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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Charlie
    Apr 30, 2021 at 12:02
  • $\begingroup$ Indeed - but there's then an operator at every one of those infinitely many points in spacetime $\endgroup$
    – HenryH
    Apr 30, 2021 at 12:24
  • $\begingroup$ 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 $f$. $\endgroup$
    – Blazej
    Apr 30, 2021 at 12:27
  • $\begingroup$ 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? $\endgroup$
    – jacob1729
    Apr 30, 2021 at 12:31
  • $\begingroup$ @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. $\endgroup$
    – Charlie
    Apr 30, 2021 at 12:33

1 Answer 1


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.


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