I know that the spectral theorem holds for unbounded normal linear operators on infinite dimensional Hilbert-spaces. We usually employ it in Quantum mechanics to explain the role of self-adjoint operators.

However, I'm not sure wether the theorem also applies to the observables of QFT, the reason being (for an interactive QFT) that we don't even know how the Hilbert space of the QFT looks like, or that the fields in QFT are operator valued distributions, and not operators. Hence the question: Is there a version of the spectral theorem that still holds in QFT (possibly with some restrictions to the used operators).


1 Answer 1


I am not sure to understand the nature of the problem.

The spectral theorem, as it is a mathematical fact, holds also in QFT. It does not matter if we do not know how the Hilbert space is made, it is sufficient to know that it is a Hilbert space and that the used operator is selfadjoint. Regarding operator valued distributions $\phi$, the spectral theorem applies to (usually the closures of) the images of these distributions $\phi(f)$ when they are selfadjoint operators.

If the theorem did not hold, then we would conclude that the space of states is not Hilbert or the operator is not selfadjoint (more generally normal).

  • $\begingroup$ I have seen different spectral theorems that did differ in wether the operator is bounded or compact, (and I haven't seen a theorem for unbounded operators), so I wasn't sure wether there might be an issue of that kind. $\endgroup$ Sep 2, 2022 at 10:46
  • $\begingroup$ Actually there is only one spectral theorem in Hilbert space. If rhe operator is bounded or compact, then the ststement specialises. But the rheorem is one. I wrote several books on the subject... $\endgroup$ Sep 2, 2022 at 11:31

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