University Physics V · The Quantum Harmonic Oscillator · 7.8

Coherent States & Squeezing

Once you have the ladder, the interesting states are not the rungs. Displacing the vacuum gives the closest a quantum oscillator comes to a classical orbit; squeezing it buys precision in one quadrature by paying for it in the other. Both are exponentials of operators, and which operator decides everything.

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The Fock states diagonalise H, but they describe nothing that moves: ⟨n|x|n⟩ = 0 on every rung. The states that do move are the eigenvectors of the operator that is not an observable. Because a is not Hermitian, a|α⟩ = α|α⟩ has a normalisable solution for every complex α, and in the Fock basis that solution is forced: cₙ = αⁿ c₀/√(n!), normalised by c₀ = e^{(−|α|²/2)}. The same state is the vacuum displaced, |α⟩ = D(α)|0⟩ with D(α) = exp(α a† − α* a), and because D†(α) a D(α) = a + α the displacement moves the mean and leaves every variance at its vacuum value.

What follows is the closest a quantum oscillator comes to a classical orbit: under H the label simply turns, α → α e^(−iωt), so the centre traces the classical path at fixed width, with Poissonian number spread Δn = |α| that shrinks relative to n̄ = |α|² as the state brightens. The price is threefold. The noise floor is exactly vacuum, not zero, so shot noise is built in. The rigid orbit is an accident of the quadratic H and dies the moment a quartic term appears.

And the family is not a basis — no two coherent states are orthogonal, and the resolution of identity needs an integral over the whole plane. Squeezing reaches below the floor in one quadrature by exponentiating a quadratic rather than a linear function of a, and pays for it in the conjugate one.

Simple definition: A coherent state is the normalisable eigenvector of the annihilation operator, a|α⟩ = α|α⟩ — equivalently the vacuum displaced in phase space by D(α), carrying vacuum-sized uncertainty in both quadratures.
Example: For α = 2 the mean occupation is |α|² = 4 with Δn = |α| = 2, the mode is found empty with probability e⁻⁴ = 1.8%, and ΔX stays at 1/√2 = 0.707 — exactly the vacuum's, whatever α is.

Eigenvector of a non-Hermitian operatora|α⟩ = α|α⟩, |α⟩ = e^{(−|α|²/2)} Σₙ αⁿ/√(n!) |n⟩

Every expectation of a and a† collapses to a power of α — moments without a single integral.

α is any complex number; the sum runs over all Fock states, so |α⟩ has no definite energy

Displacement operatorD(α) = exp(α a† − α* a), D†(α) a D(α) = a + α, |α⟩ = D(α)|0⟩

It slides the vacuum across phase space without deforming it, which is why every variance stays at its vacuum value.

Unitary; D(α)D(β) = e^{(i Im αβ*)} D(α + β), so displacements compose only up to a phase

Poissonian number statisticsP(n) = e^(−|α|²) |α|⁽²ⁿ⁾/n!, n̄ = |α|², Δn = |α|

This is shot noise: 10⁸ detected quanta scatter by 10⁴, one part in 10⁴, and no stabilisation removes it.

Fano factor (Δn)²/n̄ = 1 exactly; the relative spread Δn/n̄ = 1/|α| falls as the state brightens

Quadratures and free evolution⟨X⟩ = √2 Re α(t), α(t) = α e^(−iωt), ΔX = ΔP = 1/√2

The centre runs the classical orbit at constant width — true only because H is quadratic in a.

X = (a + a†)/√2 with [X, P] = i; in length units x = √(ħ/mω) X, so Δx = √(ħ/2mω)

Squeeze operatorS(ξ) = exp[½(ξ* a² − ξ a†²)], ξ = r e^(iθ), S†aS = a cosh r − a† e^(iθ) sinh r

Quadratic in a, so it rescales the quadratures instead of shifting them: 8.686 r dB below vacuum in one, as much above in the other.

θ = 0 gives ΔX = e^(−r)/√2 and ΔP = e^(r)/√2; the squeezed vacuum holds ⟨n⟩ = sinh²r quanta

Overlap and resolution of identity|⟨β|α⟩|² = e^{(−|α − β|²)}, (1/π) ∫ d²α |α⟩⟨α| = 1

The set resolves the identity while being linearly dependent: overcomplete, so expansions in it are not unique.

d²α = d(Re α) d(Im α); the overlap decays but never reaches zero, so no two coherent states are orthogonal

