The Canonical Commutator and Compatible Observables
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You should be comfortable applying linear operators to wavefunctions and calculating expectation values using the inner product formalism introduced in the previous lecture.
Why this matters
The commutator is the algebraic engine that dictates what can and cannot be known simultaneously in quantum mechanics.
In classical mechanics, the order in which we measure position and momentum is irrelevant; they are just numbers. In quantum mechanics, observables are operators, and the order of operations matters. The canonical commutator \([ \hat{x}, \hat{p} ]\) is the fundamental non-zero commutator in the theory. It is not merely a mathematical curiosity; it is the direct source of the Heisenberg uncertainty principle and the generator of spatial translations. Understanding which operators commute (compatible observables) and which do not (incompatible observables) allows us to classify states into simultaneous eigenstates or superpositions, forming the backbone of the Hilbert space structure.
The Order of Operations
Imagine two machines in a factory. Machine A stamps a date on a box, and Machine B paints the box. If you run Machine A first, then Machine B, the date is painted over. If you run Machine B first, then Machine A, the date is stamped on the paint. The final result depends on the order. In quantum mechanics, operators act on wavefunctions. If the result of applying operator \(\hat{A}\) then \(\hat{B}\) is different from applying \(\hat{B}\) then \(\hat{A}\), the operators do not commute. The commutator \([\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}\) measures this difference. If the commutator is zero, the order doesn't matter, and the observables are "compatible." If it is non-zero, they are "incompatible," meaning we cannot have a state that is a precise eigenstate of both simultaneously.
The Canonical Commutator
The most fundamental pair of incompatible observables is position \(\hat{x}\) and momentum \(\hat{p}\). In the position representation, \(\hat{x}\) acts by multiplication, while \(\hat{p}\) acts as a derivative. Because differentiation and multiplication do not commute (think of the product rule in calculus), their operators do not commute either.
Using the definition of the momentum operator \(\hat{p} = -i\hbar \frac{\partial}{\partial x}\), we can evaluate the commutator acting on an arbitrary differentiable wavefunction \(\psi(x)\). The derivation (see D-142) shows that the derivative term from the product rule cancels out, leaving a constant multiple of the original wavefunction.
Since this holds for any \(\psi\), we write the operator identity:
Here \(\hat{I}\) is the identity operator. This result is profound: the non-commutativity is exactly one quantum of action, \(\hbar\). It is a universal constant, independent of the specific system or state.
Compatible Observables and Simultaneous Eigenstates
Two observables \(\hat{A}\) and \(\hat{B}\) are said to be compatible if their commutator vanishes: \([\hat{A}, \hat{B}] = 0\). A key theorem in quantum mechanics states that if two Hermitian operators commute, they share a complete set of simultaneous eigenfunctions. This means we can find states that are eigenstates of both operators at once.
For example, consider the Hamiltonian \(\hat{H}\) and the angular momentum operator \(\hat{L}^2\) for a central potential. Since the potential depends only on \(r\), it is rotationally symmetric, and \([\hat{H}, \hat{L}^2] = 0\). Therefore, we can label states by energy \(E\) and angular momentum quantum numbers \(l, m\) simultaneously. These are "good quantum numbers."
Conversely, because \([\hat{x}, \hat{p}] \neq 0\), there are no states that are exact eigenstates of both position and momentum. A state with a definite position is a superposition of all momenta, and a state with a definite momentum is spread out over all space. This is the operational meaning of incompatibility.
Physical Consequences: Translations and Uncertainty
The canonical commutator has two major physical implications. First, it identifies momentum as the generator of spatial translations. The unitary operator that shifts a wavefunction by a distance \(a\) is \(U(a) = e^{-i \hat{p} a / \hbar}\). This is a direct consequence of \([\hat{x}, \hat{p}] = i\hbar\). If you expand the exponential, the commutator ensures that the position operator shifts correctly: \(U^\dagger(a) \hat{x} U(a) = \hat{x} + a\), so that \( \langle x \rangle \) in the shifted state \( U(a)|\psi\rangle \) is larger by \( a \).
Second, the commutator sets the scale for the uncertainty principle. The generalised uncertainty relation states that for any two observables \(\hat{A}\) and \(\hat{B}\):
Applying this to \(\hat{x}\) and \(\hat{p}\), and noting that the commutator is a constant \(i\hbar\), we recover the Heisenberg uncertainty principle:
This inequality is not a limitation of measurement technology; it is a fundamental property of the quantum state space. The non-zero commutator is the algebraic reason why position and momentum cannot both be sharply defined.
The derivation
The Canonical Commutator [x,p] = i hbar
Working in the position representation, where the momentum operator acts on a differentiable wavefunction as p̂ = −iℏ ∂/∂x and position acts multiplicatively as x̂ ψ = xψ, the two operators fail to commute by exactly one quantum of action: [x̂, p̂] = x̂p̂ − p̂x̂ = iℏ, understood as the operator that multiplies any state by the constant iℏ.
Students often confuse the commutator \([\hat{x}, \hat{p}]\) with the anti-commutator \(\{\hat{x}, \hat{p}\} = \hat{x}\hat{p} + \hat{p}\hat{x}\). The anti-commutator is not a multiple of the identity: it is a genuine operator (Hermitian, because \( \hat{x} \) and \( \hat{p} \) are), and its expectation value depends on the state. Only the commutator yields the universal constant \(i\hbar\). Be careful not to mix up the signs or the order of operators when expanding products; the derivative acts on everything to its right, including the \(x\) in \(\hat{x}\hat{p}\psi\). C
Worked example
Problem. Consider a particle in one dimension. Calculate the commutator \([\hat{x}, \hat{p}^2]\) using the canonical commutator \([\hat{x}, \hat{p}] = i\hbar\). Then, determine if \(\hat{x}\) and \(\hat{p}^2\) are compatible observables.
- Use the identity \([\hat{A}, \hat{B}\hat{C}] = \hat{B}[\hat{A}, \hat{C}] + [\hat{A}, \hat{B}]\hat{C}\). Here \(\hat{A}=\hat{x}\), \(\hat{B}=\hat{p}\), \(\hat{C}=\hat{p}\).
- Substitute the known commutator \([\hat{x}, \hat{p}] = i\hbar\).
- Sum the terms to find the final operator.
- Check if the result is zero to determine compatibility.
Units check. \(\hat{x}\) has units of \(\mathrm{m}\), \(\hat{p}^2\) has units of \(\mathrm{kg^2 \ m^2 / s^2}\). The LHS has units \(\mathrm{kg^2 \ m^3 / s^2}\). The RHS: \(\hbar\) is \(\mathrm{kg \ m^2 / s}\), \(\hat{p}\) is \(\mathrm{kg \ m / s}\). Product is \(\mathrm{kg^2 \ m^3 / s^2}\). Units match.
Three quick tests
- Does the commutator \([\hat{p}, \hat{p}]\) vanish?
Answer
Yes, any operator commutes with itself, so \([\hat{p}, \hat{p}] = 0\). - If \([\hat{A}, \hat{B}] = 0\), can we find a state that is an eigenstate of both \(\hat{A}\) and \(\hat{B}\)?
Answer
Yes, provided the operators are Hermitian, they share a complete set of simultaneous eigenfunctions. - What is the physical significance of the constant \(i\hbar\) in \([\hat{x}, \hat{p}] = i\hbar\)?
Answer
It sets the scale for the minimum uncertainty in position and momentum and identifies momentum as the generator of translations.
Problem set for PU-202 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.