Vacuum Rabi Oscillations and Splitting

Point a single atom at empty space and it decays: the excited electron dumps its energy into the continuum of vacuum modes and never gets it back. Trap that same atom inside a high-finesse cavity tuned to its transition and the story inverts. Now there is only one field mode to talk to — the atom emits its photon into the mode, the mode hands it straight back, and the single quantum shuttles between matter and light indefinitely. That reversible sloshing is the vacuum Rabi oscillation; in the frequency domain it shows up as a doublet where the atom's single spectral line should be. The word vacuum is the whole point. Ordinary Rabi flopping needs an applied laser whose amplitude sets the rate; kill the laser and it stops. Here the oscillation runs with the cavity in its ground state $|0\rangle$ — nothing classical drives it. The drive is the zero-point field that aq-optics-vacuum-zero-point-energy shows never switches off, coupled to the dipole through the machinery of aq-optics-field-quantization and the photon-number ladder of aq-optics-fock-states. The frequency is fixed entirely by the coupling $g$ , built from the atomic dipole moment and the vacuum field per photon. The governing model is the Jaynes–Cummings Hamiltonian, the exactly solvable cornerstone of cavity quantum electrodynamics. Under the rotating-wave approximation the interaction conserves total excitation, so the Hilbert space falls into two-dimensional blocks whose diagonalization gives an anharmonic ladder split by $2g\sqrt{n+1}$ . The lowest rung splits by $2g$ , and observing that cleanly is the strong-coupling regime. This article stays with that vacuum rung: the physics, the numbers, and why a doublet where a single line should be is a photograph of the quantization of light.

Key takeaways

A single two-level atom resonantly coupled to a single quantized cavity mode does not decay smoothly: even in the vacuum, one quantum oscillates coherently between atom and field at $\Omega_0=2g$ .
The Jaynes–Cummings Hamiltonian block-diagonalizes into $2\times2$ manifolds; the degeneracy of $|e,n\rangle$ and $|g,n+1\rangle$ is lifted by $2g\sqrt{n+1}$ , giving the dressed states.
On the vacuum rung ( $n=0$ ) the empty mode alone splits the excited manifold by $2g$ — the vacuum Rabi doublet, where the bare atom shows one line.
Seeing the oscillation or resolved doublet requires strong coupling $g\gg\kappa,\gamma$ : coherent exchange must beat cavity loss $\kappa$ and atomic decay $\gamma$ , i.e. $C=g^2/\kappa\gamma\gg1$ .
The $\sqrt{n+1}$ scaling is the fingerprint of field quantization — a classical field gives an intensity-proportional response with no fixed vacuum splitting, so the anharmonic ladder is photons made visible.
Key idea: one atom and one empty cavity mode form a coupled oscillator; the coupling splits $|e,n\rangle$ and $|g,n+1\rangle$ by $2g\sqrt{n+1}$ , so the vacuum ( $n=0$ ) alone opens a $2g$ doublet and drives oscillation at $2g$ .

The core insight

A single atom and a single empty cavity mode form one coupled oscillator; the Jaynes–Cummings coupling lifts the degeneracy of $|e,n\rangle$ and $|g,n+1\rangle$ by $2g\sqrt{n+1}$ , so the vacuum ( $n=0$ ) alone splits the atom's line into a $2g$ doublet and drives reversible oscillation at frequency $2g$ .

Where this lives today

The vacuum Rabi splitting is now the standard yardstick for any light–matter interface. In optical cavity QED, single atoms or trapped ions coupled to high-finesse resonators show resolved doublets whose spacing calibrates $g$ directly — atom–photon entanglement, quantum repeaters, and single-photon sources all begin with a measured $2g$ . The same physics runs in circuit QED, where a superconducting qubit couples to a resonator; the dipoles are so large that $g/2\pi$ reaches tens to hundreds of MHz, and the dispersive limit provides the qubit readout in essentially every superconducting processor. The anharmonicity $2g\sqrt{n+1}$ is itself a resource: absorbing one photon detunes the system from the second rung, so a strongly coupled atom–cavity acts as a photon blockade emitting antibunched photons on demand. Even in the solid state, semiconductor-microcavity excitons show the collective $2g\sqrt{N}$ splitting as polariton doublets.

