Expectation values and the σₓ σₚ product

For a normalised state the mean of a long run of A measurements is ⟨A⟩ = ⟨ψ|Â|ψ⟩, and the spread of that run is σ_A = √(⟨²⟩ − ⟨A⟩²). In a stationary state the time dependence is the global phase e^{(−iEₙ t/ħ)}, which cancels between bra and ket, so every moment computed here is a constant: nothing in |n⟩ evolves. Now watch which object each moment needs. Position moments read straight off the density, ⟨x^k⟩ = ∫₀^L x^k |ψₙ(x)|² dx, so |ψₙ|² is sufficient. Momentum moments are not like that: ⟨p²⟩ = ∫₀^L ψₙ*(−ħ²ψₙ″) dx requires the operator acting on the state, and |ψₙ|² is identical for ψₙ and for ψₙ e^(iθ(x)), two states with completely different momentum content. So the working habit is to name the operator and its domain before integrating anything — here p̂ = −iħ d/dx on square-integrable functions with square-integrable first derivative vanishing at both walls, and Ĥ = −(ħ²/2m) d²/dx² on those whose second derivative is square-integrable too — and only then to reach for a table of integrals.

Let Π be reflection about the centre, (Πf)(x) = f(L − x). It commutes with H because the box is symmetric, and it maps the eigenbasis onto itself: Πψₙ = (−1)⁽ⁿ⁺¹⁾ ψₙ, so ψ₁ is even about L/2, ψ₂ is odd, and so on. Split the position operator as x̂ = L/2 + (x̂ − L/2). The second piece is odd under Π while |ψₙ|² is even, so its integral vanishes and ⟨x⟩ = L/2 for every n, with no integration performed. The same argument kills ⟨p⟩: p̂ is odd under Π, so ⟨n|p̂|n⟩ = ⟨n|Π†p̂Π|n⟩ = −⟨n|p̂|n⟩, and the only number equal to minus itself is zero. There is a blunter second route worth knowing, because it generalises to any real bound state: ψₙ is real up to a constant phase, so ⟨p⟩ = −iħ∫₀^L ψₙ ψₙ′ dx = (−iħ/2)[ψₙ²]₀^L = 0, killed by the boundary condition itself. Both routes say a real standing wave carries no net momentum. Neither says the particle is at rest.

Write ⟨x²⟩ = (2/L)∫₀^L x² sin²(nπx/L) dx and split sin² = (1 − cos(2nπx/L))/2. The first half gives (2/L)(L³/6) = L²/3. The second needs ∫₀^L x² cos(2nπx/L) dx; two integrations by parts leave only the term in cos(2nπ) = 1, giving L³/(2n²π²). So ⟨x²⟩ = L²/3 − L²/(2n²π²), and σₓ² = ⟨x²⟩ − (L/2)² = L²(1/12 − 1/(2n²π²)). Recognise the two pieces. L²/3 and L²/12 are exactly the second moment and the variance of a uniform density 1/L on [0, L] — the classical bouncer. All the quantum content is the 1/(2n²π²) deficit, which removes 60.8% of the classical variance in the ground state, leaving σₓ = 0.1808L, then shrinks fast: 0.2658L at n = 2, 0.2788L at n = 3, 0.2862L at n = 6, against the classical 0.2887L. Note what converges. σₓ is an average over the whole box and it does converge; |ψₙ|² itself still swings between 0 and 2/L at every n and never converges pointwise to 1/L.

Inside the box V = 0, and one integration by parts, its boundary term killed by ψₙ(0) = ψₙ(L) = 0, turns ⟨p²⟩ = ‖p̂ψₙ‖² = ħ²∫₀^L |ψₙ′|² dx into 2m⟨H⟩ = 2mEₙ = (nπħ/L)². No integral has to be evaluated. With ⟨p⟩ = 0 that makes σₚ = nπħ/L — exactly |p| for a classical particle of the same energy bouncing between the walls, whose momentum distribution is two spikes at ±nπħ/L with mean zero and standard deviation nπħ/L. The box's momentum spread is already classical; only its position spread is not. The quantum version smears those spikes: |φ(p)|² ∝ n²[1 − (−1)ⁿ cos(pL/ħ)] / [(nπ)² − (pL/ħ)²]², two peaks of width of order ħ/L, unresolved at n = 1 and clearly separate by n = 4. Their tails fall only as p⁻⁴, and that is a warning. ⟨p²⟩ converges, and Parseval applied to ψₙ′ confirms ħ²(nπ/L)². But ∫p⁴|φ(p)|² dp diverges, while squaring 2mH returns a finite ħ⁴(nπ/L)⁴. The two disagree because ψₙ has a kink at each wall: ψₙ lies in the domain of p̂, but p̂ψₙ does not, so p̂ cannot legally be applied twice. Topic 5.1's remark that p̂ is symmetric but not self-adjoint here surfaces as an actual number.

Multiply the two spreads: σₓ σₚ = L√(1/12 − 1/(2n²π²)) × nπħ/L. The L cancels — the product cannot depend on the box size, which is why it is the clean invariant of this problem — and tidying gives (ħ/2)√(n²π²/3 − 2). At n = 1 that is 0.5679ħ, 13.6% above the Robertson floor; at n = 2, 1.6703ħ; at n = 6, 5.3953ħ; asymptotically nπħ/√12 = 0.9069nħ. Two expectations students bring are false. First, the ground state does not saturate the bound. Saturation requires (x̂ − ⟨x⟩)|ψ⟩ = iλ(p̂ − ⟨p⟩)|ψ⟩ with λ real, a first-order differential equation whose only solution is a Gaussian; a Gaussian is nowhere zero, so no state obeying ψ(0) = ψ(L) = 0 can ever reach ħ/2. Second, high n does not creep toward the bound — it runs away, because σₚ grows linearly in n while σₓ is capped below L/√12. One loose end: Schrödinger's sharper inequality adds a covariance term ½⟨(Δx̂, Δp̂)⟩, but ψₙ is real, so x̂ψₙ is real and p̂ψₙ is purely imaginary, that term vanishes, and Robertson's bound is the whole story here.

E₁ = σₚ²/(2m) = π²ħ²/(2mL²) is not an approximation and not a thermal residue: it is the price of the Dirichlet condition. An electron in L = 1.00 nm has E₁ = h²/(8mL²) = 0.376 eV, about fifteen times k_BT at 300 K (0.0259 eV), and cooling the sample does not touch it. Halve the box and E₁ quadruples while σₚ doubles and σₓ halves. Two exact statements fix its status. First, nothing in the domain goes lower: expand any normalised state as |ψ⟩ = Σₙ cₙ|n⟩ and ⟨H⟩ = Σₙ |cₙ|² Eₙ ≥ E₁ Σₙ |cₙ|² = E₁, with equality only for |1⟩ up to a phase, so cooling toward T = 0 can at most pile the whole population into |1⟩ and stop there. Second, the uncertainty relation alone proves that confinement costs energy, though not how much. No distribution on [0, L] has a variance above L²/4, so σₓ ≤ L/2, σₚ ≥ ħ/(2σₓ) ≥ ħ/L, and ⟨H⟩ = ⟨p²⟩/(2m) ≥ σₚ²/(2m) ≥ ħ²/(2mL²) = E₁/π², only 0.0381 eV for the same electron. Both steps are slack in the ground state, and the slack is fully accounted for: its product is 0.5679ħ rather than ħ/2, a factor 1.1357, and its σₓ is 0.1808L rather than L/2, a factor 2.7662. Those factors multiply to π, so the bound undershoots σₚ by π and E₁ by π² ≈ 9.87.