obtain $E\phi=E^*\phi$. Hence, $E=E^*$ and the eigenvalue $E$ is real.

The crucial step in the argument above is the assumption that $\phi$ is simultaneously an eigenstate of $H$ and $\mathcal{PT}$. In quantum mechanics if a linear operator $X$ commutes with the Hamiltonian $H$, then the eigenstates of $H$ are also eigenstates of $X$. However, we emphasize that the operator $\mathcal{PT}$ is not linear (it is antilinear) and thus we must make the extra assumption that the $\mathcal{PT}$ symmetry of $H$ is unbroken; that is, that $\phi$ is simultaneously an eigenstate of $H$ and $\mathcal{PT}$. This extra assumption is nontrivial because it is difficult to determine [*a priori*]{} whether the $\mathcal{PT}$ symmetry of a particular Hamiltonian $H$ is broken or unbroken. For the Hamiltonian $H$ in (\[e1\]) the $\mathcal{PT}$ symmetry is unbroken when $\epsilon\geq0$ and it is broken when $\epsilon<0$. The conventional Hermitian Hamiltonian for the quantum mechanical harmonic oscillator lies at the boundary of the unbroken and the broken regimes. Recently, Dorey [*et al.*]{} proved rigorously that the spectrum of $H$ in (\[e1\]) is real and positive [@rr4] in the region $\epsilon\geq0$. Many other $\mathcal{PT}$-symmetric Hamiltonians for which space-time reflection symmetry is not broken have been investigated, and the spectra 