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I noticed the following:

$$[L_{+},L^2]=0,\qquad [L_{+},L_3]\neq 0,\qquad [L^2,L_3]=0.$$

This would suggest, that $L^2,L_+$ have a common system of eigenfunctions, and so do $L^2,L_3$, but $L_+,L_3$ don't. How is that possible?

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Commutativity is not a transitive relation: If operator $A$ commutes with $B$ and $C$,

$$AB=BA \quad\text{and}\quad AC=CA,$$

then there is no reason that $B$ and $C$ should commute.

Example: Take $A=y$, $B=x$, and $C=p_x$.

In particular, if commuting selfadjoint operators $A$ and $B$ have a common basis of orthonormal eigenvectors, and if commuting selfadjoint operators $A$ and $C$ have a common basis of orthonormal eigenvectors, then these two bases need not be the same if the spectrum of $A$ is degenerate.

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  • $\begingroup$ Yes, that's what I noticed above. But the question is the following: There is the interpretation that they commute iff they have a common system of eigenfunctions. If you look at it from that perspective, then this should hold. So where does this reasoning lack? $\endgroup$ – Xin Wang Jan 26 '14 at 13:50
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    $\begingroup$ The answer is that (in the mentioned examples) the set $\{A,B,C\}$ of all three operators don't have a common basis of orthonormal eigenvectors. Only a subset of two operators have in certain cases a common basis of orthonormal eigenvectors. $\endgroup$ – Qmechanic Jan 26 '14 at 13:53
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    $\begingroup$ Another problem (in the mentioned example) is that $L_{+}$ is not a normal operator, so there doesn't exist an orthonormal basis of eigenvectors for $L_{+}$. $\endgroup$ – Qmechanic Jan 26 '14 at 14:17
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    $\begingroup$ Yes, but the eigenvectors are not orthogonal. See also this and this Phys.SE posts. $\endgroup$ – Qmechanic Jan 26 '14 at 14:30
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    $\begingroup$ Since (i) $L_{\pm}= L_x\pm i L_y$ and (ii) $L^2$, $L_x$, $L_y$ are all Hermitian, one may view the statement $[L^2,L_{+}]=0$ as equivalent to $[L^2,L_{-}]=0$, and, in turn, equivalent to the pair of statements about Hermitian operators: $[L^2,L_x]=0$ and $[L^2,L_y]=0$. The only caveat is that $L_x$ and $L_y$ do not commute! $\endgroup$ – Qmechanic Jan 26 '14 at 16:21
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NOTE. Since $L_+$ is not normal (normal means $A^\dagger A = AA^\dagger$) it does not admit a basis of orthonormal eigenvectors. However your question can be safely restated replacing $L_+$ for $L_2$ and I will assume it henceforth.

The most elementary case of this phenomenon is given by a triple of normal matrices in $\mathbb C^n$:

$$cI,A,B$$

with $[A,B]\neq 0$ and where $c\in \mathbb C$ is an arbitrarily fixed number. $A$ has a common basis of eigenvectors with $cI$: Every basis of eigenvectors of $A$ is such basis. Similarly, every basis of eigenvectors of $B$ is also a basis of eigenvectors of $cI$. However, though it could happen for some vector, there cannot exist a whole basis of eigenvectors in common with $A$ and $B$, otherwise referring to that basis $A$ and $B$ would be in diagonal form and thus $[A,B]=0$, which is forbidden by hypotheses.

All that is possible thanks to the fact that the eigenspaces of $cI$ are (maximally) degenerate. Two vectors $u$ and $v$ with the same eigenvalue ($c$) of respect to $cI$ remain eigenvectors of $cI$ with the same eigenvalue even if linearly composed: $au+bv$. Nevertheless if $u$ and $v$ are eigenvectors of $A$, in general $au+bu$ is not, but it could be an eigenvector of $B$ (remaining, as said, an eigenvector of $cI$)

The situation is essentially the same when dealing with $L^2$ and $L_2,L_3$. The eigenspaces $\cal H_l$ of $L^2$ are degenerate and, in each eigenspace, $L^2$ is represented by $l(l+1)I$. Moreover, as $[L^2,L_i]=0$, each eigenspace $\cal H_l$ is invariant under the action of $L_i$. I mean $L_i({\cal H}_l)\subset \cal H_l$.

Restricting to $\cal H_l$, we find the situation I outlined above: $L$ is represented by $cI$ and $L_2, L_3$ are represented by non commuting operators $A$ and $B$.

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  • $\begingroup$ Nice answer. A few questions if you don't mind. You state that since the operator is not normal it can'tt have orthononormal basis of eigenvectors. One of the postulates of QM assumes that the observables are Hermitian and then from this we have a theorem which states that every Hermitian operator has a basis of orthonormal eigenvectors. But you are saying that the this is not enough? So does the postulate actually state that every observable operator is in fact normal (hence also Hermitian)? $\endgroup$ – Alex Sep 9 '16 at 10:44
  • $\begingroup$ Also, why is it important in your answer to note that each eigenspace $\mathcal{H_l}$ is invariant under the action of $L_i$ as you stated "$L_i(\mathcal{H}) \subset \mathcal{H}$"? Thanks a lot. $\endgroup$ – Alex Sep 9 '16 at 10:44
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    $\begingroup$ QM assumes that every observable is self-adjoint (not simply Hermitian) and thus it is also normal. In general, in non-finite dimensions, even normal operators may have no orthonormal basis of (proper) eigenvectors. What is true is that a normal operator has a spectral decomposition. However, at the beginning of my answer the point was another. If an operator is not normal, then it cannot have an orthonormal basis of eigenvectors just because otherwise it would be normal! Since $L_+$ is not normal, it cannot have an orthonormal basis of eigenvectors. $\endgroup$ – Valter Moretti Sep 9 '16 at 11:21
  • $\begingroup$ Regarding your second question, the answer appears in my last sentence: Restricting to $\cal H_l$, we find the situation I outlined above: $L$ is represented by $cI$ and $L_2, L_3$ are represented by non commuting operators $A$ and $B$. If $\cal H_l$ were not invariant I would not be allowed to restrict the discussion to $\cal H_l$ and use the already discussed theory. $\endgroup$ – Valter Moretti Sep 9 '16 at 11:24
  • $\begingroup$ Okay I see. You state " Moreover, as $[L_2,L_i]=0$, each eigenspace $\mathcal{H}_l$ is invariant under the action of Li". As I understand it, the commutativity $[L^2,L_i]=0$ implies there exists a common basis of orthonormal eigenvectors, not that every eigenbasis is shared. You assume that $\mathcal{H}_{l}$ is an eigenspace of operator $L^2$, why does it follow then that $\mathcal{H}_{l}$ is invariant under $L_{i}$, what if $\mathcal{H}_{l}$ is some eigenbasis which is not shared. $\endgroup$ – Alex Sep 9 '16 at 15:09

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