The canonical commutation relation

You cannot evaluate x̂p̂ − p̂x̂ by staring at it, because neither symbol is a number; the expression is an instruction, and an instruction is defined by what it does. Supply a test function ψ(x) that is differentiable and square-integrable. First ordering: x̂p̂ψ = x(−iħ dψ/dx) = −iħxψ′, because p̂ differentiates and only then does x multiply what comes out. Second ordering: p̂x̂ψ = −iħ d(xψ)/dx, and now x sits inside the derivative, so the product rule returns −iħ(ψ + xψ′). Subtract, and the two −iħxψ′ pieces are identical and vanish: [x̂, p̂]ψ = −iħxψ′ + iħψ + iħxψ′ = iħψ. Nothing about ψ was used except differentiability, so the result is the operator identity [x̂, p̂] = iħ Î, valid on every state in the domain. Reverse the order and the sign reverses with it: [p̂, x̂] = −iħ. The domain is not decoration — ψ must be smooth enough to differentiate and decay fast enough that xψ and ψ′ stay square-integrable, which is why Schwartz functions are the standard home for this relation.

Run the derivation in the other basis and watch where the sign comes from. Write ψ(x) as a superposition of momentum eigenfunctions, ψ(x) = (2πħ)^(−1/2) ∫ φ(p) exp(ipx/ħ) dp. Acting with p̂ = −iħ ∂/∂x on the kernel gives −iħ(ip/ħ) exp(ipx/ħ) = p exp(ipx/ħ), so in this basis p̂ is plain multiplication by p. Position is the interesting one. Since ∂/∂p exp(ipx/ħ) = (ix/ħ) exp(ipx/ħ), multiplying by x is the same as acting with −iħ ∂/∂p on the kernel — but the kernel is not where φ lives. Integrate by parts, the boundary term dies because φ vanishes as |p| → ∞, and the derivative lands on φ with its sign reversed: x̂φ = +iħ dφ/dp. Now test the commutator: iħ d(pφ)/dp − p(iħ dφ/dp) = iħφ + iħp φ′ − iħp φ′ = iħφ. Same relation, same iħ, and no reference to x at all. Guessing x̂ = −iħ d/dp by false symmetry would have returned −iħ instead, and with it momentum eigenfunctions running the wrong way.

Read the derivation again and note what it assumed: a coordinate x running over all of ℝ, with p̂ = −iħ d/dx self-adjoint on that line. Change either premise and the conclusion goes. Put the particle on a circle and try [φ̂, L̂_z] = iħ with L̂_z = −iħ ∂/∂φ. Single-valuedness forces L̂_z|m⟩ = mħ|m⟩ with m an integer, so take the diagonal element in that state: ⟨m|φ̂L̂_z|m⟩ − ⟨m|L̂_zφ̂|m⟩ = mħ⟨m|φ̂|m⟩ − mħ⟨m|φ̂|m⟩ = 0, while the relation demands iħ. That is the trace argument again in miniature, and it kills the relation outright: there is no self-adjoint angle operator on a circle that obeys it. The repair is to use single-valued functions of the angle, [L̂_z, cos φ̂] = iħ sin φ̂ and [L̂_z, sin φ̂] = −iħ cos φ̂, from which an honest uncertainty bound does follow. The half-line bites too. Radial momentum p̂ᵣ = −iħ(∂/∂r + 1/r) does satisfy [r̂, p̂ᵣ] = iħ, but it has no self-adjoint extension on the half-line, so it is not an observable and the bound it appears to imply means nothing.

Because x̂ and p̂ are unbounded, the relation as written is sensitive to domains: Nelson built a pair satisfying it on a dense domain that is nevertheless not the standard one. The sharp version exponentiates both sides into bounded unitaries. With Û(a) = exp(−iap̂/ħ) and V̂(b) = exp(ibx̂/ħ), the Weyl form reads Û(a)V̂(b) = exp(−iab/ħ) V̂(b)Û(a), and Stone and von Neumann prove that on a separable Hilbert space every irreducible representation of it is unitarily equivalent to the position representation. The position picture, the momentum picture and the oscillator's ladder picture are therefore one theory in three sets of clothes. Make that concrete: put â = √(mω/2ħ)(x̂ + ip̂/mω). Then [â, â†] = (−i/2ħ)[x̂, p̂] + (i/2ħ)[p̂, x̂] = ½ + ½ = 1, which came from the canonical relation and nothing else — and it is why the oscillator spectrum ħω(n + ½) can be derived without ever writing down a wavefunction.