# 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?

• The operator field isn't a function of infinitely many variables, in $D$ spacetime dimensions the "function" $\phi(\vec x,t)$ is $\phi(x_1,x_2,...,x_{D-1},t)$ and gives you the operator at that point in spacetime. Apr 30, 2021 at 12:02
• Indeed - but there's then an operator at every one of those infinitely many points in spacetime Apr 30, 2021 at 12:24
• Each $\phi(x,t)$ is acting in the same Hilbert space, so you can think of them as an infinite set of operators labeled by points $(x,t)$. Caveat: this is not quite accurate, $\phi(x,t)$ is typically not a well-defined operator, one has to first smear it, e.g. form $\phi(f) = \int \phi(x,t) f(x,t) dx dt$ for some smooth rapidly decaying function $f$. Apr 30, 2021 at 12:27
• What do you mean by the product $\hat{\phi}(x_0,t)\dots|\psi\rangle$ in your post? What are the points $x_0,\dots$ meant to be? Apr 30, 2021 at 12:31
• @HenryH Sure, so $\phi(\vec x,t)|\psi\rangle=\phi(x_0,...,x_{D-1},t)|\psi\rangle$ is just one operator (the one at that point) acting on a vector in the Fock space. Apr 30, 2021 at 12:33

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.