Input–Output Theory Overview

Input–output theory connects a localized optical degree of freedom to the traveling fields that enter and leave its ports. For a cavity, it answers three questions in one framework:

1. How does an incident field drive the intracavity mode?
2. How does coupling to external modes damp that intracavity excitation?
3. What reflected, transmitted, or emitted field reaches a detector?

The central distinction is between a dimensionless intracavity annihilation operator $\hat a$ and a traveling-field operator $\hat b_{\rm in}(t)$ or $\hat b_{\rm out}(t)$, whose squared amplitude is a photon flux. A cavity may contain one photon on average while emitting much less than one photon in a short time bin; conversely, a bright reflected beam can result from interference between the incident field and a weak field radiated by the cavity.

For one Markovian port, the laboratory dictionary is

$$\begin{gathered} \text{intracavity damping} \quad\longleftrightarrow\quad \kappa, \\ \text{incident drive} \quad\longleftrightarrow\quad \hat b_{\rm in}(t), \\ \text{detected field} \quad\longleftrightarrow\quad \hat b_{\rm out}(t), \\ \hat b_{\rm out} = \hat b_{\rm in} + \sqrt\kappa\,\hat a. \end{gathered}$$

The final plus sign is a phase convention. This page consistently pairs it with a minus input term in the cavity Langevin equation. Other references use $\hat b_{\rm out}=\hat b_{\rm in}-\sqrt\kappa\,\hat a$ and a corresponding opposite drive phase. Replacing $a$ by $-a$ changes both signs and leaves the reflection and transmission amplitudes unchanged. Separately rephasing external port bases changes their amplitude representations; a consistently computed interference measurement is invariant.

Required background. Quantized modes supplies mode normalization; coherent light supplies field amplitudes; thermal and vacuum noise explains reservoir correlations. The general input–output derivation treats arbitrary coupling operators; here we specialize to one linear optical resonance with Markovian ports.

Cavity fields and rate conventions

The following conventions are used throughout:

• $\omega_c$ is the cavity resonance angular frequency, including the reservoir-induced frequency shift absorbed into the effective model.
• A coherent probe has angular frequency $\omega_d$.
• The drive detuning is $\Delta=\omega_d-\omega_c$.
• $\kappa_j$ is the energy-decay rate through port $j$.
• $\kappa_{\rm int}$ is the total unmonitored internal-loss rate.
• $\kappa=\sum_j\kappa_j+\kappa_{\rm int}$ is the total energy-decay rate.
• The intracavity field amplitude decays at rate $\kappa/2$.
• Traveling fields obey $[\hat b_j(t),\hat b_k^\dagger(t')]=\delta_{jk}\delta(t-t')$.
• $|\beta_{\rm in}|^2$ is an incident photon flux, not an intracavity photon number.

Decay rates have units of inverse time. Linewidths are stated in angular-frequency units; the ordinary-frequency linewidth is obtained by dividing by $2\pi$.

The Internal Cavity Mode

An ideal closed cavity has discrete electromagnetic normal modes. Restricting attention to one resonance gives

$$\hat H_c = \hbar\omega_c \left( \hat a^\dagger\hat a+\frac12 \right), \qquad [\hat a,\hat a^\dagger]=1.$$

The occupation

$$\hat n_c=\hat a^\dagger\hat a$$

is dimensionless. Its expectation is the mean number of excitations stored in that selected cavity mode.

A real cavity is open. A partially transmitting mirror, waveguide coupler, or radiation aperture turns the closed-cavity normal mode into a resonance with a finite linewidth. Input–output theory represents it as a localized quasimode $\hat a$ coupled to one or more continua. This is accurate when one isolated resonance dominates and the external coupling varies slowly across its linewidth.

Quantized Electromagnetic Modes explains conservative normal-mode quantization and field normalization. The present page begins where radiation leakage makes a closed normal-mode description insufficient.

Stored energy and photon number

Ignoring the zero-point offset, the stored cavity energy is

$$E_c = \hbar\omega_c \langle\hat n_c\rangle.$$

For a coherent cavity state $|\alpha\rangle$,

$$\langle\hat a\rangle=\alpha, \qquad \langle\hat n_c\rangle=|\alpha|^2.$$

The complex amplitude $\alpha$ has no units in this normalization. It should not be compared directly with $\beta_{\rm in}$, which has units $\mathrm{s}^{-1/2}$.

