From Axioms of relativistic quantum field theory, I find the following theorem :

We conclude this section with the following result of Wightman which demonstrates that in QFT it is necessary to consider operator-valued distributions instead of operator-valued mappings :

Proposition 8.15. Let $\Phi$ be a field in a Wightman QFT which can be realized as a map $\Phi : M \to \mathcal O$ and where $\Phi^*$ belongs to the fields. Moreover, assume that $\Omega$ is the only translation-invariant vector (up to scalars). Then $\Phi(x) = c \Omega$ is the constant operator for a suitable constant $c \in \Bbb C$.

First the notation is a bit confusing since $\Phi(x)$ is an operator and $c \Omega$ is a Hilbert vector. I suspect it is supposed to mean that every field operator just maps every vector to the vacuum, but it is not really explicit.

I tried finding the original Wightman theorem, but this does not include any bibliographical reference for this statement. The only Wightman book referenced is "PCT, Spin and Statistics, and All That", which doesn't seem to include it.

This is disregarding any concerns for the canonical commutation relation, as the Wightman axioms do not include them as a requirement.

What is the theorem exactly and how does one go about to prove it?

  • $\begingroup$ it looks like a typo, IIRC, it should read $\Phi(x)\Omega=c\Omega$. $\endgroup$ – AccidentalFourierTransform Nov 27 '16 at 13:39
  • $\begingroup$ Would that make every operator-valued function theory useless? I'm not quite sure where the problem lies. $\endgroup$ – Slereah Nov 27 '16 at 13:41
  • $\begingroup$ that's the point: if $\Phi(x)$ is an operator valued function, then it is trivial. Therefore, we conclude that we want it to be an operator valued distribution. This bypasses the theorem and we get a useful theory $\endgroup$ – AccidentalFourierTransform Nov 27 '16 at 13:42

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