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Parity anomaly in a mathcal{P}mathcal{T}-symmetric quartic Hamiltonian
Authors:Carl M. Bender
Affiliation:(1) Physics Department, Washington University, St. Louis, MO 63130, USA
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 2gx 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.
Keywords:  IE8"  >   /content/357887202457150q/10582_2006_Article_399_TeX2GIFIE8.gif"   alt="     $$mathcal{P}mathcal{T}$$   "   align="  middle"   border="  0"  > symmetry  anomaly  non-Hermitian  parity reflection  time reversal
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