Skip to main content

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.
Starting model: Coherent states of a mode and the classical fieldLeft: a coherent state's mean field at the field maximum, 2|α|E₁ sin(ωt − φ), with dashed edges ± E₁ that never widen, over the shaded ± rms band √(2n + 1) E₁ of the number state with n = 4, whose mean is zero. Right: P(n) for coherent (Poisson) and thermal light at n̄ = 4, rescaled by √(2πn̄); faint line P = 0.1, arrow the number state, P = 1.⟨E⟩/E₁ over ωt = 0 to 4πdashed ± E₁, shaded |n⟩P(n) at n̄ = 4solid: coherentdashed: thermal|n⟩, P = 10.1photon number n, 0 to 29
Start with a picture. Change the model in the lab below.
1

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.

2

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; derived

The 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; derived

Vacuum-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; derived

Photon-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; derived

Second-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; derived

The 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; derived
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

4
1 to 16; steps of 1.
0 °
0 to 360 °; steps of 15.

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.

Interactive Advanced Quantum Mechanics modelLeft: a coherent state's mean field at the field maximum, 2|α|E₁ sin(ωt − φ), with dashed edges ± E₁ that never widen, over the shaded ± rms band √(2n + 1) E₁ of the number state with n = 4, whose mean is zero. Right: P(n) for coherent (Poisson) and thermal light at n̄ = 4, rescaled by √(2πn̄); faint line P = 0.1, arrow the number state, P = 1.⟨E⟩/E₁ over ωt = 0 to 4πdashed ± E₁, shaded |n⟩P(n) at n̄ = 4solid: coherentdashed: thermal|n⟩, P = 10.1photon number n, 0 to 29
Left: a coherent state's mean field at the field maximum, 2|α|E₁ sin(ωt − φ), with dashed edges ± E₁ that never widen, over the shaded ± rms band √(2n + 1) E₁ of the number state with n = 4, whose mean is zero. Right: P(n) for coherent (Poisson) and thermal light at n̄ = 4, rescaled by √(2πn̄); faint line P = 0.1, arrow the number state, P = 1.

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 °

Baseline 4; choose 1 to 16, in steps of 1.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

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.

5

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.

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.

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.

6

Quick check

Select an answer and explain it before checking.

Four light sources are tested. Which single observation shows that the light cannot be described by any classical electromagnetic field, however noisy?
7

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.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.