External Continua

Each open port supports traveling modes over a continuum of frequencies. Let $\hat b_j(\omega)$ annihilate a continuum excitation in port $j$, with

$$[ \hat b_j(\omega), \hat b_k^\dagger(\omega') ] = \delta_{jk}\delta(\omega-\omega').$$

Factors of $2\pi$ sometimes appear on the right-hand side and then move into the Fourier transform. A consistent temporal normalization is

$$\hat b_{j,\rm in}(t) = \frac{1}{\sqrt{2\pi}} \int d\omega\, e^{-i(\omega-\omega_d)t} \hat b_{j,\rm in}(\omega),$$

where the narrowband envelope approximation extends the detuning integral to the whole real line. Physical optical frequencies remain positive: this extension is accurate when the carrier greatly exceeds the relevant bandwidth. It gives the white-field commutator

$$[ \hat b_{j,\rm in}(t), \hat b_{k,\rm in}^\dagger(t') ] = \delta_{jk}\delta(t-t').$$

The corresponding normally ordered photon flux is

$$\Phi_{j,\rm in}(t) = \left\langle \hat b_{j,\rm in}^\dagger(t) \hat b_{j,\rm in}(t) \right\rangle.$$

The field is an operator-valued distribution. Fluxes require an appropriate normally ordered or finite-bandwidth measurement, not arbitrary products of white noise at one instant. The units of $\hat b(t)$ are $\mathrm{s}^{-1/2}$. An incoming wave-packet mode

$$\hat B_f = \int dt\, f^*(t)\hat b_{\rm in}(t).$$

With normalization

$$\int dt\,|f(t)|^2=1,$$

$\hat B_f$ is dimensionless and obeys $[\hat B_f,\hat B_f^\dagger]=1$. This distinction between a flux field and an integrated wave-packet mode prevents many dimensional mistakes.

Vacuum is still an input field

An undriven port is not deleted from the equations. When its occupation is negligible at the optical frequency, it carries vacuum input with

$$\begin{aligned} \langle \hat b_{\rm in}^\dagger(t) \hat b_{\rm in}(t') \rangle_{\rm vac} &=0, \\ \langle \hat b_{\rm in}(t) \hat b_{\rm in}^\dagger(t') \rangle_{\rm vac} &=\delta(t-t'). \end{aligned}$$

The first line says the normally ordered incident photon flux vanishes. The second preserves field commutators and supplies the fluctuations associated with damping. Replacing an unused port by the number zero while retaining its loss rate is not a valid quantum model.

Coupling a Mode to a Continuum

In a rotating-wave and narrowband description, one port can be represented by a coupling Hamiltonian of the form

$$\hat H_{\rm int} ={} i\hbar \int d\omega\, \sqrt{ \frac{\kappa(\omega)}{2\pi} } \times \left[ \hat b^\dagger(\omega)\hat a - \hat a^\dagger\hat b(\omega) \right].$$

The phase of $\hat b$ has been chosen to match this page’s input–output sign convention. The important physics is the coupling magnitude and its frequency dependence.

Formally solving the continuum equations and substituting them into the cavity equation produces

• a frequency shift from the principal-value part of the reservoir response;
• a memory kernel from the finite reservoir correlation time;
• a fluctuating incident-field operator.

The cavity frequency $\omega_c$ below is understood to include the relevant renormalized shift. The Markov approximation replaces the coupling by its value near resonance,

$$\kappa(\omega)\simeq\kappa(\omega_c),$$

and extends the narrow frequency integral so the memory kernel becomes local in time.

This requires

• a coupling density that is nearly flat across the cavity linewidth;
• a reservoir correlation time short compared with cavity dynamics;
• no resolved propagation delay or coherent return path;
• a rotating-wave description adequate near the selected resonance;
• ports that can be treated as independent input channels.

Photonic band edges, narrow external resonances, long feedback loops, frequency-dependent mirror response, and overlapping quasimodes can violate these assumptions.

Damping Rate and Linewidth

Suppose the cavity couples to ports with rates $\kappa_j$ and to unmonitored internal-loss channels with total rate $\kappa_{\rm int}$. The total rate is

$$\kappa = \sum_j\kappa_j + \kappa_{\rm int}.$$

With vacuum inputs, in a frame rotating at the cavity frequency,

$$\langle\hat a(t)\rangle = e^{-\kappa t/2} \langle\hat a(0)\rangle,$$

whereas

$$\langle\hat n_c(t)\rangle = e^{-\kappa t} \langle\hat n_c(0)\rangle.$$

Therefore

• the field-amplitude decay rate is $\kappa/2$;
• the energy-decay rate is $\kappa$;
• the photon or energy lifetime is $\tau_E=1/\kappa$;
• the field-amplitude lifetime is $\tau_A=2/\kappa$.

The isolated intracavity intensity response is Lorentzian, as is transmission through a second port without a direct bypass. Reflection need not be a Lorentzian peak; a lossless one-port cavity has constant reflected power. The intracavity intensity full width at half maximum is

$$\delta\omega_{\rm FWHM}=\kappa.$$

In ordinary frequency,

$$\delta\nu_{\rm FWHM} = \frac{\kappa}{2\pi}.$$

The loaded quality factor is

$$Q_{\rm loaded} = \frac{\omega_c}{\kappa}.$$

Some communities define $\kappa$ as the field-amplitude decay rate or quote a half width rather than a full width. A fitted number is incomplete unless the definition is stated.

