# Commutator not transitive

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 $C$,

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

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

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

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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? –  Xin Wang Jan 26 '14 at 13:50
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. –  Qmechanic Jan 26 '14 at 13:53
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 $L_{+}$. –  Qmechanic Jan 26 '14 at 14:17
Yes, but the eigenvectors are not orthogonal. See also this and this Phys.SE posts. –  Qmechanic Jan 26 '14 at 14:30
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! –  Qmechanic Jan 26 '14 at 16:21

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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