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I understand that if a Hamiltonian remains invariant under the following transformations then it is PT invariant,

\begin{eqnarray} \mathrm{Parity \; reversal:} \; \; \hat{p} \to -\hat{p} \; \; \mathrm{ and } \; \; \hat{x} \to -\hat{x} \\ \mathrm{Time \; reversal:} \; \; \hat{p} \to -\hat{p}, \; \hat{x} \to \hat{x}, \; \; \mathrm{ and } \; \; i \to -i. \end{eqnarray}

However, if during parity reversal we change the sign on momentum, and during time reversal, we change it back, what's the use of performing these rotations on $\hat{p}$? Similarly, the momentum operator contains a first space derivative. Does changing $\hat{p} \to -\hat{p}$ imply that we take $\hat{x} \to -\hat{x}$ (or does the effect cancel out when both $\hat{p}$ and $\hat{x}$ are reversed)?

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  • $\begingroup$ I don't think I quite understand what you are asking here. Of course $PT$ has no effect on ${\hat p}$ but it still affects ${\hat x}$ and in general all of the other operators as well as states in the theory and is therefore still non-trivial. $\endgroup$ – Prahar Mar 23 '18 at 20:44
  • $\begingroup$ @Prahar I was under the impression that only the $\hat{p}$ and $\hat{x}$ operators were involved in the change of variables for the PT symmetry. Is it not the case? $\endgroup$ – Buddhapus Mar 24 '18 at 5:40
  • $\begingroup$ I mean all other operators that are made out of ${\hat p}$ and ${\hat x}$. For instance, the Hamiltonian, ${\hat H}$. $\endgroup$ – Prahar Mar 24 '18 at 15:46
  • $\begingroup$ @Prahar I see what you mean. But if $\hat{p} = -i \hbar \partial_x$, then for parity reversal $\hat{p}$ is invariant, right? (Because both $\hat{p}$ and $\hat{x}$ change sign). Same for time reversal, but now because we change $i \to -i$. I have the simplest Hamiltonian and it just seems that there's an overkill happening with changing of the variables for all these inversions, especially since $\hat{p}$ is defined in terms of $x$. $\endgroup$ – Buddhapus Mar 24 '18 at 17:37
  • $\begingroup$ ${\hat p}$ is not generally defined in terms of $x$. It only takes the form of a derivative when acting on position eigenstates, namely wavefunctions. For parity reversal, ${\hat p} \to - {\hat p}$ as you have written in your question. All you need to do is take $x \to - x$ in $- i \hbar \partial_x$ to see this. Under time reversal also ${\hat p} \to - {\hat p}$ since $i \to - i$. $\endgroup$ – Prahar Mar 24 '18 at 17:46

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