Why a has eigenvectors and a† has none

Nothing requires a non-Hermitian operator to have eigenvectors, but nothing forbids it either. Put |α⟩ = Σ cₙ|n⟩ into a|α⟩ = α|α⟩ and use a|n⟩ = √n |n−1⟩: matching the coefficient of |n⟩ gives √(n+1) c_(n+1) = α cₙ, so cₙ = αⁿ c₀/√(n!). Normalisation Σ|cₙ|² = |c₀|² e^(|α|²) = 1 then fixes c₀ = e^{(−|α|²/2)}. Every complex α gives exactly one normalised solution, so the spectrum of a is the whole complex plane and each eigenvalue is non-degenerate. Now run the same argument for a†. Since a†|ψ⟩ = Σₙ cₙ √(n+1)|n+1⟩ has no |0⟩ component at all, the |0⟩ line of a†|β⟩ = β|β⟩ reads 0 = β c₀, the |1⟩ line then gives c₀ = β c₁, and the whole vector collapses to zero. a† has no normalisable eigenvector, and the asymmetry is the ladder's own: lowering terminates at |0⟩, raising never terminates.

Displacement: |α⟩ = D(α)|0⟩

The same state has a constructive definition. D(α) = exp(α a† − α* a) has an anti-Hermitian exponent, so it is unitary. Because [α a†, −α* a] = |α|² is a c-number, the Baker–Campbell–Hausdorff identity e^(A+B) = e^A e^B e^{(−[A, B]/2)} splits it into D(α) = e^{(−|α|²/2)} e^{(α a†)} e^{(−α* a)}. Act on the vacuum: the rightmost factor is the identity there because a|0⟩ = 0, and e^{(α a†)}|0⟩ = Σₙ (αⁿ/n!)(a†)ⁿ|0⟩ = Σₙ (αⁿ/√(n!))|n⟩, reproducing the Fock sum exactly. The operator statement is the useful one: [a, α a† − α* a] = α is a number, so the commutator series for D†(α) a D(α) terminates after one term at a + α. A displacement adds a constant to a. The means move to ⟨X⟩ = √2 Re α and ⟨P⟩ = √2 Im α; every variance, and every commutator, is untouched. Draw it and the vacuum disc has slid to a new centre without changing shape.

Poissonian statistics and the shot-noise floor

Project onto the Fock basis: P(n) = |⟨n|α⟩|² = e^(−|α|²)|α|⁽²ⁿ⁾/n!, a Poisson distribution of parameter |α|². The moments come from the eigenvalue property with no sum at all. ⟨n⟩ = ⟨α|a†a|α⟩ = |α|², and normal-ordering gives ⟨n²⟩ = ⟨α|a†(a†a + 1)a|α⟩ = |α|⁴ + |α|², so (Δn)² = |α|² and Δn = |α|. The Fano factor (Δn)²/⟨n⟩ is exactly 1. That is the definition of shot noise, and it is why a perfectly stabilised laser still has a noise floor. A 1 ms sample of a 1 mW beam at 633 nm carries n̄ = 3.2 × 10¹² photons, so Δn = 1.8 × 10⁶ and the fractional noise is 5.6 × 10⁻⁷ — small, but not removable by improving the source. A Fano factor below 1, sub-Poissonian light, is impossible for any classical field and demands a state outside this family.

Free evolution: the label turns, the width does not

Evolve with U(t) = e^(−iHt/ħ) = e^(−iωt/2) e^(−iωtN). On the Fock sum it multiplies the |n⟩ amplitude by e^(−inωt), and since that amplitude is αⁿ/√(n!), the entire effect is α → α e^(−iωt): U(t)|α⟩ = e^(−iωt/2)|α e^(−iωt)⟩. A coherent state stays coherent, and its label runs clockwise on a circle of radius |α|. Writing x = √(ħ/2mω)(a + a†) then gives ⟨x⟩(t) = 2√(ħ/2mω)|α| cos(ωt − φ) and ⟨p⟩(t) = −√(2mħω)|α| sin(ωt − φ): the classical orbit, exactly, with no spreading, because Δx = √(ħ/2mω) at every instant. Do not generalise this. Ehrenfest's equations close on the means only when the force is linear in x; add a quartic term and ⟨V′(x)⟩ ≠ V′(⟨x⟩), the phase-space disc shears into a crescent, and the state stops being coherent within a few periods.

