
Advanced Quantum Mechanics course sequence · Interactive lesson
One atom in one mode: the Jaynes-Cummings model
This is the problem where an atom meets a quantised field and everything can still be solved by hand. Work it to see exactly what quantising the field adds to the Rabi oscillation of Quantum Mechanics II: an excited atom trading its energy with an empty mode, frequencies that step as √(n + 1), and revivals no classical field can make.
UAQM-11.09 · Lesson 9 of 9 in this unit. Position is not a completion score.Build the model
One conserved number turns an atom in a cavity into a ladder of 2 × 2 problems.
The key idea
The Jaynes-Cummings model is a two-level atom coupled to one quantised field mode with the counter-rotating terms dropped, H = ħωc a†a + ħωₐσ₊σ₋ + ħg(aσ₊ + a†σ₋), whose conserved excitation number splits it into independent 2 × 2 blocks.
Example. In a cavity with g/2π = 25 kHz and no photons, an excited atom is back in |e⟩ every π/g = 20 µs, since Pₑ = cos²(gt), and the lowest doublet is split by 2g/2π = 50 kHz; one photon in the mode widens the next doublet by √2, to 70.7 kHz.
The complete physical model and assumptions
Topic 11.08 ordered the atom-field Hamiltonian in powers of e, and its leading term, which changes the photon number by one, acts on resonance and in the dipole approximation as −d⋅E. Keep one mode of a cavity and two levels of an atom and that term becomes ħg(σ₊ + σ₋)(a + a†), with ħg = d√(ħω/(2ε₀V)): the dipole times the vacuum field of the mode, so a smaller mode couples harder. Of its four products, aσ₊ and a†σ₋ nearly conserve energy while the other two oscillate at about 2ω; dropping that pair, the rotating-wave approximation, leaves a Hamiltonian that conserves the excitation number a†a + σ₊σ₋. That one symmetry solves the model. H splits into 2 × 2 blocks, each spanned by |e, n⟩ and |g, n + 1⟩ and each a two-level problem with coupling g√(n + 1), and the √(n + 1) is where quantisation shows: it splits the doublets by 2ħg√(n + 1), makes an excited atom in an empty mode pass its energy back and forth at 2g instead of decaying, and makes the inversion driven by a coherent field collapse and then revive, which no classical field can do. The model stops where its assumptions do. Near g/ω ≈ 0.1 the counter-rotating terms return and only parity is conserved; real atoms have more than two levels; real cavities leak, and atoms radiate into modes outside them. Replace the one mode by a continuum and the reversible exchange becomes Unit 12's irreversible decay.
1From the dipole coupling to one number, g
SI units throughout; the course's metric signature is (+, −, −, −), though no four-vector appears here. Quoted from topic 11.08: the leading atom-field term changes the photon number by one, and in the dipole approximation its resonant matrix elements equal those of −d⋅E (Quantum Mechanics II). Topic 11.03's construction, applied to one standing-wave mode of a cavity, gives E(r) = Evac f(r) ε̂ (a + a†), with f real, |f| ≤ 1, mode volume V = ∫f² d³r, and Evac = √(ħω/(2ε₀V)) the vacuum field of topic 11.05 (quoted). Keep two atomic levels; parity kills the diagonal elements, so d⋅ε̂ = d(σ₊ + σ₋) with d = ⟨e|d⋅ε̂|g⟩ made real by a choice of phase. Then −d⋅E = −ħg f(r₀)(σ₊ + σ₋)(a + a†) with ħg = dEvac at an antinode: derived, and the sign is removed by relabelling |e⟩ → −|e⟩. Dimensions: C m × V m⁻¹ = J. An atom with d = ea₀ = 8.48 × 10⁻³⁰ C m in a mode of wavelength 1.00 µm and volume λ³ feels Evac = 1.06 × 10⁵ V m⁻¹ and has g/2π = 1.36 GHz; a mode a thousand times larger gives 42.9 MHz.
2Drop two terms, and an excitation number is conserved
Multiply out (σ₊ + σ₋)(a + a†). In the interaction picture σ₊a carries e(i(ωₐ − ωc)t), slow near resonance, and σ₋a† its conjugate, while σ₊a† and σ₋a carry e(±i(ωₐ + ωc)t) and turn through a cycle every π/ω. Over times of order 1/g they average away when g ≪ ωₐ + ωc, and second-order perturbation theory puts their leftover effect at a shift of order g²/(ωₐ + ωc), the Bloch-Siegert shift. Dropping them is the rotating-wave approximation. Now [a†a, aσ₊] = −aσ₊ and [σ₊σ₋, aσ₊] = +aσ₊, so N = a†a + σ₊σ₋ commutes with aσ₊, with a†σ₋ and with both free terms: derived. The dropped pair changes N by ±2, which is exactly why it had to go. For the circuit of Wallraff et al. (2004), g/2π = 5.8 MHz on a 6.04 GHz resonator (quoted): g/ω = 9.6 × 10⁻⁴, and the neglected shift, about g²/(2ω), is 2π × 2.8 kHz, 2.4 × 10⁻⁴ of the splitting 2g. Circuits built to reach g/ω above 0.1 (quoted) break the approximation; then only the parity (−1)N survives and the blocks are no longer 2 × 2.
