
Advanced Quantum Mechanics course sequence · Interactive lesson
Coherent states of a mode and the classical field
This is where the photon picture has to give back the classical wave. Drive one mode with a prescribed current and you get a coherent state: a wave whose mean is Maxwell's and whose noise is the vacuum's. Then we find the one photon-counting number, g⁽²⁾(0), that says when no classical field will do.
UAQM-11.07 · Lesson 7 of 9 in this unit. Position is not a completion score.Build the model
A classical current makes a coherent state; no classical field gives g⁽²⁾(0) below 1.
The key idea
A coherent state |α⟩ of a field mode is the state a prescribed classical current leaves the vacuum in: an eigenstate of the annihilation operator, a|α⟩ = α|α⟩, whose mean field is the classical wave of complex amplitude α and whose noise is the vacuum's.
Example. A coherent 1.00 mW beam at 633 nm delivers 3.19 × 10¹⁵ photons s⁻¹. A 1.00 μs window then holds n̄ = 3.19 × 10⁹ photons, Poisson-spread by √n̄ = 5.6 × 10⁴, a relative 1.8 × 10⁻⁵, with g⁽²⁾(0) = 1.
The complete physical model and assumptions
A prescribed classical current couples to one mode through −∫J⋅A d³r, which adds to ħω a†a a term linear in a and a†. Drive terms at different times commute to a number, so the whole evolution is a displacement: the vacuum is carried, exactly and for any pulse shape, to a coherent state |α⟩, an eigenstate of the annihilation operator. In the Heisenberg picture the same fact reads a(t) = e(−iωt)[a(0) + α(t)]: the current adds the classical solution as a c-number and leaves the vacuum's operator, and hence its fluctuations, untouched. That is the whole quantum account of a classical wave. ⟨E⟩ is Maxwell's plane wave of amplitude 2E₁|α|, the noise on it stays the vacuum's E₁, so the field looks classical once |α| ≫ 1; photon counts are Poisson with spread √n̄; and g⁽²⁾(0) = 1. Thermal light gives 2 and a number state 1 − 1/n. The model stops in two places. The current must be prescribed, unaffected by the field it makes; a laser, whose gain medium does feel its own field, only approximates this, and its phase slowly diffuses. And any randomness in the current only mixes coherent states with non-negative weights, which can raise g⁽²⁾(0) but never push it below 1. Light with g⁽²⁾(0) below 1 therefore cannot come from any classical current, however noisy; it needs a quantum emitter, which is Unit 14's antibunching.
1A prescribed current drives one mode
Work in SI units; nothing here is written covariantly, so the course metric diag(1, −1, −1, −1) never enters. Keep one box mode of frequency ω and polarisation ε, with A = √(ħ/(2ε₀ωV))(a ε e(ik⋅r) + a† ε* e(−ik⋅r)) quoted from topic 11.03. A classical current density J(r, t), prescribed and never itself a dynamical variable, couples through −∫J⋅A d³r, so H = ħω a†a + f(t) a† + f*(t) a with f(t) = −√(ħ/(2ε₀ωV)) ∫J⋅ε* e(−ik⋅r) d³r. In the interaction picture HI(t) = f(t) e(iωt) a† + f*(t) e(−iωt) a, and [HI(t), HI(t′)] = f*(t) f(t′) e(−iω(t−t′)) − c.c., a number. Slice the evolution into short steps exp(−iHI δt/ħ); because every pair commutes to a number, the quoted identity eX eY = e(X+Y) e([X, Y]/2) merges the product into one exponential and a phase: UI = e(iχ) exp(α a† − α* a), with α(t) = −(i/ħ)∫f(t′) e(iωt′) dt′, derived. Dimensions: f is an energy and dt′/ħ an inverse energy, so α is a pure number. Limit: no current, α = 0, and the vacuum stays the vacuum.
