Quantum Mechanics – Common Eigenfunctions of Commuting Operators in Degeneracy Cases

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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?

Best Answer

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.

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