The Compatible Observables Theorem

Theorem 2.3.2 shows one direction: if two operators share an eigenbasis, they commute. The converse is what makes commutators useful, and the boards use it without proof. Collected here, for Hermitian $\hat{A}$ and $\hat{B}$ the following are equivalent:

$[\hat{A}, \hat{B}] = 0$ (Definition 2.3.1).
$\hat{A}$ and $\hat{B}$ have a common orthonormal basis of eigenstates.
Measuring $\hat{A}$ , then $\hat{B}$ , then $\hat{A}$ again gives the first result again: the second measurement does not disturb the first.
$\hat{A}$ and $\hat{B}$ can be known together: every state is a sum of states with $Δ A = Δ B = 0$ , and Theorem 2.5.1 puts no lower bound on $Δ A Δ B$ .

Such observables are called compatible.

TheoremCommuting Hermitian operators share an eigenbasisbeyond the boards

If $\hat{A}$ and $\hat{B}$ are Hermitian and $[\hat{A}, \hat{B}] = 0$ , there is an orthonormal basis whose every vector is an eigenvector of both.

Proof

Let $\hat{A} u = a u$ . Then $\hat{A} (\hat{B} u) = \hat{B} \hat{A} u = a (\hat{B} u)$ : $\hat{B} u$ is again an eigenvector of $\hat{A}$ with the same $a$ , or zero. == $\hat{B}$ maps each eigenspace of $\hat{A}$ into itself.==

Non-degenerate $a$ . The eigenspace is one line, so $\hat{B} u$ is a multiple of $u$ : $\hat{B} u = b u$ , and $u$ is already a common eigenvector.

Degenerate $a$ . The eigenspace $E_{a}$ has dimension $g > 1$ , and inside it $\hat{A}$ is just $a$ times the identity: any vector of $E_{a}$ is an eigenvector of $\hat{A}$ . Restricted to $E_{a}$ , $\hat{B}$ is still Hermitian, since $⟨ f ∣ \hat{B} g ⟩ = ⟨ \hat{B} f ∣ g ⟩$ holds for all $f, g$ and so for those in $E_{a}$ . A Hermitian operator on a $g$ -dimensional space has $g$ orthonormal eigenvectors (Theorem 1.4.2). They are eigenvectors of $\hat{B}$ by construction and of $\hat{A}$ because they lie in $E_{a}$ .

Doing this in every eigenspace of $\hat{A}$ , and recalling that eigenspaces of a Hermitian operator for different eigenvalues are orthogonal, gives the common orthonormal basis.

The degenerate case is the one that matters. An eigenvector of $\hat{A}$ chosen at random inside a degenerate eigenspace is in general not an eigenvector of $\hat{B}$ ; only the right combinations are. That is why a degenerate level is labelled by the eigenvalues of a second, commuting operator:

the free particle: $e^{\pm ik x}$ have the same energy, and $\overset{p}{^}_{x}$ picks them apart (The free particle);
angular momentum: $\hat{J}^{2}$ and $\hat{J}_{z}$ commute, and $∣ j, m ⟩$ is their common basis (Theorem 4.2.2);
two spins: $∣ s_{1}, m_{1} ⟩ ∣ s_{2}, m_{2} ⟩$ and $∣ s, m ⟩$ are two different common bases, of two different sets of commuting operators (Definition 4.4.1);
identical particles: $\hat{H}$ commutes with exchange, so its eigenstates can be chosen symmetric or antisymmetric (Theorem 5.1.2).

Measurement

With a common basis ${u_{n}}$ , $\hat{A} u_{n} = a_{n} u_{n}$ and $\hat{B} u_{n} = b_{n} u_{n}$ (here $b_{n}$ are the eigenvalues of $\hat{B}$ , not components). Measuring $\hat{A}$ and getting $a$ leaves the state in $E_{a}$ ; measuring $\hat{B}$ then picks a basis vector of $E_{a}$ , still in $E_{a}$ , so a second measurement of $\hat{A}$ gives $a$ again. If $\hat{A}$ and $\hat{B}$ do not commute, the measurement of $\hat{B}$ leaves an eigenstate of $\hat{B}$ that is in general a mixture of eigenstates of $\hat{A}$ , and the first result is lost (Example 2.2.9).