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  Theorem of QFT as an operator-valued function theory

+ 3 like - 0 dislike

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?

This post imported from StackExchange Physics at 2017-10-11 16:33 (UTC), posted by SE-user Slereah
asked Nov 27, 2016 in Theoretical Physics by Slereah (540 points) [ no revision ]
it looks like a typo, IIRC, it should read $\Phi(x)\Omega=c\Omega$.

This post imported from StackExchange Physics at 2017-10-11 16:33 (UTC), posted by SE-user AccidentalFourierTransform
Would that make every operator-valued function theory useless? I'm not quite sure where the problem lies.

This post imported from StackExchange Physics at 2017-10-11 16:33 (UTC), posted by SE-user Slereah
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

This post imported from StackExchange Physics at 2017-10-11 16:33 (UTC), posted by SE-user AccidentalFourierTransform

1 Answer

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By Poincare invariance, the 1-point functions are constant, hence wlog zero (by subtracting the constant). After this subtraction, the 2-point functions are necessarily given by the Kallen-Lehmann formula (see, e.g., the treatise by Reed and Simon). Since these are distributions only, the fields must be distribution valued, too. 

answered Oct 14, 2017 by Arnold Neumaier (15,787 points) [ revision history ]

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