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!

  • $\begingroup$ 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, $\langle L_z\rangle=m\neq 0$). $\endgroup$ – AccidentalFourierTransform Nov 10 '16 at 19:57
  • $\begingroup$ 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? $\endgroup$ – Noam Chai Nov 10 '16 at 20:10
  • $\begingroup$ I checked again the quetion and it looks like the original question that I copied from my exercise... maybe the question is wrong.... $\endgroup$ – Noam Chai Nov 10 '16 at 20:13
  • $\begingroup$ 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...) $\endgroup$ – AccidentalFourierTransform Nov 10 '16 at 20:24
  • $\begingroup$ 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. $\endgroup$ – pppqqq Nov 10 '16 at 21:11

I guess that "time-symmetric" here means that the system is invariant under time-reversal symmetry:


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: $$(f,Jf)=(Tf,JTf)=(f,T^{\dagger}JTf)^*.$$
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  • $\begingroup$ If the system is symmetric under $T$, then there cannot be any non-degenerate state (as per Kramers theorem) $\endgroup$ – AccidentalFourierTransform Nov 10 '16 at 21:14
  • $\begingroup$ 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 $\endgroup$ – pppqqq Nov 10 '16 at 21:22
  • $\begingroup$ 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. $\endgroup$ – AccidentalFourierTransform Nov 10 '16 at 21:25
  • $\begingroup$ Actually youre right: I found the proof (page 80) and it indeed is only valid for systems with total spin a half-integer. $\endgroup$ – AccidentalFourierTransform Nov 10 '16 at 21:38
  • $\begingroup$ 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) $\endgroup$ – Noam Chai Nov 11 '16 at 8:24

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