The physical picture

Take two identical pendulums tuned to the same frequency and join them with a weak spring. The pair has two normal modes, in-phase and out-of-phase, split by the spring stiffness, and energy started in one drains completely into the other and back at a beat frequency equal to that splitting. The atom-in-a-cavity is exactly this: the atom is one pendulum, the field mode the other, the spring is $g$ , and "atom excited, cavity empty" is one pendulum holding all the energy. It hands its quantum to the mode and takes it back at the beat frequency $2g$ — the vacuum Rabi oscillation — while the two normal-mode frequencies form the doublet. The spring is the vacuum itself. The mode carries a zero-point field $E_{\mathrm{vac}}=\sqrt{\hbar\omega_c/2\varepsilon_0 V}$ even with no photons in it, and that field couples to the dipole $d$ to give $g=dE_{\mathrm{vac}}/\hbar$ . Shrinking the mode volume $V$ concentrates the field and raises $g$ — why the game is played in tiny high-finesse optical cavities and confined microwave resonators. The first figure shows the coupling at work in time: the excitation is not lost, it oscillates.

Figure

Vacuum Rabi oscillation of one quantum between atom and empty cavity, g/2π ≈ 26 MHz. Excited-atom probability P_e = cos²(gt) (blue) and one-photon probability P = sin²(gt) (amber) trade places completely: at t ≈ 9.5 ns the photon sits entirely in the cavity, the cycle repeats every 2π/2g ≈ 19 ns, so the beat is 2g/2π ≈ 53 MHz

Because the coupling scales as $\sqrt{n+1}$ , the rungs are not equally spaced — the ladder is anharmonic. A classical field would give a splitting proportional to its amplitude, tunable with no special value at zero drive. Quantum mechanics instead hands you a fixed, drive-free splitting of $2g$ on the vacuum rung, then $2g\sqrt{2}$ , then $2g\sqrt{3}$ — a spacing no classical field can fake. The second figure lays the ladder next to its spectroscopic shadow: scan a weak probe across the bare frequency and find no absorption on resonance but two peaks at $\pm g$ — the single line replaced by a doublet.

Figure

Left: the anharmonic Jaynes–Cummings ladder — each manifold n splits by 2g√(n+1), so the vacuum rung opens by 2g and higher rungs by 2g√2 and 2g√3 (split-exaggerated, not to vertical scale). Right: the vacuum-rung probe spectrum is a doublet with peaks at ±g separated by 2g/2π ≈ 53 MHz and a dip on the bare resonance

The core equations

The whole subject is one Hamiltonian solved on one small subspace; each equation below follows from the one before it. The Jaynes–Cummings Hamiltonian has a free-field term, a free-atom term, and a coupling that exchanges one atomic excitation for one photon. Its exact solubility comes from the rotating-wave approximation, which discards terms that would create or destroy energy in pairs.

\hat H = \hbar\omega_c\,\hat a^\dagger\hat a + \tfrac{1}{2}\hbar\omega_a\,\hat\sigma_z + \hbar g\left(\hat a\,\hat\sigma_+ + \hat a^\dagger\hat\sigma_-\right) \qquad (1)

where: $\omega_c,\ \omega_a$ — cavity-mode and atomic-transition angular frequencies; $\hat a,\ \hat a^\dagger$ — photon annihilation/creation operators; $\hat\sigma_z,\ \hat\sigma_\pm$ — atomic inversion and raising/lowering operators; $g$ — atom–field coupling, half the vacuum Rabi splitting. Physical reading: $\hat a\hat\sigma_+$ destroys a photon and excites the atom, $\hat a^\dagger\hat\sigma_-$ does the reverse, so total excitation $\hat N=\hat a^\dagger\hat a+\tfrac12(\hat\sigma_z+1)$ is conserved. The coupling $g$ is not free — it is the dipole's energy in the mode's zero-point field, divided by $\hbar$ .