Cavity Langevin Equation

Move to a frame rotating at the drive frequency $\omega_d$, with $\Delta=\omega_d-\omega_c$. For a linear cavity,

$$\dot{\hat a} ={} \left( i\Delta-\frac{\kappa}{2} \right)\hat a - \sum_j \sqrt{\kappa_j}\, \hat b_{j,\rm in}(t) - \sqrt{\kappa_{\rm int}}\, \hat\xi_{\rm int}(t).$$

The operator $\hat\xi_{\rm int}$ collects internal-loss noise. If internal loss contains physically distinct reservoirs, each should have its own rate and input operator.

Dropping every input operator would give damped deterministic motion, but it would also make the canonical commutator decay:

$$[ \hat a(t), \hat a^\dagger(t) ] \stackrel{\rm no\ noise}{=} e^{-\kappa t}.$$

The vacuum inputs restore the missing part and keep $[\hat a(t),\hat a^\dagger(t)]=1$. Dissipation and quantum noise are two aspects of the same port coupling.

Input and Output Fields

For each monitored Markovian port,

$$\hat b_{j,\rm out}(t) = \hat b_{j,\rm in}(t) + \sqrt{\kappa_j}\,\hat a(t).$$

The output is a coherent sum of a prompt incident field and the field radiated by the cavity. Reflection minima and phase flips arise from interference between these amplitudes. They cannot be obtained by adding incident and emitted intensities.

Top: a localized mode $\hat a$ couples to two monitored ports and one unmonitored loss bath. Bottom: for a lossless single port, a normalized rising one-photon pulse loads the cavity completely at $t=0$, followed by free release. The plotted occupation is $e^{-\kappa|t|}$; Exercise 1 derives it.

What the boundary condition means

The relation is local at the coupling plane. It does not include propagation phases between the coupler and a remote detector. Those phases should be added to the transfer path, not hidden in $\kappa_j$.

The input field is defined before interaction with the cavity; the output field is defined after interaction. Quantum causality determines which input–system correlations vanish at equal-time limits and is handled carefully in the full stochastic derivation. The compact boundary condition should not be manipulated as though every operator at time $t$ were an independent commuting number.

Internal loss is another port

Absorption, scattering into uncollected modes, and imperfect confinement can be modeled as additional bath ports. Their outputs exist physically even if no detector monitors them. Treating internal loss as a rate without its noise input is acceptable for mean classical amplitudes but incomplete for quantum fluctuations, commutators, and output states.

Coherent Drive and Intracavity Build-Up

Drive port 1 with a coherent state, separating its amplitude from vacuum fluctuations:

$$\hat b_{1,\rm in}(t) = \beta_1+\delta\hat b_{1,\rm in}(t).$$

The incident photon flux is

$$\Phi_{\rm in}=|\beta_1|^2.$$

Let $\alpha=\langle\hat a\rangle$ and take every other input to have zero mean. Then

$$\dot\alpha = \left( i\Delta-\frac{\kappa}{2} \right)\alpha - \sqrt{\kappa_1}\,\beta_1.$$

The steady-state amplitude is

$$\alpha_{\rm ss} = - \frac{ \sqrt{\kappa_1}\,\beta_1 }{ \kappa/2-i\Delta }.$$

Define the cavity susceptibility

$$\chi_c(\Delta) = \frac{1}{\kappa/2-i\Delta}.$$

Then

$$\alpha_{\rm ss} = -\sqrt{\kappa_1}\, \chi_c(\Delta)\beta_1.$$

If every input fluctuation is vacuum, the steady cavity state is coherent and its mean photon number is

$$\bar n_c = \frac{ \kappa_1|\beta_1|^2 }{ (\kappa/2)^2+\Delta^2 }.$$

This Lorentzian build-up is largest on resonance. It scales with the coupling rate through the driven port but is limited by the total linewidth, including every monitored and unmonitored decay channel.

Reflection and Transmission

Assume port 1 is coherently driven, while port 2 and the internal-loss ports have zero mean input. The output means are

$$\begin{aligned} \beta_{1,\rm out} &= \beta_1+\sqrt{\kappa_1}\,\alpha_{\rm ss}, \\ \beta_{2,\rm out} &= \sqrt{\kappa_2}\,\alpha_{\rm ss}. \end{aligned}$$

Define complex amplitude coefficients relative to the incident field:

$$r(\Delta) = \frac{\beta_{1,\rm out}}{\beta_1}, \qquad t(\Delta) = \frac{\beta_{2,\rm out}}{\beta_1}.$$

Substituting the steady-state cavity amplitude gives

$$r(\Delta) = 1 - \frac{ \kappa_1 }{ \kappa/2-i\Delta },$$

and

$$t(\Delta) = - \frac{ \sqrt{\kappa_1\kappa_2} }{ \kappa/2-i\Delta }.$$

The direct term $1$ in $r$ is the prompt reflected input in this port convention. The second term is radiation from the cavity. A deep reflection dip is destructive interference between them, not evidence that the cavity emits no field.

