University Physics V · Quantum Uncertainty and Commutation Relations · 8.1
The Commutator as an Algebraic Object
Before any uncertainty relation, learn to compute [A, B] the way you differentiate — by rule, not by grinding out matrix elements. Four identities do nearly all the work, the Pauli matrices are where you test them, and one line about the trace proves that x̂ and p̂ can never be matrices.
Build the model
Connect the measurement to the mechanism.
The commutator [A, B] = AB − BA is bookkeeping for the one property operator multiplication has that ordinary multiplication does not: order. Treat it as an algebraic object in its own right and it obeys a small closed set of rules — linear in each slot, antisymmetric, a Leibniz product rule that peels factors off one at a time, and a Jacobi identity binding any three operators. Those four make [⋅,⋅] a Lie bracket, and they let you evaluate [x̂, p̂²] or [σₓ, σy σz] without writing a single matrix element or integral.
Two consequences arrive at once. The Pauli matrices close under the bracket, [σᵢ, σⱼ] = 2i εᵢⱼₖ σₖ, so the whole of angular-momentum algebra already sits inside three 2×2 matrices. And the trace, being cyclic, annihilates every commutator: Tr(AB) = Tr(BA) forces Tr[A, B] = 0 for any two d×d matrices, while Tr(iħ Id) = iħd is never zero.
So [x̂, p̂] = iħ has no finite-dimensional solution at all — not a poor one, none. That is what the canonical relation costs: an infinite-dimensional space, operators with no finite norm, and a domain you are obliged to state, because Wintner and Wielandt extend the same verdict to every pair of bounded operators.
- Simple definition
- The commutator of two operators, [A, B] = AB − BA, is the operator measuring how much their order matters; it is linear in each argument, antisymmetric, and obeys the product rule [A, BC] = [A, B]C + B[A, C].
- Example
- σₓ σy = iσz while σy σₓ = −iσz, so [σₓ, σy] = 2iσz — entries of modulus 2, not zero. Their anticommutator (σₓ, σy) is the zero matrix: these two are as non-commuting as a pair of Hermitian unitaries can be.
The bracket is an operator on the same space, never a number, and it carries the units of the product AB.
α, β are complex scalars; [A, A] = 0 and [A, cI] = 0 for every A.
Peels one factor at a time, so [x̂, p̂²] = [x̂, p̂]p̂ + p̂[x̂, p̂] = 2iħp̂ needs no matrix elements at all.
Order is preserved: the spectator factor stays on the side it started.
The consistency condition that lets conserved quantities close into an algebra, as [Lᵢ, Lⱼ] = iħ εᵢⱼₖ Lₖ does.
Cyclic in A, B, C; with antisymmetry it makes [⋅,⋅] a Lie bracket.
Any product of Paulis collapses to one Pauli times a phase, and Sᵢ = (ħ/2)σᵢ turns the 2 into [Sₓ, Sy] = iħSz.
Each σᵢ is Hermitian, unitary, traceless and dimensionless; σᵢ² = I.
For Ĥ = p̂²/2m + V(x̂) this gives [x̂, Ĥ] = iħp̂/m and [p̂, Ĥ] = −iħV′(x̂) in one line each.
f, g given by convergent power series; ħ = 1.0546 × 10⁻³⁴ J s.
No finite matrices satisfy [x̂, p̂] = iħ, which is why the canonical pair needs an infinite-dimensional space.
Tr(AB) = Tr(BA) for any d×d matrices, so the first holds at every finite d.
The bracket is an operator, and it has units
[A, B] = AB − BA is defined only where both products are, and what it returns is another operator on the same space — not a number, not a scalar measure of disagreement. It inherits units: [x̂, p̂] carries J s, the units of x times p, which is why ħ and not a pure number stands on the right. Hermiticity constrains the answer sharply. If A and B are both Hermitian, then [A, B]† = (AB)† − (BA)† = B†A† − A†B† = BA − AB = −[A, B], so the commutator of two observables is anti-Hermitian and can never itself be an observable. Multiply by i and it becomes Hermitian: i[A, B] is measurable, and this is exactly why the canonical relation reads iħ rather than ħ. A real constant on the right would contradict Hermiticity before any physics was done. The partner object, the anticommutator {A, B} = AB + BA, is Hermitian for Hermitian A and B, and the pair split any product: AB = ½{A, B} + ½[A, B].
Four rules, and you stop multiplying matrices
Bilinearity lets you expand sums slot by slot; antisymmetry gives [A, A] = 0 and lets you swap arguments at the cost of a sign; the Leibniz rule [A, BC] = [A, B]C + B[A, C] peels one factor off a product; the Jacobi identity ties three operators together. The Leibniz rule is the workhorse, and its one trap is that the spectator factor must stay on the side it started: writing [A, BC] = [A, B]C + [A, C]B is wrong unless everything in sight happens to commute. Applied to the canonical pair it gives [x̂, p̂²] = [x̂, p̂]p̂ + p̂[x̂, p̂] = iħp̂ + iħp̂ = 2iħp̂, and induction extends that to [x̂, p̂ⁿ] = iħ n p̂ⁿ⁻¹. In other words the map B ↦ [x̂, B] acts on products exactly as iħ d/dp does — it is a derivation. That single structural fact is why the classical Poisson bracket and the quantum commutator obey the same identities, and why {x, p} = 1 and [x̂, p̂] = iħ sit opposite each other in the correspondence.