3Each block is a two-level problem: the dressed ladder
For N = n + 1 the block holds |e, n⟩, energy ħ(nωc + ωₐ), and |g, n + 1⟩, energy ħ(n + 1)ωc, joined by ħg√(n + 1) from a†|n⟩ = √(n + 1)|n + 1⟩. A 2 × 2 Hermitian matrix with diagonal difference ħδ, δ = ωₐ − ωc, and off-diagonal ħg√(n + 1) has eigenvalues at its mean ± (ħ/2)√(δ² + 4g²(n + 1)): derived. On resonance the dressed states are (|e, n⟩ ± |g, n + 1⟩)/√2 and the splittings run 2g, 2.83g, 3.46g, 4g; the lone ground state |g, 0⟩ is unshifted. Three limits check it. As g → 0 the bare levels return. For |δ| ≫ g√(n + 1) the shifts become ±ħg²(n + 1)/δ, Quantum Mechanics II's second-order perturbation theory: at δ = 10g the exact n = 0 shift is 0.0990g against 0.1g. For n ≫ 1 the splitting tends to 2g√n, which is dE₀/ħ for a classical field of amplitude E₀ = 2Evac√n, the semiclassical Rabi frequency; at n = 100 they differ by 0.5%. Two coupled harmonic oscillators, which is what a classical Lorentz atom in a cavity is, give an evenly spaced ladder, so the √2 step from the first doublet to the second is a quantum mark, resolved in a circuit by Fink et al. (2008) (quoted).
4Vacuum Rabi oscillation: emission that comes back
Start the atom in |e⟩ with the mode empty. |e, 0⟩ lies in the N = 1 block, so on resonance |ψ(t)⟩ = e(−iωt)[cos(gt)|e, 0⟩ − i sin(gt)|g, 1⟩] and Pₑ = cos²(gt) = (1 + cos 2gt)/2: derived. By t = π/(2g) the photon is in the mode with certainty and by π/g it is back in the atom; at gt = π/4 atom and mode are maximally entangled and the atom alone is in the mixed state ½(|e⟩⟨e| + |g⟩⟨g|). Nothing decays because the block has only two energies: every amplitude is a sum of two phases e(−iEt/ħ), and so periodic. A semiclassical atom cannot do this at all: in a classical field of zero amplitude the coupling vanishes, and an atom in |e⟩, with ⟨d⟩ = 0, radiates no field of its own, so it stays excited for ever. In the spectrum the same physics is two lines 2g apart, the vacuum Rabi splitting, resolved once g exceeds the cavity and atomic linewidths. Measured (quoted): Brune et al. (1996) saw circular Rydberg atoms oscillate in an empty 51 GHz cavity at 2g/2π ≈ 47 kHz, and Wallraff et al. (2004) resolved 2g/2π = 11.6 MHz in a circuit, a vacuum Rabi period of 86 ns.
5A coherent field: the collapse
Let the mode start in a coherent state with n̄ = |α|², whose photon numbers are Poisson, Pₙ = e(−n̄)n̄ⁿ/n!, with spread Δn = √n̄ (topic 11.07, quoted). Each component |e, n⟩ Rabi-oscillates in its own block, so the inversion is W = Pₑ − Pg = Σₙ Pₙ cos(2g√(n + 1)t), exactly. For n̄ ≫ 1 expand the frequency about n̄: 2g√(n + 1) ≈ 2g√(n̄ + 1) + g(n − n̄)/√(n̄ + 1). Averaging over a near-Gaussian spread of variance n̄ multiplies the cosine by e(−g²t²n̄/(2(n̄ + 1))), so W ≈ cos(2g√(n̄ + 1)t) e(−g²t²/2): derived. The collapse time tc = √2/g does not depend on n̄, because the spread of Rabi frequencies, Δn times their slope g/√(n̄ + 1), stays close to g. For n̄ = 10, P₉ = P₁₀ = 0.125, the mean Rabi frequency is 6.63g, and the exact sum falls to 1/e at gt = 1.47, so only about one and a half Rabi cycles are seen. The collapse is dephasing: nothing has left the atom and the mode. Nor does it prove quantisation. A classical field of fixed amplitude 2Evac√n̄ would Rabi-oscillate at 2g√n̄ for ever, but one whose intensity is spread as widely would collapse in the same way.
