# Need help understanding the quantization procedure in QFT for the Klein-Gordon equation

I am having some conceptual issues with the quantization process in QFT. So we define a particle as a collection of functions $\mathcal{H}$ that satisfy some differential equation (more specifically a unitary representation of the double cover of the restricted Poincare group). Then, by construction time evolution of these states is defined by the Lie algebra element corresponding to time translations. We then use these single particle states to generate a Fock space to represent multi-particle states.

So why isn't this the end of the story? For example, quantization in QFT for the Klein-Gordon field goes further by decomposing an element of the single particle Hilbert space

$$\phi(\textbf{x},t) = \sum q_{i}(t)f_{i}(t)$$

where $\{f_{i}\}$ is a basis for $\mathcal{H}$. We then find that the coefficients $\{q_{i}(t)\}$ evolve like an infinite collection of independent (classic) harmonic oscillators. We then go on to treat these coefficients quantum mechanically, then end up with a "quantum field" $\Phi$.

So my confusion is why do we quantize the coefficients. We already have our Hilbert space and time/space translation evolution defined. More specifically why are we "quantizing" elements of the single particle Hilbert space? Technically speaking, the quantum field $\Phi$ is an operator on the Fock space and the path to its construction could in some sense be irrelevant. So it in itself is quantum mechanical and consistent with the standard quantum framework. But what makes this operator so essential in QFT? Why does this operator play a privileged role? Is there a way to define $\Phi$ via a more algebraic property by saying "$\Phi$ is the unique operator on the Fock space that satisfies..." ?

The point of view you outlined is, in my opinion, quite involved. The quantization of free theories is quite well understood. I will concentrate on bosons.

We start with a classical structure, the "phase space" or single particle structure. This structure is a symplectic one. So let $H$ be a complex Hilbert space with scalar product $(\cdot,\cdot)_c$; denote by $\Sigma(H)=(K,B(\cdot,\cdot))$ the symplectic structure where $K$ is $H$ viewed as a real Hilbert space with scalar product $(\cdot,\cdot)=\mathrm{Re}(\cdot,\cdot)_c$ and $B(\cdot,\cdot)=\mathrm{Im}(\cdot,\cdot)_c$ is a symplectic form. On finite dimensions, $\Sigma(H)$ is the phase space with the symplectic form given by the Poisson brackets.

A (Segal) quantization of $\Sigma(H)$ is a (strongly) linear map $R(\cdot)$ from $K$ to the self-adjoint operators on some Hilbert space such that:

1) $e^{iR(\cdot)}$ is weakly continuous when restricted to any finite dimensional subspace of $K$;

2) $e^{iR(z_1)}e^{iR(z_2)}=e^{-iB(z_1,z_2)/2}e^{iR(z_1+z_2)}$ for all $z_1,z_2\in K$.

$R(\cdot)$ is an operator valued distribution, the quantum field. It also preserves the correct symplectic structure of the classical space, i.e. the canonical commutation relation (in their exponentiated Weyl form).

When $H$ is finite dimensional, there is only one irreducible unitarily inequivalent quantization $R(\cdot)$ (Stone-von Neumann theorem); when $H$ is infinite dimensional they are uncountably many. One of them is the Fock representation but it is only one (very important, but just one). The problem is that for the moment dynamics has not entered into play. Dynamics would be given by the suitable quantization of a group of symplectomorphisms on $\Sigma(H)$.

With linear symplectomorphisms (free theories), we can characterize the dynamics (and also symmetries) by means of Shale's theorem. Let $T$ be a linear symplectomorphism of $\Sigma(H)$. Then there exists a unitary operator $Y(T)$ on the Fock representation such that $R(Tz)=Y(T)R(z)Y(T)^*$ if and only if $(T^*T)^{1/2}-1$ is Hilbert-Schmidt.

It remains the problem, and is still unsolved, to quantize interacting dynamics, i.e. nonlinear symplectomorphisms of $\Sigma(H)$. For example it is commonly believed that a change of Hilbert space, from the Fock one, is necessary (Haag's theorem).