University Physics IV · The Quantum Harmonic Oscillator · 8.8
Coherent States & the Classical Limit
Stationary states are frozen: nothing in a single ψₙ moves. Attach the phases to a sum of them and the ladder's even spacing does what no other potential can — the packet returns exactly once per classical period. This is where the oscillator finally reproduces Newton, and where you see what that costs.
Build the model
Connect the measurement to the mechanism.
Every stationary state of the oscillator is frozen: |ψₙ|² carries no time dependence, so nothing in a single eigenstate resembles a swinging mass. Motion appears only when you add eigenstates and let each carry its own phase e(−iEₙ t/ℏ). Because Eₙ = (n + ½)ℏω, those phases split into a common factor e(−iωt/2) and a relative part e(−inωt): the half-quantum is a global phase with no physical effect, and every relative phase turns at an integer multiple of the single frequency ω.
That one fact — uniform level spacing — carries the whole topic. It makes the probability density of any superposition whatsoever exactly periodic with the classical period T = 2π/ω, with no spreading, no dephasing, and none of the fractional revivals a spectrum spaced as n² produces. Push it further and you reach the coherent state, the ground state simply displaced off centre, whose middle obeys Newton's equation exactly while its width never changes — the closest a quantum state comes to an orbit.
The cost is that all of this belongs to the exactly quadratic potential. Give a real bond its anharmonic term and the spacing stops being uniform, the packet dephases after tens of periods, and the tidy revival is gone.
- Simple definition
- A coherent state is the particular superposition of oscillator eigenstates you get by displacing the ground state off centre: it spreads over n with Poisson statistics, its centre obeys the classical equation of motion, and its width stays at the ground-state value for ever.
- Example
- For a 1.0 g mass swinging at 1.0 Hz with amplitude 1.0 cm, E = 1.97 μJ against ℏω = 6.63 × 10⁻³⁴ J, so ⟨n⟩ = 3.0 × 10²⁷, and the packet width 9.2 × 10⁻¹⁷ m is one part in 10¹⁴ of the swing.
Eₙ t/ℏ = (n + ½)ωt, so the half-quantum is a global factor and only e(−inωt) survives into |ψ|².
cₙ = ∫ ψₙ* ψ(x,0) dx, dimensionless, with Σ|cₙ|² = 1; ω = √(k/m) in rad s⁻¹
Every relative phase is an integer multiple of ωt, so they all come back together — a spectrum spaced as n² cannot.
T is the classical period in s; ψ itself returns only after 2T, picking up a factor −1 at T.
No neighbouring pair in the sum, no motion: (ψ₀ + ψ₂)/√2 keeps ⟨x⟩ = 0 for ever, though its density still breathes at 2ω.
x = √(ℏ/2mω)(a + a†) raises or lowers n by one, so ⟨x⟩ is a sum of nearest-neighbour beats at ω.
It is an eigenstate of the lowering operator, never of H: an energy measurement is spread over about √⟨n⟩ levels.
α is a dimensionless complex number; ⟨n⟩ = |α|² and Δn = |α|, so the occupation is Poissonian.
The Gaussian is carried rigidly along the classical path; spreading would need a force that is not linear in x.
Δx is the ground-state width in m, the same at every α and every t; Δx Δp = ℏ/2 at all times.
The local average of |ψₙ|² tends to this at large n — the n nodes between the peaks never go away.
Normalised on |x| < A, where A is the classical amplitude in m; it diverges as the speed v → 0.
Attach the phases, then see what survives
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.
Even spacing buys an exact revival
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 its is the one spectrum whose gaps are all equal.
Only neighbouring levels move the centre
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.
In a quadratic well, Ehrenfest is exact
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.
The coherent state: the ground state, pushed
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 the relative energy spread is 1/√⟨n⟩. 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⟩.
Large n is a shape, not a motion
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.
Change one variable at a time
Make the relationship visible.
Sweep the phase and watch the packet ride the classical path without changing shape; then push α from 0.25 to 3 and watch the swing grow while the bar across the packet stays exactly as long — that ratio, 2α, is the entire classical limit.
MEAN OCCUPATION α²3.06
AMPLITUDE A2.47 x₀
PACKET CENTRE ⟨x⟩1.24 x₀
SWING IN PACKET WIDTHS3.50
Live interpretationMEAN OCCUPATION α²: 3.06. AMPLITUDE A: 2.47 x₀. PACKET CENTRE ⟨x⟩: 1.24 x₀. SWING IN PACKET WIDTHS: 3.50
Catch the common trap
Explain before calculating.
A particle in a harmonic well of angular frequency ω is prepared in the equal superposition (ψ₀ + ψ₂)/√2 of the ground state and the second excited state. What happens next?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron sits in a harmonic well with ω = 1.0 × 10¹⁵ rad s⁻¹ and is prepared in the equal superposition (ψ₀ + ψ₁)/√2. Find the amplitude of ⟨x⟩, the period of its oscillation, and the mean energy. Take m = 9.11 × 10⁻³¹ kg.
