ω
p
−[a
p
, a
†
q
]e
ip·x−iq·y
+ [a
†
p
, a
q
]e
−ip·x+iq·y
=
Z
d
3
p d
3
q
(2π)
6
(−i)
2
r
ω
q
ω
p
(2π)
3
−δ
3
(p − q)e
ip·x−iq·y
− δ
3
(q − p)e
−ip·x+iq·y
=
(−i)
2
Z
d
3
p
(2π)
3
−e
−ip·(x−y)
− e
ip·(y−x)
= iδ
3
(x − y).
Note that to prove the inverse direction, we have to invert the relation between
φ(x), π(x) and a
p
, a
†
p
and express a
p
and a
†
p
in terms of φ and π by using
Z
d
3
x φ(x) e
ip·x
=
1
p
2ω
p
a
−p
+ a
†
p
Z
d
3
x π(x) e
ip·x
= (−i)
r
ω
p
2
a
−p
− a
†
p
.
So our creation and annihilation operators do satisfy commutation relations
similar to the case of a simple harmonic oscillator.
The next thing to do is to express
H
in terms of
a
p
and
a
†
p
. Before we plunge
into the horrendous calculations that you are probably going to skip, it is a
good idea to stop and think what we are going to expect. If we have enough
faith, then we should expect that we are going to end up with infinitely many
decoupled harmonic oscillators. In other words, we would have
H =
Z
d
3
p
(2π)
3
(a harmonic oscillator of frequency ω
p
).
But if this were true, then we would have a serious problem. Recall that each
harmonic oscillator has a non-zero ground state energy, and this ground state
energy increases with frequency. Now that we have infinitely many harmonic
oscillators of arbitrarily high frequency, the ground state energy would add up to
infinity! What’s worse is that when we derived the ground state energy for the
harmonic oscillator, the energy is
1
2
ω
[
a, a
†
]. Now our [
a
p
, a
†
p
] is (2
π
)
3
δ
3
(
p − q
),
i.e. the value is infinite. So our ground state energy is an infinite sum of infinities!
This is so bad.
These problems indeed will occur. We will discuss how we are going to avoid
them later on, after we make ourselves actually compute the Hamiltonian.
As in the classical case, we have
H =
1
2
Z
d
3
x (π
2
+ (∇φ)
2
+ m
2
φ
2
).
For the sake of sanity, we will evaluate this piece by piece. We have
Z
d
3
x π
2
= −
Z
d
3
x d
3
p d
3
q
(2π)
6
√
ω
p
ω
q
2
(a
p
e
ip·x
− a
†
p
e
−ip·x
)(a
q
e
iq·x
− a
†
q
e
−iq·x
)
= −
Z
d
3
x d
3
p d
3
q
(2π)
6
√
ω
p
ω
q
2
(a
p
a
q
e
i(p+q)·x
− a
†
p
a
q
e
i(q−p)·x
− a
p
a
†
q
e
i(p−q)·x
+ a
†
p
a
†
q
e
−i(p+q)·x
)
= −
Z
d
3
p d
3
q
(2π)
3
√
ω
p
ω
q
2
(a
p
a
q
δ
3
(p + q) − a
†
p
a
q
δ
3
(p − q)
− a
p
a
†
q
δ
3
(p − q) + a
†
p
a
†
q
δ
3
(p + q))
=
Z
d
3
p
(2π)
3
ω
p
2
((a
†
p
a
p
+ a
p
a
†
p
) − (a
p
a
−p
+ a
†
p
a
†
−p
)).
That was tedious. We similarly compute
Z
d
3
x (∇φ)
2
=
Z
d
3
x d
3
p d
3
q
(2π)
6
1
2
√
ω
p
ω
q
(ipa
p
e
ip·x
− ipa
†
p
e
−ip·x
)
(iqa
q
e
iq·x
− iqa
†
q
e
−iq·x
)
=
Z
d
3
p
(2π)
3
p
2
2ω
p
((a
†
p
a
p
+ a
p
a
†
p
) + (a
p
a
−p
+ a
†
p
a
†
−p
))
Z
d
3
x m
2
φ
2
=
Z
d
3
p
(2π)
3
m
2
2ω
p
((a
†
p
a
p
+ a
p
a
†
p
) + (a
p
a
−p
+ a
†
p
a
†
−p
)).