The Pauli matrices are the smallest place to test it
Three traceless Hermitian 2×2 matrices carry the whole product table σᵢ σⱼ = δᵢⱼ I + i εᵢⱼₖ σₖ. Multiply σₓ = [[0,1],[1,0]] by σy = [[0,−i],[i,0]] and you get [[i,0],[0,−i]] = iσz; reverse the order and you get −iσz. So [σₓ, σy] = 2iσz and (σₓ, σy) = 0, the two halves of the product table read off separately. The bracket closes — a commutator of Paulis is a Pauli times a number — so these three span the Lie algebra su(2), and rescaling to Sᵢ = (ħ/2)σᵢ turns 2i εᵢⱼₖ σₖ into iħ εᵢⱼₖ Sₖ. Jacobi is worth checking here because it is cheap: [σₓ, [σy, σz]] = [σₓ, 2iσₓ] = 0, and the same for the two cyclic partners, so the sum vanishes term by term. Note also what the right-hand sides look like: 2iσz is traceless. Nothing in this algebra ever produces a non-zero multiple of the identity, which is the loophole the next section closes.
The trace kills [x̂, p̂] = iħ in finite dimensions
For finite matrices Tr(AB) = Σⱼ Σₖ Aⱼₖ Bₖⱼ = Tr(BA), because the double sum is symmetric under relabelling. Therefore Tr[A, B] = Tr(AB) − Tr(BA) = 0 for every pair of d×d matrices, with no exceptions and no dependence on d. Now take the trace of both sides of [x̂, p̂] = iħ I. The left gives 0; the right gives iħd, which is non-zero for every d ≥ 1 — for d = 1000 it is 1.05 × 10⁻³¹ J s, not a rounding error. The canonical commutation relation therefore has no finite-dimensional matrix solution: not an inaccurate one, none. Be precise about what is being forbidden. Non-commuting finite matrices are ordinary, and [σₓ, σy] = 2iσz is one. What no finite dimension permits is a commutator equal to a non-zero multiple of the identity, because the identity has trace d and commutators have trace zero.
Bounded is not enough either: the operators must be unbounded
Infinite dimensions remove the trace argument but not the obstruction. Suppose A and B are bounded operators with [A, B] = cI. The Leibniz rule and induction give [A, Bⁿ] = n c Bⁿ⁻¹, so n|c| ‖Bⁿ⁻¹‖ = ‖[A, Bⁿ]‖ ≤ 2‖A‖ ‖Bⁿ‖ ≤ 2‖A‖ ‖B‖ ‖Bⁿ⁻¹‖. If no power of B vanishes, cancel ‖Bⁿ⁻¹‖ and read off n|c| ≤ 2‖A‖‖B‖ for every n, which fails as soon as n exceeds 2‖A‖‖B‖/|c|; if some power does vanish, the smallest such n forces c = 0. That is the Wintner–Wielandt theorem, and its verdict is that x̂ and p̂ have no finite norm at all. Unbounded operators are not defined on the whole Hilbert space, only on a dense domain, so [x̂, p̂]ψ = iħψ is a statement about the vectors ψ for which both x̂p̂ψ and p̂x̂ψ exist — the Schwartz functions, say. Quoting a commutator without its domain is quoting half a theorem, and it is the half that discretised code discovers the hard way.
What the algebra buys once you have it
Almost every later result in the course is an evaluated commutator. The Heisenberg equation reads dÂ/dt = (i/ħ)[Ĥ, Â] + ∂Â/∂t, so anything commuting with Ĥ is conserved: for a free particle [Ĥ, p̂] = 0 and momentum is constant, while adding V(x̂) gives [Ĥ, p̂] = [V(x̂), p̂] = iħV′(x̂) and hence dp̂/dt = −V′(x̂), the operator form of F = −dV/dx. The Robertson bound takes ⟨[A, B]⟩ as its only input, so the bracket is the entire content of the uncertainty relation rather than a step towards it. Vanishing commutators do work too: two commuting Hermitian operators share a complete eigenbasis, which is what makes a complete set of commuting observables a usable labelling scheme for degenerate states. And Jacobi is not decoration — it is what makes A ↦ [A, ·] respect the algebra, so that conserved quantities close into one rather than generating an endless list of new operators.
Change one variable at a time
Make the relationship visible.
Push d from 2 to 10: the plateau of +1s lengthens while the single top-rung bar deepens to {1-d|0}, so the defect concentrates rather than dilutes. Then drag n up the ladder — every rung obeys [a, a†] = 1 until the last, which pays the whole trace bill alone.
ENTRY AT RUNG n1
TOP-RUNG DEFECT 1 − d-5
TRACE OF THE COMMUTATOR0
RUNGS OBEYING THE ALGEBRA83 %
Live interpretationENTRY AT RUNG n: 1. TOP-RUNG DEFECT 1 − d: −5. TRACE OF THE COMMUTATOR: 0. RUNGS OBEYING THE ALGEBRA: 83 %
Catch the common trap
Explain before calculating.
