Expectation values and the virial split

Define â = (1/√2)(x̂/b + i b p̂/ħ) with b = √(ħ/mω), and invert it: x̂ = (b/√2)(â + â†) and p̂ = i(ħ/b√2)(↠− â). Now ⟨n|x̂|n⟩ is a sum of ⟨n|â|n⟩ and ⟨n|â†|n⟩, and â|n⟩ is proportional to |n−1⟩ while â†|n⟩ is proportional to |n+1⟩. Neither overlaps |n⟩, so both terms are zero: ⟨x̂⟩ = 0, and by the identical argument ⟨p̂⟩ = 0, for every n. Parity gets you the first of those — |ψₙ|² is even, so the odd integrand ∫x|ψₙ|²dx vanishes — but parity says nothing about ⟨p²⟩, and the ladder says everything. The payoff is immediate: since Δx² = ⟨x²⟩ − ⟨x⟩² and ⟨x⟩ = 0, the variance and the second moment are the same number, and the same holds for momentum.

Square the ladder forms. x̂² = (b²/2)(â² + ↲ + â↠+ â†â). The first two terms connect |n⟩ to |n∓2⟩, so their diagonal elements vanish; the last two combine as â↠+ â†â = 2N̂ + 1, whose eigenvalue is 2n + 1. Hence ⟨x²⟩ = (b²/2)(2n + 1) = (n + ½)ħ/mω. The momentum square is p̂² = (ħ²/2b²)(2N̂ + 1 − â² − ↲), giving ⟨p²⟩ = (n + ½)mωħ — the same factor, with b² traded for ħ²/b². Put numbers on it. An electron (m = 9.11 × 10⁻³¹ kg) in a parabolic quantum dot with ħω = 10 meV has b = ħ/√(mħω) = 2.76 nm, so in the ground state Δx = b/√2 = 1.95 nm and Δp = (ħ/b)/√2 = 2.70 × 10⁻²⁶ kg m s⁻¹.

For any stationary state the virial theorem reads 2⟨T⟩ = ⟨x dV/dx⟩. A quadratic potential has x dV/dx = mω²x² = 2V, so 2⟨T⟩ = 2⟨V⟩ and the energy divides exactly in half: ⟨T⟩ = ⟨V⟩ = ½(n + ½)ħω. Check it against the moments already in hand — ⟨p²⟩/2m = ½(n+½)ħω and ½mω²⟨x²⟩ = ½(n+½)ħω — and they agree. The same theorem tells you where the halving stops. For a homogeneous potential V ∝ xᵏ it gives 2⟨T⟩ = k⟨V⟩, so ⟨T⟩/E = k/(k+2), and only k = 2 splits evenly. The infinite well is the k → ∞ limit: V = 0 wherever the particle is, so ⟨V⟩ = 0 and ⟨T⟩ = E. Hydrogen is k = −1: in the ground state ⟨T⟩ = −E = 13.6 eV and ⟨V⟩ = 2E = −27.2 eV. The classical oscillator shares the 50:50 split, because cos²ωt and sin²ωt both time-average to ½.

With ⟨x⟩ = ⟨p⟩ = 0, Δx = √((n+½)ħ/mω) and Δp = √((n+½)mωħ), so Δx Δp = (n + ½)ħ exactly. Read that carefully. At n = 0 the product is ħ/2, sitting precisely on the Heisenberg bound; the ground state is a Gaussian, and a Gaussian is the only wavefunction that saturates Δx Δp ≥ ħ/2. At n = 1 the product is already 3ħ/2, three times the floor, and at n = 10 it is twenty-one times. So "the oscillator is a minimum-uncertainty system" is true of one state and false of all the rest, and what disqualifies the excited states is their nodes: a function with nodes cannot be Gaussian. Geometrically the product is an area in phase space. The classical orbit of energy Eₙ is an ellipse of semi-axes Aₙ and Pₙ enclosing πAₙPₙ = (n + ½)h — the Bohr–Sommerfeld rule ∮p dx = nh plus the half quantum the old theory missed, arriving here as a consequence rather than a postulate.

Run the logic backwards and the ground state falls out without solving anything. In any state with ⟨x⟩ = ⟨p⟩ = 0 the mean energy is ⟨H⟩ = ⟨p²⟩/2m + ½mω²⟨x²⟩ = Δp²/2m + ½mω²Δx². The uncertainty principle forbids Δp from dropping below ħ/2Δx, so ⟨H⟩ ≥ ħ²/(8mΔx²) + ½mω²Δx². Differentiate with respect to Δx and set the derivative to zero: −ħ²/(4mΔx³) + mω²Δx = 0, giving Δx⁴ = ħ²/4m²ω², so Δx² = ħ/2mω — exactly the ground-state variance found above — and ⟨H⟩ = ħω/4 + ħω/4 = ħω/2. Squeezing the particle inward raises the kinetic term faster than it lowers the potential term, and letting it spread does the reverse. A ⁴⁰Ca⁺ ion trapped at ω/2π = 1.0 MHz therefore keeps 2.1 neV of motion, equal to k_B × 24 μK, which is the temperature the trap must beat.

The classical turning point of the n-th state is Aₙ = √((2n+1)ħ/mω), and Δx = √((n+½)ħ/mω), so Δx = Aₙ/√2 at every n without exception. That is precisely the rms displacement of a classical oscillator of amplitude Aₙ, since the time average of cos²ωt is ½. The second moments therefore match the classical values even at n = 0, where no classical motion of that energy is ever observed. The shapes do not match. The ground state's probability density is a Gaussian peaked at the centre, while the classical density, proportional to 1/v(x), diverges at the turning points where the particle moves slowest — and about 16% of the ground-state probability lies beyond ±A₀, in the classically forbidden region. Correspondence in this topic is a claim about moments and about large-n averages, never about the shape of the distribution.