In QFT is there a single Hilbert space or a fiber bundle of Hilbert spaces? In QFT, I understand that we have field operators $\hat \phi(\underline{x},t)$ acting on a Hilbert space $\mathcal{H}$. Operators (e.g. creation/annihilation operators) can change the state in $\mathcal{H}$ so $\hat \phi(\underline{x},t)|\psi\rangle \to|\psi'\rangle$
What I don't understand is whether there is a copy of $\mathcal{H}$ at every point in spacetime - ie a fiber bundle so that $|\psi\rangle$ describes the state at $(\underline x,t)$ or whether there is one single $\mathcal{H}$ for the whole universe. In other words, is the field an infinite collection of operators, each acting on its own $\mathcal{H}$, or a single $\mathcal{H}$ that's acted on by an infinite collection of operators?
If (as I suspect) it's more like the latter, I'm confused about what this even means, mathematically. Does $\hat \phi(\underline{x},t)|\psi\rangle$ basically mean $\hat \phi(\underline{x_0},t_0)\hat \phi(\underline{x_1},t_1)\hat \phi(\underline{x_2},t_2)...|\psi\rangle$? And if it's something like that, what does that actually mean given that this isn't actually a countable infinity so we can't apply the operators sequentially like this? Or does it mean something completely different?
 A: There is only one Hibert space.
I like to think of a QFT as just being the ordinary quantum mechnics of a system with many degrees of freedom.  For example consider a bunch of beads of mass $m$ sliding along the $x$ axis so that the $x$ coordinate of the $i$-th bead is $\eta_i$. Adjacent beads   connected by springs with energy $E=k(\eta_{i+1}-\eta_i-a)^2/2$, so the equilibrium separation is $a$.
If there are $N$ masses there are $N$ degrees of freedom.
We quantize this system, as we would any system of $N$  particles, by setting $\pi_i= m\dot \eta_i$ and setting the  commutators to  $[\eta_i, \pi_j]= i\hbar \delta_{ij}$. The   resulting   Hilbert space  is
$$
{\mathcal H}=\bigotimes_{i=1}^N L^2[{\mathbb R}_i]= L^2[\otimes_{i=1}^N {\mathbb R}_i]
$$
where $\eta_i\in {\mathbb R}_i$ is the position of the $i$-th bead. The wavefunctions are therefore $\psi(\eta_1,\ldots \eta_N)$ and the inner product is
$$
\langle\chi|\psi\rangle = \int_{{\mathbb R}^N} \chi^*(\eta_1,\ldots \eta_N)\psi(\eta_1,\ldots \eta_N) d\eta_1\cdots d\eta_N.
$$
The classical-mechanics normal modes labelled by their wavenumber $k$.  As in any "small vibrations" problem   each normal mode  can be regarded as an independent   harmonic oscillator and when we quantize the system these oscillators  are quantized. If the oscillator with frequency $\omega(k)$  is in  its $n$-the excited state the system has $n$ "phonons" of momentum $k$. The phonons are the "elementary particles" of the system and the  quantum fields are the $\eta_i$.
If you make the masses small and $a$ small (and hence $N$ large) so the mass density remains the same, you get a model of a one-dimensional elastic body.  We can relabel   $\eta_i\to\eta(x)$ where $x\equiv  ia$ labels the equilibrium position of the bead and now you have a continuum QFT.
