$\mathcal{PT}$ symmetry, we say that the $\mathcal{PT}$ symmetry is unbroken. Conversely, if the solution does not possess $\mathcal{PT}$ symmetry, we say that the $\mathcal{PT}$ symmetry is broken.

It is extremely easy to prove that if the $\mathcal{PT}$ symmetry of a Hamiltonian $H$ is unbroken, then the spectrum of $H$ is real: Assume that (i) $H$ possesses $\mathcal{PT}$ symmetry (that is, that $H$ commutes with the $\mathcal{PT}$ operator), and (ii) if $\phi$ is an eigenstate of $H$ with eigenvalue $E$, then it is simultaneously an eigenstate of $\mathcal{PT}$ with eigenvalue $\lambda$ (it is this second assumption of [*unbroken*]{} $\mathcal{P
T}$ symmetry that is crucial): $$\begin{aligned}
H\phi=E\phi\quad{\rm and}\quad\mathcal{PT}\phi=\lambda\phi.
\label{e2}\end{aligned}$$ We first show that the eigenvalue $\lambda$ is a pure phase. We multiply $\mathcal{PT}\phi=\lambda\phi$ on the left by $\mathcal{PT}$ and use the fact that $\mathcal{P}$ and $\mathcal{T}$ commute and that $\mathcal{P}^2=\mathcal{
T}^2=1$ to conclude that $\phi=\lambda^*\lambda\phi$ and thus $\lambda=e^{i
\alpha}$ for some real $\alpha$. Next, we introduce a convention used throughout this paper. Without loss of generality we replace the eigenstate $\phi$ by $e^{
-i\alpha/2}\phi$ so that its eigenvalue under the operator $\mathcal{PT}$ is unity: $\mathcal{PT}\phi=\phi$. Next, we multiply the eigenvalue equation $H
\phi=E\phi$ on the left by $\mathcal{PT}$ and use $[\mathcal{PT},H]=0$ to 