Common eigenfunctions of commuting operators: case of degeneracy

As proved in the answer to this post, if the operators $$\hat A$$ and $$\hat B$$ commute, then they have the same eigenstates.

Let $$\hat A\psi_{A_i}=A_i\psi_{A_i}\qquad \Rightarrow\qquad \hat B\hat A\psi_{A_i}=\hat B(A_i\psi_{A_i})=A_i\hat B\psi_{A_i}\equiv > A_i\phi .$$ Now, due to the vanishing of the commutator we have that $$\hat B\hat A\psi_{A_i}=\hat A\hat B\psi_{A_i}=\hat A\phi$$ From the RHS of the last equations, we have that $$\hat A\phi=A_i\phi,$$ meaning that $$\phi$$ is also an eigenstate of $$\hat A$$ with eigenvalue $$A_i$$. This could happen for the following reasons:

1. $$\phi=c\psi_{A_i}$$, with $$c$$ a constant. Hence, commuting operators have simultaneous eigenstates.
2. $$\phi\neq c\psi_{A_i}$$. In this case the operator $$\hat A$$ must have degenerate eigenstates, namely $$\phi$$ and $$\psi_{A_i}$$. Even at this case, the non-degenerate eigenstates of $$\hat A$$ are simultaneously eigenstates of $$\hat B$$.

However, what happens in the second case, where the eigenvalue $$A_i$$ is degenerate? Can we say that the eigenfunctions corresponding to the degenerate eigenvalue $$A_i$$ are not eigenfunctions of $$\hat B$$? Do we know something else about them?

If $$[\hat A,\hat B]=0$$ and they are both non-degenerate, then every eigenstate of $$\hat A$$ is an eigenstate of $$\hat B$$ and vice versa.

If $$[\hat A,\hat B]=0$$ and $$\hat A$$ has a degenerate spectrum, then you are guaranteed the existence of one common eigenbasis. However, you are not guaranteed that every eigenstate of $$\hat A$$ will be an eigenstate of $$\hat B$$.

As a simple counterexample to illustrate that last statement, take the operators $$\hat A = \begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix} \quad\text{and}\quad \hat B = \begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix},$$ for which $$(1,0,0)^T$$ is an eigenstate of $$\hat{A}$$ but not $$\hat B$$ even though $$\hat A\hat B=\hat B\hat A=\hat B$$.

If the information you have is that $$[\hat A,\hat B]=0$$, $$\hat A$$ has a degenerate spectrum and $$v$$ is an eigenstate of $$\hat A$$ in a space with degenerate eigenvalue, then you cannot make any inferences about its relationship to $$\hat B$$ $$-$$ it might be an eigenstate, or it might not.

• Is this true even if the operators aren't Hermitian? Aug 28, 2020 at 13:21
• @Philip In principle, yes, but you need additional external guarantees that they both have eigenstates to begin with. That pulls you into a game of provide-the-minimal-assumptions which is for mathematicians and people with too much time on their hands to play. Aug 28, 2020 at 13:25
• That's what I thought, I wasn't completely sure :) Aug 28, 2020 at 13:26

In your case you seem to have defined $$\phi_i = \hat{B}\psi_i$$, where $$i=1,2,3,\dots N$$ is the degree of degeneracy.

It should be clear to you that the states $$\phi_i$$ are still eigenstates of $$\hat{A}$$. However, there is no reason for them to, a priori, be eigenstates of $$\hat{B}$$. In fact, since every $$\phi_i$$ is an eigenstate of $$\hat{A}$$, you can write it as a linear combination of the "degenerate" eigenstates of $$\hat{A}$$, $$\psi_i$$. The action of $$\hat{B}$$ could then be for example to take one eigenstate to a different one. (You could have, say, $$\hat{B}\psi_1 = \psi_2$$, for example.)

Therefore in general I don't think there is anything special that can be said in this case without any further information. However, if the operators $$\hat{A}$$ and $$\hat{B}$$ are Hermitian, then we are guaranteed that we can diagonalise $$\hat{B}$$ within this subspace spanned by $$\psi_i$$, and therefore there exists at least $$N$$ linear combinations of the $$\psi_i$$s that are also eigenstates of $$\hat{B}$$.

In other words, in the case of Hermitian operators, at least one simultaneous eigenbasis can be found.

Example: Consider the Hamiltonian for a free particle: $$\hat{H} = \frac{\hat{p}^2}{2m}.$$

Clearly, $$\hat{H}$$ and $$\hat{p}$$ commute, but not all states of definite energy are states of definite momentum. For example, a state $$|E_1\rangle \propto |p\rangle + |-p\rangle$$ would have the same energy as the state $$|E_2\rangle \propto |p\rangle - |-p\rangle$$ and so on. However, clearly there is one basis (the basis of $$|p_i \rangle$$) which is a simultaneous eigenbasis of both $$\hat{H}$$ and $$\hat{p}$$.

When one of the two commuting operators has degenerate eigenfunctions, one can always construct their linear combinations which will be the eigenfunctions of the other operator.