2Displacing the vacuum gives an eigenstate of a
Call D(α) = exp(α a† − α* a) the displacement operator; it is unitary because its exponent is anti-Hermitian. With X = α a† and Y = −α* a, [X, Y] = |α|², so the quoted identity gives D(α) = e(−|α|²/2) e(α a†) e(−α* a). The last factor leaves |0⟩ alone, and with a†ⁿ|0⟩ = √(n!)|n⟩ the state is |α⟩ = D(α)|0⟩ = e(−|α|²/2) Σ αⁿ/√(n!) |n⟩, derived. Since Σ |α|(2n)/n! = e(|α|²), it is normalised. Apply a: a|n⟩ = √n|n − 1⟩ shifts the sum by one and returns α|α⟩, so the coherent state is an eigenstate of the annihilation operator, derived; equivalently D†(α) a D(α) = a + α. Free evolution only turns the label, e(−iωt a†a)|α⟩ = |α e(−iωt)⟩, so a coherent state stays coherent and its complex amplitude rotates at ω like the classical mode amplitude. The Heisenberg equation da/dt = −iωa − (i/ħ)f(t) says it in one line: a(t) = e(−iωt)[a(0) + α(t)], the free operator plus a c-number, derived.
3Maxwell's wave as the mean, the vacuum's noise on top
Take the free field E = iE₁(a ε e(i(k⋅r − ωt)) − a† ε* e(−i(k⋅r − ωt))) of topic 11.03, with E₁ = √(ħω/(2ε₀V)), and write α = |α|e(iφ) with ε real. The eigenvalue property turns every a into α, so ⟨α|E|α⟩ = −2E₁|α| sin(k⋅r − ωt + φ) ε, derived: a classical plane wave of amplitude 2E₁|α| whose phase is the phase of α. Now square the component along ε. E² contains a a† + a†a, and a a† = a†a + 1, so ⟨E²⟩ = ⟨E⟩² + E₁², derived: the variance is ħω/(2ε₀V), the vacuum value of topic 11.05, at every amplitude and every instant. The Heisenberg form says why: the current adds a number to a and leaves its fluctuations alone. The noise-to-signal ratio is therefore 1/(2|α|), 5 × 10⁻⁴ for |α|² = 10⁶. Hold the amplitude fixed and |α| = amplitude/(2E₁) grows as 1/√ħ, which is what the classical limit means here. A number state is the opposite: ⟨n|a|n⟩ = 0 gives ⟨E⟩ = 0 everywhere, while ⟨E²⟩ = (2n + 1)E₁² is all fluctuation.
4Photon counting sees a Poisson distribution
Square the amplitudes: P(n) = |⟨n|α⟩|² = e(−n̄) n̄ⁿ/n! with n̄ = |α|², a Poisson distribution, derived. The moments come from the eigenvalue property without any summing: ⟨a†a⟩ = |α|², and (a†a)² = a†a†aa + a†a gives ⟨(a†a)²⟩ = |α|⁴ + |α|², so the variance equals the mean and Δn/n̄ = 1/|α|. A coherent state with n̄ = 100 holds 100 ± 10 photons, and even the likeliest counts, 99 and 100, each turn up only 4.0% of the time. Physically the current emits each photon independently of the others. Limit: |α| → 0 leaves P(0) = 1, the vacuum. Measured: in 1965 Arecchi counted photons from a He-Ne laser running well above threshold and found the Poisson distribution, while the same light scattered from a moving ground-glass disc, a pseudo-thermal source, gave the Bose–Einstein distribution P(n) = n̄ⁿ/(1 + n̄)(n+1), quoted from statistical mechanics. Run near threshold, a laser gives counts broader than Poisson, because its intensity itself still fluctuates: the classical excess the final section bounds.
5Second-order coherence: 1, 2 and 1 − 1/n
For one mode, g⁽²⁾(0) = ⟨a†a†aa⟩/⟨a†a⟩² compares the rate of coincidences at zero delay with the rate for independent arrivals. Since a†a†aa = a†a(a†a − 1) = n(n − 1), it equals 1 + (⟨(Δn)²⟩ − n̄)/n̄², derived. Coherent: the eigenvalue property gives |α|⁴/|α|⁴ = 1 at once. Thermal: ρ = Σ (1 − x)xⁿ|n⟩⟨n| with x = exp(−ħω/kB T) is quoted from statistical mechanics; then n̄ = x/(1 − x) and ⟨n(n − 1)⟩ = 2x²/(1 − x)² = 2n̄², so g⁽²⁾(0) = 2 at every temperature, derived. Number state: n(n − 1)/n² gives g⁽²⁾(0) = 1 − 1/n, derived, which is 0 for one photon, because one photon cannot be detected twice. Limits: as n → ∞ the number state tends to 1, and thermal light keeps 2 however dim, since bunching is a ratio. Measured: Hanbury Brown and Twiss saw the thermal excess in the correlated photocurrents of two detectors in 1956, diluted by detectors slower than the coherence time; pseudo-thermal sources with long coherence times show the full factor of 2.