g = \frac{d\,E_{\mathrm{vac}}}{\hbar}, \qquad E_{\mathrm{vac}} = \sqrt{\frac{\hbar\omega_c}{2\varepsilon_0 V}} \qquad (2)

where: $d$ — atomic transition dipole matrix element on the mode polarization; $E_{\mathrm{vac}}$ — rms vacuum (zero-point) field; $V$ — cavity mode volume; $\varepsilon_0$ — vacuum permittivity. Physical reading: $g\propto d/\sqrt{V}$ , so strong coupling wants a big dipole (Rydberg atoms, superconducting qubits) and a small mode volume. Because the coupling only links $|e,n\rangle$ to $|g,n+1\rangle$ , the Hamiltonian is block-diagonal in $2\times2$ manifolds. On resonance those states are degenerate, so the eigenvectors are symmetric and antisymmetric combinations.

|\pm,n\rangle = \frac{1}{\sqrt 2}\left(|e,n\rangle \pm |g,n+1\rangle\right) \qquad (3)

where: $|e,n\rangle$ — atom excited with $n$ photons; $|g,n+1\rangle$ — ground-state atom with $n+1$ photons; $|\pm,n\rangle$ — the two dressed eigenstates of manifold $n$ . Physical reading: the stationary states are equal superpositions of both — the excitation is genuinely shared, an atom–photon polariton. Diagonalizing each manifold shifts the dressed states symmetrically by $\hbar g\sqrt{n+1}$ — the anharmonic ladder of the second figure. A splitting is a Rabi frequency, so each rung both splits and oscillates at $\Omega_n$ .

E_{\pm,n} = \hbar\omega_c\left(n+\tfrac{1}{2}\right) \pm \hbar g\sqrt{n+1}, \qquad \Omega_n = 2g\sqrt{n+1}, \quad \Omega_0 = 2g \qquad (4)

where: $E_{\pm,n}$ — dressed-state energies of manifold $n$ ; $\Omega_n=(E_{+,n}-E_{-,n})/\hbar$ — the manifold's Rabi frequency and splitting; $\Omega_0=2g$ — the vacuum Rabi frequency. Physical reading: the splitting grows with $n$ but not linearly, and even the $n=0$ rung splits and oscillates — the irreducible quantum of the effect, with an irrational spacing no classical drive reproduces. Putting the atom in $|e\rangle$ with the cavity empty makes an equal superposition of $|+,0\rangle$ and $|-,0\rangle$ ; those dressed states beat and the excitation flops fully between atom and mode.

P_e(t) = \cos^2\!\left(g t\right) = \tfrac{1}{2}\left[1 + \cos(2g t)\right] \qquad (5)

where: $P_e(t)$ — probability the atom is excited at time $t$ ; $gt$ — accumulated half-Rabi phase. Physical reading: fully excited at $gt=0,\pi,\dots$ and fully de-excited at $gt=\pi/2,3\pi/2,\dots$ ; the population oscillates at $2g$ , not $g$ , because probability is amplitude squared. Away from resonance the manifold states are no longer degenerate, and the splitting takes the hyperbolic-Rabi form. Detuning always increases the splitting and suppresses the flop amplitude.

E_{\pm,n} = \hbar\omega_c\!\left(n+\tfrac12\right) + \tfrac{\hbar\Delta}{2} \pm \tfrac{\hbar}{2}\sqrt{\Delta^2 + 4g^2(n+1)} \qquad (6)

where: $\Delta=\omega_a-\omega_c$ — atom–cavity detuning; $\sqrt{\Delta^2+4g^2(n+1)}$ — the generalized Rabi frequency (times $\hbar$ , the splitting). Physical reading: the resonant splitting $2g$ is the minimum — sweeping the atom through the cavity traces an avoided crossing of closest approach $2g$ . Finally, observability is a contest between the coherent rate $g$ and the loss rates. The dimensionless number that decides it is the cooperativity.

C = \frac{g^2}{\kappa\gamma}, \qquad C \gg 1 \ \text{for strong coupling} \qquad (7)

where: $C$ — single-atom cooperativity; $\kappa$ — cavity field decay rate; $\gamma$ — atomic dipole decay rate. Physical reading: $C$ counts roughly how many coherent exchanges happen before the quantum is lost; the doublet resolves when $2g$ exceeds the widths $\kappa+\gamma$ — the same as $C\gg1$ .

Derivation AThe coupling term and the rotating-wave approximation

The goal: show where the exchange coupling $\hbar g(\hat a\hat\sigma_+ + \hat a^\dagger\hat\sigma_-)$ in equation (1) comes from, and why we discard half the terms the dipole interaction $-\hat{\mathbf d}\cdot\hat{\mathbf E}$ produces — the rotating-wave approximation.

