# Is this definition of the Fourier Transform of a quantum field operator rigorous?

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?

• 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. Commented Oct 14, 2021 at 14:55

• 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. Commented Oct 14, 2021 at 12:30
• 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. Commented Oct 14, 2021 at 14:58
• @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. Commented Oct 14, 2021 at 15:03