The question is this:

Does $$L_+ L_- Y_{lm} $$ ,where $Y_{lm}$ is a spherical harmonic function, equals to zero. If so, why?

The two operators above are defined as $$L_+ ={L_x + iL_y } $$ $$L_-={L_x -iL_-} $$ where $L_x and L_y$ are components of the angular momentum operator . We can also prove that: $$L_+ L_- = L^2 - {L_z}^2 +\hbar L_z $$ $$L_- L_+ = L^2 - {L_z}^2 -\hbar L_z $$ and the the eigenvalues of $L^2$ and $L_z$ are $l(l+1)\hbar ^2 $ and $m \hbar $

Note: Maybe it' s my fault by I can't figure out why or find the proof somewhere. Note2: I am posting this question because I saw a problem where it is asked to prove that $$-l<m<l$$ using the operators $L_+ and L_- $ and in the solution the above equation of the question is used.

Thank you.

  • 4
    $\begingroup$ I'll answer your question if you tell my if $A^\sigma C\rho^\dagger$ is negative or positive. Protip: Write down the definitions of the three symbols and write down that expression in using the definitions. $\endgroup$
    – Nikolaj-K
    Apr 25, 2015 at 16:24
  • $\begingroup$ Would you tell me what these symbols are or should I search them? $\endgroup$ Apr 25, 2015 at 16:28
  • $\begingroup$ @NikolajK: I would be happy to try your suggestion if I knew what the symbols mean. $\endgroup$ Apr 25, 2015 at 16:35
  • 1
    $\begingroup$ Nikolaj wasn't being serious - he meant to imply that you should define what $L_+$ and $L_-$ are and how they act on $Y_{lm}$ for this question to become answerable. $\endgroup$
    – ACuriousMind
    Apr 25, 2015 at 16:47
  • $\begingroup$ Okay, sorry. I have made an edition to the post. $\endgroup$ Apr 25, 2015 at 16:59

1 Answer 1


Does $$L_+ L_- Y_{lm} $$ ,where $Y_{lm}$ is a spherical harmonic function, equals to zero. If so, why?

It may or may not equal zero depending on the value of $m$. If $m$ is equal to $-l$ then yes, otherwise no.

If $m=-l$ then applying the lowering operator annihilates the state (i.e., gives zero) since there is no state with an $m$ lower than $-l$.

Otherwise, applying the lowering operator gives you the state with $m=m-1$ (times an overall factor which you can look up) and then applying the raising operator gives you back the original state (again, up to a combined overall factor).


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