Putting all these together, we have
H =
1
2
Z
d
3
x (π
2
+ (∇φ)
2
+ m
2
φ
2
)
=
1
4
Z
d
3
p
(2π)
3
−ω
p
+
p
2
ω
p
+
m
2
ω
p
(a
p
a
−p
+ a
†
p
a
†
−p
)
+
ω
p
+
p
2
ω
p
+
m
2
ω
p
(a
p
a
†
p
+ a
†
p
a
p
)
Now the first term vanishes, since we have ω
2
p
= p
2
+ m
2
. So we are left with
=
1
4
Z
d
3
p
(2π)
3
1
ω
p
(ω
2
p
+ p
2
+ m
2
)(a
p
a
†
p
+ a
†
p
a
p
)
=
1
2
Z
d
3
p
(2π)
3
ω
p
(a
p
a
†
p
+ a
†
p
a
p
)
=
Z
d
3
p
(2π)
3
ω
p
a
†
p
a
p
+
1
2
[a
p
, a
†
p
]
=
Z
d
3
p
(2π)
3
ω
p
a
†
p
a
p
+
1
2
(2π)
3
δ
3
(0)
.
Note that if we cover the
R
d
3
p
(2π)
3
, then the final three lines are exactly what we
got for a simple harmonic oscillator of frequency ω
p
=
p
p
2
+ m
2
.
Following the simple harmonic oscillator, we postulate that we have a vacuum
state |0i such that
a
p
|0i = 0
for all p.
When
H
acts on this, the
a
†
p
a
p
terms all vanish. So the energy of this ground
state comes from the second term only, and we have
H |0i =
1
2
Z
d
3
p
(2π)
3
ω
p
(2π)
3
δ
3
(0) |0i = ∞|0i.
since all ω
p
are non-negative.
Quantum field theory is always full of these infinities! But they tell us
something important. Often, they tell us that we are asking a stupid question.
Let’s take a minute to explore this infinity and see what it means. This is
bad, since what we want to do at the end is to compute some actual probabilities
in real life, and infinities aren’t exactly easy to calculate with.
Let’s analyze the infinities one by one. The first thing we want to tackle is
the δ
3
(0). Recall that our δ
3
can be thought of as
(2π)
3
δ
3
(p) =
Z
d
3
x e
ip·x
.
When we evaluate this at
p
=
0
, we are then integrating the number 1 over all
space, and thus the result is infinite! You might say, duh, space is so big. If
there is some energy everywhere, then of course the total energy is infinite. This
problem is known as infrared divergence. So the idea is to look at the energy
density, i.e. the energy per unit volume. Heuristically, since the (2
π
)
3
δ
3
(
p
) is
just measuring the “volume” of the universe, we would get rid of it by simply
throwing away the factor of (2π)
3
δ
3
(0).
If we want to make this a bit more sophisticated, we would enclose our
universe in a box, and thus we can replace the (2
π
)
3
δ
3
(
0
) with the volume
V
.
This trick is known as an infrared cutoff . Then we have
E
0
=
E
V
=
Z
d
3
p
(2π)
3
1
2
ω
p
.
We can then safely take the limit as
V → ∞
, and forget about this
δ
3
(
0
) problem.
This is still infinite, since
ω
p
gets unbounded as
p → ∞
. In other words, the
ground state energies for each simple harmonic oscillator add up to infinity.
These are high frequency divergences at short distances. These are called
ultraviolet divergences. Fortunately, our quantum field theorists are humble
beings and believe their theories are wrong! This will be a recurring theme —
we will only assume that our theories are low-energy approximations of the real
world. While this might seem pessimistic, it is practically a sensible thing to do
— our experiments can only access low-level properties of the universe, so what
we really see is the low-energy approximation of the real theory.
