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Is it the case that if you have simultaneous eigenkets say $|l m \rangle$ then we have both $$\sum_{l} |l m \rangle \langle l m \rangle = I ~~\text{ and }~~ \sum_{m} |l m \rangle \langle l m \rangle = I ?$$ Also, is there always degeneracy when having shared eigenstates? Consider the simultaneous eigenkets of the angular momentum operators $J^{2}$ and $J_{z}$, since we have that $|l, m \rangle$ and $|l', m \rangle$ are both eigenstates of the angular momentum operator $J_{z}$ with $J_{z}|l, m \rangle = m \hbar |l, m \rangle$ and $J_{z}|l', m \rangle = m \hbar |l', m \rangle$ clearly the operators $J_{z}$ and $J^2$ are degenerate.

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2 Answers 2

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If you apply, for example, your first sum to the state $|l^\prime,m^\prime\rangle$ you get $$ \sum_{l}|l,m\rangle\langle l,m|l^\prime,m^\prime\rangle = \langle l^\prime,m|l^\prime,m^\prime\rangle |l^\prime,m\rangle = \delta_{m,m^\prime}|l^\prime,m\rangle $$ i.e. this is $0$ unless $m^\prime$ happens to match the $m$ you chose for your sum. The problem is that you ignoring the degeneracy in the spectrum of $J^2$ and not summing over a complete basis. To obtain the identity, therefore, you must sum over both $l$ and $m$ $$ I = \sum_{l,m}|l,m\rangle\langle l,m| $$

As for the second part of your question $p$ and $p^3$ are operators with a basis of mutual eigenstates and neither of which is degenerate

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  • $\begingroup$ Thanks for your answer. My question was motivated by the following query. Please have a look if you have a chance. $\endgroup$
    – user101311
    May 6, 2017 at 16:41
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No. The unit operator is $$ \hat 1= \sum_{\ell m}\vert \ell m\rangle\langle \ell m\vert\, . $$ The simplest intuition into this is to compare with spherical harmonics, for which the completeness relation involves a sum on both $\ell$ and $m$. Of course you may have a priori knowledge of a state $\vert\psi\rangle$ and know it's an eigenstate of $\hat L_z$ with eigenvalue $M\hbar$. Then of course \begin{align} \vert\psi\rangle&=\sum_{\ell m}\vert \ell m\rangle\langle \ell m\vert\psi\rangle=\sum_{\ell m}\vert \ell m\rangle\langle \ell m\vert\psi\rangle\,\delta_{Mm}\, ,\\ &=\sum_{\ell}\vert \ell M\rangle\langle \ell M\vert\psi\rangle \end{align} just like you may know that $F(\theta,\phi)=f(\theta)e^{iM\phi}$ and so restrict the sum to the appropriate $m$ but this is because you are implicitly using additional information on the eigenvalue to simplify your expansion. The same general type of argument holds for restricting to a single $\ell$. Thus type of simplification will not work in general.

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