Step 1. Write $\hat H_{\mathrm{int}}=-\hat{\mathbf d}\cdot\hat{\mathbf E}\propto(\hat\sigma_++\hat\sigma_-)(\hat a+\hat a^\dagger)$ , since the dipole operator $\propto\hat\sigma_++\hat\sigma_-$ and the field $\propto\hat a+\hat a^\dagger$ . Why: the product has four cross terms to sort.

Step 2. Expand into $\hat\sigma_+\hat a + \hat\sigma_-\hat a^\dagger + \hat\sigma_+\hat a^\dagger + \hat\sigma_-\hat a$ . The first two conserve excitation (excite the atom while absorbing a photon, or the reverse); the last two do not. Why: only energy-conserving processes drive a sustained resonant oscillation.

Step 3. In the interaction picture the kept terms rotate slowly at $\omega_a-\omega_c\approx0$ , while the "counter-rotating" terms oscillate at $\omega_a+\omega_c\approx2\omega_c$ and average to zero, leaving $\hat H_{\mathrm{int}}\approx\hbar g(\hat\sigma_+\hat a + \hat\sigma_-\hat a^\dagger)$ . Why: a term oscillating millions of times per Rabi cycle contributes nothing net — the rotating-wave approximation, valid when $g\ll\omega_c$ , which every cavity-QED experiment satisfies. *Verified: the surviving coupling conserves $\hat N=\hat a^\dagger\hat a+\tfrac12(\hat\sigma_z+1)$ , which block-diagonalizes the Hamiltonian into the $2\times2$ manifolds used below; consistent with Loudon Ch. 7 and Haroche and Raimond.*

Go deeper · Derivation B · 1 minVacuum Rabi splitting and oscillation from the lowest manifold

Derivation BVacuum Rabi splitting and oscillation from the lowest manifold

The goal: solve the single-excitation problem by hand — diagonalize the $2\times2$ block spanned by $|e,0\rangle$ and $|g,1\rangle$ for the splitting $2\hbar g$ , then time-evolve $|e,0\rangle$ for the flop $P_e(t)=\cos^2(gt)$ . Both fall out of the same tiny matrix — a splitting in frequency and a beat in time are one physics.

Step 1. In the basis $\{|e,0\rangle,|g,1\rangle\}$ at resonance $\omega_a=\omega_c\equiv\omega$ : $|e,0\rangle$ carries $+\tfrac12\hbar\omega$ and $|g,1\rangle$ carries $-\tfrac12\hbar\omega+\hbar\omega=+\tfrac12\hbar\omega$ , so the diagonal entries are equal and $H=\hbar\left(\begin{smallmatrix}\omega/2 & g\\ g & \omega/2\end{smallmatrix}\right)$ . Why: the states are exactly degenerate before coupling — atom and empty cavity "in tune."

Step 2. Diagonalize: eigenvalues $E_\pm=\hbar(\omega/2\pm g)$ with eigenvectors $|\pm,0\rangle=(|e,0\rangle\pm|g,1\rangle)/\sqrt2$ , so the splitting is $E_+-E_-=2\hbar g$ . Why: off-diagonal $\hbar g$ splits a degenerate pair by twice the coupling — a weak probe absorbs at $\omega\pm g$ , the doublet.

Step 3. Evolve in time: decompose $|e,0\rangle=(|+,0\rangle+|-,0\rangle)/\sqrt2$ , let each dressed state pick up its phase $e^{-iE_\pm t/\hbar}$ , and recombine to $|\psi(t)\rangle=e^{-i\omega t/2}[\cos(gt)\,|e,0\rangle-i\sin(gt)\,|g,1\rangle]$ . Why: their relative phase $2gt$ periodically reassembles the excitation on atom or photon.