Lossless single-sided cavity

For one external port and no internal loss,

$$\kappa=\kappa_1, \qquad \kappa_2=\kappa_{\rm int}=0.$$

The reflection coefficient becomes

$$r(\Delta) = \frac{ -\kappa/2-i\Delta }{ \kappa/2-i\Delta }.$$

Its magnitude is unity at every frequency:

$$|r(\Delta)|=1.$$

The cavity cannot absorb or transmit energy, so it acts as a frequency-dependent phase shifter. On resonance $r(0)=-1$ in this convention. Far from resonance, $r\rightarrow1$.

Measuring only reflected power would miss this lossless cavity completely: $R=|r|^2=1$ everywhere. Phase-sensitive detection or interferometric reference reveals the resonance.

Two-sided cavity and internal loss

For drive through port 1, define

$$\mathcal D(\Delta) = \left(\frac{\kappa}{2}\right)^2+\Delta^2.$$

The reflected and transmitted power fractions are

$$R(\Delta) = \frac{ \left[ (\kappa_2+\kappa_{\rm int}-\kappa_1)/2 \right]^2 + \Delta^2 }{ \mathcal D(\Delta) },$$

and

$$T(\Delta) = \frac{ \kappa_1\kappa_2 }{ \mathcal D(\Delta) }.$$

The fraction leaving through internal-loss channels is

$$A(\Delta) = \frac{ \kappa_1\kappa_{\rm int} }{ \mathcal D(\Delta) }.$$

For passive vacuum loss,

$$R(\Delta)+T(\Delta)+A(\Delta)=1.$$

This identity is a strong convention and algebra check. If a passive empty-cavity model predicts a sum larger than one, a sign, linewidth, or normalization has been mixed incorrectly.

On resonance,

$$r(0) = \frac{ \kappa_2+\kappa_{\rm int}-\kappa_1 }{ \kappa },$$

and

$$t(0) = - \frac{ 2\sqrt{\kappa_1\kappa_2} }{ \kappa }.$$

A symmetric lossless two-sided cavity has $\kappa_1=\kappa_2=\kappa/2$. It gives $r(0)=0$ and $t(0)=-1$: all resonant power exits port 2, with a convention-dependent phase.

Critical coupling and impedance matching

For a one-sided cavity with external coupling $\kappa_{\rm ext}$ and internal loss $\kappa_{\rm int}$,

$$r(0) = \frac{ \kappa_{\rm int}-\kappa_{\rm ext} }{ \kappa_{\rm int}+\kappa_{\rm ext} }.$$

Three regimes follow:

• undercoupled: $\kappa_{\rm ext}<\kappa_{\rm int}$;
• critically coupled: $\kappa_{\rm ext}=\kappa_{\rm int}$;
• overcoupled: $\kappa_{\rm ext}>\kappa_{\rm int}$.

At critical coupling, $r(0)=0$. All resonant incident power enters the cavity and leaves through internal-loss channels in this one-port model. The zero reflection results from destructive interference, while the energy disposition is fixed by the available decay channels.

For a two-sided cavity driven through port 1, zero resonant reflection instead requires

$$\kappa_1=\kappa_2+\kappa_{\rm int}.$$

Calling both situations “critical coupling” is common, but the transmitted and absorbed fractions differ. The port topology must accompany the phrase.

Coupling efficiencies

For undriven empty-cavity decay into vacuum with constant rates, the escape probability for one initial excitation through port $j$ is

$$\eta_j^{\rm esc} = \frac{\kappa_j}{\kappa}.$$

The total monitored escape efficiency is

$$\eta_{\rm mon}^{\rm esc} = \frac{ \sum_{j\in{\rm monitored}}\kappa_j }{ \kappa }.$$

Escape efficiency is not the same as propagation transmission, spatial mode matching, or detector quantum efficiency. Those later losses multiply the probability that an escaped excitation produces the intended record.

Frequency-Domain Scattering Matrix

For several linear ports, the empty-cavity amplitude scattering matrix has the compact form

$$S_{jk}(\omega) = \delta_{jk} - \frac{ \sqrt{\kappa_j\kappa_k} }{ \kappa/2-i(\omega-\omega_c) }.$$

The diagonal term contains the prompt path. The rank-one resonant term comes from entering and leaving through the cavity mode.

If every loss bath is included as an explicit port, the full passive scattering matrix is unitary on the real-frequency axis. The submatrix restricted to experimentally monitored ports is generally not unitary when internal loss is traced out. Apparent nonunitarity is then missing information and energy in unobserved channels, not a violation of quantum mechanics.