6Revivals: the discreteness of n made visible
The exact sum is over integers, and neighbouring terms differ in frequency by 2g(√(n + 2) − √(n + 1)) ≈ g/√(n̄ + 1). They come back into phase when that difference times t reaches 2π: tᵣ ≈ 2π√(n̄ + 1)/g ≈ 2π√n̄/g, derived. For n̄ = 10 the formula gives gt = 20.8 and the exact sum peaks at 20.9 with 0.57 of full contrast; later revivals near ktᵣ come lower and broader, because √n is curved, and in the end overlap. Collapse and revival separate cleanly since tᵣ/tc ≈ π√(2n̄), 14 at n̄ = 10. No classical field does this: a fixed amplitude gives no collapse, and a continuously spread one gives Rabi frequencies that fill a band and never rephase. The revival is the photon's discreteness read off the atom alone. Measured (quoted): Rempe, Walther and Klein (1987) sent rubidium Rydberg atoms through a superconducting cavity on a 21.5 GHz transition, chose their speed to set the interaction time, and saw the excitation collapse and revive, in a field built up by earlier atoms rather than a coherent one. It needs both lifetimes long compared with tᵣ, and a single mode.
Key equations
Read the units, assumptions and limits before substituting.
Single-photon coupling
ħg = d √(ħω/(2ε₀V))
Derived from −d⋅E with one mode's vacuum field Evac = √(ħω/(2ε₀V)). g grows as √(ω/V), so shrinking the mode is the lever; at a node g = 0.
d = |⟨e|d⋅ε̂|g⟩| in C m; ω the mode's angular frequency in s⁻¹; V its mode volume in m³; ε₀ in F m⁻¹; g in s⁻¹, at an antinode.Jaynes-Cummings Hamiltonian
H = ħωc a†a + ħωₐσ₊σ₋ + ħg(aσ₊ + a†σ₋)
Derived. It commutes with N = a†a + σ₊σ₋, so it never mixes different excitation numbers: the model is a stack of independent 2 × 2 problems.
σ₊ = |e⟩⟨g|, σ₋ = |g⟩⟨e|; ωc, ωₐ the mode and atomic angular frequencies in s⁻¹; aσ₋, a†σ₊ dropped, needing g, |δ| ≪ ωₐ + ωc.Dressed doublets
Eₙ,± = ħ(n + 1)ωc + ħδ/2 ± (ħ/2)√(δ² + 4g²(n + 1))
Derived. On resonance the splitting is 2ħg√(n + 1): 2g, 2.83g, 3.46g, 4g. That uneven ladder is the photon's graininess written into a spectrum.
δ = ωₐ − ωc in s⁻¹; n = 0, 1, 2, …; each doublet comes from the block of |e, n⟩ and |g, n + 1⟩; |g, 0⟩ stays alone at energy 0.Rabi oscillation inside one block
Pₑ(t) = cos²(g√(n + 1) t)
Derived. At n = 0 it is the vacuum Rabi oscillation at angular frequency 2g: the photon emitted into the empty mode is reabsorbed, and two energies cannot make a decay.
Atom starts in |e, n⟩, δ = 0, t in s. Detuned: Pₑ = 1 − (4g²(n + 1)/Ωₙ²) sin²(Ωₙt/2), with Ωₙ = √(δ² + 4g²(n + 1)).Inversion for a coherent field, and its collapse
W(t) = Σₙ Pₙ cos(2g√(n + 1) t) ≈ cos(2g√(n̄ + 1) t) e(−g²t²/2)
Derived. The envelope falls to 1/e at tc = √2/g whatever n̄ is: the spread of Rabi frequencies dephases the sum. It is a dephasing, not a decay.
W = Pₑ − Pg; Pₙ = e(−n̄)n̄ⁿ/n! with n̄ ≫ 1 and δ = 0; the envelope comes from linearising √(n + 1) about n̄.Revival time
tᵣ ≈ 2π√(n̄ + 1)/g ≈ 2π√n̄/g
Derived: neighbouring photon numbers differ in Rabi frequency by about g/√(n̄ + 1) and come back into phase. A continuous spread never does, so revivals mark discrete n.
n̄ ≫ 1, δ = 0, tᵣ in s. Detuned, tᵣ ≈ πΩ̄/g² with Ω̄ = √(δ² + 4g²(n̄ + 1)); the k-th revival sits near ktᵣ, lower and broader.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at n̄ = 0: the curve is the vacuum Rabi oscillation cos(2gt), and it never decays. Raise n̄ to 10: it collapses by gt ≈ 1.5 and revives near gt ≈ 21, where the dashed envelope is flat. Then add detuning: the contrast drops and the revival moves later.
