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Let us consider a Wightman field theory for the free scalar neutral field $\phi$, and let $O\mapsto\mathfrak F(O)$ be the net of local von Neumann field algebras. If we take a non-empty bounded open subset $O$ of $\mathbb R^4$, then according to the Reeh-Schlieder theorem the vacuum vector $\Omega_0$ of the Fock construction is cyclic. This implies that the set of vectors $\{W\Omega_0\ |\ W\in\mathfrak F(O)\}$ is total in the Fock Hilbert space $\Gamma(\mathscr H)$, i.e. it is dense.

Why is $\mathfrak F(O)$ not irreducible in this case? If we take $C$ in the commutant of $\mathfrak F(O)$, i.e. $C\in\mathfrak F(O)'$, then we can show that $\Omega_0$ is an eigenvector of $C$, i.e. $\exists\lambda_C\in\mathbb R$ s.t. $C\Omega_0=\lambda_C\Omega_0$. Moreover, using commutativity, we deduce that $(\psi,CW\Omega_0)=\lambda_C(\psi,W\Omega_0)$ for any $\psi\in\Gamma(\mathscr H)$ and $W\in\mathfrak F(O)$. But since the $W\Omega_0$s are elements of a dense subset of the Fock Hilbert space, then we must conclude that $C=\lambda_C\text{Id}$ for any $C\in\mathfrak F(O)'$, i.e. $\mathfrak F(O)$ is irreducible.

What's wrong with the above proof?

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I am not familiar with the objects in your question, so I might be completely missing the point, but it seems that you are considering a specific representation of the local algebra. Namely the action on the Fock space, which is irreducible. But an algebra is irreducible if the regular representation is irreducible. Consider a reducible algebra and a irreducible quotient with an irreducible action of it on a Hilbert space. Then the composition through the quotient will give you an irreducible representation of the irreducible algebra. –  MBN Jul 3 '11 at 0:06
    
This may not be the case in your example, may be the actions is fateful. –  MBN Jul 3 '11 at 0:06
    
Yes I'm considering the Fock representation of the local field algebras. As I have said in the question, it is known that when $O$ is bounded the field algebra $\mathfrak F(O)$ in the Fock representation is not irreducible, but I fail to see this. –  Phoenix87 Jul 3 '11 at 14:50
    
It might be helpful to have a reference for the known statement that when $O$ is bounded the field algebra $\mathfrak F(O)$ in the Fock representation is not irreducible. –  Qmechanic Jul 6 '11 at 18:20
    
One such reference may be the book of Bogoliubov et al., General Principles of Quantum Field Theory, specifically the discussion that follows Proposition 8.1. –  Phoenix87 Jul 7 '11 at 9:18
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1 Answer 1

up vote 3 down vote accepted

The question (v1) deals with the axiomatic QFT formulation of the real Klein-Gordon field $\phi$ in Minkowski spacetime $M$ with a vacuum state $\Omega_0$.

Let $O\subseteq M$ be a fixed non-empty bounded open subset. I would say that the error in OP's proof is the sentence If we take $C$ in the commutant of ${\mathfrak F}(O)$, i.e. $C\in{\mathfrak F}(O)'$, then we can show that $\Omega_0$ is an eigenvector of $C$.

My counterexample goes roughly like this. Since $O\subseteq M$ is bounded, we can choose another non-empty open subset $U$ in $M$ that is causally disconnected from $O$, i.e. such that every pair of points $x\in O$ and $y\in U$ are space-like separated. Consider next a test-function $f\in S(M)$ with support ${\mathrm supp}(f)\subseteq U$ in such a way that the one-particle state $\phi[f]\Omega_0$ does not vanish.

From one of the Wightman axioms, we know that $C:=\phi[f]$ commutes with the operators in ${\mathfrak F}(O)$. However, a one-particle state $\phi[f]\Omega_0$ cannot be proportional to the vacuum state $\Omega_0$.

As a consequence, ${\mathfrak F}(O)$ is not irreducible.

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This is exactly what I was looking for! Thank you! –  Phoenix87 Jul 8 '11 at 7:42
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