6What no classical field can do
Let the current be random: amplitude jitter, phase drift, added noise. Each realisation leaves a coherent state, so the field is a mixture ρ = ∫P(α)|α⟩⟨α| d²α with P a probability density, and every normal-ordered moment is an average of numbers: g⁽²⁾(0) = ⟨|α|⁴⟩/⟨|α|²⟩² = 1 + Var(|α|²)/⟨|α|²⟩² ≥ 1, derived, because a variance cannot be negative. The same step for the intensity I of any classical stochastic wave gives ⟨I²⟩ ≥ ⟨I⟩², so g⁽²⁾(0) ≥ 1 for every classical field, with equality only for a stable intensity. Glauber's theory, named here, writes every state in this P form; light is nonclassical when no non-negative P exists, and a measured g⁽²⁾(0) below 1 proves it. Poisson counts, discrete clicks and ⟨E⟩ = 0 prove nothing: a stable classical wave on a square-law detector gives Poisson clicks, quoted from the semiclassical theory of photodetection. Measured: Kimble, Dagenais and Mandel first saw antibunching in 1977, in sodium resonance fluorescence, as coincidences rising away from zero delay, which the quoted classical bound g⁽²⁾(τ) ≤ g⁽²⁾(0) forbids; with a fluctuating number of atoms in view their g⁽²⁾(0) stayed near 1, and single quantum dots now reach g⁽²⁾(0) below 0.01.
Key equations
Read the units, assumptions and limits before substituting.
A classical current displaces the mode
α(t) = −(i/ħ) ∫ f(t′) e(iωt′) dt′, integrated from −∞ to t
Drive terms at different times commute to a number, so the evolution is exactly D(α) = exp(αa† − α*a) times a phase, carrying |0⟩ to |α⟩.
f(t) = −√(ħ/(2ε₀ωV)) ∫ J⋅ε* e(−ik⋅r) d³r, an energy in J; current density J in A m⁻²; α dimensionless; derivedThe mean field is the classical wave
⟨α|E(r, t)|α⟩ = −2E₁|α| sin(k⋅r − ωt + φ) ε
Because a|α⟩ = α|α⟩, the operator field averages to Maxwell's plane wave of amplitude 2E₁|α|, and the phase of α is the phase of the wave.
E₁ = √(ħω/(2ε₀V)) in V m⁻¹; V the quantisation volume in m³; α = |α|e(iφ); ε real; derivedVacuum-sized noise at every amplitude
⟨α|E²|α⟩ − ⟨α|E|α⟩² = E₁² = ħω/(2ε₀V)
Signal grows as |α|, noise does not: ΔE over the amplitude is 1/(2|α|), which at fixed amplitude scales as √ħ. That is the classical limit.
E the field component along ε, in V m⁻¹; topic 11.05's vacuum value, the same for every α and t; derivedPhoton-number amplitudes
⟨n|α⟩ = e(−|α|²/2) αⁿ/√(n!)
Squared, P(n) = e(−n̄) n̄ⁿ/n! with n̄ = |α|²: Poisson, variance n̄, so Δn/n̄ = 1/|α|. A mean of 100 photons is 100 ± 10.
n = 0, 1, 2, …; from D(α)|0⟩ with the quoted identity e(X+Y) = eX eY e(−[X, Y]/2), [X, Y] a number; derivedSecond-order coherence from number moments
g⁽²⁾(0) = ⟨a†a†aa⟩/⟨a†a⟩² = 1 + (⟨(Δn)²⟩ − n̄)/n̄²
Coherent light gives 1, thermal light 2 (variance n̄² + n̄), a number state 1 − 1/n (variance 0). Below 1 needs a variance below the mean.
one mode, any state; a†a†aa = n(n − 1) because [a, a†] = 1; dimensionless; derivedThe classical bound
g⁽²⁾(0) = ⟨I²⟩/⟨I⟩² = 1 + Var(I)/⟨I⟩² ≥ 1
Equality needs a stable intensity. A measured g⁽²⁾(0) below 1, Unit 14's antibunching, rules out every classical field at once.
