# Wave-function solution in time-reversal transformation

1. Jan 3, 2016

### Happiness

In obtaining (5.362) from (5.359), we first get

$U_{\tau}i\hbar\frac{\partial}{\partial t}\Psi(t) = U_{\tau}H^*\Psi^*(-t)$

In order to obtain the LHS of (5.362), $U_{\tau}$ must commute with $\frac{\partial}{\partial t}$. But how do we know that they commute?

2. Jan 4, 2016

### ShayanJ

The Hamiltonian is assumed to be time-independent, so its reasonable to expect U to be time-independent too. So we just assume it and continue. If it was a bad assumption, it would give us non-sense. Then we could go back and relax this assumption. But if it works(as it did here), good for us!

3. Jan 4, 2016

### Happiness

But it appears I could get non-sense out of this assumption:

Since $U_{\tau}$ commutes with $i\hbar\frac{\partial}{\partial t}$, it also commutes with $H$ by the time-dependent Schrondinger equation. From (5.361), $H^*=U_{\tau}^{\dagger}HU_{\tau}$. Since $U_{\tau}$ commutes with $H$, $H^*=U_{\tau}^{\dagger}U_{\tau}H=H$, which contradicts the assumption that $H$ is not real.

4. Jan 4, 2016

### ShayanJ

I don't see how that follows!

5. Jan 4, 2016

### Happiness

$UH\Psi=Ui\hbar\frac{\partial}{\partial t}\Psi=i\hbar\frac{\partial}{\partial t}U\Psi=HU\Psi$

Since this is true for any $\Psi$, $UH=HU$.

6. Jan 4, 2016

### ShayanJ

Here you're assuming that $U\Psi$ is a solution to Schrodinger's equation only because $\Psi$ is! Can you prove this?