# Sakurai's derivation of the time-reversal operator

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.

• 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. Feb 21, 2020 at 11:11
• 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. Feb 21, 2020 at 11:24
• @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.
– user245141
Feb 21, 2020 at 12:03

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.