# Why is time evolution unitary in $PT$-symmetry?

I have a question on the time evolution for a $PT$-symmetric Hamiltonian. So far I have only read that time evolution was unitary because $H$ commutes with $PT$ and the newly constructed operator $C$ but I don't really understand why this implicates that time evolution is unitary. Can someone explain that to me?

Time evolution never becomes unitary under naive inner product $\langle\psi|\phi\rangle$ you start with for which the Hamiltonian is not Hermitian. Instead (and here you need that $\mathcal{C}$ operator) you construct a new norm $\langle\psi|\mathcal{CPT}|\phi\rangle$ and in terms of that new norm the Hamiltonian becomes Hermitian and it generates the unitary evolution.
The best way to look at "$\mathcal{PT}$-symmetric quantum theory" is the idea of pseudo-Hermitian operators. You start with Hermitian $h$ and use non-unitary operator $\eta$ to construct non-Hermitian operator $H$, \begin{equation} H=\eta h \eta^{-1},\quad H\neq H^\dagger \end{equation} However you may take the norm $\langle\psi|\phi\rangle_{\eta}\equiv\langle \psi |\eta^\dagger\eta|\phi\rangle$ and for that norm $H$ is Hermitian. However if you try to construct $\eta$ and $h$ for most of the known nice $\mathcal{PT}$-symmetric Hamiltonians like $p^2+x^2(ix)^\epsilon$ it's only possible in the perturbation theory and the resulting $h$ is ugly and non-local so working with $H$ may be still useful.
I would add that no real application of $\mathcal{PT}$-symmetric Hamiltonians that I know actually use this special norm for probability measure. Those applications usually involve open quantum systems and simply use this relation with Hermitian Hamiltonians to explain certain peculiar properties that arise when you have $\mathcal{PT}$-symmetry.