Under this assumption, we would want to cut off the integral at high mo-
mentum in some way. In other words, we just arbitrarily put a bound on the
integral over
p
, instead of integrating over all possible
p
. While the cut-off point
is arbitrary, it doesn’t really matter. In (non-gravitational) physics, we only care
about energy differences, and picking a different cut-off point would just add a
constant energy to everything. Alternatively, we can also do something slightly
more sophisticated to avoid this arbitrariness, as we will see in the example of
the Casimir effect soon.
Even more straightforwardly, if we just care about energy differences, we can
just forget about the infinite term, and write
H =
Z
d
3
p
(2π)
3
ω
p
a
†
p
a
p
.
Then we have
H |0i = 0.
While all these infinity cancelling sound like bonkers, the theory actually fits
experimental data very well. So we have to live with it.
The difference between this
H
and the previous one with infinities is an
ordering ambiguity in going from the classical to the quantum theory. Recall
that we did the quantization by replacing the terms in the classical Hamiltonian
with operators. However, terms in the classical Hamiltonian are commutative,
but not in the quantum theory. So if we write the classical Hamiltonian in a
different way, we get a different quantized theory. Indeed, if we initially wrote
H =
1
2
(ωq − ip)(ωq + ip),
for the classical Hamiltonian for a single harmonic operator, and then did the
quantization, we would have obtained
H = ωa
†
a.
It is convenient to have the following notion:
Definition (Normal order). Given a string of operators
φ
1
(x
1
) ···φ
n
(x
n
),
the normal order is what you obtain when you put all the annihilation operators
to the right of (i.e. acting before) all the creation operators. This is written as
:φ
1
(x
1
) ···φ
n
(x
n
): .
So, for example,
:H: =
Z
d
3
p
(2π)
3
ω
p
a
†
p
a
p
.
In the future, we will assume that when we quantize our theory, we do so in a
way that the resulting operators are in normal order.
Applications — The Casimir effect
Notice that we happily set
E
0
= 0, claiming that only energy differences are mea-
sured. But there exists a situation where differences in the vacuum fluctuations
themselves can be measured, when we have two separated regions of vacuum. In
this example, we will also demonstrate how the infrared and ultraviolet cutoffs
can be achieved. We are going to enclose the world in a box of length
L
, and
then do the ultraviolet cutoff in a way parametrized by a constant
a
. We then
do some computations under these cutoffs, derive an answer, and then take the
limit as
L → ∞
and
a →
0. If we asked the right question, then the answer will
tend towards a finite limit as we take these limits.
To regulate the infrared divergences, we put the universe in a box again. We
make the
x
direction periodic, with a large period
L
. So this is a one-dimensional
box. We impose periodic boundary conditions
φ(x, y, z) = φ(x + L, y, z).
We are now going to put two reflecting plates in the box at some distance
d L
apart. The plates impose φ(x) = 0 on the plates.
d
L
The presence of the plates means that the momentum of the field inside them is
quantized
p =
πn
d
, p
y
, p
z
for
n ∈ Z
. For a massless scalar field, the energy per unit area between the
plates is
E(d) =
∞
X
n=1
Z
dp
y
dp
z
(2π)
2
1
2
r
πn
d
2
+ p
2
y
+ p
2
z
.
The energy outside the plates is then E(L − d). The total energy is then
E = E(d) + E(L − d).
This energy (at least naively) depends on
d
. So there is a force between the
plates! This is the Casimir effect, predicted in 1945, and observed in 1958. In
the lab, this was done with the EM field, and the plates impose the boundary
conditions.
Note that as before, we had to neglect modes with
|p|
too high. More precisely,
we pick some distance scale
a d
, and ignore modes where
|p| a
−1
. This is
known as the ultraviolet cut-off. This is reasonable, since for high momentum
modulus, we would break through the plates. Then we have
E(d) = A
X
n
Z
dp
y
dp
z
(2π)
2
1
2
|p|e
−a|p|
.
