symmetry in quantum physics and angular momentum I'm having trouble proving this argument in quantum physics:

On a stationary state (with no degeneracy), if the system is time-symmetric but not necessarily symmetric under rotations, prove that the expectation value of the angular momentum is zero.

As I understood:


*

*Stationary state: the wave function is an eigenvector of the Hamiltonian.

*Symmetry in time means that the Hamiltonian is constant.

*If the system is not necessarily symmetric under rotations then the Hamiltonian does not necessarily commute with the angular momentum operator.
But how can I prove the argument with that? can anyone help me?
Thank you!
 A: I guess that "time-symmetric" here means that the system is invariant under time-reversal symmetry:
$$T^{\dagger}HT=H$$
Recall that the angular momentum $\boldsymbol J$ transforms under time
reversal as $$T^\dagger \boldsymbol J T=-\boldsymbol J.$$
First, as AccidentalFourierTransform points out, let us note that the state in question cannot be an half-integer spin, by Kramers degeneracy (see comments below).
Now, suppose that $\vert E\rangle$ is a non degenerate eigenstate of $H$. Since $H$ commutes with $T$ we have: $$HT\vert E\rangle =T H \vert E \rangle =ET\vert E\rangle. $$
Since $\vert E\rangle$ is non degenerate, and $T$ is antiunitary, this means that: $$T\vert E\rangle = e^{2i\alpha}\vert E\rangle. $$
Using the antiunitary character of $T$ you can easily see that WLOG we may put $\alpha=0$.
Now, using $T\vert E\rangle =\vert E \rangle$, the antiunitarity of $T$ and the hermiticity of $\boldsymbol J$ we have: $$\langle E\vert \boldsymbol J \vert E \rangle = (\langle E \vert T^{\dagger}\boldsymbol J T\vert E\rangle )^* = -(\langle E\vert \boldsymbol J \vert \boldsymbol E\rangle)^*=-\langle E\vert \boldsymbol J \vert  E\rangle$$
which implies:$$\langle E\vert \boldsymbol J \vert  E\rangle=0.$$

In response to Noam Chai's comment:

*

*The fact that $HT\vert E\rangle = ET\vert E \rangle$, together with the assumption that the eigenvalues $E$ is nondegenerate, allows me to conclude that $T\vert E \rangle =c\vert E \rangle$ for some complex number $c$.

*Since $T$ is in particular an isometry, we must have $\vert c \vert =1$, so $c=e^{2i\alpha}$ for some real number $2\alpha$ beetween $0$ and $2\pi$, say (also $\alpha$ doesn't depend on time, since nor $T$ nor $\vert E \rangle$ does, by assumption). Now, multiplying the equation $T\vert E\rangle =e^{2i\alpha}\vert E\rangle$ by $e^{-i\alpha}$, and using the fact that $T$ is antilinear, we obtain: $$e^{i\alpha}\vert E\rangle = e^{-i\alpha}T\vert E\rangle = T(e^{i\alpha}\vert E\rangle),$$ so that the phase factor can actually be absorbed in the definition of $\vert E \rangle $.

*The $*$ comes from the definition of the adjoint of an antilinear operator. Let me switch to the mathematician's Hilbert space notation: $$\langle g \vert f\rangle \longrightarrow (g,f).$$
This notation is clearer when dealing with antilinear operators. Now, neglecting domain's issues (which in fact do not occur in the case of $T$), the adjoint of an antilinear operator $A$ is defined by the equation: $$(A^{\dagger}g,f)=(g,Af)^*.$$ You see that the LHS is linear in $f$, so it must be the RHS, hence the $*$.
In this notation, with $\vert E \rangle \to f$, my last equation reads: $$(f,Jf)=(Tf,JTf)=(f,T^{\dagger}JTf)^*.$$
