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I have an hamiltonian operator

$$\hat{H}=-\frac{\hbar^2}{2m}\nabla^2+V(\hat{r})$$

for which there exists an antiunitary operator $\hat{O}$ such as

$$OH^*O^{-1}=H$$

If $\psi(r,t)$ is a solution of the time dependent Schrödinger equation I have to show that the function

$$\tilde{\psi}(r,t)=O \psi^{*}(r,-t)$$ is also a solution of that equation. My attempt was this: I start from

$$i\hbar\frac{\partial}{\partial t}\psi(r,t)=H\psi(r,t)$$

and I firstly change the variable $t\rightarrow -t=t'$ then I took the complex conjugate of both sides of the equation so as to obtain

$$i\hbar \frac{\partial}{\partial t'}\psi^*(r,-t')=H^{*}\psi^*(r,-t')$$

Then I apply the operator $O$ to both sides from the left:

$$ O \left(i\hbar \frac{\partial}{\partial t'}\psi^*(r,-t')\right)=OH^{*}\psi^*(r,-t')=HO\psi^*(r,-t')$$ Now I don't understand how to proceed because in my opinion the antiunitarity property and the other requirement for $O$ are not sufficient to prove the assertion. My question arises reading a paragraph of the following book: Quantum Mechanics: A New Introduction, Konishi,Paffuti, pag 121, in which the authors state that the property $OH^*O^{-1}=H$ and the antiunitarity of O are sufficient to claim that $\tilde{\psi}$ is a solution, without giving other details.

Is this procedure correct? If so is there a way to express the symmetry property (time reversal) of the Hamiltonian with a commutator? I know for example that if the symmetry is expressed by an unitary operator $S$ then $H$ is invariant under the symmetry if $[S,H]=O$, and I wonder if there is a similar property for the antiunitary operator $\hat{O}$ alternative to $OH^{*}=HO$. Update: if I assume instead that $[O,H]=O$ then maybe I can exploit the antilinearity of O and the fact that O anticommute with the time derivative operator(Does the time inversion operator commute or anticommute with the total time derivative) to say that: $$O\left(i\hbar \frac{\partial}{\partial t'}\right)=-i\hbar O\left( \frac{\partial}{\partial t'}\right)=i\hbar \left( \frac{\partial}{\partial t'}\right)O$$ But anyway I don't understand why it appears H* in the book.

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2 Answers 2

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I think that you are confusing the unitary matrix $O$ that appears in the first quantized formula $$ OH^*O^{-1} =H $$ with the antilinear operator $T$ that acts as on the second quantised field operator $\psi$ as $$ T\psi T^{-1}= O^\dagger\psi $$ See, for example, appendix C2 in https://arxiv.org/abs/2009.00518.

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$O$ should be a unitary operator. The anti-unitary part of time-reversal already puts the $*$ in $H$, so $O$ is unitary. Then you can move $O$ inside the time derivative on the left-hand side and the assertion follows.

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