Note that as we set a → 0, we get the previous expression.
Since we are scared by this integral, we consider the special case where we
live in the 1 + 1 dimensional world. Then this becomes
E
1+1
(d) =
π
2d
∞
X
n=1
ne
−anπ/d
= −
1
2
d
da
X
n
e
−anπ/d
!
= −
1
2
d
da
1
1 − e
−aπ/d
=
π
2d
e
−aπ/d
(1 − e
−aπ/d
)
2
=
d
2πa
2
−
π
24d
+ O(a
2
).
Our total energy is
E = E(d) + E(L − d) =
L
2πa
2
−
π
24
1
d
+
1
L − d
+ O(a
2
).
As
a →
0, this is still infinite, but the infinite term does not depend on
d
. The
force itself is just
∂E
∂d
=
2π
24d
2
+ O
d
2
L
2
+ O(a
2
),
which is finite as
a →
0 and
L → ∞
. So as we remove both the infrared and UV
cutoffs, we still get a sensible finite force.
In 3 + 1 dimensions, if we were to do the more complicated integral, it turns
out we get
1
A
∂E
∂d
=
π
2
480d
4
.
The actual Casimir effect for electromagnetic fields is actually double this due
to the two polarization states of the photon.
Recovering particles
We called the operators
a
p
and
a
†
p
the creation and annihilation operators.
Let us verify that they actually create and annihilate things! Recall that the
Hamiltonian is given by
H =
1
2
Z
d
3
q
(2π)
3
ω
q
a
†
q
a
q
.
Then we can compute
[H, a
†
p
] =
Z
d
3
q
(2π)
3
ω
q
[a
†
q
a
q
, a
†
p
]
=
Z
d
3
q
(2π)
3
ω
q
a
†
q
(2π)
3
δ
3
(p − q)
= ω
p
a
†
p
.
Similarly, we obtain
[H, a
p
] = −ω
p
a
p
.
which means (like SHM) we can construct energy eigenstates by acting with
a
†
p
.
We let
|pi = a
†
p
|0i.
Then we have
H |pi = ω
p
|pi,
where the eigenvalue is
ω
2
p
= p
2
+ m
2
.
But from special relativity, we also know that the energy of a particle of mass
m
and momentum p is given by
E
2
p
= p
2
+ m
2
.
So we interpret
|pi
as the momentum eigenstate of a particle of mass
m
and
momentum
p
. And we identify
m
with the mass of the quantized particle. From
now on, we will write E
p
instead of ω
p
.
Let’s check this interpretation. After normal ordering, we have
P =
Z
π(x)∇φ(x) d
3
x =
Z
d
3
p
(2π)
3
pa
†
p
a
p
.
So we have
P |pi =
Z
d
3
q
(2π)
3
qa
†
q
a
q
(a
†
p
|0i)
=
Z
d
3
q
(2π)
3
qa
†
q
(a
†
p
a
q
+ (2π)
3
δ
3
(p − q)) |0i
= pa
†
p
|0i
= p |pi.
So the state has total momentum p.
What about multi-particle states? We just have to act with more
a
†
p
’s. We
have the n-particle state
|p
1
, p
2
, ··· , p
n
i = a
†
p
1
a
†
p
2
···a
†
p
n
|0i.
Note that
|p, qi
=
|q, pi
for any
p, q
. So any two parts are symmetric under
interchange, i.e. they are bosons.
Now we can tell what our state space is. It is given by the span of particles
of the form
|0i, a
†
p
|0i, a
†
p
a
†
q
|0i, ···
This is known as the Fock space. As in the case of SHM, there is also an operator
which counts the number of particles. It is given by
N =
Z
d
3
p
(2π)
3
a
†
p
a
p
.
So we have
N |p
1
, ··· , p
n
i = n |p
1
, ··· , p
n
i.
It is easy to compute that
[N, H] = 0
So the particle number is conserved in the free theory. This is not true in general!
