Relation between the Fourier transform and the Fock Space

Question

First consider the classical Klein-Gordon field. The equation is $(\Box +m^2)\phi = 0$ which upon using the Fourier transform becomes (here I denote the Fourier transform with the $\hat{}$ as usual.

$$(\partial_t^2+\omega_{k}^2)\hat{\phi}=0$$

this equation has solution (with the condition of reality imposed)

$$\hat{\phi}(\mathbf{k},t)=a_{\mathbf{k}}e^{-i\omega_k t}+a_\mathbf{k}^\ast e^{i\omega t}$$

which in turns end up giving the general solution

$$\phi(\mathbf{x},t)=\int \dfrac{d^3 \mathbf{k}}{(2\pi)^3\sqrt{2\omega_k}}(a_\mathbf{k}e^{-ik_\mu x^\mu}+a_{-\mathbf{k}}^\ast e^{k_\mu x^\mu})$$

There is nothing fancy here. It is the standard method for solving a differential equation with Fourier transform.

Now, when quantizing the field, the natural way to lift this expression is

$$\phi(\mathbf{x},t)=\int \dfrac{d^3 \mathbf{k}}{(2\pi)^3\sqrt{2\omega_k}}(a_\mathbf{k}e^{-ik_\mu x^\mu}+a_{-\mathbf{k}}^\dagger e^{k_\mu x^\mu})$$

which is again just the Fourier transform of operators.

Now comes my question: in one way we can deduce from this that the canonical commutation relations for $\phi$ are equivalent to the commutation relations for the $a_\mathbf{k}$ being $[a_{\mathbf{k}},a_{\mathbf{k'}}^\dagger]=(2\pi)^3\delta(\mathbf{k}-\mathbf{k}')$ and this is quite straightforward.

The problem I have is: one can then just present a way to build all this, which is the Fock space. If one defines the Fock space it comes naturaly with a pair of operators $a$ and $a^\dagger$ which obeys exactly that commutation relations and can be used to define the fields.

In this approach one knows beforehand the solution - use the Fock space - and just works with it. The operators of creation and annihilation as well as the Fock space are considered to be built first, and then the quantum fields are defined.

What I want to know is how does one arrive at this conclusion, to use the Fock space. It seems that always when one expand one operator in a Fouirer transform, the coefficients of the Fourier transform are creation and annihilation operators in a Fock space. I've even saying some people saying that "it is obvious from the expansion $\phi(x,t)$ that the Fourier coefficients are creation and annihilation operators in a Fock space".

What is thus really the relation between Fourier transform and Fock space? Why "it is obvious" that when expanding one operator into Fourier modes, the Fourier coefficients are creation/annihilation operators in a certain Fock space? How does one arrive at the Fock space by the Fourier transform?

Answer 1

This was (is) one of my biggest bugbears whilst learning QFT. The reasoning behind using a Fock space is actually really simple and intuitive for a scalar field (provided you are comfortable with standard QM) but it's always masked by the horrible concept of 'canonical quantisation'.

Take the Fourier transform of the Klein-Gordon equation:

$$ \partial_t^2\hat{\phi} = -\omega_{k}^2\hat{\phi} $$

this is the classical equation of motion for a harmonic oscillator. There's one of these for every possible momentum (this corresponds to $\phi$ being a field over space transforming into $\hat{\phi}$ which is a field over momentum).

It's harder to transform the Lagrangian (since it includes terms like $\left(\frac{\partial\phi}{\partial t}\right)^2$), but one can assume it similarly describes $\hat{\phi}$ as an infinite spectrum of harmonic oscillators. Given this, it's reasonable to assume that the quantum Lagrangian/Hamiltonian for $\hat{\phi}$ similarly corresponds to an infinite spectrum of quantum harmonic oscillators. This we do know the form of:

$$ H = \int \frac{d^{3}p}{2E_{p}}a_{p}^{\dagger}a_{p}, $$

where I've dropped the zero point energy because you can (it corresponds to reordering the fields, which is an ambiguity that exists in going to the non-commuting quantum case from the commuting classical case) and it avoids the usual infinite-energy problems.

Now we just claim that $\phi$ has the same quantum Lagrangian as in the classical case and that the above Hamiltonian is the Fourier transform (of the Legendre transform) of the Lagrangian. If you work through you find that you get out the canonical form of the field. David Tong does this on page 24 of his notes, though he does it by essentially proposing the canonical form as an ansatz.

Then you just use your infinite set of annihilation and creation operators that arose naturally to generate the infinite set of QHO number states (one for each momentum). This is identical to the Fock space generated by the momentum operator, so you just treat it as a Fock space.

Answer 2

The Fourier transform of the field yields to creation/annihilation operators which create a particle with a given momentum. You can for example also define creation/annihilation operators in position space simply via $$a_x = \frac{1}{\sqrt{2}}(\phi(x) + i \pi(x))\text{ and }a_x^\dagger = \frac{1}{\sqrt{2}}(\phi(x) - i \pi(x))$$ that fulfill $[a_x,a_y^\dagger] = \delta(x-y)$ (you can scale them by $\sqrt{2\pi}$ if you want).

Now you can write $$\phi(x) = \int \mathrm d^3x\,u_xa_x + u_x^*a_x^\dagger$$ where the $u_x$ are the solutions that are a delta peak at $t=0$.

Then you can basically Fourier transform the field $\phi(x)$ and obtain the creation/annihilation operators for the momentum modes.