Angular momentum eigenfunctions and commutators

Question

If two operators commute, they have a common set of eigenfunctions (right?). As $[L^{2}, L_{x}] = 0$, $[L^{2}, L_{z}] = 0$, $L^{2}$ and $L_{x}$ have a common set and $L^{2}$ and $L_{z}$ have a common set. But $[L_{z}, L_{x}] = i\hbar L_{y}$, therefore, they don't have a common set. Can someone explain why there is no actual inconsistency here?

Answer 1

Your reasoning is valid only if the eigenvalues of the operators in question are non-degenerate.

The meaning of two operators commuting is that they preserve each other's eigensubspaces. That is, we can construct eigensubspaces such that any linear combination of vectors from a given eigensubspace of one of the operators is also an eigenvector of the other operator. If all your operators only had one eigenvector corresponding to each eigenvalue, your line of thinking would be valid. However, consider the following explicit operators

$ A = \begin{pmatrix}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \\ \end{pmatrix}, B = \begin{pmatrix}\lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \gamma \\ \end{pmatrix}, C = \begin{pmatrix}p & q & 0 \\ r & s & 0 \\ 0 & 0 & t \\ \end{pmatrix} $

You can see that $[A,B] = [B, C] = 0$ but $[A,C]\neq 0$. What is happening here is that because $A$ has different eigenvalues for $(1,0,0)$ and $(0,1,0)$ and $C$ "mixes together" these eigenvectors, it does not commute with $A$.

However, $C$ does commute with $B$ because there exists a different basis where $C$ is diagonal and $B$ remains diagonal too. Essentially, any basis change where you use linear combinations of $(1,0,0)$ and $(0,1,0)$ to write new basis vectors is one where $B$ stays diagonal.

$L^2$ has degenerate eigenvalues and this resolves your perceived inconsistency.

Answer 2

You can always select a set which is either an eigenfunction of $L^{2}$and $L_{z}$ or eigenfunction of $L^{2}$and $L_{x}$ or of $L^{2}$and $L_{y}$. These three sets are differnt one can be obtained from another by rotation of the axis. However you cannot select one that is eigenfunction of $L^{2}$, $L_{z}$ and $L_{y}$ for instance because $L_{z}$ and $L_{y}$ do not commute. So there is no problem.

Answer 3

Zoe likes both chocolate ice cream, and vanilla ice cream, but not mixtures.

Xavier doesn't like either chocolate or vanilla on their own, but when they are combined, he loves that.

Linda likes all combinations of chocolate and vanilla flavours.

Therefore

Zoe and Linda have tastes in common; they can both enjoy the same ice cream.

Xavier and Linda have tastes in common; they can both enjoy the same ice cream.

There is no flavour that Zoe and Xavier both like (except the special case of no ice cream, when they are not hungry).