Coherent states and the classical limit

Expand the initial state on the eigenbasis, ψ(x,0) = Σₙ cₙ ψₙ(x) with cₙ = ∫ ψₙ* ψ(x,0) dx, then evolve it by attaching one phase per term: ψ(x, t) = Σₙ cₙ ψₙ(x) e^{(−iEₙ t/ℏ)}. Substituting Eₙ = (n + ½)ℏω factorises every phase into e^(−iωt/2) times e^(−inωt). The first factor is common to all terms, so it is a global phase: it cancels out of |ψ|² and out of every expectation value, and the zero-point half-quantum is therefore invisible here. What is left is the relative phases. Squaring the sum gives diagonal terms |cₙ|²|ψₙ|², which never move, plus cross terms carrying e^(−ikωt) with k = n − m an integer — so every moving part of the density oscillates at an integer multiple of the single frequency ω. An eigenstate has no cross terms at all, which is exactly why its density is static however large n is.

Every relative phase turning at an integer multiple of ω has a sharp consequence: after T = 2π/ω each factor e^(−inωt) has advanced by exactly 2πn and come back to 1. So |ψ(x, t + T)|² = |ψ(x, t)|² for any initial state whatsoever — no approximation, no spreading, no dephasing. The wavefunction itself picks up the global e^(−iωT/2) = −1 and needs 2T to return. Nothing else in this course behaves like that. In the infinite well Eₙ ∝ n², the phases advance as n², and the state only rebuilds after Tᵣₑᵥ = 4mL²/πℏ, which for an electron in a 1.0 nm box is 11 fs — a time with no classical counterpart at all — while in between the packet passes through fractional revivals that split it into copies of itself. The oscillator revives at the classical period because every gap in its spectrum is the same ℏω.

Write x = √(ℏ/2mω)(a + a†). Since a and a† shift n by one, ⟨n|x|m⟩ vanishes unless m = n ± 1, so the only surviving cross terms in ⟨x⟩ join neighbours — and every neighbouring gap is ℏω, so ⟨x⟩ can oscillate at ω and nothing else. Take ψ = (ψ₀ + ψ₁)/√2. With ⟨0|x|1⟩ = √(ℏ/2mω) the two cross terms add to ⟨x⟩ = √(ℏ/2mω) cos ωt = 0.71 x₀ cos ωt, where x₀ = √(ℏ/mω) is the oscillator's natural length. The centre does move at the classical frequency, but the swing is 0.71 x₀ while the packet's own width runs between 0.71 x₀ and 1.0 x₀ over the cycle: a blob wobbling by less than its own size. Now take (ψ₀ + ψ₂)/√2 instead. No neighbouring pair, so ⟨x⟩ = 0 for ever — yet the density is far from static, because its cross term carries e^(−2iωt) and makes the blob breathe twice per period.

The general result d⟨p⟩/dt = −⟨dV/dx⟩ becomes something stronger here. For V = ½mω²x² the slope dV/dx = mω²x is linear, so ⟨dV/dx⟩ = mω²⟨x⟩ with no approximation, and the pair d⟨x⟩/dt = ⟨p⟩/m and d⟨p⟩/dt = −mω²⟨x⟩ closes on itself. The centre of any state at all — one eigenstate, two, a lopsided mess — obeys the classical equation of motion exactly. Add a cubic term to V and this fails at once, because the force then goes as x², ⟨x²⟩ ≠ ⟨x⟩², and the mean position no longer obeys a closed equation. But notice what the theorem does not say. It fixes the centre, not the shape: the (ψ₀ + ψ₁)/√2 packet satisfies it perfectly while looking nothing like a particle. Classical behaviour needs a second condition, that the width be small compared with the swing.

Take the ground-state Gaussian and displace it: ψ(x,0) = ψ₀(x − d). Expanded on the eigenbasis that is |α⟩ = e^{(−|α|²/2)} Σₙ αⁿ|n⟩/√(n!), with α fixed by d = √(2ℏ/mω) α. Three things follow. The occupation is Poissonian, P(n) = e^(−⟨n⟩)⟨n⟩ⁿ/n!, with mean ⟨n⟩ = |α|² and spread Δn = |α|, so Δn/⟨n⟩ = 1/√⟨n⟩, which is also the relative energy spread once ⟨n⟩ ≫ 1. The centre follows ⟨x⟩ = √(2ℏ/mω)|α| cos(ωt − φ), which is Ehrenfest at work. And the width stays at the ground-state value √(ℏ/2mω) at every instant, so the shape never changes: a rigid Gaussian sliding along the classical trajectory, at minimum uncertainty the whole way. The number that decides how classical it looks is A/Δx = 2|α| = 2√⟨n⟩.

The classical particle spends most of its time where it moves slowest, which gives the density P(x) = 1/(π√(A² − x²)) — piled up at the turning points ±A, thin in the middle. A high-n eigenstate does look like that: its envelope grows towards the turning points at ±√(2n+1) x₀ and its wavelength lengthens there. But it matches only after you average over its n nodes, at each of which the exact probability is zero, and it still leaks past the turning points (15.7% for n = 0, shrinking with n but never vanishing). More to the point, |ψₙ|² is stationary — the same picture at every instant, with ⟨x⟩ = 0 for ever. The correspondence at large n is between a smoothed eigenstate density and a time-averaged classical density. What actually swings is a superposition spanning many neighbouring n.