Commuting operators

Consider two generic operators $\hat{A}$ and $\hat{B}$, and for simplicity assume that they both have a set of non-degenerate eigenstates, $\{\vert\psi^a\rangle \}$ and $\{\vert\psi^b\rangle\}$, so that we have $\hat{A} \vert\psi_j^a\rangle = a_j \vert\psi_j^a\rangle$ and $\hat{B} \vert\psi_j^b\rangle = b_j \vert\psi_j^b\rangle$, where the $a_j$ and $b_j$ are the corresponding eigenvalues. Any wavefunction can be expressed as a linear combination of either set of eigenstates, e.g.:

$$\vert \psi \rangle = \sum_j c_j \vert\psi^a_j \rangle,$$ (12.1)

with the sum running over all states and $c_j = \langle \psi_j^a \vert \psi \rangle$ ^12.1. Consider the commutator:

$$[\hat{A},\hat{B}] \vert\psi \rangle = [\hat{A}\hat{B} - \hat{B}\h... ...i \rangle = \sum_j [\hat{A}\hat{B} - \hat{B}\hat{A}] c_j \vert\psi^a_j \rangle.$$ (12.2)

This quantity is zero if every term in the sum is zero, which is the case if and only if $\{\vert\psi^a\rangle \}$ are also eigenstates of $\hat{B}$ and so $\{\vert\psi^b\rangle\} \equiv \{\vert\psi^a\rangle\}$. Let us prove the two if's of this statement, and let us begin with the assumption and the eigenstates of $\hat{A}$ are also eigenstates of $\hat{B}$:

$$\hat{A}\hat{B} \vert\psi^a_j \rangle - \hat{B}\hat{A} \vert\psi^a... ..._j \rangle = a_j b_j \vert\psi^a_j \rangle - b_j a_j \vert\psi^a_j \rangle = 0,$$ (12.3)

which proves it. Now consider:

$$0 = \hat{A}\hat{B} \vert\psi^a_j \rangle - \hat{B}\hat{A} \vert\p... ...gle = \hat{A} \hat{B} \vert\psi^a_j \rangle - \hat{B}a_j \vert\psi^a_j \rangle,$$ (12.4)

which shows that $\hat{B} \vert\psi^a_j \rangle$ must be proportional to an eigenstate of $\hat{A}$. Since we assumed that the states are not degenerate $\hat{B} \vert\psi^a_j \rangle$ must be proportional to $\vert\psi^a_j\rangle$, meaning that $\vert\psi^a_j\rangle$ is also an eigenstate of $\hat{B}$. For the degenerate case the equivalent statement is that is always possible to find a common set of eigenstates for two commuting operators. This is achieved by forming appropriate linear combinations of eigenstates in the respective degenerate subsets (any linear combination of eigenstates in the degenerate set is still an eigenstate with the same eigenvalue). The proof of this statement is only slightly more involved but will not be given here.