Under time reversal T, the momentum operator changes sign but the position operator remains the same. So if you have a Hamiltonian of the form H(X,P)=P^2 + V(X) , then it's invariant under time reversal since momentum is squared. This means H and T commute, so that if a state has eigenvalue E of the operator H, then T operated on the state also has the same eigenvalue E under the operator H. However, since T^2=-1 and not +1, this implies that T operated on the state is not the same state. In other words, whenever the Hamiltonian is of the form P^2+V(X), every state is twice degenerate?(adsbygoogle = window.adsbygoogle || []).push({});

For the electrons, this kind of makes sense because spin isn't specified by the Hamiltonian of the form P^2+V(X), and spin can offer a degeneracy of two in this case. But what about for spinless particles?

Also is it sloppy to say that if the Hamiltonian doesn't depend on time, then it is invariant under time reversal? What if the Hamilotnian were H=P^2+P? Here the Hamiltonian seems to me to not depend on time, but the way time flows?

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# Question on Time Reversal and Degeneracy

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