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The operator $L^2$ commutes with each of the operators $L_x$, $L_y$ and $L_z$, yet $L_x$, $L_y$ and $L_z$ do not commute with each other. From linear algebra, we know that if two hermitian operators commute, they admit complete sets of common/simultaneous eigenfunctions. The way I understand this statement is that the eigenfunctions of both operators are the same. So, if that were the case, that would mean that $L_x$ has the same eigenfunctions as $L^2$. The same goes for $L_y$ and $L_z$. That would mean that $L_x$, $L_y$ and $L_z$ all have the same eigenfunctions, which doesn't seem to be true since they do not commute with each other. How is this resolved?

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6 Answers 6

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Be very careful about what the theorem is saying! You posted in a comment:

If two Hermitian operators, A and B, commute and if A has no degenerate eigenvalue, then each eigenvector of A is also an eigenvector of B.

This statement is not symmetric in A and B! How does it apply to our situation? $L^2$ and $L_z$ commute, and $L_z$ has no degenerate eigenvalues. The statement above means that every eigenvector of $L_z$ is also an eigenfunction of $L^2$. No problem there.

But $L^2$ does have degenerate eigenvalues, and so there are some eigenvectors of $L^2$ that are not eigenvalues of $L_z$. (In particular, the eigenvectors of $L_y$ and $L_x$ are eigenvectors of $L^2$ but not of $L_z$.)

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  • $\begingroup$ Just a minor correction: L_{z} does have degenerate eigenvectors. So, it goes both ways. $\endgroup$
    – Arnab
    Apr 5, 2021 at 23:44
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Your formulation is correct:

From linear algebra, we know that if two hermitian operators commute, they admit complete sets of common/simultaneous eigenfunctions.

However, if two hermitian operators commute, it's not true that every set of eigenfunctions for either of them will be one of these sets of common eigenfunctions.

Thus, $L^2$ has shared eigenbases with each of $L_x$, $L_y$ and $L_z$, but those are different eigenbases. The only state that's common to all three is the spherically-symmetric $L^2=0$ state.

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Commuting operators do not necessarily share ALL eigenstates, just some set.

An eigenstate shared by $L^2$ and $L_x$ will not be the same as that shared by $L^2$ and $L_y$

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  • $\begingroup$ What about commuting "hermitian" operators? They share ALL their eigenfunctions, as far as I know. I'm reading Griffith's introduction to quantum mechanics and Zettili's Quantum mechanics: concepts and applications. They both state that commuting hermitian operators share all their eigenfunctions. $\endgroup$
    – user626542
    Apr 4, 2021 at 16:24
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    $\begingroup$ @user626542 Nope, with Hermitian operators you can always find at least one set of eigenfunctions that are the same. For example, consider the free particle Hamiltonian $p^2/2m$ and the momentum operator $p$. The eigenbasis $|p\rangle$ is clearly a simultaneous eigenbasis. However, consider the following eigenbasis $|p\rangle + |-p\rangle$. This is a complete set, but it is an eigenstate of $H$ but not of $p$! So you can see that $H$ has an eigenbasis that is not an eigenbasis of $p$, even though they are both Hermitian and commuting. $\endgroup$
    – Philip
    Apr 5, 2021 at 4:30
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Since there's probably a mathematical development of this property in your textbook/notes, I'm guessing you want an intuitive approach on it. If this doesn't help you I'll develop the mathematics:

Imagine $L^2$ has having three properties, let's call them colors. $L^2$ is green, red, and blue. $L_x , L_y , L_z$ are respectively just green, red and blue. If two operators have the same color, they'll commute, and if they don't have any common color, they won't commute. You can easily see the commutation relationships you describe above are true for this "colorful" approach.

In fact, think of this funny experiment. The identity matrix surely commutes with any matrix you can think of, let's say, A and B. If it worked as you think, this would mean that any arbitrary pair of matrices A and B must commute!

