A solved problem for time reversal symmetry

Source: Problem 4-10,'Modern Quantum Mechanics'; J.J. Sakurai

problem: suppose a spinless particle is bound to a fixed center by a potential V(x) so asymmetrical that no energy level is degenerate. Using time-reversal invariance prove <L>=o

for any energy eigenstate. (This is known as quenching of orbital angular momentum.) If the wave function of such a non degenerate eigenstate is expanded as $\sum _{l}\sum _{m}F_{lm}(r)Y_{l}^{m}(\theta ,\varphi )$

what kind of phase restriction do we obtain on $F_{lm}(r)$?

Solution: Under reversal time $P\to -P$ and $r\to -r$, [H,K]=0 Assume $HK|\ n\rangle =EK|\ n\rangle$ where $|\ n\rangle$ is the eigenket of Hamiltonian. So, $K|\ n\rangle$ is also a eigenket of H with the same eigenvalue.

If we do not have any degeneracy, so $K|\ n\rangle =e^{i\delta }|\ n\rangle =|\ m\rangle$

that $\delta$ is a real number. $\langle \ n|L|\ n\rangle =\langle \ m|KLK^{-1}|\ m\rangle =-\langle \ m|L|\ m\rangle =-e^{-i\delta }\langle \ n|L|\ n\rangle e^{i\delta }=-\langle \ n|L|\ n\rangle$

$\Rightarrow \langle \ n|L|\ n\rangle =0\Rightarrow <L>=o$ $\psi _{n}(x)=\langle \ x|\ n\rangle =\sum _{lm}\langle \ x|\ l,m\rangle \langle \ l,m|\ n\rangle =\sum _{l}\sum _{m}F_{lm}(r)Y_{l}^{m}(\theta ,\varphi )$

Also

$\psi _{n}(x)=e^{-i\delta }\langle \ x|\ m\rangle =e^{-i\delta }\psi _{n}^{*}(x)=e^{-i\delta }\sum _{m}F_{lm}^{*}(r)Y_{l}^{m}(\theta ,\varphi )=e^{-i\delta }\sum _{m}F_{lm}^{*}(r)(-1)^{m}Y_{l}^{-m}(\theta ,\varphi )=e^{-i\delta }\sum _{m}F_{l,-m}^{*}(r)(-1)^{-m}Y_{l}^{m}(\theta ,\varphi )$

with comparison two different form of function, we get

$F_{lm}(r)=F_{l,-m}^{*}(r)(-1)^{m}e^{-i\delta }$