Squeezing needs a quadratic, not a linear, exponent

To get below the vacuum floor you need an operator quadratic in a, not linear. S(ξ) = exp[½(ξ* a² − ξ a†²)] with ξ = r e^(iθ) is again unitary, and its action is a Bogoliubov transformation, S†(ξ) a S(ξ) = a cosh r − a† e^(iθ) sinh r, mixing a with a† instead of adding a number to it. Take θ = 0 and the dimensionless quadratures rescale cleanly: X → X e^(−r), P → P e^(+r). The variances become e^(−2r)/2 and e^(+2r)/2, so their product stays at ¼ and the ellipse has exactly the vacuum disc's area. In decibels the squeezed quadrature sits 8.686 r below vacuum. The cost is real: S(ξ)|0⟩ carries ⟨n⟩ = sinh²r photons and only even Fock components, because a†² creates them in pairs. Ten decibels means r = 1.151 and 2.03 photons, and it takes a two-photon process — parametric down-conversion, or a spring constant modulated at 2ω — since no linear drive can produce it.

Overcomplete, and the phase you measure at

Two coherent states are never orthogonal. ⟨β|α⟩ = exp(−½|α|² − ½|β|² + β*α), so |⟨β|α⟩|² = e^{(−|α − β|²)}: at |α − β| = 2 that is e⁻⁴ = 1.8%, at 3 it is 1.2 × 10⁻⁴, and it never reaches zero. Yet the set still resolves the identity, (1/π)∫d²α |α⟩⟨α| = 1 with d²α = d(Re α)d(Im α). A linearly dependent family that resolves unity is overcomplete: any one |α⟩ can be expanded in the others, so expansion coefficients are not unique and phase space is described by quasi-distributions (Q, Wigner, P) rather than by amplitudes. Overcompleteness also fixes what an experiment reads. Homodyne detection measures the rotated quadrature X_φ = (a e^(−iφ) + a† e^(iφ))/√2, whose variance in a θ = 0 squeezed state is ½(e^(−2r)cos²φ + e^(2r)sin²φ): get the local-oscillator phase 5° wrong and 10 dB of squeezing reads as 7.6 dB.

Change one variable at a time

Make the relationship visible.

Interactive model

Coherent amplitude |α|1.6

Squeeze parameter r0.40

Phase ωt25 °

Set r = 0 and drag ωt: the dot runs clockwise round the orbit while the noise circle never changes size. Now raise r: the ellipse turns with the dot, and the measured ΔX on the right dips below the vacuum line twice a turn and rises above it twice.

MEAN NUMBER |α|²2.56 quanta

ΔX AT PHASE ωt0.619

ΔX⋅ΔP (bound is 0.5)0.605

SQUEEZE COST sinh²r0.17 quanta

Live interpretationMEAN NUMBER |α|²: 2.56 quanta. ΔX AT PHASE ωt: 0.619. ΔX⋅ΔP (bound is 0.5): 0.605. SQUEEZE COST sinh²r: 0.17 quanta

Catch the common trap

Explain before calculating.

Misconception check

The common error is to read "coherent" as "classical, therefore noiseless". A coherent state is not noise-free: it carries exactly the vacuum's noise, Δx Δp = ħ/2, in every quadrature and at every amplitude. What makes it look classical is a ratio, not an absence — the signal grows as |α| while the spread stays put, so Δn/n̄ = 1/|α| falls, and by n̄ = 10⁸ the fractional noise is 10⁻⁴. Two corollaries get missed. Squeezing does not evade the Robertson bound: S(ξ) holds ΔX ΔP at ½ and merely moves variance between conjugate quadratures, so 10 dB below vacuum in X buys 10 dB above it in P. And a† has no coherent states — ⟨α|a† = α*⟨α| is a statement about the bra, while a†|α⟩ is not proportional to |α⟩. The raising ladder has no top rung to terminate on, so a† has no normalisable eigenvector at all.

A single mode of frequency ω is prepared in the coherent state |α⟩ with α = 3 (real and positive), then left to evolve freely under H = ħω(a†a + ½). Which statement describes the state a quarter period later, at ωt = π/2?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA single mode is prepared in the coherent state |α⟩ with α = 2 (real and positive). Find the mean occupation, the number uncertainty, the probability of detecting exactly zero quanta and exactly four, and the quadrature uncertainty ΔX.

n̄ = ⟨α|a†a|α⟩ = |α|² = 4. Because a|α⟩ = α|α⟩ and ⟨α|a† = α*⟨α|, this is just a product of numbers — no Fock sum, no integral.
Normal-order the second moment: ⟨n²⟩ = ⟨α|a†(a†a + 1)a|α⟩ = |α|⁴ + |α|² = 16 + 4 = 20, so (Δn)² = 20 − 16 = 4 and Δn = |α| = 2. The Fano factor (Δn)²/n̄ = 4/4 = 1 exactly: Poissonian.
P(n) = e⁻⁴ 4ⁿ/n!. P(0) = e⁻⁴ = 0.0183, so the mode reads empty 1.8% of the time; P(4) = e⁻⁴ × 256/24 = 0.0183 × 10.67 = 0.195.
D(α) is unitary and does not deform the vacuum, so ΔX = ΔP = 1/√2 = 0.707 whatever α is. Only the centre moved, to ⟨X⟩ = √2 Re α = 2.83 and ⟨P⟩ = 0.

