Light held beside an atom in a band gap

A Photon Cloud That Cannot Escape

Why does an excited atom tuned into a waveguide's band gap keep part of its excitation instead of emitting it all? Because light at the atom's frequency cannot travel along the waveguide. Some excitation runs away at first, but the rest forms a bound state: the atom stays partly excited, wrapped in a photon cloud that thins out exponentially with distance. The deeper the frequency sits in the gap, the tighter the cloud and the more excitation remains. Back in the band, almost all of it leaks away.

Move the atom's transition frequency across a band edge of the waveguide. Follow the wave that leaves and the photon cloud that stays localized, then compare what remains in the atom with bound states calculated separately.

A single two-level atom is coupled to one site of an ideal resonator array. The Hamiltonian uses the rotating-wave approximation and a single excitation, with no disorder, absorption or other decay channels. The bound states of an infinite chain serve as analytic references. The local cloud at a finite time does not by itself measure the total weight of the bound state.

At the edge of the propagation band

Excitation still in the atom

P_e=0

TOTAL BOUND-STATE WEIGHT · INFINITE CHAIN

0

ATOM

0\%

NEAR ATOM

0\%

OUTSIDE REGION

0\%

Schematic resonator array. Ribbon height: fixed square-root probability gain; atom glow: excited population. It is a diagnostic over the waveguide, not a mean field or particle path.

Jt=0

01 / Where the excitation goes

Scrub through the whole journey.

Full 401-site model, time increasing downward. Drag vertically or use arrows to select a time. Fixed square-root color scale saturates at probability 0.04 per site; white marks the current snapshot.

$x/a\ \longrightarrow\qquad Jt\ \downarrow$

02 / A band with two edges

Drag the atom across the edge.

Cyan: propagation band. Gold: bare atom frequency; drag vertically or use arrows. Coral: the two dressed bound energies, including the weak branch. All are model references.

$ka/\pi\ \longrightarrow\qquad (\omega-\omega_c)/J$

What stays in the atom?

Cyan: complete time evolution. Dashed coral: the bound-state contribution alone, including interference between both bound states. The continuum is absent from that reference, so it need not agree at early times.

Each preparation gives one outcome

Find the excitation.

0 / 10000

Each trial starts again in the excited atom, evolves to the selected time, then measures either the atom or one waveguide site. These are ideal independent destructive snapshots, not repeated monitoring of one photon. Fixed seed; every outcome retained.

Recorded atom fraction

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95% Wilson interval

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Largest norm error

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Largest boundary probability

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Count the atom and keep every position.

Gold: acquired frequencies. Cyan: integrated model bins, with a separate atom outcome. All 401 sites are included. Square-root vertical scale; tick labels remain probabilities.

Latest 20 records · export all

Trial	Outcome	Site position

The model, units and limits

\frac{H}{\hbar J}=\frac{\Delta}{J}|e\rangle\langle e|-\sum_x(|x+1\rangle\langle x|+\mathrm{h.c.})+\sum_j\frac{g_j}{J}(|e\rangle\langle x_j|+\mathrm{h.c.})

Positive position is rightward along the array. Spacing sets length, nearest-neighbor hopping sets frequency, and its inverse sets time. Frequencies are relative to the array center, not negative absolute photon energies. The atom starts with one excitation, the rotating-wave approximation is used, and the waveguide starts in vacuum. There is no drive after preparation, and there are no counter-rotating terms, loss, disorder or readout backaction before the final snapshot.

401 open sites, fourth-order Runge–Kutta at a time step of 0.01, duration 64. The state is never clipped or renormalized. The displayed norm and the probability in the outer eight sites show any numerical loss and boundary contamination. Sampling normalizes only the cumulative measurement probabilities by the displayed norm.

E_\pm-\Delta=\frac{g^2\operatorname{sgn}(E_\pm)}{\sqrt{E_\pm^2-4J^2}},\qquad Z_\pm=\left[1+\frac{g^2|E_\pm|}{(E_\pm^2-4J^2)^{3/2}}\right]^{-1}

Both infinite-chain bound states are retained. The total bound overlap is the sum of their atomic weights; the long-time averaged atomic contribution is the sum of their squares. A neighborhood of twelve sites on either side is a finite observation region, not an exact projector onto bound states.

Primary sources

The model follows the single-excitation resonator-chain treatment; the cited circuit experiment has additional device geometry and two-qubit physics not simulated here.

Start with the essentials

Focus question: Why does an excited atom tuned into a waveguide's band gap keep part of its excitation instead of emitting it all?
One-sentence intuition: Outside the propagation band, the atom and a localized field together form a dressed eigenstate.

Core mathematical model

Total probability of the single excitation

P_e(t)+\sum_x|\psi_x(t)|^2=1

In an ideal snapshot readout, finding the excitation in the atom and finding it in the waveguide are mutually exclusive outcomes.

A finite propagation band

\omega(k)-\omega_c=-2J\cos(ka)

The band edges lie two hopping units above and below the band center. The same dispersion drives the animation.

Common difficulties

A cloud is not a trajectory

Typical misconceptionThe excitation splits into many particles that can each be measured on their own.

Better mental modelEach preparation holds exactly one excitation in total. Repeated snapshots show how it is distributed. The ribbon height displays the square root of the probability.

Where the reference curve comes from

Typical misconceptionThe reference curve was reconstructed from the saved samples.

Better mental modelA single two-level atom is coupled to one site of an ideal resonator array. The Hamiltonian uses the rotating-wave approximation and a single excitation, with no disorder, absorption or other decay channels. The bound states of an infinite chain serve as analytic references. The local cloud at a finite time does not by itself measure the total weight of the bound state.

Run the experiment