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I found a quite challenge quantum mechanics problem in a preparation sample test for a midterm. The sample test does not have a solution, so it is bothering.

The question reads as follows:

Consider an electron moving in a central potential. Suppose that we know the matrix element of the $z$-position operator between two states: $$\langle j',m'|\,z\,|j,m\rangle $$ (i) Justify that $m'=m$ for this matrix element to be non-zero;

(ii) What are the constraints on $j$ and $j'$ for non-zero matrix elements?

(iii) Given a none-zero matrix element $\langle j',m'|\,z\,|j,m\rangle$ (with $m=m'$ and suitable constraints on $j'$ and $j$), give a general formula to compute the matrix elements: $$\langle j',m'''|\,x\,|j,m''\rangle $$

I am stuck at the (i) question. I tried to use algebraic methods but it seemed useless. By algebraic methods, I mean inserting $J_z$ and noticing $[J_z,z]=0$, but this only gives a relation of $m$. Then I tried to use spherical harmonic wavefunction $Y_m^j$. However, if $\hat{z}=\hat{r}\cos(\hat{\theta})$, then one will eventually compute the following integral:

$$\int_{-1}^{1}P^m_j(x)P^{m'}_{j'}(x)xdx $$ where we have $P_j^m(-x)=(-1)^{(m+j)}P_j^m(x)$. Hence, (i) statement may not be true.

Can someone give me a hint or some detailed calculation? Appreciated in advance.

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Hint: There is a useful recursion formula, $$(2l+1)xP^m_l(x) = (l + m)P^m_{l-1}(x)+(l-m+1)P^m_{l+1}(x).$$

But this is a bit... awkward. I can't see how someone could see this through on an exam.

There is another way (which may be better), where one can exploit the relationships between the commutators of $L_z$. In fact, you've already done that, but it seems you've missed something, so maybe go back and check your work again.

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  • $\begingroup$ Does this mean that the statement $(i)$ is not true in general? According to the recursion formula, the integral is non-vanishing if $j$ and $j'$ deviate 1. $\endgroup$ – Hamio Jiang Nov 25 '18 at 19:37
  • $\begingroup$ Sorry, what does "deviate 1" mean? $\endgroup$ – Hanting Zhang Nov 25 '18 at 19:46
  • $\begingroup$ I mean when $|j-j'|=1$ $\endgroup$ – Hamio Jiang Nov 25 '18 at 19:48
  • $\begingroup$ Ah, I think part (ii) handles your question $\endgroup$ – Hanting Zhang Nov 25 '18 at 19:49
  • $\begingroup$ No, the $(ii)$ assumes that $m\neq m'$, so they are different. $\endgroup$ – Hamio Jiang Nov 25 '18 at 19:50

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