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Question on Time Reversal and Degeneracy

  1. Sep 2, 2008 #1
    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?

    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?
  2. jcsd
  3. Sep 3, 2008 #2
    Never mind. Wikipedia explains it:


    So it turns out that T^2 can be +1 or -1, the latter for spin 1/2, the former for spin 0. So there is no degeneracy if T^2=1.

    Still, it's a little weird. If you charge conjugate twice, you should get the same state. If you do parity twice, you may or may not get the same state depending on your phase convention? And if you do time reversal twice, you definitely don't get the same state (for a spin 1/2 particle).

    I suppose this is not too weird since you have to rotate an electron around two circles to get it back the same. But one would normally expect that if you reversed time twice, nothing would happen, and you'd be back where you were. Same with parity.
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