Live results
- MEAN RABI FREQUENCY Ω̄/g
- 6.63
- PREDICTED REVIVAL g tᵣ
- 20.8
- PHOTON SPREAD Δn = √n̄
- 3.16
- RABI CONTRAST
- 1.00
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: 10.0; Detuning δ/g: 0.0
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Fermi's golden rule invites the wrong answer. Take aσ₊ + a†σ₋ as the perturbation and |g, 1⟩ as the final state, and the atom seems to decay at a rate 2πg²ρ. But the golden rule needs a continuum: final states whose density is smooth over a frequency range much wider than the rate it predicts, so that amplitude leaking into them never comes back in phase. One mode is one final state. Its density is a δ-function, the 'rate' is zero off resonance and infinite on it, and both answers are wrong. The exact solution has no rate at all: Pₑ = cos²(gt) empties the atom by t = π/(2g) and refills it by π/g. Irreversibility needs a continuum restored. A cavity whose photons leak at κ ≫ g turns the oscillation into decay at 4g²/κ, and an atom with no cavity decays at the Weisskopf-Wigner rate; Unit 12 derives both.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
An atom with transition dipole d = 2.5 × 10⁻²⁹ C m on a 780 nm line sits at an antinode of a resonant standing-wave mode of waist w₀ = 20 µm and length L = 100 µm, so V = πw₀²L/4. Find Evac, g/2π, the vacuum Rabi splitting and the time the atom takes to hand its energy to the empty mode. The cavity loses photons at κ/2π = 4.0 MHz and the atom radiates into other modes at γ/2π = 6.0 MHz: is the coupling strong?
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.
- Mode volume and frequency: V = π(20 × 10⁻⁶ m)²(100 × 10⁻⁶ m)/4 = 3.14 × 10⁻¹⁴ m³, and ω = 2πc/λ = 2.415 × 10¹⁵ s⁻¹, so ħω = 2.547 × 10⁻¹⁹ J.
- Vacuum field of the mode (topic 11.05, quoted): 2ε₀V = 5.563 × 10⁻²⁵ F m², so Evac = √(2.547 × 10⁻¹⁹ J ÷ 5.563 × 10⁻²⁵ F m²) = √(4.578 × 10⁵ V² m⁻²) = 677 V m⁻¹. Units: J/(F m²) = V² m⁻².
- Coupling (derived in this topic): ħg = dEvac = 2.5 × 10⁻²⁹ C m × 677 V m⁻¹ = 1.69 × 10⁻²⁶ J, so g = 1.60 × 10⁸ s⁻¹ and g/2π = 25.5 MHz. Units: C m × V m⁻¹ = J.
- Vacuum Rabi splitting: 2g/2π = 51.1 MHz. Starting from |e, 0⟩, Pₑ = cos²(gt) first reaches zero at t = π/(2g) = 9.79 ns.
- Checks: g/κ = 6.4 and g/γ = 4.3, so the exchange completes before either loss acts and the two lines are resolved: strong coupling. The rotating-wave approximation needs g ≪ 2ω, and g/ω = 6.6 × 10⁻⁸; the dipole approximation needs an atom much smaller than λ, about 0.5 nm against 780 nm.
Answer. Evac = 677 V m⁻¹, g/2π = 25.5 MHz, vacuum Rabi splitting 51.1 MHz, and the energy is in the mode after 9.8 ns. g exceeds κ and γ by factors of 6.4 and 4.3: strong coupling.
MediumExample 2
A two-level artificial atom is resonant with a 5.0 GHz superconducting resonator, g/2π = 4.0 MHz. A short pulse leaves the resonator in a coherent state with n̄ = 9 and the atom in |e⟩. Find the spread of photon numbers, the mean Rabi frequency, and the collapse and revival times. Do a photon lifetime 1/κ = 50 µs and an atomic lifetime 1/γ = 30 µs let the revival be seen, and what would a classical field of the same amplitude do instead?
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.
- Photon numbers (topic 11.07, quoted): Poisson with Δn = √9 = 3.0; the most likely values are P₈ = P₉ = 0.132, and P₀ = e(−9) = 1.2 × 10⁻⁴, so the vacuum term is negligible.
