Expectation values in a box

Every |ψₙ|² is symmetric about the midpoint: sin(nπ(L − x)/L) = ±sin(nπx/L), and squaring removes the sign, so the density mirrors across x = L/2. A symmetric density has its first moment at the centre, giving ⟨x⟩ = L/2 for every n with no integration at all. Do it the long way as a check: ⟨x⟩ = (2/L)∫₀ᴸ x sin²(nπx/L) dx = (1/L)∫₀ᴸ x[1 − cos(2nπx/L)] dx, and the cosine term vanishes — by parts, its boundary term carries sin(2nπ) = 0 and what remains is a sine integrated over whole periods — leaving (1/L)(L²/2) = L/2. Then read the result carefully. It is the mean of a distribution, not a place the particle prefers: in the n = 2 state |ψ₂(L/2)|² = 0 exactly, so the single point the mean names is the one point a measurement never returns.

The second moment needs real work: (2/L)∫₀ᴸ x² sin²(nπx/L) dx = L²(1/3 − 1/(2n²π²)), where 1/3 is what a uniform density would give and the n-dependent term is the correction the nodes impose. Subtract ⟨x⟩² = L²/4 and the position spread is σx = L√(1/12 − 1/(2n²π²)). Put numbers in: 0.1808L at n = 1, 0.2658L at n = 2, 0.2788L at n = 3, and 0.2887L as n → ∞. That limit is exactly L/√12, the standard deviation of a uniform distribution on [0, L] — the classical answer for a particle bouncing at constant speed between the walls, recovered rather than assumed. The ground state is the tightest because its single hump piles density near the middle; each excitation pushes weight outwards, but σx can never exceed 0.2887L.

Neither momentum moment needs quadrature. For the mean, ⟨p⟩ = −iħ∫₀ᴸ ψₙ ψₙ′ dx = −iħ[ψₙ²/2]₀ᴸ = 0, because ψₙ is real and vanishes at both walls — true of every real bound state, not a quirk of the box. For the mean square, use the equation you already solved: V = 0 inside, so p̂²ψₙ = 2mEₙ ψₙ, giving ⟨p²⟩ = 2mEₙ = (nπħ/L)². Since ⟨p⟩ = 0, that mean square is the entire variance, so σp = nπħ/L. The physical reading is the standing wave: √(2/L) sin(nπx/L) = (1/2i)√(2/L)[e^(inπx/L) − e^(−inπx/L)], equal weights on +nπħ/L and −nπħ/L, which cancel in the first moment and reinforce in the second. One caution: those exponentials do not satisfy the wall conditions, so the true momentum density is broadened, not two spikes: for n ≥ 2 it is a pair of peaks near ±nπħ/L, but for n = 1 the broadening is as wide as the separation and the two merge into a single hump centred on p = 0.

Now combine them: σx σp = L√(1/12 − 1/(2n²π²)) × nπħ/L = ħ√(n²π²/12 − 1/2). The width cancels, so the product depends on n alone — a 0.20 nm well and a 2.0 nm well give the same answer. It runs 0.568ħ, 1.670ħ and 2.627ħ for n = 1, 2, 3, and reaches 9.04ħ at n = 10. Three readings follow. The bound holds everywhere, since even the smallest value is 1.136 × ħ/2. The bound is never saturated, because saturation demands a Gaussian and a sine hump clipped at two walls is not one — the ground state misses by 14%. And excited states are worse, not better: σp grows exactly as n while σx is already near its ceiling, so the product climbs towards nπħ/√12 = 0.907nħ, without limit.

Because ⟨p⟩ = 0, every joule of Eₙ = ⟨p²⟩/2m is momentum variance — and variance is precisely what the uncertainty bound puts a floor under. Anything trapped in [0, L] has σx ≤ L/2, so σp ≥ ħ/(2σx) ≥ ħ/L and therefore E ≥ ħ²/(2mL²). The exact ground state is π² times that, E₁ = π²ħ²/(2mL²): the estimate has the right form and is loose by a factor 9.87, which is what an inequality is for. The scale it sets is the whole point. An electron in L = 0.50 nm gets E₁ = 2.41 × 10⁻¹⁹ J = 1.50 eV, the energy of a photon of wavelength about 825 nm, just past the red end of the visible: confinement lands on the optical scale, which is why a quantum dot's diameter fixes its colour. A 1.0 g bead in a 10 cm box gets E₁ = 5.5 × 10⁻⁶³ J, a speed of 3.3 × 10⁻³⁰ m s⁻¹. Same formula; only one of them is measurable.

|Ψₙ(x, t)|² = |ψₙ(x)|² for all t, so nothing here moves: ⟨x⟩ sits at L/2 forever, σx never changes, and Ehrenfest's d⟨x⟩/dt = ⟨p⟩/m = 0 agrees. There is no bouncing and no orbit to picture — motion appears only once two energies are superposed, which is a different lesson. Nor is σx a blur around one particle: it is the spread of outcomes across many identically prepared measurements, each of which returns a sharp position. It is not a Gaussian σ either — for n = 1 only 65.0% of results land within one σx of L/2, against 68.3% for a normal distribution. Finally the idealisation sends a bill: the wall kinks ψ′, so the momentum density falls only as p⁻⁴. ⟨p²⟩ survives that tail, ⟨p⁴⟩ diverges, and p̂ is not self-adjoint on this domain at all.