Nonuniqueness of the C operator in PT-symmetric quantum mechanics

The C operator in PT-symmetric quantum mechanics satisfies a system of three simultaneous algebraic operator equations, $C^2=1$, $[C,PT]=0$, and $[C,H]=0$. These equations are difficult to solve exactly, so perturbative methods have been used in the past to calculate C. The usual approach has been to express the Hamiltonian as $H=H_0+εH_1$, and to seek a solution for C in the form $C=e^Q P$, where $Q=Q(q,p)$ is odd in the momentum p, even in the coordinate q, and has a perturbation expansion of the form $Q=εQ_1+ε^3 Q_3+ε^5 Q_5+\ldots$. [In previous work it has always been assumed that the coefficients of even powers of $ε$ in this expansion would be absent because their presence would violate the condition that $Q(p,q)$ is odd in p.] In an earlier paper it was argued that the C operator is not unique because the perturbation coefficient $Q_1$ is nonunique. Here, the nonuniqueness of C is demonstrated at a more fundamental level: It is shown that the perturbation expansion for Q actually has the more general form $Q=Q_0+εQ_1+ε^2 Q_2+\ldots$ in which {\it all} powers and not just odd powers of $ε$ appear. For the case in which $H_0$ is the harmonic-oscillator Hamiltonian, $Q_0$ is calculated exactly and in closed form and it is shown explicitly to be nonunique. The results are verified by using powerful summation procedures based on analytic continuation. It is also shown how to calculate the higher coefficients in the perturbation series for Q.

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