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My understanding is that in classical field theory, we study a classical field $\phi(x,t)$ where for each $x\in\mathbb{R}^3$, $t\in\mathbb{R}$, $\phi(x,t)$ is a scalar. In quantum field theory, we promote each $\phi(x,t)$ to an operator $\hat{\phi}(x,t)$ on an Hilbert space.

My questions are:

  1. Given a problem, how do we know what that Hilbert space is? Is it some sort of Fock space?
  2. Does every $\hat{\phi}(x,t)$ act on the same Hilbert space?
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Quantum fields (as defined by the Wightman axioms) are operator-valued distributions. We must smear them with a test function $f$ (usually a Schwartz space function) to obtain an (in general unbounded) operator on a Hilbert space $\mathcal{H}$: $$\phi(f) := \int_{\mathbb{R}^{d+1}} f(x,t) \phi(x,t) \ \mathrm{d} x \mathrm{d} t.$$ The Hilbert space $\mathcal{H}$ is part of the data that defines a QFT model. This Hilbert space does not need to be a Fock space.

To answer your second question, the Wightman axiom W1 in the linked Wikipedia entry demands that a dense subspace $D \subset \mathcal{H}$ exists such that, for each test function $f$, the smeared quantum field $\phi(f)$ is an operator with domain $D$. Thus, the smeared quantum fields act on the same Hilbert space with common dense domain $D$.

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  • $\begingroup$ What do you mean by: "$\mathcal{H}$ is usually a Fock space"? As far as I recall, a Fock space is the infinite direct sum of direct sums of Hilbert spaces. $\endgroup$ Apr 16 at 13:01
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    $\begingroup$ A Fock space $\mathcal{F}$ is an infinite direct sum of tensor products of a (one-particle) Hilbert space $\mathfrak{h}$, i.e. $\mathcal{F} = \bigoplus_{n\in\mathbb{N}} \mathfrak{h}^{\otimes n}$, and therefore $\mathcal{F}$ is also a Hilbert space. $\endgroup$
    – Janik
    Apr 16 at 20:27
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    $\begingroup$ @Janik Could you expand on why one cannot assume that an interacting theory has an associated Fock space? $\endgroup$
    – Martin C.
    Apr 21 at 14:39
  • $\begingroup$ @MartinC. I think it's a reference to Haag's theorem. But in the Schrodinger formulation of QFT, a wavefunctional evolves in time according to Schrodinger equation. I don't see how the states of interacting theories aren't wavefunctionals. Is the Schrodinger formulation of QFT wrong somehow? $\endgroup$
    – Ryder Rude
    Apr 21 at 15:54
  • $\begingroup$ In principle, you could assume that the Hilbert space is a Fock space as all Hilbert spaces are isomorphic. However, these isomorphisms are not compatible with a Wightman QFT. For example, consider representations of the canonical commutation relations (CCRs). According to Haag's theorem infinitely many unitarily inequivalent representations of CCRs exist. One of these representations is a Fock representation, but there are infinitely many more representations which may not be a Fock representation. See also physics.stackexchange.com/q/264279 $\endgroup$
    – Janik
    Apr 22 at 16:33

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