# 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!

• Unless I have misunderstood the question, I don't think you can prove that, because that statement is false (consider, e.g., a stationary state of the Hydrogen atom: there, $\langle L_z\rangle=m\neq 0$). – AccidentalFourierTransform Nov 10 '16 at 19:57
• Hi, I don't know a lot about the Hydrogen atom, but I forgot to mention that the stationary state is not degenerate, does it help? – Noam Chai Nov 10 '16 at 20:10
• I checked again the quetion and it looks like the original question that I copied from my exercise... maybe the question is wrong.... – Noam Chai Nov 10 '16 at 20:13
• I have the feeling that the statement is still false, but I could be wrong (what if we consider a free particle, i.e., a plane wave: there, the states are non-degenerate, but the angular momentum is still non-zero...) – AccidentalFourierTransform Nov 10 '16 at 20:24
• hi Noam, I'm quite sure that "time-symmetric" here means "symmetric under time-reversal". In the sense you are considering (the hamiltonian doesn't explicitly depend on time) the proposition is false, consider the hamiltonian $H=gS_z$. I wrote my answer under this assumption, let me know if it's not what you was looking for. – pppqqq Nov 10 '16 at 21:11

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:

1. 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 $$c$$.
2. 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 $$\vert E \rangle$$.
3. 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: $$(f,Jf)=(Tf,JTf)=(f,T^{\dagger}JTf)^*.$$
• If the system is symmetric under $T$, then there cannot be any non-degenerate state (as per Kramers theorem) – AccidentalFourierTransform Nov 10 '16 at 21:14
• Kramers theorem applies to half integer spins, right? Also, I don't know why I did such a mess to prove that $HT\vert E\rangle = ET\vert E \rangle$ XD – pppqqq Nov 10 '16 at 21:22
• Kramers theorem is much more general: it applies to any system that is invariant under $T$ (the general proof can be found somewhere in Weinberg's QFT, Vol I). Initially, I also thought that the time-symmetry of the OP was related to $T$, but this cannot be, because in such a case there would be no non-degenerate states. – AccidentalFourierTransform Nov 10 '16 at 21:25
• Actually youre right: I found the proof (page 80) and it indeed is only valid for systems with total spin a half-integer. – AccidentalFourierTransform Nov 10 '16 at 21:38
• Thank you ! few questions: 1. where did you use HT|E⟩=ET|E⟩? I don't see any use of this fact. 2. can you explain more about alpha? does the exponential power suppose to be time dependent? (the eigenvalue of T) – Noam Chai Nov 11 '16 at 8:24