I a classical field's intensity, averaged over its randomness; holds for light whose Glauber P function is non-negative; derivedInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Raise n̄: the wave grows as 2√n̄ while its dashed edges stay ± E₁ from it, and the Poisson peak moves out as n̄ but widens only as √n̄, while the thermal spread stays as wide as its mean. Turn φ: the wave slides along t, but no bar moves, because photon counting cannot see the phase.
Live results
- MEAN-FIELD AMPLITUDE 2|α|
- 4.00 E₁
- COHERENT Δn/n̄ = 1/|α|
- 0.500
- THERMAL Δn/n̄
- 1.118
- g⁽²⁾(0) OF |n⟩, n = n̄
- 0.750
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Mean photon number n̄ = |α|²: 4; Phase φ of α: 0 °
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The tempting picture is that a classical wave carrying energy nħω simply is n photons, the number state |n⟩, so that a brighter laser is a bigger number state and the most classical light has a definite photon count. It is the reverse. In |n⟩, ⟨n|a|n⟩ = 0, so ⟨E⟩ vanishes at every point and instant: the state has energy but no phase and no oscillating field for a dipole to follow. An oscillating mean field needs ⟨a⟩ ≠ 0, which needs a superposition of neighbouring photon numbers; the coherent state spreads over n̄ ± √n̄ and buys a phase fixed to about 1/(2|α|) radians. At |α|² = 10 the two states have the same time-averaged ⟨E²⟩ = 21E₁², yet one is a wave and the other is not, and only the number state, with g⁽²⁾(0) = 0.90, lies beyond every classical field. A definite photon number is the most quantum light there is, not the most classical.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A cavity mode at ν = 10.0 GHz with mode volume V = 5.0 cm³ is driven from the vacuum into a coherent state with n̄ = 25. Find the vacuum field E₁, the amplitude of ⟨E⟩ at the field maximum, its noise ΔE, the photon-number spread, and the probability that a count finds exactly 25 photons.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- ω = 2π × 1.00 × 10¹⁰ s⁻¹ = 6.28 × 10¹⁰ s⁻¹, so ħω = 1.0546 × 10⁻³⁴ J s × 6.283 × 10¹⁰ s⁻¹ = 6.63 × 10⁻²⁴ J.
- E₁ = √(ħω/(2ε₀V)) = √(6.626 × 10⁻²⁴ J ÷ (2 × 8.854 × 10⁻¹² F m⁻¹ × 5.0 × 10⁻⁶ m³)) = √(7.48 × 10⁻⁸ V² m⁻²) = 2.74 × 10⁻⁴ V m⁻¹. Dimensions: J/(F m⁻¹ × m³) = V² m⁻², since J/F = V².
- |α| = √25 = 5, so ⟨E⟩ oscillates with amplitude 2E₁|α| = 10 × 2.74 × 10⁻⁴ = 2.74 × 10⁻³ V m⁻¹, while ΔE = E₁ = 2.74 × 10⁻⁴ V m⁻¹. Their ratio is 1/(2|α|) = 0.10.
- Poisson: Δn = √25 = 5, so Δn/n̄ = 0.20. P(25) = e(−25) 25²⁵/25! = 0.0795; P(24) is the same, so 24 and 25 are the joint likeliest counts.
- Limit: at |α| → 0 the amplitude vanishes but ΔE stays E₁, which is the vacuum of topic 11.05.
Answer. E₁ = 2.7 × 10⁻⁴ V m⁻¹; ⟨E⟩ has amplitude 2.7 × 10⁻³ V m⁻¹ with ΔE = 2.7 × 10⁻⁴ V m⁻¹, a ratio of 0.10; n = 25 ± 5, and P(25) ≈ 0.080.
MediumExample 2
A drive f(t) = F cos(ωd t), with F/ħ = 2.00 × 10⁵ s⁻¹, acts from t = 0 to T = 100 μs on a vacuum mode with ω/2π = 10.0 GHz. Find α(T) and n̄ on resonance, then |α| and n̄ for detunings (ω − ωd)/2π = 2.50 kHz and 10.0 kHz, and justify dropping the counter-rotating term.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- α(T) = −(iF/ħ) ∫ cos(ωd t′) e(iωt′) dt′ from 0 to T = −(iF/2ħ) ∫ [e(iΔt′) + e(i(ω + ωd)t′)] dt′, with Δ = ω − ωd.
