Parity anomaly in a $\mathcal{P}\mathcal{T}$-symmetric quartic Hamiltonian

Abstract

In this paper, two independent methods are used to show that the non-Hermitian $\mathcal{P}\mathcal{T}$-symmetric wrong-sign quartic Hamiltonian H = (1/2m)p ² − gx ⁴ is exactly equivalent to the conventional Hermitian Hamiltonian $\tilde H = ({1 \mathord{\left/ {\vphantom {1 {2m}}} \right. \kern-\nulldelimiterspace} {2m}})p^2 + 4gx^4 - \hbar ({{2g} \mathord{\left/ {\vphantom {{2g} m}} \right. \kern-\nulldelimiterspace} m})^{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-\nulldelimiterspace} 2}} x$. First, this equivalence is demonstrated by using elementary differential-equation techniques and second, it is demonstrated by using functional-integration methods. As the linear term in the Hermitian Hamiltonian $\tilde H$ is proportional to ℏ, this term is anomalous; that is, the linear term in the potential has no classical analog. The anomaly is a consequence of the broken parity symmetry of the original non-Hermitian $\mathcal{P}\mathcal{T}$-symmetric Hamiltonian. The anomaly term in $\tilde H$ remains unchanged if an x ² term is introduced into H. When such a quadratic term is present in H, this Hamiltonian possesses bound states. The corresponding bound states in $\tilde H$ are a direct physical measure of the anomaly. If there were no anomaly term, there would be no bound states.