  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 $H=H_0+\epsilon H_1$, 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 $Q=\epsilon
Q_1+\epsilon^3 Q_3+\epsilon^5 Q_5+\ldots$. [In previous work it has always been
assumed that the coefficients of even powers of $\epsilon$ 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
$Q=Q_0+\epsilon Q_1+\epsilon^2 Q_2+\ldots$ in which {\it all} powers and not
just odd powers of $\epsilon$ 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.
