Compatible and Incompatible Observables

Can two observables be measured simultaneously, both yielding definite values? The answer is encoded in a single algebraic object: the commutator. This lesson links the order in which measurements can be performed to whether their operators commute.

The commutator

For two operators $\hat A$ and $\hat B$, the commutator is

If $[\hat A, \hat B] = 0$ the operators commute; otherwise they do not. Order of operator application is exactly like order of physical operations, so the commutator measures how much the result depends on which observable you handle first.

Compatible observables commute

Two observables are compatible if they can have a complete set of simultaneous eigenstates — states with a definite value for both at once. The fundamental theorem is:

Two Hermitian observables share a complete orthonormal basis of common eigenvectors if and only if they commute, $[\hat A, \hat B] = 0$.

One direction is quick. If a common eigenbasis $\{|n\rangle\}$ exists, with $\hat A|n\rangle = a_n|n\rangle$ and $\hat B|n\rangle = b_n|n\rangle$, then on each basis vector

so $[\hat A,\hat B] = 0$ on a complete basis, hence everywhere. The converse (commuting implies a shared eigenbasis) is the spectral content of the theorem.

Physically, compatible observables can be measured in either order with the same statistics, and measuring one does not disturb the value of the other. They define good quantum numbers that label states jointly — for instance energy, total angular momentum, and its $z$-component for a hydrogen atom.

Incompatible observables do not commute

When $[\hat A, \hat B] \neq 0$ the observables are incompatible: no state can have a sharp value of both. Measuring one disturbs the other. The canonical qubit example is the pair $\hat X$ and $\hat Z$. Compute the commutator directly:

so

The Pauli operators obey the cyclic relations $[\hat X,\hat Y] = 2i\hat Z$, $[\hat Y,\hat Z] = 2i\hat X$, $[\hat Z,\hat X] = 2i\hat Y$ — none of them commute, so no single qubit state has definite $X$, $Y$, and $Z$ spin components simultaneously. This is why measuring $\hat Z$ on the eigenstate $|+\rangle$ of $\hat X$ gives a random result: $|+\rangle$ has a sharp $X$ value but a completely uncertain $Z$ value.

The uncertainty principle

Incompatibility is quantified by the Robertson uncertainty relation: for any state,

where $\Delta A = \sqrt{\langle\hat A^2\rangle - \langle\hat A\rangle^2}$ is the standard deviation. When the commutator is nonzero, the product of uncertainties has a positive lower bound — you cannot make both arbitrarily small. For position and momentum, $[\hat x,\hat p] = i\hbar$, which gives the Heisenberg relation $\Delta x\,\Delta p \geq \hbar/2$. The familiar uncertainty principle is just this theorem applied to a particular incompatible pair.

Summary

Compatibility is not a vague notion — it is the algebraic statement $[\hat A, \hat B] = 0$. Commuting observables share an eigenbasis and can be jointly determined; non-commuting observables obey an uncertainty relation that forbids simultaneous sharpness. The commutator thus controls both the structure of states and the limits of measurement.

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