that $\int dy\,\delta(x-y)\delta(y-z)=\delta(x-z)$.

Second, we show how to reconstruct the parity operator $\mathcal{P}$ in terms of the eigenstates. In coordinate space the parity operator is given by $\mathcal{P
}(x,y)=\delta(x+y)$, so from (\[e6\]) we get $$\mathcal{P}(x,y)=\sum_n(-1)^n\phi_n(x)\phi_n(-y).
\label{e9}$$ By virtue of (\[e8\]) the square of the parity operator is unity: $\mathcal{P}
^2=1$.

Third, we reconstruct $H$ in coordinate space: $$H(x,y)=\sum_n(-1)^nE_n\phi_n(x)\phi_n(y).
\label{e10}$$ Using (\[e6\]) – (\[e8\]) we can see that $H$ satisfies $H\phi_n(x)=E_n
\phi_n(x)$.

Fourth, we construct the coordinate-space Green’s function $G(x,y)$: $$G(x,y)=\sum_n(-1)^n\frac{1}{E_n}\phi_n(x)\phi_n(y).
\label{e11}$$ The Green’s function is the functional inverse of $H$; that is, $G$ satisfies $$\int dy\,H(x,y)G(y,z)=\left[-\frac{d^2}{dx^2}+x^2(ix)^\epsilon\right]G(x,z)=
\delta(x-z).
\label{e12}$$ The time-independent Schrödinger equation (\[e3\]) cannot be solved analytically; it can only be solved numerically or perturbatively. However, the differential equation for $G(x,z)$ in (\[e12\]) [*can*]{} be solved exactly and in closed form because it is a Bessel equation [@rr9]. The technique is to consider the case $0<\epsilon<2$ so that we may treat $x$ as real and then to decompose the $x$ axis into two regions, $x>z$ and $x<z$. We solve the differential equation in each region in terms of Bessel functions and patch the solutions together at $x=z$. Then, using this coordinate-space representation of the Green’s function, we construct an exact 