Commuting Operators and Common Eigenbases

When can two observables both have definite values at once? The answer is one of the cleanest results in quantum mechanics: exactly when they commute. Commutation is the algebraic condition; sharing a common eigenbasis is its geometric meaning. This single equivalence underlies good quantum numbers, complete sets of commuting observables, and the uncertainty principle.

The commutator

The commutator of two operators measures their failure to commute:

[A, B] = AB - BA.

If $[A,B] = 0$ we say $A$ and $B$ commute; then $AB = BA$ and the order of application does not matter. The commutator is itself an operator, antisymmetric in its arguments ( $[A,B] = -[B,A]$ ) and linear in each. For the Pauli operators, direct multiplication gives the non-trivial relations $[X, Y] = 2iZ$ , $[Y, Z] = 2iX$ , $[Z, X] = 2iY$ — none of the three commute with each other, which is why a qubit cannot have definite $x$ -, $y$ -, and $z$ -spin simultaneously.

The theorem

Two Hermitian operators are simultaneously diagonalizable if and only if they commute.

That is, $[A,B] = 0$ exactly when there exists a single orthonormal basis $\{|n\rangle\}$ in which both are diagonal:

A|n\rangle = a_n |n\rangle, \qquad B|n\rangle = b_n |n\rangle.

Such a basis is a common eigenbasis. Each basis vector carries a definite value of $A$ and a definite value of $B$ at the same time.

Why commuting forces a shared basis

One direction is quick. Suppose a common eigenbasis exists. Then on each $|n\rangle$ ,

AB|n\rangle = A(b_n|n\rangle) = a_n b_n |n\rangle = b_n a_n |n\rangle = B(a_n|n\rangle) = BA|n\rangle.

Since $AB$ and $BA$ agree on every vector of a basis, they agree everywhere: $[A,B] = 0$ .

The converse takes a little more work. Assume $[A,B]=0$ and let $|a\rangle$ be an eigenvector of $A$ with eigenvalue $a$ . Then

A\big(B|a\rangle\big) = B\big(A|a\rangle\big) = a\,\big(B|a\rangle\big),

so $B|a\rangle$ is also an eigenvector of $A$ with the same eigenvalue $a$ — it stays inside the $a$ -eigenspace. Thus $B$ maps each eigenspace of $A$ into itself. Restricting the Hermitian operator $B$ to that subspace and diagonalizing it there produces vectors that are eigenvectors of both $A$ and $B$ . Doing this in every eigenspace assembles a common eigenbasis.

Compatible observables and good quantum numbers

Commuting observables are called compatible: they can be measured together without one disturbing the other's value, and their joint eigenvalues $(a_n, b_n)$ serve as quantum numbers labeling the states. A maximal set of mutually commuting observables — a complete set of commuting observables (CSCO) — has a common eigenbasis in which each state is uniquely tagged by its list of eigenvalues. This is how, for example, the hydrogen atom's states are labeled by $(n, \ell, m)$ : the corresponding operators all commute.

The link to uncertainty

When operators do not commute, no common eigenbasis exists, so no state can have a sharp value of both. The Robertson uncertainty relation quantifies this,

\Delta A\,\Delta B \ge \tfrac{1}{2}\,\big|\langle [A,B]\rangle\big|,

with the lower bound set by the expectation of the commutator. For position and momentum, $[\hat x, \hat p] = i\hbar$ , giving the Heisenberg bound $\Delta x\,\Delta p \ge \hbar/2$ . The non-vanishing commutator is the precise reason the two quantities cannot be simultaneously definite.

Try it

Compute the commutator $[X, Z] = XZ - ZX$ of the Pauli operators and return its entry in row $0$ , column $1$ . Since the result is nonzero, $X$ and $Z$ do not share a common eigenbasis.

Run your code to see the quantum state.