Some pairs of quantities can be known together and some cannot. The dividing line is the commutator: observables that commute share a basis of eigenvectors, so a state can be sharp for both, measuring one does not disturb the other, and the order of measuring them does not matter. A set of commuting observables whose values pin down a single state is the fullest description quantum mechanics allows.

The latitude and longitude of a city can both be read off one map, and reading one does not smudge the other. Commuting observables are like that: coordinates of the same grid.

Two observables are compatible when they commute. Then they have a common orthonormal basis of eigenvectors, and conversely. In a common eigenvector both have definite values, and measuring one leaves the result of the other unchanged.

• The common eigenvectors are labelled by both eigenvalues: $\ket{a,b}$.
• Measuring $\hat{A}$ then $\hat{B}$ gives the same statistics as then , and a repeat of the first still gives its first result.

Apply $\hat{A}$ to $\hat{B}\ket{a}$ and swap the order, which commuting allows. The result says $\hat{B}\ket{a}$ is again an eigenvector of with the same eigenvalue . Since is not degenerate, that eigenvector is a multiple of — which makes an eigenvector of . The degenerate case needs one more step, diagonalising inside each eigenspace of .

The two operators commute, so their order can be exchanged.

Use $\hat{A}\ket{a}=a\ket{a}$ and pull the number out.

So $\hat{B}\ket{a}$ is either zero or an eigenvector of $\hat{A}$ with the same eigenvalue.

Non-degenerate means those eigenvectors are multiples of $\ket{a}$; zero is the multiple $b=0$.

Two observables can be sharp together in a whole basis of states exactly when they commute. Such observables are called compatible.

Commuting observables have a common orthonormal basis of eigenvectors, in which both are diagonal. In each of those states both have definite values.

If $\hat{A}$ and $\hat{B}$ commute and $v$ is an eigenvector of $\hat{A}$ with eigenvalue , then is also an eigenvector of with eigenvalue , or zero.

A complete set of commuting observables labels every basis state by its list of eigenvalues, with no two states sharing a list.

Great work! You now know how to:

• test two observables for compatibility with their commutator
• prove that commuting observables share eigenvectors
• label states by a complete set of commuting observables