- Counter-rotating piece: its size is at most F/(ħ(ω + ωd)) = 2.00 × 10⁵ ÷ 1.26 × 10¹¹ = 1.6 × 10⁻⁶, negligible beside the resonant 10.
- Resonant term: α = −(F/2ħΔ)(e(iΔT) − 1), so |α| = (F/ħΔ)|sin(ΔT/2)|. As Δ → 0 this tends to FT/(2ħ) = 2.00 × 10⁵ × 1.00 × 10⁻⁴ ÷ 2 = 10.0, with α = −10.0i, so n̄ = 100 ± 10. Dimensions: F/ħ in s⁻¹ times T in s is a pure number.
- At 2.50 kHz: Δ = 1.571 × 10⁴ s⁻¹ and ΔT/2 = π/4, so |α| = 12.73 × 0.7071 = 9.00 and n̄ = 81.1. Here α = 6.37(1 − i).
- At 10.0 kHz: ΔT/2 = π, sin π = 0, so α = 0. The current pumped photons in for the first half of the pulse and took them back in the second, and the mode ends exactly in the vacuum, as a detuned classical oscillator returns to rest.
Answer. On resonance α = −10.0i and n̄ = 100; at 2.50 kHz |α| = 9.00 and n̄ = 81.1; at 10.0 kHz α = 0, the vacuum. The counter-rotating term changes α by at most 1.6 × 10⁻⁶.
HardExample 3
A pulsed classical current drives one mode from the vacuum. Its phase is random from pulse to pulse, and its amplitude takes one of two values with equal probability, leaving |α|² = 4.0 or 16.0. Find n̄, g⁽²⁾(0), the photon-number variance two ways, the probability of an empty pulse, and the ensemble ⟨E⟩ and time-averaged ⟨E²⟩ at the field maximum in units of E₁. Which of these separate this light from the number state |10⟩?
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Each pulse leaves a coherent state, so ρ is an average of |α⟩⟨α| with non-negative weights, a classical mixture. Normal-ordered moments average as numbers: n̄ = ⟨|α|²⟩ = (4.0 + 16.0)/2 = 10.0, and ⟨a†a†aa⟩ = ⟨|α|⁴⟩ = (16 + 256)/2 = 136.
- g⁽²⁾(0) = 136/10.0² = 1.36, at least 1 as every classical mixture must be.
- Variance from g⁽²⁾: ⟨(Δn)²⟩ = n̄ + (g⁽²⁾ − 1)n̄² = 10.0 + 0.36 × 100 = 46. Directly: the mean Poisson variance, 10.0, plus the spread of |α|², [(4 − 10)² + (16 − 10)²]/2 = 36, gives 46 again, so Δn = 6.8.
- P(0) = (e(−4) + e(−16))/2 = 9.16 × 10⁻³, against e(−10) = 4.54 × 10⁻⁵ for a steady coherent beam of the same mean, where empty pulses are 202 times rarer, and exactly 0 for |10⟩.
- Field: the random phase averages ⟨E⟩ to 0 at every instant. Each pulse gives a time-averaged ⟨E²⟩ = (2|α|² + 1)E₁², so the ensemble gives (2n̄ + 1)E₁² = 21E₁². The number state |10⟩ also has ⟨E⟩ = 0 and ⟨E²⟩ = (2 × 10 + 1)E₁² = 21E₁².
- Verdict: ⟨E⟩ and ⟨E²⟩ cannot tell them apart. g⁽²⁾(0) of 1.36 against 0.90, a variance of 46 against 0, and P(0) of 9.2 × 10⁻³ against 0 can, and only the 0.90 lies beyond every classical field.
Answer. n̄ = 10.0, g⁽²⁾(0) = 1.36, variance 46 (Δn ≈ 6.8), P(0) = 9.2 × 10⁻³; ⟨E⟩ = 0 and ⟨E²⟩ = 21E₁², identical to |10⟩, which only g⁽²⁾(0) = 0.90 and the photon statistics distinguish.
Quick check
Select an answer and explain it before checking.
Review & continue
A coherent state |α⟩ of a field mode is the state a prescribed classical current leaves the vacuum in: an eigenstate of the annihilation operator, a|α⟩ = α|α⟩, whose mean field is the classical wave of complex amplitude α and whose noise is the vacuum's.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.