The simple matrix assumes no nonresonant background except the identity prompt path. Real devices can have cable delay, mirror propagation phase, mode mismatch, Fano interference, or a direct bypass channel. Those effects should be modeled explicitly rather than absorbed into a complex, frequency-dependent “$\kappa$.”

Time-Domain Response and Ringdown

After initial transients have decayed, the stable linear cavity field is a causal convolution of the past input with the impulse response:

$$\hat a(t) ={} - \sum_j\sqrt{\kappa_j} \times \int_{-\infty}^{t}ds\, e^{(i\Delta-\kappa/2)(t-s)} \hat b_{j,\rm in}(s) - \sqrt{\kappa_{\rm int}} \times \int_{-\infty}^{t}ds\, e^{(i\Delta-\kappa/2)(t-s)} \hat\xi_{\rm int}(s).$$

For evolution from a finite initial time $t_0$, start the integrals at $t_0$ and add $e^{(i\Delta-\kappa/2)(t-t_0)}a(t_0)$.

A drive varying slowly compared with $1/\kappa$ approximately follows the instantaneous steady state. A pulse with bandwidth comparable to or larger than $\kappa$ excites transient dynamics and is reshaped by the cavity.

Cavity ringdown

Prepare a coherent cavity state with amplitude $\alpha_0$, leave every bath in vacuum, and turn the coherent drive off at $t=0$. Then

$$\alpha(t) = \alpha_0 e^{(i\Delta-\kappa/2)t},$$

and

$$\bar n_c(t) = |\alpha_0|^2e^{-\kappa t}.$$

With vacuum incident fields, the emitted photon flux through port $j$ is

$$\Phi_{j,\rm out}(t) = \kappa_j\bar n_c(t).$$

Integrating the entire ringdown gives

$$N_j = \int_0^\infty dt\, \Phi_{j,\rm out}(t) = \frac{\kappa_j}{\kappa} |\alpha_0|^2.$$

The initial photons divide among channels according to their rates. Summing over every external and internal port returns $|\alpha_0|^2$.

Ringdown directly measures the energy-decay rate when the detector response is faster than the decay and no slow mode, thermal repopulation, or coherent feedback distorts the exponential. A field-amplitude measurement decays with time constant $2/\kappa$; direct intensity decays with time constant $1/\kappa$.

What a Detector Sees

Detectors act on traveling output fields, not directly on $\hat a$. Input–output theory supplies the operator entering each measurement model.

Output photon flux

At port $j$,

$$\Phi_{j,\rm out}(t) = \left\langle \hat b_{j,\rm out}^\dagger(t) \hat b_{j,\rm out}(t) \right\rangle.$$

Using the boundary relation,

$$\Phi_{j,\rm out} ={} \langle \hat b_{j,\rm in}^\dagger \hat b_{j,\rm in} \rangle + \kappa_j \langle\hat a^\dagger\hat a\rangle + \sqrt{\kappa_j} \langle \hat b_{j,\rm in}^\dagger\hat a + \hat a^\dagger\hat b_{j,\rm in} \rangle.$$

The final line is interference. For vacuum input and spontaneous cavity emission, the emitted flux is $\kappa_j\langle\hat n_c\rangle$. For a coherently driven reflection experiment, omitting the interference terms destroys the reflection dip and gives the wrong energy balance.

Photon Counting explains the detector POVM, efficiency, dark counts, dead time, timing response, and number-resolution limits.

Output quadratures

Balanced homodyne detection measures a mode of $\hat b_{j,\rm out}(t)$ selected by the local oscillator. The prompt coherent mean is often subtracted electronically or displaced optically. The remaining fluctuations still contain incident noise, cavity radiation, and their correlations. Subtracting a mean neither isolates cavity fluctuations nor changes the output-field commutator.

Homodyne and Heterodyne Detection explains local-oscillator mode selection, quadrature calibration, heterodyne noise, and optical tomography. Conditional state updates are described in quantum trajectories.

Spectra and correlations

Output spectra and correlation functions follow by replacing each output operator with its input–output expression and evaluating the required multi-time averages. An empty linear cavity filters Gaussian inputs linearly. A quadratic parametric Hamiltonian can squeeze a Gaussian input but preserves Gaussianity. Saturable emitters, higher-order nonlinearities, or conditional measurements can instead produce non-Gaussian output.

The boundary condition alone does not compute internal multi-time correlators. Those come from the cavity’s Hamiltonian and open-system dynamics, often through the quantum regression theorem or a stochastic trajectory calculation.