Answern̄ = 4, Δn = 2 (a 50% relative spread), P(0) = 1.8%, P(4) = 19.5%, and ΔX = ΔP = 0.707 — vacuum-sized, and independent of α.

MediumA nanomechanical beam has m = 1.0 × 10⁻¹⁵ kg and ω/2π = 1.0 MHz. It is driven into a coherent state whose displacement oscillates with amplitude 1.0 nm. Find the zero-point length, the coherent label |α|, the mean occupation, and the fractional position noise. Take ħ = 1.055 × 10⁻³⁴ J s.

x = √(ħ/2mω)(a + a†) defines the zero-point length x_zpf = √(ħ/2mω). With ω = 2π × 1.0 × 10⁶ = 6.283 × 10⁶ rad s⁻¹, 2mω = 1.257 × 10⁻⁸ kg s⁻¹, so x_zpf = √(1.055 × 10⁻³⁴ / 1.257 × 10⁻⁸) = √(8.395 × 10⁻²⁷) = 9.16 × 10⁻¹⁴ m.
⟨x⟩(t) = x_zpf⟨a + a†⟩ = 2x_zpf|α| cos(ωt − φ), so the swing amplitude is 2x_zpf|α|. Setting 2x_zpf|α| = 1.0 × 10⁻⁹ m gives |α| = 1.0 × 10⁻⁹ / 1.833 × 10⁻¹³ = 5.46 × 10³.
n̄ = |α|² = 2.98 × 10⁷ quanta. Cross-check against the classical energy: ½mω²A² = ½(1.0 × 10⁻¹⁵)(6.283 × 10⁶)²(1.0 × 10⁻⁹)² = 1.974 × 10⁻²⁰ J, and n̄ħω = 2.98 × 10⁷ × 6.63 × 10⁻²⁸ = 1.974 × 10⁻²⁰ J. They agree.
Δx = x_zpf for every coherent state, so Δx/A = x_zpf/(2x_zpf|α|) = 1/(2|α|) = 9.2 × 10⁻⁵, and Δn/n̄ = 1/|α| = 1.8 × 10⁻⁴.

Answerx_zpf = 9.2 × 10⁻¹⁴ m, |α| = 5.5 × 10³, n̄ = 3.0 × 10⁷ quanta, Δx/A = 9.2 × 10⁻⁵. The quantum blur is one part in 10⁴ of the orbit, which is why the motion reads as classical.

HardA squeezed-vacuum source is aligned so the measured quadrature sits on the short axis of the noise ellipse, giving 10.0 dB below the vacuum level. Find r and the mean photon number of the squeezed vacuum, state the anti-squeezed quadrature in dB, then find the noise if the local-oscillator phase is 5.0° off, and the misalignment at which all advantage is lost.

Noise power relative to vacuum is (ΔX)²/(ΔX_vac)² = e^(−2r), and 10.0 dB below vacuum means e^(−2r) = 10^(−1.00) = 0.100. So 2r = ln 10 = 2.303 and r = 1.151.
⟨n⟩ = ⟨0|S†(ξ) a†a S(ξ)|0⟩ = sinh²r = (e^(2r) − 2 + e^(−2r))/4 = (10.0 − 2 + 0.100)/4 = 2.03 photons. Ten decibels costs about two photons per mode, and they arrive in pairs — S(ξ)|0⟩ has only even Fock components.
The orthogonal quadrature carries e^(+2r) = 10.0, that is +10.0 dB. The product ΔX ΔP = ½ is untouched, so the ellipse has exactly the vacuum disc's area — nothing was gained overall, only moved.
At a local-oscillator misalignment φ the measured variance is ½(e^(−2r)cos²φ + e^(2r)sin²φ). With φ = 5.0°: 0.100 × 0.9924 + 10.0 × 0.007596 = 0.0992 + 0.0760 = 0.1752, and 10 log₁₀(0.1752) = −7.56 dB.
The advantage vanishes when that ratio reaches 1: 0.100 + 9.90 sin²φ = 1 gives sin²φ = 0.900/9.90 = 0.0909, so φ = 17.5°.

Answerr = 1.15, ⟨n⟩ = 2.03 photons, with +10.0 dB in the anti-squeezed quadrature. A 5.0° phase error already throws away 2.4 of the 10 dB, and beyond 17.5° the squeezed beam is noisier than vacuum.