matrix elements of $\hat{z}$ operator under the angular momentum basis

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.

Hint: There is a useful recursion formula, $$(2l+1)xP^m_l(x) = (l + m)P^m_{l-1}(x)+(l-m+1)P^m_{l+1}(x).$$
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.
• 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. – Hamio Jiang Nov 25 '18 at 19:37
• I mean when $|j-j'|=1$ – Hamio Jiang Nov 25 '18 at 19:48
• No, the $(ii)$ assumes that $m\neq m'$, so they are different. – Hamio Jiang Nov 25 '18 at 19:50