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  • $\begingroup$ While that does make sense, the intuitive approach was not what I'm looking for. My problem is that there seems to me to be two results that contradict each other. The first is that if L<sup>2</sup> commutes with L<sub>i</sub>, then L<sup>2</sup> and L<sub>i</sub> have the same eigenfunctions. The second is that the L<sub>i</sub> s do not commute with each other and thus do not have the same eigenfunctions. I hope this is a better phrasing. $\endgroup$
    – user626542
    Apr 4, 2021 at 16:09
  • $\begingroup$ They do have common eigenfunctions, but not enough to form a basis for the desired space. $\endgroup$ Apr 4, 2021 at 16:20
  • $\begingroup$ But if every eigenfunction of one of them is also an eigenfunction of the other, how are they not enough to form a basis? $\endgroup$
    – user626542
    Apr 4, 2021 at 16:22
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If two observables $A$ and $B$ commute, i.e. if $[A,B]=0$, then there exists a common eigenbasis. In other words, there is a basis $\{|\phi_n\rangle\}_n$ for which $$A|\phi_n\rangle= a_n\, |\phi_n\rangle \quad\text{and}\quad B|\phi_n\rangle= b_n\, |\phi_n\rangle \quad.$$

Now consider the case where $A$ also commutes with another observable $C$. Then this does not imply that $C|\phi_n\rangle= c_n\, |\phi_n\rangle$ for all $n$: The basis $\{|\phi_n\rangle\}_n$ is, in general, not an eigenbasis of $C$.

The fact that $L^2$ commutes with all $L_x$, $L_y$ and $L_z$ does not imply that the e.g. $L_x$ and $L_y$ commute. Indeed, as you pointed out, they do not commute and hence do not share a common eigenbasis, although each of them shares a common eigenbasis with $L^2$.

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  • $\begingroup$ This is how I interpreted your answer. Lx and Ly each have a common eigenbasis with L2. Yet, they do not have a common eigenbasis among themselves. If that is indeed what you mean, let me try another phrasing. As stated in Zettili's Quantum mechanics: concepts and applications, If two Hermitian operators, A and B, commute and if A has no degenerate eigenvalue, then each eigenvector of A is also an eigenvector of B. If true, then every eigenvector of L2 is an eigenvector of Lx , Ly and Lz. Hence, Lx , Ly and Lz share common basis sets. $\endgroup$
    – user626542
    Apr 4, 2021 at 16:44
  • $\begingroup$ For completeness, the statement says that one of the operators does not have any degenerate eigenvalues. While this is not the case for L2, it is indeed the case for Lz and the statement only requires one of them to not have degenerate eigenvalues $\endgroup$
    – user626542
    Apr 4, 2021 at 16:45
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    $\begingroup$ Well, if the eigenvalues of one operator are non-degenerate, then there exists only one eigenbasis for that operator. Therefore, if it commutes with multiple operators, they must all share the same eigenbasis. However, if the eigenvalues are degenerate, then there can be multiple eigenbasis for the operator. That way it can share one set of eigenbasis with one operator and another set with another operator without the two operators sharing an eigenbasis. That's what degeneracy does, as far as I understand. $\endgroup$
    – user626542
    Apr 4, 2021 at 20:03
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    $\begingroup$ @user626542: Your idea that "the statement only requires one of them to not have degenerate eigenvalues" is a bit misleading, since the theorem statement doesn't imply that every eigenvector of $A$ is an eigenvector of $B$ *and vice versa; it's not bidirectional in that way. Rather, it is true that every eigenvector of $L_x$ is an eigenvector of $L^2$, and similarly every eigenvector of $L_z$ is an eigenvector of $L^2$. However, it does not imply that these two sets of eigenvectors (for $L_x$ and $L_z$) are the same sets of eigenvectors; ... $\endgroup$ Apr 5, 2021 at 0:00
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    $\begingroup$ ... and it is not the case that every eigenvector of $L^2$ is an eigenvector of either $L_x$ or $L_z$, due to the degeneracy of $L^2$. $\endgroup$ Apr 5, 2021 at 0:09
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This is possible precisely because $L^2$ is degenerate: for an eigenvalue $l(l + 1)$, it has an eigenspace of dimension $2l + 1$ (i.e., it has this many linearly independent eigenstates). The choice of basis in a degenerate eigenspace is not unique -- thus explaining how it can be that the eigenstates $L^2$ shares with $L_x$ are not the same as the ones $L^2$ shares with $L_y$.

It is possible to simultaneously diagonalize $L^2$ and $L_x$, or $L^2$ and $L_y$, or $L^2$ and $L_z$. But the most you can do for $L^2$, $L_x$, $L_y$, and $L_z$ together is simultaneous block diagonalization: $L_x$, $L_y$, and $L_z$ are nonzero only in each $(2l + 1) \times (2l + 1)$ block corresponding to each $L^2$ eigenvalue. And $L^2$ commutes with all of them because it is proportional to the identity matrix in each block.

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