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Let there be a a quantum field operator $\hat\phi(t,\vec{x})$ which, because it acts (pointwise) on a separable Hilbert space, I expect I can write as $$\hat\phi(t,\vec{x}) = \sum_n\sum_m\phi^n_m(t,\vec{x})\,|n\rangle\langle m|.$$

I am told in Quantum Field theory that I can define the Fourier Transform of $\hat\phi(t,\vec{x})$, which I will call $\hat{\tilde{\phi}}(t,\vec{p})$, by doing

$$\hat\phi(t,\vec{x}) = \sum_{n,m}\phi^n_m(t,\vec{x})\,|n\rangle\langle m| = \sum_{n,m}\left(\iiint_{\mathbb{R}^3}\mathrm{e}^{\mathrm{i}\vec{p}\cdot\vec{x}}\tilde\phi{}^n_m(t,\vec{p})\,\mathrm{d}^3p\right)|n\rangle\langle m| = \iiint_{\mathbb{R}^3}\mathrm{e}^{\mathrm{i}\vec{p}\cdot\vec{x}}\underbrace{\sum_{n,m}\tilde\phi{}^n_m(t,\vec{p})\,|n\rangle\langle m|}_{\hat{\tilde\phi}(t,\vec{p})}\,\mathrm{d}^3p.$$ up to some factor that fixes the units of the differential and the $(2\pi)^3$ that comes with FTs.

Is this actually rigorous? Does the expression in components for $\hat{\tilde\phi}(t,\vec{p})$ always converge by virtue of some theorem (Paserval's, Riezs' or some other one, I tried but didn't succeed) or am I making any additional assumption when I do this?

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    $\begingroup$ I'm not sure why this question is being downvoted. OP is asking a valid question, even though their assumption is wrong (as is explained in mike stone's answer), it is a reasonable thing to assume and ask here about. $\endgroup$ Commented Oct 14, 2021 at 14:55

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As with all quantum fields you need to remember that they are operator-valued distributions. So convergence is only ever in the sense of distributions.

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  • $\begingroup$ Any clue, then, on how would I be able to prove that? Functions or distributions, I have no starting point :( $\endgroup$
    – Pablo T.
    Commented Oct 14, 2021 at 12:03
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    $\begingroup$ Use the Schwartz class as test functions. Then the unitary isometry is guaranteed by the way the Forier transform works with that class: whatever properties the eperator has in $x$ space it has in $p$ space. $\endgroup$
    – mike stone
    Commented Oct 14, 2021 at 12:30
  • $\begingroup$ I find it strange, nonetheless, that these $\phi^n_m(t,\vec{x})$ are distributions and not proper functions, because they should (I thought) yield a proper operator when evaluated at a point, and evaluating at a point is not something you usually want to do with distributions. I mean, it makes sense for them to be so, but I don't quite understand the full deets. $\endgroup$
    – Pablo T.
    Commented Oct 14, 2021 at 14:58
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    $\begingroup$ @PabloT. it is just one of those strange things in QFT you have to deal with. Quantum fields are really operator-valued distributions and not functions. It is easy to convince yourself of this: if they were functions, the propagator $\left< 0 \right| \phi(x) \phi(y) \left| 0 \right>$ wouldn't have singularities when $x = y$ or when they are lightlike separated. $\endgroup$ Commented Oct 14, 2021 at 15:03

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