Compatible Observables

In classical mechanics you can, in principle, measure every property of a system — position, momentum, energy, angular momentum — all at once and with arbitrary precision. Quantum mechanics denies that privilege: certain pairs of observables are incompatible, meaning measuring one necessarily disturbs the other. Other pairs are compatible, meaning they can be measured simultaneously without any fundamental trade-off. The key to telling these two cases apart is the commutator.

The commutator

Given two operators $\hat{A}$ and $\hat{B}$ , their commutator is defined as

$[\hat{A},\hat{B}] \equiv \hat{A}\hat{B} - \hat{B}\hat{A}.$

The commutator is itself an operator. If $[\hat{A},\hat{B}] = 0$ (the zero operator), the operators commute. If $[\hat{A},\hat{B}] \neq 0$ , they do not.

The order of operator products matters because operators act on states, and applying $\hat{A}$ after $\hat{B}$ can leave the state in a different place than applying $\hat{B}$ after $\hat{A}$ . For ordinary numbers $ab - ba = 0$ always, so commutativity is automatic. For operators it is a non-trivial constraint.

Compatible observables share eigenstates

Two Hermitian operators $\hat{A}$ and $\hat{B}$ are called compatible if and only if they commute: $[\hat{A},\hat{B}] = 0$ . The deep reason this matters is the following theorem:

Simultaneous eigenstate theorem. If $[\hat{A},\hat{B}] = 0$ and both operators are Hermitian, there exists a complete orthonormal basis of states $|n\rangle$ satisfying $\hat{A}|n\rangle = a_n|n\rangle$ and $\hat{B}|n\rangle = b_n|n\rangle$ simultaneously.

A state that is simultaneously an eigenstate of both $\hat{A}$ and $\hat{B}$ has a definite value for both observables: measuring either quantity returns the corresponding eigenvalue with probability one, and neither measurement disturbs the other. This is the precise meaning of "simultaneously measurable."

The proof proceeds in one direction by acting with $\hat{A}$ on the eigenvalue equation for $\hat{B}$ : if $\hat{B}|n\rangle = b_n|n\rangle$ and $[\hat{A},\hat{B}] = 0$ , then $\hat{B}(\hat{A}|n\rangle) = \hat{A}(\hat{B}|n\rangle) = b_n(\hat{A}|n\rangle)$ , so $\hat{A}|n\rangle$ is again an eigenstate of $\hat{B}$ with the same eigenvalue $b_n$ . Within each eigenspace of $\hat{B}$ , the operator $\hat{A}$ acts and can be diagonalised, yielding the simultaneous eigenstates.

Incompatible observables and the uncertainty principle

When $[\hat{A},\hat{B}] \neq 0$ the two observables are incompatible: no complete basis of simultaneous eigenstates exists, so they cannot be jointly diagonalised (individual states may still happen to be eigenstates of both — for instance any state annihilated by both operators). The Robertson uncertainty relation makes this quantitative. For any state $|\psi\rangle$ ,

$\sigma_A \,\sigma_B \;\geq\; \frac{1}{2}\left|\langle[\hat{A},\hat{B}]\rangle\right|,$

where $\sigma_A$ and $\sigma_B$ are the standard deviations of the two observables in the state $|\psi\rangle$ , and the angle brackets denote the expectation value. If $[\hat{A},\hat{B}] = 0$ the right-hand side vanishes for every state, recovering the compatible case; if the commutator is a nonzero operator the right-hand side can still vanish in particular states (those where $\langle[\hat{A},\hat{B}]\rangle = 0$ ), so the bound is informative only where that expectation is nonzero.

The canonical example is position $\hat{x}$ and momentum $\hat{p}$ in one dimension. Their commutator is

$[\hat{x},\hat{p}] = i\hbar,$

where $\hbar$ is the reduced Planck constant. Inserting this into the Robertson relation gives

$\sigma_x\,\sigma_p \;\geq\; \frac{\hbar}{2},$

the famous Heisenberg uncertainty principle: there is no quantum state in which position and momentum are simultaneously sharp. The non-commutativity of $\hat{x}$ and $\hat{p}$ is not a statement about experimental imprecision — it is a structural property of the operators.

A complete set of commuting observables

In practice, specifying a quantum state completely requires enough compatible observables to fix every degree of freedom of the system. A complete set of commuting observables (CSCO) is a collection of mutually commuting Hermitian operators whose joint eigenstates are unique (up to phase). For the hydrogen atom, for instance, the Hamiltonian $\hat{H}$ , the total angular momentum squared $\hat{L}^2$ , and the $z$ -component of angular momentum $\hat{L}_z$ all commute with one another and together constitute a CSCO: the quantum numbers $n$ , $\ell$ , $m_\ell$ label a unique energy eigenstate.