# Time Inversion Symmetry and Angular Momentum

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1. Nov 17, 2016

### Yoni V

1. The problem statement, all variables and given/known data
Let $\left|\psi\right\rangle$ be a non-degenerate stationary state, i.e. an eigenstate of the Hamiltonian. Suppose the system exhibits symmetry for time inversion, but not necessarily for rotations. Show that the expectation value for the angular momentum operator is zero.

2. Relevant equations

3. The attempt at a solution
I'm trying to write the mathematical implications for each of the above statements, e.g. $$T(-iH)T^{-1}=iH,\; R(iH)R^{-1}=iH$$ where R,T are the corresponding unitary and anti unitary operators, and H is the Hamiltonian.
But I really don't see where this leads me. This is the beginning of the semester, so I still have very little intuition about how to take advantage of different properties such as unitarity, symmetries and commutation relations...

2. Nov 17, 2016

### strangerep

When asked to show that an expectation value for some observable $J$ is 0 in the state $|\psi\rangle$, one approach is to try proving that $$\langle\psi|J|\psi\rangle ~=~ - \langle\psi|J|\psi\rangle ~.$$ In your case, you can replace $|\psi\rangle$ by $T|\psi\rangle$ (where $T$ is the operator of time inversion). But then you must also think about how $T$ acts on angular momentum. A bit of googling should reveal the answer, or you can use the trick of thinking of time reversal as motion reversal.