One line of algebra replaces every case-by-case argument about which quantities can be sharp together. Two observables are compatible exactly when their operators commute.

§7.2 found that position and momentum have eigenfunctions that cannot both be sharp — but only by inspecting each in turn. This section supplies the general test, and it is one line of algebra.

The answer will be the sharpest statement in the book: $[\hat x,\hat p] = i\hbar$, from which the uncertainty principle follows as a theorem rather than an experimental fact.

7.3 How much has to be specified?

Classically, a particle in one dimension needs two numbers — position and momentum — and from them every other dynamical quantity follows. In three dimensions it needs six.

Quantum mechanically the count is different, and smaller:

Observables that can be specified together are called compatible observables compatible observables Two observables that can be sharply defined at the same time. Equivalently, two whose operators commute, and which therefore share a complete set of simultaneous eigenfunctions. defined in ch. 7 — open in glossary , and a set of them large enough to fix the state uniquely is a complete set of compatible observables complete set of compatible observables The smallest set of mutually compatible observables whose simultaneous eigenvalues label a quantum state uniquely — one for a particle in 1-D, three in 3-D. defined in ch. 7 — open in glossary . What §7.3 does not yet supply is a way to tell which observables are compatible. That is §7.4.

7.4 The commutator

The commutator commutator [Â,B̂] = ÂB̂ − B̂Â. It vanishes exactly when two observables are compatible, so it is the test for whether they can be known together. defined in ch. 7 — open in glossary of two operators is defined as

The claim — proved for $\hat x$ and $\hat p$ below, and true in general — is that $\hat A$ and $\hat B$ describe compatible observables if $[\hat A,\hat B] = 0$ and incompatible observables if $[\hat A,\hat B] \ne 0$.

The canonical commutation relation

Apply $(\hat x\hat p - \hat p\hat x)$ to an arbitrary $\Psi(x,t)$ and use the product rule:

This is the canonical commutation relation canonical commutation relation [x̂,p̂] = iħ. The single algebraic fact that forbids a state of definite position and momentum, and from which the uncertainty principle can be derived. defined in ch. 7 — open in glossary , and the book says outright that it is “so important in quantum mechanics” that it earns a name of its own.

Why no state has both

Assume the impossible: a simultaneous eigenfunction $\psi_{x'p'}$ with

Act on it with the commutator two ways. Using Eq. (7.18):

because $x'$ and $p'$ are numbers, and numbers commute. Using Eq. (7.17):

Both are correct, so $i\hbar\psi_{x'p'} = 0$, so $\psi_{x'p'} = 0$ everywhere.

And the uncertainty principle follows

Define the deviation operators $\widehat{\Delta x} = \hat x - \langle x\rangle$ and $\widehat{\Delta p} = \hat p - \langle p\rangle$. Since $\langle x\rangle$ and $\langle p\rangle$ are numbers, they commute with everything, so

Then the Schwarz inequality schwarz inequality ∫|α|²∫|β|² ≥ |∫α*β|², for any two square-integrable functions. The purely mathematical step behind the uncertainty principle: with α = ÂΨ and β = B̂Ψ it becomes ⟨A²⟩⟨B²⟩ ≥ |∫Ψ*ÂB̂Ψ|². It is Cauchy–Schwarz, and it is an equality exactly when α and β are parallel — which is why the oscillator ground state saturates ΔxΔp = ħ/2. defined in ch. 7 — open in glossary — built in problems 5 and 6 — gives

and therefore

Three dimensions, and a set that does commute

The book’s example: $\hat x$, $\hat y$ and $\hat p_z$. Since $\hat p_z$ differentiates only $z$, and $x$ and $y$ are constants as far as it is concerned, the product rule generates nothing:

so $[\hat x,\hat y] = [\hat x,\hat p_z] = [\hat y,\hat p_z] = 0$ — three mutually compatible observables, with simultaneous eigenfunctions

The pattern is the one that matters: $\hat x$ and $\hat p_x$ fail to commute, but $\hat x$ and $\hat p_z$ commute perfectly. Incompatibility is a property of a pair, not of position and momentum as categories. You may know exactly where a particle is along $x$ while knowing exactly how fast it moves along $z$.

Where this goes next

§7.5 asks what happens when an observable commutes with the Hamiltonian in particular — and the answer is that its expectation value never changes. Conservation laws turn out to be commutators that vanish, and the vanishing traces back to a symmetry of $\hat H$.