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Do commuting operators always share the same eigenfunctions? Could you kindly explain to me if this is true or false and examples for this, as I am a little confused.

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    $\begingroup$ Related: math.stackexchange.com/questions/1227031/… $\endgroup$ – Tyberius Jun 4 at 22:25
  • $\begingroup$ i'm still confused? $\endgroup$ – Ezra Jun 5 at 16:42
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    $\begingroup$ as a quick summary, no. This is only true if both operators are diagonalizable (can be transformed such that they form a diagonal matrix of the eigenvalues). A common subcase of being diagonalizable is when there are no repeated eigenvalues. Some operators with repeated eigenvalues can be diagonalized, but this is a more involved proof. $\endgroup$ – Tyberius Jun 5 at 16:47
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If two operators commute, then there exists a basis for the space that is simultaneously an eigenbasis for both operators.

However, if one of the operators has two eigenvectors with the same eigenvalue, any linear combination of those two eigenvectors is also an eigenvector of that operator, but that linear combination might not be an eigenvector of the second operator.

As an example consider the states $|l,m\rangle = |1,1\rangle$ and $|l,m\rangle = |1,-1\rangle$. These are both eigenvectors of $\hat{L}^2$ and $\hat{L}_z$ (the angular momentum operators).

The eigenvalues of $\hat{L}_z$ are $\hbar$ and $-\hbar$, respectively, but the state $$|1,1\rangle + |1,-1\rangle$$ is clearly not an eigenvector of $\hat{L}_z$, but it is still an eigenvector of $\hat{L}^2$. This is probably why we use the term "mutual" rather than "same". Finally, we form linear combinations of the $|1,m\rangle$ states to get the $l=1$ states that are eigenvectors of, say, $\hat{L}_y$.

As a straight-forward example, consider the following two matrices: $$L = \left[ {\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}} \right]$$ and $$Z = \left[ {\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}} \right].$$ The vectors $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ are eigenvectors of both operators. The latter two are eigenvectors of $L$ with the same eigenvalue (namely $1$), but they are eigenvectors of $Z$ with different eigenvalues (namely, $1$ and $-1$). If we instead use the vectors $(0,1,1)$ and $(0,1,-1)$, these are still eigenvectors of $L$ with eigenvalue $1$, but they are no longer eigenvectors of $Z$, as you can verify.

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Let operators $\hat{A}$ and $\hat{B}$ commute, $[\hat{A},\hat{B}]=0$, and consider the eigenvalue equation for $\hat{A}$: $$ \hat{A}|\psi\rangle=\lambda|\psi\rangle. $$ The first thing to prove is that $\hat{B}|\psi\rangle$ is also an eigenstate of $\hat{A}$ with the same eigenvalue. To do this: $$ \hat{A}(\hat{B}|\psi\rangle)=\hat{B}(\hat{A}|\psi\rangle)=\hat{B}(\lambda|\psi\rangle)=\lambda(\hat{B}|\psi\rangle). $$ If $\lambda$ is a non-degenerate eigenvalue, this means that $\hat{B}|\psi\rangle\propto|\psi\rangle$ (as for non-degenerate $\lambda$ all eigenstates must be the same up to a constant), so we can write: $$ \hat{B}|\psi\rangle=\mu|\psi\rangle, $$ where $\mu$ is the proportionality constant. This shows that $|\psi\rangle$ is also an eigenstate of $\hat{B}$ (with eigenvalue $\mu$). So commuting observables share the same eigenstates if the associated eigenvalues are non-degenerate.

If $\lambda$ is degenerate, then $|\psi\rangle$ need no longer be an eigenstate of $\hat{B}$. To see this, let us have $g$ linearly independent eigenstates, $|\psi^i\rangle$ for $i=1,\ldots,g$, with the same eigenvalue $\lambda$. Construct an arbitrary linear superposition within the $g$-fold degenerate subspace:

$$ |\chi\rangle=\sum_{i=1}^gc_i|\psi^i\rangle $$ for arbitrary coefficients $c_i$. Then, $$ \hat{A}|\chi\rangle=\hat{A}\left(\sum_{i=1}^gc_i|\psi^i\rangle\right)=\sum_{i=1}^gc_i\hat{A}|\psi^i\rangle=\sum_{i=1}^gc_i\lambda|\psi^i\rangle=\lambda\left(\sum_{i=1}^gc_i|\psi^i\rangle\right)=\lambda|\chi\rangle. $$

This means that any linear combination within the degenerate subspace is also an eigenstate of $\hat{A}$ in that subspace with the same eigenvalue $\lambda$. In turn, this implies that all we can say about $\hat{B}|\psi\rangle$ is that it belongs to this $g$-fold degenerate subspace. This means that $\hat{B}|\psi\rangle$ could be proportional to the result of one of these linear combinations rather than one of the original eigenstates, showing that for degenerate eigenvalues the eigenstates of $\hat{A}$ and $\hat{B}$ need not be the same.

However, as the action of $\hat{B}$ on $|\psi\rangle$ is restricted to the degenerate subspace, then you can always diagonalize $\hat{B}$ within that subspace to find the eigenstates of $\hat{B}$. These eigenstates will also be eigenstates of $\hat{A}$ because they are simply linear combinations of the degenerate eigenstates, and as we have seen any such linear combination is also an eigenstate of $\hat{A}$. This means that you can always choose an appropriate linear combination of the original eigenstates of $\hat{A}$ to create a new set of eigenstates that are eigenstates of $\hat{B}$ too.

This answer is adapted from one I wrote in the Physics Stack Exchange which asked specifically about the example in which $\hat{A}$ is the Hamiltonian and $\hat{B}$ the parity operator.

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