- Mean Rabi frequency (derived): 2g√(n̄ + 1)/2π = 2 × 4.0 MHz × √10 = 25.3 MHz, a period of 39.5 ns. A classical field of amplitude 2Evac√9 drives 2g√9/2π = 24.0 MHz at full contrast for ever.
- Collapse (derived): tc ≈ √2/g = √2 ÷ (2π × 4.0 × 10⁶ s⁻¹) = 56 ns. The exact Poisson sum falls to 1/e at gt = 1.48, which is 59 ns, about 1.5 Rabi periods.
- Revival (derived): tᵣ ≈ 2π√(n̄ + 1)/g = √10 ÷ (4.0 × 10⁶ s⁻¹) = 0.79 µs, and the exact sum peaks at gt = 19.8, also 0.79 µs, at 0.58 of full contrast. The cruder 2π√n̄/g gives 0.75 µs.
- Losses: κtᵣ = 0.79 µs ÷ 50 µs = 0.016 and γtᵣ = 0.79 µs ÷ 30 µs = 0.026, and on average n̄κtᵣ = 0.14 photons leak out before tᵣ, so the revival should be clearly visible (an estimate). Checks: g/ω = 8.0 × 10⁻⁴, and tᵣ/tc = 14, close to π√(2n̄) = 13.3.
Answer. Δn = 3.0; mean Rabi frequency 25.3 MHz; collapse after about 56 to 59 ns; revival near 0.79 µs, well inside both lifetimes. A classical field of the same amplitude oscillates at 24.0 MHz for ever, with no collapse and no revival.
HardExample 3
An atom entering an empty resonant cavity in |e⟩ obeys Pₑ = cos²(gt) with g/2π = 20.0 kHz. The cavity is now fed a weak coherent field, n̄ = 0.50. Find the frequencies and amplitudes of the oscillating terms in Pₑ(t) and the value of Pₑ at t = 25.0 µs, and compare both with a classical field of the same amplitude. Then show how two measured amplitudes would determine n̄.
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.
- Each photon number contributes Pₙcos²(g√(n + 1)t) = (Pₙ/2)[1 + cos(2g√(n + 1)t)] (derived), so Pₑ(t) = ½ + ½Σₙ Pₙ cos(2g√(n + 1)t): lines at 2g√(n + 1)/2π = 40.0, 56.6, 69.3 and 80.0 kHz, in the ratios 1 : √2 : √3 : 2.
- Weights (Poisson, topic 11.07, quoted): P₀ = e(−0.50) = 0.607, P₁ = 0.303, P₂ = 0.076, P₃ = 0.013, so the line amplitudes ½Pₙ are 0.303, 0.152, 0.038 and 0.006.
- At t = 25.0 µs = π/g: n = 0 gives cos²(π) = 1, n = 1 gives cos²(π√2) = 0.071, n = 2 gives cos²(π√3) = 0.444, n = 3 gives cos²(2π) = 1 and n = 4 gives cos²(π√5) = 0.544. Weighted: Pₑ = 0.6065 + 0.0215 + 0.0336 + 0.0126 + 0.0009 = 0.675.
- A classical field of amplitude 2Evac√n̄ gives one line, 2g√n̄/2π = 28.3 kHz, which is absent from the quantum signal, and Pₑ(π/g) = cos²(π√0.50) = 0.367. The quantum value is 1.8 times larger because the n = 0 component, weight 0.61, has completed a full vacuum Rabi cycle and is back in |e⟩.
- Inverting: the 56.6 kHz amplitude over the 40.0 kHz one is P₁/P₀ = n̄ = 0.50, and the next ratio is P₂/P₁ = n̄/2 = 0.25; successive ratios n̄/(n + 1) test that the field is Poissonian. Limits: as n̄ → 0 only the 40.0 kHz line remains, the vacuum Rabi oscillation that the classical model misses entirely; for n̄ ≫ 1 the lines crowd towards 2g√n̄ and the early-time classical result returns.
Answer. Lines at 40.0, 56.6, 69.3 and 80.0 kHz with amplitudes 0.303, 0.152, 0.038 and 0.006; Pₑ(25.0 µs) = 0.675, against 0.367 from the classical field's single 28.3 kHz line; n̄ = P₁/P₀ = 0.50.
Quick check
Select an answer and explain it before checking.
Review & continue
The Jaynes-Cummings model is a two-level atom coupled to one quantised field mode with the counter-rotating terms dropped, H = ħωc a†a + ħωₐσ₊σ₋ + ħg(aσ₊ + a†σ₋), whose conserved excitation number splits it into independent 2 × 2 blocks.
- 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.