# Hamiltonian invariant under time reversal symmetry

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.

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.
$$O$$ 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.