Relation to Open Systems

The same physical port appears in several equivalent descriptions. For cavity port $j$, define

$$\hat L_j = \sqrt{\kappa_j}\,\hat a.$$

For a vacuum bath, the port contributes

$$\mathcal D[\hat L_j]\rho = \kappa_j\mathcal D[\hat a]\rho$$

to the unconditional master equation, where

$$\mathcal D[\hat a]\rho = \hat a\rho\hat a^\dagger - \frac12 \left\{ \hat a^\dagger\hat a,\rho \right\}.$$

For vacuum inputs and a linear cavity in the drive frame,

$$\dot\rho ={} - \frac{i}{\hbar} [ \hat H_{\rm rot}+\hat H_{\rm drive}, \rho ] + \sum_j \kappa_j\mathcal D[\hat a]\rho + \kappa_{\rm int}\mathcal D[\hat a]\rho.$$

The Hamiltonian terms are

$$\hat H_{\rm rot} = -\hbar\Delta\, \hat a^\dagger\hat a,$$

and, for drive through port 1,

$$\hat H_{\rm drive} = i\hbar\sqrt{\kappa_1} \left( \beta_1^*\hat a - \beta_1\hat a^\dagger \right).$$

They reproduce the mean Langevin equation used above.

The dictionary is

• master equation: discard all output records and retain $\mathcal D[L_j]\rho$;
• Langevin equation: retain the incident quantum fields as operator noise;
• input–output relation: retain the outgoing fields;
• quantum trajectory: condition the state on a specified measurement of one or more outgoing fields.

Combining all vacuum loss terms into $\kappa\mathcal D[a]\rho$ is sufficient for unconditional cavity dynamics. It is insufficient for predicting which port carries an emitted photon or how efficiently that output is monitored. Port-resolved operators must be retained for those questions.

Coherent input versus a classical force

Moving a coherent input amplitude into $\hat H_{\rm drive}$ is a displacement of the bath description. Vacuum fluctuations remain in the input channel. Replacing the whole quantum input by a deterministic force gives the correct mean for a linear cavity but loses shot noise and cannot predict output quantum statistics.

Thermal, Squeezed, and Nonclassical Inputs

The boundary condition does not require vacuum input. For a stationary thermal Markov field with occupation $\bar n$,

$$\begin{aligned} \langle \hat b_{\rm in}^\dagger(t) \hat b_{\rm in}(t') \rangle &= \bar n\,\delta(t-t'), \\ \langle \hat b_{\rm in}(t) \hat b_{\rm in}^\dagger(t') \rangle &= (\bar n+1)\delta(t-t'). \end{aligned}$$

This ideal white thermal field has no finite total photon flux over an unlimited bandwidth. Real detection selects a finite band. In the Markov master equation a thermal port contributes $\kappa_j[(\bar n_j+1)\mathcal D[a]+\bar n_j\mathcal D[a^\dagger]]\rho$. For independent broad thermal reservoirs (including internal ports), the driven steady cavity occupation is

$$\langle a^\dagger a\rangle_{\rm ss} =|\alpha_{\rm ss}|^2+\frac{\sum_j\kappa_j\bar n_j}{\kappa}.$$

Zero mean input therefore does not imply zero incident noise or zero stored photons. A squeezed input also has anomalous correlations such as $\langle\hat b_{\rm in}(t)\hat b_{\rm in}(t')\rangle$. A finite-bandwidth squeezed source is generally not perfectly delta correlated and may require an explicit auxiliary mode or colored-noise model.

For nonclassical wave packets containing one or a few photons, the input state is not summarized by a coherent amplitude. One can retain temporal input modes explicitly, use Fock-state master equations, or solve a scattering problem. The empty-cavity transfer amplitude remains useful, but a classical mean-field calculation does not determine the output state.

When the Simple Cavity Model Fails

The one-mode Markov model should be enlarged when

• several cavity resonances overlap within a linewidth;
• external coupling varies appreciably across the signal bandwidth;
• a nearby waveguide band edge or narrow resonance creates memory;
• propagation delay and coherent feedback are dynamically resolved;
• mirror dispersion or absorption cannot be represented by a flat loss bath;
• the cavity couples ultrastrongly enough that rotating-wave port coupling becomes questionable;
• nonlinear dynamics makes a single linear susceptibility inadequate;
• the output contains spatial or polarization modes not represented by the declared ports.

A useful repair is often to promote the structured part of the environment to an explicit system mode. The remaining broad continuum can then be treated Markovian at the enlarged boundary.

Exercises

Exercise 1: Loading one photon with a rising pulse

An empty resonant cavity is driven through port 1 by a normalized one-photon envelope $f(t)=\sqrt\kappa e^{\kappa t/2}$ for $t\leq0$, and zero thereafter. All other ports are vacuum. Show that the largest cavity occupation occurs at $t=0$ and equals $\kappa_1/\kappa$. What happens for a lossless one-port cavity, whose steady reflected power is always unity?