Step 4. Read off: $P_e(t)=|\langle e,0|\psi(t)\rangle|^2=\cos^2(gt)$ , which is $1$ at $t=0$ , $0$ at $gt=\pi/2$ (photon in the cavity), back to $1$ at $gt=\pi$ . Why: the splitting $2g$ from Step 2 is the angular frequency of the flop, so doublet spacing and flop period give the same $g$ . *Verified: eigenvalues $\hbar(\omega/2\pm g)$ , splitting $2\hbar g$ , and $P_e(t)=\cos^2(gt)$ reproduced by explicit $2\times2$ evolution; numerically $g/2\pi=26.4$ MHz gives splitting $2g/2\pi=52.8$ MHz and half-transfer time $\pi/2g=9.5$ ns, matching the figures.*

Go deeper · Derivation C · 1 minCollapse and revival with a coherent field

Derivation CCollapse and revival with a coherent field

The goal: show what happens when the cavity holds a coherent state $|\alpha\rangle$ — a near-classical field with a Poisson spread of photon numbers. Each $n$ drives its own Rabi frequency $2g\sqrt{n+1}$ ; because those are incommensurate the flops first dephase (collapse) then rephase (revive). No classical field can produce the revival, so seeing it proves the field energy comes in discrete photons.

Step 1. Weight each manifold by $P_n=e^{-\bar n}\bar n^{\,n}/n!$ with $\bar n=|\alpha|^2$ , giving $P_e(t)=\tfrac12[1+\sum_n P_n\cos(2g\sqrt{n+1}\,t)]$ . Why: a coherent state superposes all photon numbers, so the signal is a Poisson-weighted sum of flops.

Step 2. The spread of frequencies $\delta\Omega\approx g$ dephases the sum after a collapse time $t_c\sim1/g$ , independent of $\bar n$ . Why: many cosines of slightly different frequency interfere destructively once their phases spread by $\sim2\pi$ , flattening the signal to $\tfrac12$ .

Step 3. But the frequencies are discrete, so the phases eventually realign. Neighbouring terms rephase when $2g(\sqrt{\bar n+2}-\sqrt{\bar n+1})\,t_r\approx2\pi$ , i.e. $t_r\approx2\pi\sqrt{\bar n}/g$ , giving a partial revival. Why: discreteness makes the dephasing reversible — a continuous classical intensity never returns, so the revival is the signature of quantization. *Verified: standard result in Haroche and Raimond, observed in the Rydberg micromaser; for $\bar n=9$ and $g/2\pi=26.4$ MHz, $t_c\approx6$ ns and $t_r\approx0.11$ μs, well separated.*

Limiting cases

The behaviour is governed by the coupling $g$ against the loss rates $\kappa,\gamma$ and by the detuning $\Delta=\omega_a-\omega_c$ .

Case	Result	Meaning
Strong coupling	$g\gg\kappa,\gamma$ ( $C\gg1$ )	Coherent exchange dominates; the oscillation runs for many cycles and the $2g$ doublet is resolved.
Weak coupling (Purcell)	$g\ll\kappa$	The lossy mode acts as a fast reservoir; the atom decays irreversibly at a Purcell-enhanced rate $\propto g^2/\kappa$ — no oscillation, no resolved splitting.
Large detuning (dispersive)	$\lvert\Delta\rvert\gg g$	No real exchange; the interaction reduces to a photon-number-dependent ac-Stark shift $\sim g^2/\Delta$ — the basis of QND photon counting and dispersive qubit readout.
Higher rungs	$\Omega_n=2g\sqrt{n+1}$	With $n$ photons present the splitting and flop grow as $\sqrt{n+1}$ ; a coherent state seeding many rungs gives collapse and revival.
Many atoms	$\Omega_N=2g\sqrt{N}$	$N$ identical atoms couple collectively; the splitting is enhanced by $\sqrt{N}$ , visible even in a classical (polariton) picture.

Common trapVacuum Rabi splitting is just the ordinary Rabi splitting of a strong laser

It is not, and conflating them erases the point. The classical Rabi frequency scales with an applied field's amplitude and tunes to any value including zero — switch the laser off and the flopping stops. The vacuum Rabi frequency $\Omega_0=2g$ has no applied field: the cavity is in its ground state $|0\rangle$ , and the splitting is fixed by the zero-point field through $g=dE_{\mathrm{vac}}/\hbar$ . That is why a nonzero splitting survives at zero drive, and why higher rungs scale as $\sqrt{n+1}$ rather than linearly in intensity — both impossible for a classical field. A bookkeeping trap rides alongside: many texts (this one included) write the coupling as $\hbar g(\hat a\hat\sigma_++\hat a^\dagger\hat\sigma_-)$ so the splitting is $2g$ , while others define $\Omega_0$ as the full splitting. Before quoting a number, check whether your symbol is the coupling or the full splitting.

