advanced32 min studyLesson 475 in Physics

Resolve the coupled atom–cavity doublet

Derive excitation blocks, detuned transfer, and the square-root photon dependence of the Jaynes–Cummings model.

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What you’ll understand

  • State dipole, single-mode, two-level, and rotating-wave assumptions.
  • Diagonalize a fixed-total-excitation block.
  • Distinguish coupling g, vacuum doublet splitting, and population oscillation frequency.

A single excited atom can exchange energy reversibly with an initially empty cavity mode. The vacuum supplies an available final photon state, so this exchange does not require a classical drive already present. The Jaynes–Cummings model describes it with a two-level atom, one quantized mode, and a rotating-wave interaction. We will use g as the off-diagonal coupling in angular-frequency units, making the vacuum doublet splitting equal to 2g on resonance rather than calling both quantities the same frequency.

Declare what was removed from the interaction

The electric dipole interaction couples the atomic dipole to the mode field at the atom’s position. A two-level approximation retains one selected transition and a single-mode approximation excludes other cavity modes. The coupling includes the transition dipole projected onto local polarization and the normalized mode amplitude. It is therefore position and orientation dependent. A small cavity mode volume can increase the coupling, but geometry alone does not guarantee strong coupling once losses and unwanted transitions are included.

Before the rotating-wave approximation, the interaction contains terms that simultaneously create an atomic and field excitation or remove both. Near resonance, when g and relevant detunings and decay scales are much smaller than optical frequencies, these terms rotate rapidly and can be neglected. The retained terms exchange one excitation between atom and field. This approximation fails in ultrastrong coupling and can fail for unsuitable drive or spectral conditions; the Jaynes–Cummings Hamiltonian is not an exact arbitrary-coupling light–matter theory.

H^ℏ=ωca^†a^+ωaσ^+σ^−+g(a^σ^++a^†σ^−),Δ=ωa−ωc\frac{\hat H}{\hbar}=\omega_c\hat a^\dagger\hat a+\omega_a\hat\sigma_+\hat\sigma_-+g(\hat a\hat\sigma_++\hat a^\dagger\hat\sigma_-),\qquad \Delta=\omega_a-\omega_c
ωc, ωa, Δ, and real g are in s⁻¹ as angular frequencies. σ+=|e⟩⟨g|; zero-point constants are omitted. A fixed atom, dipole approximation, selected two levels and mode, and near-resonant weak relative coupling are assumed.↔ Scroll wide equations horizontally.

Total excitation reduces the Hilbert space

The retained interaction conserves cavity photon number plus excited atomic occupation. For an initially excited atom and n photons, the motion stays within the pair of states consisting of that initial state and a ground-state atom with n plus one photons. The ladder matrix element supplies square root of n plus one. Diagonalizing this two-dimensional block gives two dressed states separated by the generalized angular frequency. The vacuum block remains coupled because the square root is one when n is zero.

On resonance the two dressed states are equal-weight superpositions, with energies shifted oppositely by g times square root of n plus one. Their relative phase determines the oscillation between bare states. Off resonance, the energy mismatch changes the mixing angle and prevents complete transfer from the initial excited atom. A spectral doublet and a time-domain population oscillation describe related physics but use different observables, and their interpretation in a dissipative experiment also requires linewidth information.

Hnrelℏ=12(Δ2gn+12gn+1−Δ),Ωn=Δ2+4g2(n+1),En,+−En,−=ℏΩn\frac{H_n^{\mathrm{rel}}}{\hbar}=\frac12\begin{pmatrix}\Delta&2g\sqrt{n+1}\\2g\sqrt{n+1}&-\Delta\end{pmatrix},\qquad \Omega_n=\sqrt{\Delta^2+4g^2(n+1)},\qquad E_{n,+}-E_{n,-}=\hbar\Omega_n
The basis is {|e,n⟩,|g,n+1⟩}, and a common block energy is subtracted. Ωₙ is the population oscillation angular frequency and dressed-state splitting divided by ℏ; n is an integer occupation.↔ Scroll wide equations horizontally.

