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.


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Nov 25, 2018 at 19:37
  • $\begingroup$ Sorry, what does "deviate 1" mean? $\endgroup$
    – cxx
    Nov 25, 2018 at 19:46
  • $\begingroup$ I mean when $|j-j'|=1$ $\endgroup$ Nov 25, 2018 at 19:48
  • $\begingroup$ Ah, I think part (ii) handles your question $\endgroup$
    – cxx
    Nov 25, 2018 at 19:49
  • $\begingroup$ No, the $(ii)$ assumes that $m\neq m'$, so they are different. $\endgroup$ Nov 25, 2018 at 19:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.