Worked examples

1 · Coupling and splitting from the vacuum field

A cesium atom (transition wavelength $\lambda=852$ nm, effective dipole moment $d\approx2.7\times10^{-29}$ C·m) sits in a Fabry–Pérot cavity with mode volume $V\approx3.1\times10^{-14}$ m³ (Gaussian mode, waist $w_0=20$ μm, length $L=100$ μm). Find $E_{\mathrm{vac}}$ , the coupling $g$ , the splitting $2g/2\pi$ , and the time for one photon to appear fully in the cavity.

Step 1. Mode frequency: $\omega_c=2\pi c/\lambda=2\pi(3.00\times10^8)/(852\times10^{-9})=2.21\times10^{15}$ rad/s, so $\hbar\omega_c=2.33\times10^{-19}$ J (about $1.46$ eV, the D-line photon energy — a sanity check).

Step 2. Vacuum field: $E_{\mathrm{vac}}=\sqrt{\hbar\omega_c/2\varepsilon_0V}=\sqrt{(2.33\times10^{-19})/(2\cdot8.85\times10^{-12}\cdot3.1\times10^{-14})}$ . The denominator is $5.5\times10^{-25}$ , so $E_{\mathrm{vac}}=\sqrt{4.2\times10^{5}}\approx\mathbf{6.5\times10^{2}\ V/m}$ — a few hundred V/m from a mode with zero photons.

Step 3. Coupling: $g=dE_{\mathrm{vac}}/\hbar=(2.7\times10^{-29})(647)/(1.05\times10^{-34})=1.66\times10^{8}$ rad/s, i.e. $g/2\pi\approx\mathbf{26\ MHz}$ , and the splitting $2g/2\pi\approx\mathbf{53\ MHz}$ .

Step 4. The excitation flops as $P_e=\cos^2(gt)$ , reaching zero (photon in the cavity) at $gt=\pi/2$ , so $t=\pi/2g\approx\mathbf{9.5\ ns}$ ; it returns at $t=\pi/g\approx19$ ns. Strongly coupled: $g/2\pi\approx26$ MHz, a $53$ MHz doublet, a $\sim19$ ns period.

2 · Is it actually strong coupling? Cooperativity from measured rates

A landmark single-atom optical cavity-QED experiment reports $(g,\kappa,\gamma)/2\pi=(34,\ 4.1,\ 2.6)$ MHz ( $\kappa$ cavity field decay, $\gamma$ atomic dipole decay). Compute the cooperativity, check the doublet is resolvable, and find the critical photon and atom numbers.

Step 1. Cooperativity: $C=g^2/\kappa\gamma=(34)^2/(4.1\times2.6)=1156/10.7\approx\mathbf{108}$ . Since $C\gg1$ the system is deep in strong coupling — about a hundred coherent exchanges per decoherence event.

Step 2. Resolvability: peaks at $\pm g/2\pi=\pm34$ MHz (separation $2g/2\pi=68$ MHz), each of half-width $(\kappa+\gamma)/2\pi\approx6.7$ MHz — the separation beats the linewidth tenfold, so two clean peaks resolve.

Step 3. Critical photon number $n_0=\gamma^2/2g^2=(2.6)^2/(2\cdot34^2)=6.76/2312\approx\mathbf{2.9\times10^{-3}}$ and critical atom number $N_0=2\kappa\gamma/g^2=2(4.1)(2.6)/34^2=21.3/1156\approx\mathbf{1.8\times10^{-2}}$ , both far below one. Strong coupling is reached with well under one photon and one atom — the essence of single-quantum cavity QED.

3 · Detuning and the collapse–revival clock

Using $g/2\pi=26$ MHz from Example 1: (a) if the atom is detuned by $\Delta/2\pi=30$ MHz, what is the effective splitting? (b) if instead the cavity is filled with a coherent state of mean photon number $\bar n=9$ , estimate the collapse and first-revival times.

