0
$\begingroup$

I am reading the section in Sakurai's Modern Quantum Mechanics (2nd Edition) on the time-reversal operator $\Theta$ in quantum mechanics. He provides a short derivation of the properties that the time-reversal operator $\Theta$ should have. In particular, in equation (4.4.31) on page 291 it is stated that the time-reversal operator must obey

$$ \Theta H = H \Theta \quad (*) $$

where $H$ is the Hamiltonian of the system. This result is true when acting on any states $|\psi\rangle$ in the Hilbert space and it also appears that no particular form of $H$ was assumed. At first glance this property is telling me that $[H , \Theta]=0$ for all Hamiltonians, in other words all Hamiltonians have time-reversal invariance!

This is obviously not true and in fact later in the book theorem 4.2 assumes a time-reversal invariant Hamiltonian implying that not all Hamiltonians are time-reversal invariant.

One would assume time-reversal invariance means $[H,\Theta] = 0$ so I am unsure how this differs from (*) above. Any comments would be greatly appreciated.

$\endgroup$
3
  • $\begingroup$ Generically the Hamiltonian does not need to commute with the time-reversal operator. For instance for an electron in an external magnetic field, time-reversal is broken. $[H,\Theta]=H\Theta - \Theta H$ obviously. $\endgroup$ Feb 21, 2020 at 11:11
  • $\begingroup$ From what I understand, Sakurai says that $\Theta$ ought to not be unitary for the time-reversal symmetry to make sense, and assumes it to be antiunitary, which leads to the case where $[H, \Theta] = 0$, but as I understand it's not compelled for $\Theta$ to be antiunitary, the only restriction is that it can't be unitary. This is just how I interpret what Sakurai says, so I'm not 100% sure of it. $\endgroup$
    – Mr. Nobody
    Feb 21, 2020 at 11:24
  • $\begingroup$ @Mr.Nobody the point is that previously he showed that if $\Theta$ is unitary then we get $\Theta H = - H \Theta$, and then he goes on to derive that we also get $\Theta H = H \Theta$, therefore it cannot be unitary. The equation the OP marked with $(*)$ is not the general case, and Sakurai uses it during this proof. $\endgroup$
    – user245141
    Feb 21, 2020 at 12:03

1 Answer 1

3
$\begingroup$

it also appears that no particular form of $H$ was assumed.

This is not the case. In page $290$ Sakurai says:

If the motion obeys symmetry under time reversal, we expect the preceding ket to be the same as $$\Theta|\alpha,t_0;t=-\delta t\rangle$$

So we are beginning with the assumption of time reversal invariance. So if the condition $[H,\Theta]=0$ holds, then the system is time reversal invariant. This means not all Hamiltonians are symmetric under time reversal.

In fact take the case of angular momentum in a magnetic field: $H=\mu \hat{J}\cdot B$. You can easily see that this Hamiltonian is not symmetric under time reversal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.