Usually, when particles are allowed to interact, the interactions may create or
destroy particles. It is only because we are in a free theory that we have particle
number conservation.
Although we are calling these states “particles”, they aren’t localized —
they’re momentum eigenstates. Theoretically, we can create a localized state via
a Fourier transform:
|xi =
Z
d
3
p
(2π)
3
e
−ip·x
|pi.
More generally, we can create a wave-packet, and insert ψ(p) to get
|ψi =
Z
d
3
p
(2π)
3
e
−ip·x
ψ(p) |pi.
Then this wave-packet can be both partially localized in space and in momentum.
For example, we can take ψ to be the Gaussian
ψ(p) = e
−p
2
/2m
.
Note now that neither
|xi
nor
|ψi
are
H
-eigenstates, just like in non-relativistic
quantum mechanics.
Relativistic normalization
As we did our quantum field theory, we have completely forgotten about making
our theory Lorentz invariant. Now let’s try to recover some. We had a vacuum
state
|0i
, which we can reasonably assume to be Lorentz invariant. We then
produced some 1-particle states
|pi
=
a
†
p
|0i
. This is certainly not a “scalar”
quantity, so it would probably transform as we change coordinates.
However, we can still impose compatibility with relativity, by requiring that
its norm is Lorentz invariant. We have chosen |0i such that
h0|0i = 1.
We can compute
hp|qi = h0|a
p
a
†
q
|0i = h0|a
†
q
a
p
+ (2π)
3
δ
3
(p − q) |0i = (2π)
3
δ
3
(p − q).
There is absolutely no reason to believe that this is a Lorentz invariant quantity,
as p and q are 3-vectors. Indeed, it isn’t.
As the resulting quantity is a scalar, we might hope that we can come up
with new states
|pi = A
p
|pi
so that
hp|qi = (2π)
3
A
∗
p
A
q
δ
3
(p − q)
is a Lorentz invariant quantity. Note that we write non-bold characters for
“relativistic” things, and bold characters for “non-relativistic” things. Thus, we
think of the p in |pi as the 4-vector p, and the p in |pi as the 3-vector p.
The trick to figuring out the right normalization is by looking at an object
we know is Lorentz invariant, and factoring it into small pieces. If we know that
all but one of the pieces are Lorentz invariant, then the remaining piece must be
Lorentz invariant as well.
Our first example would be the following:
Proposition. The expression
Z
d
3
p
2E
p
is Lorentz-invariant, where
E
2
p
= p
2
+ m
2
for some fixed m.
Proof. We know
R
d
4
p certainly is Lorentz invariant, and
m
2
= p
µ
p
µ
= p
2
= p
2
0
− p
2
is also a Lorentz-invariant quantity. So for any m, the expression
Z
d
4
p δ(p
2
0
− p
2
− m
2
)
is also Lorentz invariant. Writing
E
2
p
= p
2
0
= p
2
+ m
2
,
integrating over p
0
in the integral gives
Z
d
3
p
2p
0
=
Z
d
3
p
2E
p
,
and this is Lorentz invariant.
Thus, we have
Proposition. The expression
2E
p
δ
3
(p − q)
is Lorentz invariant.
Proof. We have
Z
d
3
p
2E
p
· (2E
p
δ
3
(p − q)) = 1.
Since the RHS is Lorentz invariant, and the measure is Lorentz invariant, we
know 2E
p
δ
3
(p − q) must be Lorentz invariant.
From this, we learn that the correctly normalized states are
|pi =
p
2E
p
|pi =
p
2E
p
a
†
p
|0i.
These new states satisfy
hp|qi = (2π)
3
(2E
p
)δ
3
(p − q),
and is Lorentz invariant.
We can now define relativistically normalized creation operators by
a
†
(p) =
p
2E
p
a
†
p
.
Then we can write our field as
φ(x) =
Z
d
3
p
(2π)
3
1
2E
p
(a(p)e
ip·x
+ a
†
(p)e
−ip·x
).