Theorem: No stationary finite-second-moment local state

Suppose a product extension has a negative transverse eigenvalue $-{\kappa }^2$. There is no time-translation-invariant state on its extension-selected compact-test field algebra having finite second moments for every compact smooth field test.

In particular, no regular spectral ground state or regular KMS state with that finite-moment property exists. This excludes every massless ${\mathsf{L}}_L$, every diagonal combination with some ${\lambda }_j<0$, and every boundary relation with $H_W\ngeqq 0$.

Proof. The full spatial operator is bounded below. Choose a compact smooth spatial section $v$ with nonzero spectral measure on some interval $[-b,-a]\subset (-\infty ,0)$. Such a section exists: compact tests are dense, and the negative spectral projection is nonzero. For a charged field use the underlying real phase space.

Choose a nonnegative, nonzero $\chi \in {C_c}^{\infty }(\mathbb{R})$, supported in $[-A,A]$, and let $f(t,x)=\chi (t)v(x)$. The commutator pairing of $f$ and its time translate $f_T$ is obtained by integrating

against $\chi (s)\chi (s^{\prime })$.

For $\lambda =-{\omega }^2<0$, this contribution equals

It is positive and grows at least as a positive constant times $e^{\sqrt{a}T}$ on a smaller negative spectral interval. For nonnegative spectrum, the elementary bound $|\sin (u\sqrt{\lambda })/\sqrt{\lambda }|\le |u|$ bounds the entire contribution by $C(T+2A){\Vert v\Vert }^2$. The commutator magnitude is therefore unbounded in $T$.

In an invariant state with finite $\omega (\Phi (f{)}^2)$, Cauchy–Schwarz gives

independently of $T$, a contradiction.