Step 1. Detuned splitting ( $n=0$ ): $\sqrt{\Delta^2+4g^2}\big/2\pi=\sqrt{30^2+(2\cdot26)^2}=\sqrt{900+2735}=\sqrt{3635}\approx\mathbf{60\ MHz}$ , up from $53$ MHz. Detuning widens the splitting and shrinks the flop amplitude — the $2g$ minimum sits at $\Delta=0$ .

Step 2. Collapse time: $t_c\sim1/g=1/(1.66\times10^8)\approx\mathbf{6\ ns}$ , essentially independent of $\bar n$ — the spread of Rabi frequencies $\sim g$ dephases the sum within a period.

Step 3. First revival: $t_r\approx2\pi\sqrt{\bar n}/g=2\pi(3)/(1.66\times10^8)\approx\mathbf{0.11\ μs}$ . The revival at $\sim110$ ns arrives about twenty collapse-times after the signal has apparently died — the cleanest fingerprint that the field is quantized.

Physics of the portrait

The vacuum Rabi effect lived on paper for thirty years before anyone could watch a single atom trade one photon with one mode; the person who did most to make it laboratory reality is Serge Haroche.

Origins · 1 minSerge Haroche and how this was discovered

Serge Haroche

born 1944, Casablanca and Paris · cavity quantum electrodynamics

Serge Haroche built his career on a deceptively simple ambition: to observe one atom interacting with one photon without destroying either. Working in Paris from the 1980s with Rydberg atoms — whose kilodebye dipoles make $g$ huge — and superconducting microwave cavities with photon lifetimes of tenths of a second, his group reached the predicted strong-coupling regime. In 1996 Brune, Haroche and collaborators directly recorded the quantum Rabi oscillation and its $\sqrt{n+1}$ dependence on photon number, resolving the discrete rungs of the dressed-state ladder. Their later quantum non-demolition experiments counted cavity photons one by one and watched individual quantum jumps of the field, using the dispersive limit $\lvert\Delta\rvert\gg g$ of the same model. For "ground-breaking experimental methods that enable measuring and manipulation of individual quantum systems," Haroche shared the 2012 Nobel Prize with David Wineland. His work turned the vacuum Rabi splitting into a routine diagnostic and seeded the circuit-QED architectures underpinning much of superconducting quantum computing.

History

1963

— Jaynes and Cummings publish the exactly solvable atom–single-mode model

1985

— Meschede, Walther and Müller operate the one-atom maser, single-atom cavity QED

1992

— Thompson, Rempe and Kimble resolve the vacuum Rabi doublet for atoms in an optical cavity

1996

— Brune, Haroche and co-workers observe the quantum Rabi oscillation and its √n scaling

2004

— Wallraff and co-workers measure vacuum Rabi splitting in circuit QED with a superconducting qubit

Going further

Texts. Loudon, The Quantum Theory of Light, 3rd ed., Ch. 7 · Foot, Atomic Physics, Ch. 9 · Saleh and Teich, Fundamentals of Photonics, 2nd ed., Ch. 13–14 · Haroche and Raimond, Exploring the Quantum: Atoms, Cavities and Photons (2006) · Scully and Zubairy, Quantum Optics, Ch. 6 · Jaynes and Cummings, Proc. IEEE 51, 89 (1963) · Demtröder, Laser Spectroscopy.

Provenance. The Jaynes–Cummings Hamiltonian, the rotating-wave approximation, the dressed states, and the $2g\sqrt{n+1}$ scaling follow Loudon Ch. 7, Foot Ch. 9, and Haroche and Raimond; the vacuum-field expression is recalled from aq-optics-field-quantization and aq-optics-vacuum-zero-point-energy, and the Fock ladder $|n\rangle$ from aq-optics-fock-states. A verifier should confirm: (1) the resonant $2\times2$ diagonalization gives eigenvalues $\hbar(\omega/2\pm g)$ and splitting $2\hbar g$ ; (2) $P_e(t)=\cos^2(gt)$ ; (3) the collapse–revival series with $t_r\approx2\pi\sqrt{\bar n}/g$ ; (4) the worked numbers $E_{\mathrm{vac}}\approx6.5\times10^2$ V/m, $g/2\pi\approx26$ MHz, $2g/2\pi\approx53$ MHz, and $C\approx108$ for $(g,\kappa,\gamma)/2\pi=(34,4.1,2.6)$ MHz. Device-specific magnitudes are representative laboratory values, not universal constants.

