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Cavity and waveguide quantum electrodynamics

L2 · Chapter 2

How to keep track of light after it leaves

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An open-system model compresses escaping light into a decay rate when the environment forgets quickly enough. Keeping the environment explicitly shows how earlier emission can affect the present. Only a separation of time scales turns that history into a coefficient multiplying the current amplitude.

Write the system and environment together

The dimensionless amplitudes are c(t)c(t) for the system and bq(t)b_q(t) for environmental mode qq. Frequencies Δ\Delta and ωq\omega_q are relative to a reference. Complex coupling gqg_q also has inverse-time units; its phase fixes a relative coupling convention. The conjugate pairing below makes the exchange Hamiltonian Hermitian.

Multiply the second equation by eiωqte^{i\omega_qt}. The product rule cancels the known free rotation, leaving only the source c(t)c(t). Integrating this expression from the initial time produces the history integral in the next section.

c˙=−iΔc−i∑qgqbq,b˙q=−iωqbq−igq∗c\dot c=-i\Delta c-i\sum_qg_qb_q,\qquad\dot b_q=-i\omega_qb_q-ig_q^*c
Opposite exchange directions use conjugate couplings, preserving probability in the lossless model.
ddt ⁣(eiωqtbq(t))=−igq∗eiωqtc(t)\frac d{dt}\!\left(e^{i\omega_qt}b_q(t)\right)=-ig_q^*e^{i\omega_qt}c(t)
This integrating-factor method removes a known rotation before adding the source contributions.

Eliminating a mode leaves its history behind

Let c(t)c(t) be the system excitation amplitude and bq(t)b_q(t) the amplitude of environmental mode qq. Frequencies are measured relative to a chosen reference. Starting with an empty environment, a mode receives amplitude from the system at every earlier time. Each contribution accumulates free phase after it is emitted.

Substituting the integral solution into the system equation removes the explicit mode variables but retains the past system amplitude. The memory kernel assigns a complex weight to each age of that history. A large number of modes alone does not ensure rapid cancellation.

bq(t)=−igq∗∫0te−iωq(t−s)c(s) dsb_q(t)=-ig_q^*\int_0^t e^{-i\omega_q(t-s)}c(s)\,ds
The integration variable is emission time. Initially occupied modes add a freely propagating input term.
c˙(t)=−iΔc(t)−∫0tK(t−s)c(s) ds,K(u)=∑q∣gq∣2e−iωqu\dot c(t)=-i\Delta c(t)-\int_0^t K(t-s)c(s)\,ds,\qquad K(u)=\sum_q|g_q|^2e^{-i\omega_qu}
The kernel has units of inverse seconds squared. Integrating over time produces the correct units for an amplitude derivative.

Reading: Gardiner and Collett · Sheremet et al.

When a history becomes a decay rate

If mode density and coupling vary slowly across the relevant frequencies, their phases spread apart quickly. The kernel is then appreciable only for a short time. If the system barely changes during that interval, replace its earlier amplitude inside the integral by the current one and extend the upper limit beyond the correlation time.

The real part of the resulting integral gives half the population decay rate Γ\Gamma; its imaginary part shifts the frequency by δω\delta\omega. Both come from the same environment. Absorbing the shift into detuning gives exponential amplitude decay. A band edge, narrow filter or long propagation delay can invalidate this separation of scales.

c˙≃−[i(Δ+δω)+Γ2]c,∣c(t)∣2=e−Γt∣c(0)∣2\dot c\simeq-\left[i(\Delta+\delta\omega)+\frac{\Gamma}{2}\right]c,\qquad |c(t)|^2=e^{-\Gamma t}|c(0)|^2
Squaring the amplitude doubles its decay exponent. The Markov approximation removes resolvable environmental memory.

Connect a cavity amplitude to one travelling port

For a cavity with one output port, ccc_c is dimensionless. A travelling single-photon envelope ξ(t)\xi(t) has units of inverse square-root seconds, making its squared magnitude a probability flux. Choose the port phases to obtain the equations below.

The output combines the field passing directly through the port and the field emitted by the cavity. Add their amplitudes before calculating flux. This interference changes phase and pulse shape. For this lossless one-port cavity, stationary monochromatic input and output have equal intensity. An intensity dip requires another output or loss channel. Adding an emitted intensity to the input intensity would miss it.

With no incoming light and one initial cavity photon, output flux equals the rate at which cavity population is lost. An atom adds coherent exchange: cavity population may fall and then rise as the atom feeds it again.

c˙c=−(iΔc+κ2)cc−κ ξin,ξout=ξin+κ cc\dot c_c=-\left(i\Delta_c+\frac{\kappa}{2}\right)c_c-\sqrt\kappa\,\xi_{\rm in},\qquad \xi_{\rm out}=\xi_{\rm in}+\sqrt\kappa\,c_c
These are single-excitation linear cavity equations with a memoryless port. Changing the input phase convention changes both relevant signs together.

Reading: Gardiner and Collett · Sheremet et al.

Find the probability missing from a cavity

Worked example: let one initial photon decay through an empty input port, with κ/(2π)=10 MHz\kappa/(2\pi)=10\,\mathrm{MHz}. The population lifetime is 1/κ≃15.9 ns1/\kappa\simeq15.9\,\mathrm{ns}. After three lifetimes, e−3≃0.0498e^{-3}\simeq0.0498 remains and 0.9500.950 has left. Their sum is one.

An additional uncollected loss port contributes to the total decay rate. The recorded port receives only part of that flux. A later chapter separately evaluates collection probability and the quality of the collected temporal mode.

Pcav(T)=e−κT,Pout(T)=∫0Tκe−κt dt=1−e−κTP_{\rm cav}(T)=e^{-\kappa T},\qquad P_{\rm out}(T)=\int_0^T\kappa e^{-\kappa t}\,dt=1-e^{-\kappa T}
The integral counts probability emitted within the observation window. The un-emitted remainder stays in the system.

Try the prediction

Compare system population with escaping light

  1. Predict how a stronger port coupling changes system exchange and the output pulse.
  2. Change only the port parameter. Compare atom and bath populations with the accumulated probability in each output channel.
  3. Find the clearest departure from exponential decay and explain it using memory or coherent exchange.

Exercises and explanations

  1. The population lifetime is one hundred nanoseconds. How long does amplitude take to fall by the natural exponential factor?

    Read the explanation

    Two hundred nanoseconds. Population is the squared amplitude magnitude, so amplitude decays at half its rate.

  2. Three quarters of the total decay enters the collected port. What is the collection probability after complete decay?

    Read the explanation

    It is three quarters for a single initial excitation, constant branching and two memoryless channels. This number alone says nothing about temporal shape.

  3. Why can an environment with many modes still violate the Markov approximation?

    Read the explanation

    Its spectrum and propagation determine its correlation time. A band edge or delayed path can retain memory while the system changes. Compare that correlation time with the system’s evolution time.

Sources and calculation scope

  1. Gardiner and Collett, Input and output in damped quantum systems, Phys. Rev. A 31, 3761 (1985)

    Section II specifies rotating-wave coupling to a flat bath and derives input–output boundary relations. The course states its own port phase convention and treats the finite-band chain separately from a white-noise port.

  2. Sheremet et al., Waveguide quantum electrodynamics, Rev. Mod. Phys. 95, 015002 (2023)

    The theory overview and Appendix A place single-excitation propagation and input–output relations in their model context. Our two-amplitude examples and schematic figures are original teaching calculations.

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