Read transfer probability with its prefactor

Starting in the excited bare state, the ground-atom plus extra-photon probability is a sine squared multiplied by a detuning-dependent amplitude. The generalized frequency alone does not determine how much population transfers. At large detuning oscillations become faster but their transfer amplitude becomes small. On resonance the first full swap occurs when g times square root of n plus one times time equals pi over two. Vacuum exchange is thus a finite, coherent process.

A coherent cavity state contains many n blocks with different square-root frequencies, so their population contributions can dephase and later revive in the ideal model. A single two-state geometric lab cannot reproduce that multi-block evolution. It can display equal-weight dressed superpositions within one declared excitation block and show their phase distinction. Its playback is fixed relative-phase precession, not a tunable Jaynes–Cummings exchange, and it contains neither detuning-dependent dynamics nor cavity decay.

Pg,n+1(t)=4g2(n+1)Ωn2sin⁡2(Ωnt2),Δ=0:Pg,n+1(t)=sin⁡2(gn+1 t)P_{g,n+1}(t)=\frac{4g^2(n+1)}{\Omega_n^2}\sin^2\left(\frac{\Omega_nt}{2}\right),\qquad \Delta=0:\quad P_{g,n+1}(t)=\sin^2(g\sqrt{n+1}\,t)
The atom starts in |e⟩ with exactly n cavity photons, and dynamics is unitary under the declared model. Time is in seconds; this probability excludes spontaneous emission, cavity loss, drive, and atomic motion.↔ Scroll wide equations horizontally.

Find detuned vacuum transfer

An atom begins in |e,0⟩ with g=2.0×10⁶ s⁻¹ and detuning Δ=3.0×10⁶ s⁻¹. Find the dressed splitting in angular-frequency units, maximum transfer probability, and first time at that maximum.

  1. The vacuum block has n+1=1. Its generalized frequency is √[(3.0×10⁶)²+4(2.0×10⁶)²]=5.0×10⁶ s⁻¹.

  2. The transfer prefactor is 16/25=0.64. Detuning prevents a full swap even though the coupled state oscillates.

  3. The first sine-squared maximum occurs at Ω₀t/2=π/2, giving t=π/(5.0×10⁶)≈0.628 μs. The dressed energy separation is ℏ times 5.0×10⁶ s⁻¹.

The result

The splitting is 5.0×10⁶ s⁻¹, maximum transfer is 0.64, and its first occurrence is about 0.628 μs in this lossless model.

Explore the model

Interactive lab

Interactive lab · quantum

Inspect one dressed excitation block

For fresh runs Reset and set BOTH mixing angle 90°, relative phase 0°; then Reset and set mixing 90°, phase 180°. Associate the basis with |e,n⟩ and |g,n+1⟩. Predict equal bare probabilities but opposite Bloch x components. The lab shows pure-state geometry, not g-controlled exchange, detuning, or loss.

Inspect one dressed excitation blockA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forResonant dressed states differ by relative phase even when their bare-state probabilities agree.

Check your understanding

With interaction ℏg(aσ++a†σ−), what is the resonant vacuum dressed-state splitting?

Try it yourself

On resonance an atom begins in |e,3⟩ with g=1.0×10⁶ s⁻¹. Find Ω₃ and the first complete swap time. Compare that time with the same coupling and an initially empty cavity.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use √(n+1), and distinguish the sine argument from the population oscillation frequency.

Compare with the solution

√(3+1)=2, so Ω₃=2g×2=4.0×10⁶ s⁻¹. The first complete swap is t=π/(4.0×10⁶)≈0.785 μs. For vacuum it is π/(2.0×10⁶)≈1.571 μs, twice as long. These are bare-state transfer times in the lossless fixed-number model.

Take it with you

  • The rotating-wave interaction conserves total atom-plus-mode excitation.
  • Coupling in the n block scales as g√(n+1), including vacuum exchange.
  • Detuning changes both the oscillation frequency and the maximum transfer.

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