Prerequisites. Quantization of the Electromagnetic Field · Fock States and the Photon-Number Basis · Vacuum State and Zero-Point Energy. Leads to. Photon Antibunching · Single-Photon Sources · Coherent States.

Ask the tutor: ask me why an empty cavity can still split an atom's line into two, or to work the coupling $g$ for your own cavity's mode volume and atomic dipole.

Why does the atom get its photon back instead of just decaying?

In free space the atom couples to a continuum of empty modes: it emits into one of infinitely many channels, the photon flies away, and the chance of that mode returning the excitation is negligible — irreversible spontaneous emission. A cavity removes almost all those channels and leaves one long-lived mode, so the photon has nowhere to go but back to the atom. As long as the exchange rate $g$ beats the leak rate $\kappa$ and the decay $\gamma$ , the excitation cycles many times before it is lost — strong coupling is precisely reabsorption winning over escape.

Is the vacuum Rabi splitting the same as the Lamb shift or the AC Stark shift?

They are cousins, all rooted in atom–vacuum coupling, but different quantities. The Lamb shift is a tiny energy shift from coupling to all the vacuum modes of free space; it is second-order (dispersive) and does not split a line. The vacuum Rabi splitting is a first-order, resonant effect of coupling to one mode, producing two peaks of spacing $2g$ . The AC Stark shift is what the vacuum Rabi coupling becomes in the far-detuned limit $\lvert\Delta\rvert\gg g$ : the exchange is virtual and shifts levels by $\sim g^2/\Delta$ per photon. The same Hamiltonian gives a splitting on resonance and a shift off resonance.

Why does the population oscillate at 2g but the levels split by 2g as well — where did the factor come from?

Both are $2g$ , and that is not a coincidence. The dressed states $|\pm,0\rangle$ sit at $\hbar(\omega/2\pm g)$ , so their frequency difference is $2g$ . Starting in $|e,0\rangle$ populates an equal superposition of both, and any superposition of two stationary states beats at their frequency difference. So $P_e=\cos^2(gt)=\tfrac12[1+\cos(2gt)]$ oscillates at exactly the splitting; the lone $g$ inside the cosine is a bookkeeping artefact of $\cos^2$ .

What sets the size of g in a real experiment, and how do you make it big?

From $g=dE_{\mathrm{vac}}/\hbar$ with $E_{\mathrm{vac}}=\sqrt{\hbar\omega_c/2\varepsilon_0V}$ , you get $g\propto d/\sqrt{V}$ : two levers, a bigger dipole $d$ or a smaller mode volume $V$ . Neutral atoms have modest dipoles, so optical experiments push on $V$ with tiny high-finesse cavities. Rydberg atoms have enormous dipoles, so microwave cavity QED reaches strong coupling with larger modes. Superconducting qubits in circuit QED win on both fronts — gigantic effective dipoles and micron-scale modes — giving the largest couplings of all, up to a fraction of the mode frequency in the ultrastrong regime where even the rotating-wave approximation fails.

If a probe finds no absorption on resonance, where did the atomic line go?

It did not vanish — it moved. The bare resonance at $\omega_a$ is no longer an eigenstate; the true stationary states are the dressed states at $\omega_c\pm g$ . A weak probe absorbs only at a real transition frequency, and there is none at $\omega_a=\omega_c$ , so it sails through unabsorbed on the old line and instead absorbs strongly at $\pm g$ . This transparency flanked by two peaks is the most recognizable trace of strong coupling, and its spacing $2g$ reads out the coupling — the frequency-domain face of the flop $\cos^2(gt)$ .

Does the vacuum Rabi splitting require a single atom, or works for many?

Both, with a twist. A single atom gives splitting $2g$ . Put $N$ identical atoms in the mode and they respond collectively: the field couples to a symmetric collective excitation with coupling $g\sqrt{N}$ , so the splitting grows to $2g\sqrt{N}$ . This $\sqrt{N}$ enhancement is how the effect was often first seen — with atomic beams or clouds — and why the doublet appears even in a classical linear-transmission calculation, as two polariton peaks. The single-atom, single-photon version is the true quantum limit; that is why the single-atom, $\sqrt{n+1}$ -resolved measurement is the one that certifies quantization.