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On page 336 of Shankar's 'Principles of Quantum Mechanics' the author states "The $Y_l^m$ functions are mutually orthogonal because they are nondegenerate eigenfunctions of $L^2$ and $L_z$, which are Hermitian on single-valued functions of $\theta$ and $\phi$."

I don't see how the $Y_l^m$ can be nondegenerate.

Consider $L^2|lm\rangle = l(l+1)\hbar^2|lm\rangle$. For example, the eigenvalue $6\hbar^2$ has possible eigenstates $|22\rangle, |21\rangle,...|2,-2\rangle$. So there is degeneracy for the eigenvalue $l(l+1)$.

Consider $L_z|lm\rangle = m\hbar|lm\rangle$. For example, the eigenvalue $2\hbar$ has possible eigenstates $|22\rangle, |32\rangle, |42\rangle,...$Again there is degeneracy for the eigenvalue $mh$.

So if the $Y_l^m$ are degenerate functions of $L^2$ and $L_z$ what does the author mean?

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If you consider $L^2$ and $L_z$ simultaneously then $|lm\rangle$ are non-degenerate (each has a different $\textbf{pair}$ of eigenvalues). Considering them together is enough to show that $Y_l^m$ are orthogonal: if $l\neq l'$ you have $\langle lm|L^2|l'm'\rangle = 0 $ and if $m\neq m'$ you have $\langle lm|L_z|l'm'\rangle = 0 $. Both imply $\langle lm|l'm'\rangle = 0$.

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  • $\begingroup$ I had considered that the author meant considering $L^2$ and $L_z$ simultaneously, in which case it can be shown that the $Y_l^m$ are nondegenerate. The product of $L^2$ and $L_z$ is Hermitian, which means that the $Y_l^m$ must be orthogonal. However, by the wording in the statement I thought the author wanted the operators to be considered separately, but I guess not. Thanks for the reply. $\endgroup$
    – Nitram
    Apr 2, 2021 at 16:36
  • $\begingroup$ It is perfectly fine to have orthogonal states having the same eigenvalue of some Hermitian operator as long as there is another operator for which they have different eigenvalues. So there is no difference if you consider a product of $L^2$ and $L_z$ or just two of them separately. $\endgroup$
    – Viking
    Apr 2, 2021 at 17:24
  • $\begingroup$ So what does the author mean when he says that the $Y_l^m$ functions are nondegenerate eigenfunctions of $L^2$ and $L_z$? Are they not degenerate eigenfunctions of these two operators? $\endgroup$
    – Nitram
    Apr 3, 2021 at 10:37
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    $\begingroup$ Most likely, as you said in the first comment, the author means considering $L^2$ and $L_z$ simultaneously - because it is a correct statement and it is enough to prove that $Y_l^m$ are orthogonal. $\endgroup$
    – Viking
    Apr 3, 2021 at 14:43

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