- Attach the phases: ψ(t) = (ψ₀ e(−iωt/2) + ψ₁ e(−3iωt/2))/√2. The common e(−iωt/2) is a global phase, leaving a single relative phase e(−iωt) — so whatever moves, moves at ω.
- With c₀ = c₁ = 1/√2 and ⟨0|x|1⟩ = √(ℏ/2mω), the two cross terms add to ⟨x⟩ = 2 × ½ × √(ℏ/2mω) cos ωt = √(ℏ/2mω) cos ωt.
- The natural length is x₀ = √(ℏ/mω) = √(1.055 × 10⁻³⁴ / 9.11 × 10⁻¹⁶) = √(1.158 × 10⁻¹⁹) = 3.40 × 10⁻¹⁰ m, so the amplitude is x₀/√2 = 2.41 × 10⁻¹⁰ m.
- Period T = 2π/ω = 6.28 × 10⁻¹⁵ s. Mean energy ⟨E⟩ = ½ × ½ℏω + ½ × 3/2 ℏω = ℏω = 1.05 × 10⁻¹⁹ J = 0.66 eV.
AnswerAmplitude 0.241 nm at a period of 6.28 fs, with ⟨E⟩ = ℏω = 0.66 eV. The swing, 0.71 x₀, is no bigger than the packet's own width — a two-level state is not an orbit.
MediumA 1.0 g mass on a spring oscillates at f = 1.0 Hz with amplitude 1.0 cm. Model it as a coherent state. Find the mean quantum number, the fractional spread in energy, and the packet width compared with the amplitude, then say what makes the motion look classical.
- ω = 2πf = 6.28 rad s⁻¹, so the classical energy is E = ½mω²A² = ½ × 1.0 × 10⁻³ × 39.5 × 1.0 × 10⁻⁴ = 1.97 × 10⁻⁶ J.
- One rung of the ladder is ℏω = 1.055 × 10⁻³⁴ × 6.28 = 6.63 × 10⁻³⁴ J — which is just hf, as it must be at 1.0 Hz.
- ⟨n⟩ = E/ℏω = 1.97 × 10⁻⁶ / 6.63 × 10⁻³⁴ = 3.0 × 10²⁷, so |α| = √⟨n⟩ = 5.5 × 10¹³.
- Poisson statistics give Δn = |α| = 5.5 × 10¹³, hence ΔE = 5.5 × 10¹³ × 6.63 × 10⁻³⁴ = 3.6 × 10⁻²⁰ J and ΔE/E = 1/√⟨n⟩ = 1.8 × 10⁻¹⁴.
- Δx = √(ℏ/2mω) = √(1.055 × 10⁻³⁴ / 1.26 × 10⁻²) = 9.2 × 10⁻¹⁷ m, so A/Δx = 2|α| = 1.1 × 10¹⁴.
Answer⟨n⟩ ≈ 3.0 × 10²⁷, ΔE/E ≈ 1.8 × 10⁻¹⁴, and the swing is 1.1 × 10¹⁴ packet widths wide. Nothing has stopped being quantum: the rungs and the packet are simply far below anything measurable.
HardA single ⁴⁰Ca⁺ ion of mass 6.64 × 10⁻²⁶ kg is held in a trap with ω = 2π × 1.00 MHz and prepared in a coherent state with α = 2.00, real. Find the packet width, the amplitude of the motion, the probabilities that an energy measurement returns n = 0 and n = 4, and decide whether this counts as classical motion.
- Width first: 2mω = 2 × 6.64 × 10⁻²⁶ × 6.283 × 10⁶ = 8.34 × 10⁻¹⁹ kg s⁻¹, so Δx = √(1.055 × 10⁻³⁴ / 8.34 × 10⁻¹⁹) = √(1.26 × 10⁻¹⁶) = 1.12 × 10⁻⁸ m = 11.2 nm.
- Amplitude: A = √(2ℏ/mω)|α| = 2|α|Δx = 4.00 × 11.2 nm = 45.0 nm, and ⟨x⟩ = 45.0 nm × cos ωt with period 2π/ω = 1.00 μs.
- Occupation: ⟨n⟩ = |α|² = 4.00, and ⟨E⟩ = (⟨n⟩ + ½)ℏω = 4.50 × 6.63 × 10⁻²⁸ = 2.98 × 10⁻²⁷ J.
- Poisson weights: P(0) = e(−4) = 0.018, and P(4) = e(−4) × 4⁴/4! = 0.018 × 10.67 = 0.195 — the most likely single outcome, and still under one chance in five.
- Judge it: A/Δx = 2|α| = 4.0 and Δn/⟨n⟩ = 1/|α| = 0.50. The centre follows the classical path exactly, but a swing of four packet widths with a 50% energy spread is a quantum state; getting Δx/A down to 1% needs 2√⟨n⟩ = 100, that is ⟨n⟩ = 2.5 × 10³.
AnswerΔx = 11.2 nm, A = 45.0 nm, P(0) = 0.018, P(4) = 0.195. The centre obeys the classical equation exactly, but with a swing of only four widths and a 50% energy spread this is not classical motion — that needs ⟨n⟩ ≈ 2.5 × 10³.