Solution

In the one-excitation sector the cavity excitation amplitude satisfies $\dot c=-\kappa c/2-\sqrt{\kappa_1}f$. Thus, during the pulse,

$$c(t)=-\sqrt{\kappa_1}\int_{-\infty}^t e^{-\kappa(t-s)/2}f(s)ds =-\sqrt{\frac{\kappa_1}{\kappa}}e^{\kappa t/2}.$$

The loading probability is $|c(0)|^2=\kappa_1/\kappa$. Cauchy–Schwarz applied to the integral at zero bounds this probability by $\kappa_1\int_{-\infty}^0e^{\kappa s}ds=\kappa_1/\kappa$ for any normalized pulse ending then; the rising exponential saturates the bound.

For $\kappa_1=\kappa$, the reflected envelope $f_{\rm out}=f+\sqrt\kappa c$ vanishes while the photon enters. At zero the cavity contains the full photon, then releases a decaying exponential. There is no irreversible absorption or contradiction with unit steady-state reflection: energy is stored temporarily. The ideal pulse has an infinitely long leading tail; truncation and mode mismatch reduce loading. This time-reversal principle was tested by Bader et al.

Exercise 2: Vacuum noise preserves the cavity commutator

For one resonant port, consider

$$\dot{\hat a} = -\frac{\kappa}{2}\hat a - \sqrt\kappa\,\hat b_{\rm in}(t).$$

Assume $[\hat a(0),\hat a^\dagger(0)]=1$, the initial cavity and input commute, and the input has the white-noise commutator. Solve formally and show that $[\hat a(t),\hat a^\dagger(t)]=1$.

Solution

The formal solution is

$$\hat a(t) ={} e^{-\kappa t/2}\hat a(0) - \sqrt\kappa \int_0^t ds\, e^{-\kappa(t-s)/2} \hat b_{\rm in}(s).$$

Initial system–field cross commutators vanish. Therefore

$$[ \hat a(t),\hat a^\dagger(t) ] ={} e^{-\kappa t} + \kappa \int_0^t ds\, e^{-\kappa(t-s)}.$$

The integral is

$$\kappa \int_0^t ds\, e^{-\kappa(t-s)} = 1-e^{-\kappa t}.$$

Hence

$$[ \hat a(t),\hat a^\dagger(t) ] = e^{-\kappa t} + 1-e^{-\kappa t} = 1.$$

Without the input-noise term, only the first contribution would remain and the canonical algebra would decay.

Exercise 3: Intracavity response and linewidth

A cavity is driven through port 1 with photon flux $\Phi_{\rm in}=|\beta_1|^2$. Starting from

$$\dot\alpha = \left( i\Delta-\frac{\kappa}{2} \right)\alpha - \sqrt{\kappa_1}\beta_1,$$

find the steady-state photon number. Show that its intensity full width at half maximum is $\kappa$.

Solution

Setting $\dot\alpha=0$ gives

$$\alpha_{\rm ss} = - \frac{ \sqrt{\kappa_1}\beta_1 }{ \kappa/2-i\Delta }.$$

Therefore

$$\bar n_c(\Delta) = \frac{ \kappa_1\Phi_{\rm in} }{ (\kappa/2)^2+\Delta^2 }.$$

On resonance,

$$\bar n_c(0) = \frac{ 4\kappa_1\Phi_{\rm in} }{ \kappa^2 }.$$

Half maximum occurs when

$$\left(\frac{\kappa}{2}\right)^2+\Delta^2 = 2\left(\frac{\kappa}{2}\right)^2,$$

so

$$|\Delta|=\frac{\kappa}{2}.$$

The two half-maximum points are separated by $\kappa$, so the intensity FWHM is $\kappa$ in angular-frequency units.

Exercise 4: A lossless one-port cavity can be invisible in power

For

$$r(\Delta) = \frac{ -\kappa/2-i\Delta }{ \kappa/2-i\Delta },$$

show that reflected power is independent of detuning. What observable still reveals the cavity?

Solution

The squared magnitudes of numerator and denominator are equal:

$$\left| -\frac{\kappa}{2}-i\Delta \right|^2 = \left| \frac{\kappa}{2}-i\Delta \right|^2 = \frac{\kappa^2}{4}+\Delta^2.$$

Therefore

$$R(\Delta)=|r(\Delta)|^2=1.$$

The cavity changes the phase of $r$. In this convention it changes from the far-detuned value $r\simeq1$ to $r(0)=-1$ on resonance. Interferometry, homodyne detection, or network analysis of the complex reflected amplitude reveals that phase response even though a power-only scan is flat.

Exercise 5: Power conservation for a two-sided cavity

Using

$$\mathcal D = \left(\frac{\kappa}{2}\right)^2+\Delta^2$$

and $\kappa=\kappa_1+\kappa_2+\kappa_{\rm int}$, verify that the fractions

$$\begin{aligned} R &= \frac{ [(\kappa_2+\kappa_{\rm int}-\kappa_1)/2]^2 + \Delta^2 }{ \mathcal D }, \\ T &= \frac{\kappa_1\kappa_2}{\mathcal D}, \\ A &= \frac{\kappa_1\kappa_{\rm int}}{\mathcal D} \end{aligned}$$

satisfy $R+T+A=1$.

