(Un)countability in QFT

Question

I am a mathematician self-studying physics, and a currently working on QFT with Srednicki's book.

One thing that bothers me is that for a scalar field (in the Hamiltonian version) there is a harmonic oscillator at every point of momentum space. These all commute with each other, which means that they can be observed simultaneously. This is an uncountable number of harmonic oscillators.

If a classical field is nonzero, then it is nonzero at uncountably many places. So it seems to me that a generic measurement of the field will yield uncountably many points with non-zero energy, corresponding to uncountably many bosons.

However, QFT seems to treat a field as having a countable number of particles. Each Feynmann diagram only involves a finite number of particles, and the LSZ reduction formula deals with a finite number of incoming/outgoing particles.

So my question is, Will a generic field state have countably many particles or uncountably many particles, and how does this reduce to a classical field in the classical limit?

Answer 1

The uncountable number of harmonic oscillators represent "opportunities", one-particle states into which new particles may be created.

But a generic field state has not only a "countable" number of particles but a finite number of particles. That's necessary at least when the particle is massive or charged, otherwise the total energy or charge would be infinite. We only want to consider states that are a finite energy or charge above the ground state.

For massless and neutral particles such as photons, the number of particles may be formally divergent due to loop effects and divergences but this result has nothing to do with the observation of the simple uncountability of one-particle states you have made.

I think that it is helpful for you to understand the relationship between the field theory's Fock space and the direct sum of $n$-particle Hilbert spaces for all integers $n$ – because if you understand there is no problem in the latter, you will see that there is no problem in the former because the Hilbert spaces are isomorphic.

Answer 2

A generic state of QFT has an unfixed number of particles, that are at most countable.

This is because the Fock space is constructed as a direct sum of the subspaces with any possible number of particles, from zero to infinity. Mathematicallly, let $\mathscr{H}_1$ be the one particle space, $\mathscr{H}_n$ the (symmetric or anti-symmetric) $n$-fold tensor product of $\mathscr{H}_1$, and $\mathscr{H}_0=\mathbb{C}$ the vacuum space (no particles). Then the (symmetric, just to fix the ideas) Fock space over $\mathscr{H}_1$ is $\Gamma_s(\mathscr{H}_1)=\bigoplus_{n=0}^\infty \mathscr{H}_n$.

Thus a state $\Gamma_s(\mathscr{H}_1)\ni\Psi=(\Psi_0,\Psi_1,\dotsc,\Psi_n,\dotsc)$ in general has non-zero $n$-particle components for all $n\in\mathbb{N}$, with the constraint that the norm $\lVert\Psi\rVert^2=\sum_{n=0}^\infty \lVert \Psi_n\rVert^2_{\mathscr{H}_n}$ is finite. The annihilation and creation operators let you jump from $\mathscr{H}_n$ and $\mathscr{H}_{n\pm 1}$ and are formally defined, usually, as operator-valued distributions that make sense when acting on functions in the one-particle space $\mathscr{H}_1$.

Suppose $\mathscr{H}_1=L^2(\mathbb{R}^d)$, and let $a^\#(k)$ be the annihilation/creation operators, and $\phi(k)=\frac{1}{\sqrt{2}}(a^*(k)+a(k))$ be the field operator (again that make sense only when acting on $f\in L^2(\mathbb{R}^d)$). Then the average of $\langle\sqrt{\hslash} \phi(k)\rangle$ over suitable $\hslash$-dependent states (e.g. coherent states) converges in the limit $\hslash\to 0$ to the corresponding classical field $\varphi(k)\in L^2(\mathbb{R}^d)$. This can be rigorously proved in some (sufficiently regular) system.