You truncate the harmonic-oscillator ladder to the lowest six Fock states and build 6×6 matrices for a and a†. The first five diagonal entries of [a, a†] all come out equal to 1. What is the sixth?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyUsing only the product rule and [x̂, p̂] = iħ, evaluate [x̂, p̂²] and [x̂², p̂]. Then take Ĥ = p̂²/2m for a free particle, find [x̂, Ĥ], and feed it into the Heisenberg equation dÂ/dt = (i/ħ)[Ĥ, Â].
- Leibniz in the right slot: [x̂, p̂²] = [x̂, p̂]p̂ + p̂[x̂, p̂].
- Each bracket is iħ times the identity, which commutes past anything, so [x̂, p̂²] = iħp̂ + iħp̂ = 2iħp̂.
- Leibniz in the left slot: [x̂², p̂] = x̂[x̂, p̂] + [x̂, p̂]x̂ = iħx̂ + iħx̂ = 2iħx̂. Note the two results differ by which operator survives, not by a sign.
- With Ĥ = p̂²/2m, bilinearity pulls out the constant: [x̂, Ĥ] = (1/2m)[x̂, p̂²] = iħp̂/m.
- Heisenberg needs [Ĥ, x̂] = −[x̂, Ĥ] = −iħp̂/m, so dx̂/dt = (i/ħ)(−iħp̂/m) = p̂/m.
Answer[x̂, p̂²] = 2iħp̂, [x̂², p̂] = 2iħx̂, and [x̂, Ĥ] = iħp̂/m, which the Heisenberg equation turns into dx̂/dt = p̂/m — velocity, recovered from algebra alone.
MediumEvaluate [σₓ, σy σz] using only bilinearity, antisymmetry and the product rule, with no matrix multiplication. Check the result a second way by collapsing the product first, then rescale to Sᵢ = (ħ/2)σᵢ and show where the factor of 2 in the Pauli relation goes.
- Leibniz on the product in the right slot: [σₓ, σy σz] = [σₓ, σy]σz + σy[σₓ, σz].
- From [σᵢ, σⱼ] = 2i εᵢⱼₖ σₖ: εxyz = +1 gives [σₓ, σy] = 2iσz, and εxzy = −1 gives [σₓ, σz] = −2iσy.
- Substitute and use σᵢ² = I: 2iσzσz + σy(−2iσy) = 2i I − 2i I = 0.
- Second route: the product collapses first, σy σz = i εyzx σₓ = iσₓ, so [σₓ, iσₓ] = i[σₓ, σₓ] = 0. The two routes agree, which is the check.
- Rescale: [Sₓ, Sy] = (ħ/2)²[σₓ, σy] = (ħ²/4)(2iσz) = i(ħ²/2)σz, and iħSz = iħ(ħ/2)σz is the same operator.
Answer[σₓ, σy σz] = 0 by both routes, and [Sₓ, Sy] = iħSz. The 2 in the Pauli relation is an artefact of the normalisation σᵢ = 2Sᵢ/ħ, not a physical factor.
HardA student truncates the oscillator ladder to the lowest N = 4 Fock states and builds 4×4 matrices for a and a†, expecting to confirm [a, a†] = I numerically. Compute the diagonal of [a, a†] and its trace, identify the defect operator [a, a†] − I, and say what this proves about [x̂, p̂] = iħ and whether a larger N repairs it.
- In the truncated space a|n⟩ = √n|n−1⟩ for n = 1, 2, 3 and a|0⟩ = 0, while a†|n⟩ = √(n+1)|n+1⟩ for n = 0, 1, 2 but a†|3⟩ = 0, because |4⟩ is no longer in the basis. That amputation is the whole effect.
- Diagonals: (a†a)ₙₙ = n gives 0, 1, 2, 3. And (aa†)ₙₙ = n + 1 gives 1, 2, 3 for n = 0, 1, 2, but 0 for n = 3.
- Subtracting entry by entry, the diagonal of [a, a†] is (1 − 0, 2 − 1, 3 − 2, 0 − 3) = (1, 1, 1, −3), and its trace is 3 − 3 = 0, exactly as Tr(AB) = Tr(BA) demands.
- The defect is [a, a†] − I = diag(0, 0, 0, −4) = −4|3⟩⟨3|, of operator norm 4 = N. At general N it is −N on the top rung, so refining the truncation deepens the defect; only the fraction of rungs obeying the algebra, (N − 1)/N, improves.
- In dimensionless quadratures X = x̂√(mω/ħ) and P = p̂/√(mħω), a = (X + iP)/√2 and [a, a†] = 1 is precisely [x̂, p̂] = iħ, so the verdict transfers: taking the trace of [x̂, p̂] = iħ IN gives 0 = iħN, false for every finite N.
Answerdiag[a, a†] = (1, 1, 1, −3), trace 0, defect −4|3⟩⟨3| of norm 4 = N. The failure never shrinks with N — it concentrates on the top rung — and the trace identity rules out every finite N.