2Free field theory

III Quantum Field Theory 2.1 Review of simple harmonic oscillator
Michael Peskin once famously said “Physics is the subset of human experience
that can be reduced to coupled harmonic oscillators”. Thus, to understand
quantum mechanics, it is important to understand how the quantum harmonic
oscillator works.
Classically, the simple harmonic oscillator is given by the Hamiltonian
H =
1
2
p
2
+
1
2
ω
2
q
2
,
where
p
is the momentum and
q
is the position. To obtain the corresponding
quantum system, canonical quantization tells us that we should promote the
p
and
q
into complex “operators”
ˆp, ˆq
, and use the same formula for the Hamiltonian,
namely we now have
ˆ
H =
1
2
ˆp
2
+
1
2
ω
2
ˆq
2
.
In the classical system, the quantities
p
and
q
used to satisfy the Poisson brackets
{q, p} = 1.
After promoting to operators, they satisfy the commutation relation
[ˆq, ˆp] = i.
We will soon stop writing the hats because we are lazy.
There are a few things to take note of.
(i)
We said
p
and
q
are “operators”, but did not say what they actually
operate on! Instead, what we are going to do is to analyze these operators
formally, and after some careful analysis, we show that there is a space the
operators naturally act on, and then take that as our state space. This is,
in general, how we are going to do quantum field theory (except we tend
to replace the word “careful” with “sloppy”).
During the analysis, we will suppose there are indeed some states our
operators act on, and then try to figure out what properties the states
must have.
(ii)
The process of canonical quantization depends not only on the classical
system itself, but how we decide the present our system. There is no
immediate reason why if we pick different coordinates for our classical
system, the resulting quantum system would be equivalent. In particular,
before quantization, all the terms commute, but after quantization that is
no longer true. So how we decide to order the terms in the Hamiltonian
matters.
Later, we will come up with the notion of normal ordering. From then
onwards, we can have a (slightly) more consistent way of quantizing
operators.
After we’ve done this, the time evolution of states is governed by the
Schr¨odinger equation:
i
d
dt
|ψi = H |ψi.
In practice, instead of trying to solve this thing, we want to find eigenstates
|Ei
such that
H |Ei = E |Ei.
If such states are found, then defining
|ψi = e
iEt
|Ei
would give a nice, stable solution to the Schr¨odinger equation.
The trick is to notice that in the classical case, we can factorize the Hamilto-
nian as
H = ω
r
ω
2
q +
i
2ω
p
r
ω
2
q +
i
2ω
p
.
Now
H
is a product of two terms that are complex conjugates to each other,
which, in operator terms, means they are adjoints. So we have the benefit
that we only have to deal with a single complex object
p
ω
2
q
+
i
2ω
p
(and its
conjugate), rather than two unrelated real objects. Also, products are nicer than
sums. (if this justification for why this is a good idea doesn’t sound convincing,
just suppose that we had the idea of doing this via divine inspiration, and it
turns out to work well)
We now do the same factorization in the quantum case. We would not
expect the result to be exactly the above, since that working relies on
p
and
q
commuting. However, we can still try and define the operators
a =
i
2ω
p +
r
ω
2
q, a
=
i
2ω
p +
r
ω
2
q.
These are known as creation and annihilation operators for reasons that will
become clear soon.
We can invert these to find
q =
1
2ω
(a + a
), p = i
r
ω
2
(a a
).
We can substitute these equations into the commutator relation to obtain
[a, a
] = 1.
Putting them into the Hamiltonian, we obtain
H =
1
2
ω(aa
+ a
a) = ω
a
a +
1
2
[a, a
]
= ω
a
a +
1
2
.
We can now compute
[H, a
] = ωa
, [H, a] = ωa.
These ensure that a, a
take us between energy eigenstates if
H |Ei = E |Ei,
then
Ha
|Ei = (a
H + [H, a
]) |Ei = (E + ω)a
|Ei.
Similarly, we have
Ha |Ei = (E ω)a |Ei.
So assuming we have some energy eigenstate
|Ei
, these operators give us loads
more with eigenvalues
··· , E 2ω, E ω, E, E + ω, E + 2ω, ··· .
If the energy is bounded below, then there must be a ground state
|0i
satisfying
a |0i
= 0. Then the other “excited states” would come from repeated applications
of a
, labelled by
|ni = (a
)
n
|0i,
with
H |ni =
n +
1
2
ω |ni.
Note that we were lazy and ignored normalization, so we have hn|ni 6= 1.
One important feature is that the ground state energy is non-zero. Indeed,
we have
H |0i = ω
a
a +
1
2
|0i =
ω
2
|0i.
Notice that we managed to figure out what the eigenvalues of
H
must be, without
having a particular state space (assuming the state space is non-trivial). Now we
know what is the appropriate space to work in. The right space is the Hilbert
space generated by the orthonormal basis
{|0i, |1i, |2i, ···}.