Solution

The numerator of the sum is

$$\mathcal N ={} \frac14 ( \kappa_2+\kappa_{\rm int}-\kappa_1 )^2 + \Delta^2 + \kappa_1\kappa_2 + \kappa_1\kappa_{\rm int}.$$

Let $x=\kappa_2+\kappa_{\rm int}$. Then

$$\begin{aligned} \frac14(x-\kappa_1)^2+\kappa_1x &= \frac14(x+\kappa_1)^2 \\ &= \frac{\kappa^2}{4}. \end{aligned}$$

Thus

$$\mathcal N = \frac{\kappa^2}{4}+\Delta^2 = \mathcal D,$$

and $R+T+A=1$.

Exercise 6: Coupling regime from a resonant reflection

A one-port cavity has

$$\frac{\kappa_{\rm ext}}{2\pi} = 4.0\ \mathrm{MHz}, \qquad \frac{\kappa_{\rm int}}{2\pi} = 1.0\ \mathrm{MHz}.$$

Find its loaded intensity linewidth, resonant reflection amplitude, reflected power, and internally lost power fraction. Classify the coupling regime.

Solution

The loaded rate is the sum:

$$\frac{\kappa}{2\pi} = 5.0\ \mathrm{MHz}.$$

This is the intensity FWHM in ordinary-frequency units. On resonance,

$$\begin{aligned} r(0) &= \frac{ \kappa_{\rm int}-\kappa_{\rm ext} }{ \kappa_{\rm int}+\kappa_{\rm ext} } \\ &= \frac{1-4}{1+4} = -0.6. \end{aligned}$$

Therefore

$$R(0)=|r(0)|^2=0.36.$$

There is no transmission port, so passivity gives

$$A(0)=1-R(0)=0.64.$$

Because $\kappa_{\rm ext}>\kappa_{\rm int}$, the cavity is overcoupled. The negative reflection amplitude encodes the resonant phase flip; power alone would not distinguish its sign.

Exercise 7: Ringdown branching ratios

A cavity initially contains $\bar n_c(0)=10$ photons. Its decay rates are in the ratio

$$\kappa_1:\kappa_2:\kappa_{\rm int} = 5:3:2.$$

The loaded linewidth is $\kappa/(2\pi)=10\ \mathrm{MHz}$. Find the energy lifetime, field-amplitude lifetime, and mean number of photons eventually leaving through each channel.

Solution

The total energy-decay rate is

$$\kappa = 2\pi \left( 10\ \mathrm{MHz} \right).$$

The energy lifetime is

$$\tau_E = \frac1\kappa \simeq 15.9\ \mathrm{ns},$$

and the field-amplitude lifetime is

$$\tau_A = \frac2\kappa \simeq 31.8\ \mathrm{ns}.$$

The normalized branching fractions are $0.5$, $0.3$, and $0.2$. Therefore

$$N_1=5, \qquad N_2=3, \qquad N_{\rm int}=2.$$

Their sum equals the initial mean occupation. The internal share is physically emitted into or absorbed by unmonitored bath degrees of freedom even though it does not appear in a collected optical record.

Exercise 8: Same master equation, different collection

Compare two empty cavities with the same total rate $\kappa$. Cavity A has one monitored port with $\kappa_1=\kappa$. Cavity B has $\kappa_1=\kappa/4$ and $\kappa_{\rm int}=3\kappa/4$. Both baths are vacuum.

1. Show that their unconditional undriven master equations are identical.
2. If each starts in $|1\rangle$, find the probability that the photon exits through the monitored port.
3. Explain why the total master equation alone cannot answer the second question.

Solution

For cavity A, the dissipative term is

$$\kappa\mathcal D[\hat a]\rho.$$

For cavity B, the monitored and internal terms add:

$$\begin{aligned} \frac{\kappa}{4} \mathcal D[\hat a]\rho + \frac{3\kappa}{4} \mathcal D[\hat a]\rho &= \kappa\mathcal D[\hat a]\rho. \end{aligned}$$

Thus both have the same unconditional decay of every intracavity observable.

The probability for one initial excitation to leave through port 1 is its branching ratio:

$$P_1=\frac{\kappa_1}{\kappa}.$$

Therefore

$$P_1^{(A)}=1, \qquad P_1^{(B)}=\frac14.$$

Combining channels into one total dissipator erases their physical labels. The port-resolved coupling operators $L_1=\sqrt{\kappa_1}\,a$ and $L_{\rm int}=\sqrt{\kappa_{\rm int}}\,a$ must be retained to predict collection probabilities or measurement records.

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