Why is the local field algebra $\mathfrak F(O)$ associated with a bounded non-empty open region $O$ of space-time not irreducible?

Question

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?

Answer 1

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.