Quantum Langevin Equations

A quantum Langevin equation is a Heisenberg-picture equation of motion for a system operator after environmental degrees of freedom have been eliminated but their fluctuating input operators have not been discarded. It is the operator version of the statement that dissipation and noise arrive together.

The density-matrix master equation answers:

How does the system state evolve after the environment is ignored?

A quantum Langevin equation answers:

How do system observables evolve while the incident and emitted fields remain visible?

This distinction is essential in quantum optics, circuit QED, optomechanics, quantum-limited amplification, and continuous measurement. The same physical port can appear as a Lindblad dissipator, an input–output boundary condition, or a measured field depending on what information is retained.

For the input-field normalization used here, see Input–Output Theory. For the noise taxonomy, see Quantum Noise.

The Basic Structure

The classical Langevin equation for a damped variable has the schematic form

$$\dot x = \text{deterministic drift} + \text{noise}.$$

The quantum version replaces $x$ by an operator and replaces the random force by bath or input-field operators. For a Markovian bosonic input field,

$$[b_{\mathrm{in}}(t),b_{\mathrm{in}}^\dagger(t')] = \delta(t-t').$$

For vacuum input,

$$\langle b_{\mathrm{in}}(t)b_{\mathrm{in}}^\dagger(t')\rangle = \delta(t-t'), \qquad \langle b_{\mathrm{in}}^\dagger(t)b_{\mathrm{in}}(t')\rangle = 0.$$

For thermal input with occupation $\bar n$,

$$\langle b_{\mathrm{in}}^\dagger(t)b_{\mathrm{in}}(t')\rangle = \bar n\,\delta(t-t'),$$

and

$$\langle b_{\mathrm{in}}(t)b_{\mathrm{in}}^\dagger(t')\rangle = (\bar n+1)\delta(t-t').$$

These are not ordinary functions of time. They are idealized white-noise operator distributions. Products at the same time require a convention, usually supplied by quantum stochastic calculus or by starting from a finite-bandwidth bath and taking the Markov limit.

General Markov Form

Consider a system Hamiltonian $H$ coupled to one Markov input channel by a coupling operator $L$. In the same convention as

$$b_{\mathrm{out}}(t) = b_{\mathrm{in}}(t)+L(t),$$

the Langevin equation for a system operator $X$ is commonly written

$$\begin{aligned} \dot X =& \frac{i}{\hbar}[H,X] + \frac12 L^\dagger[X,L] + \frac12[L^\dagger,X]L \\ &+ b_{\mathrm{in}}^\dagger(t)[X,L] + [L^\dagger,X]b_{\mathrm{in}}(t). \end{aligned}$$

The first line is the adjoint Lindblad drift. The second line is the input noise. Other sign conventions for the input–output boundary relation move minus signs between the coupling Hamiltonian, the input field, and the output field. The convention must be used consistently.

With several independent Markov channels, one adds a copy of this structure for each $L_j$ and $b_{j,\mathrm{in}}(t)$:

$$\begin{aligned} \dot X =& \frac{i}{\hbar}[H,X] + \sum_j \left[ \frac12 L_j^\dagger[X,L_j] + \frac12[L_j^\dagger,X]L_j \right] \\ &+ \sum_j \left[ b_{j,\mathrm{in}}^\dagger(t)[X,L_j] + [L_j^\dagger,X]b_{j,\mathrm{in}}(t) \right]. \end{aligned}$$

For vacuum input, or for a coherent input after the known mean field has been displaced into the Hamiltonian, the corresponding unconditional Schrödinger-picture generator is

$$\dot\rho = -\frac{i}{\hbar}[H,\rho] + \sum_j\mathcal D[L_j]\rho,$$

where

$$\mathcal D[L]\rho = L\rho L^\dagger - \frac12\{L^\dagger L,\rho\}.$$

The master equation is recovered when the input fields are assigned their state and the outgoing fields are not retained as records. Thermal, squeezed, or otherwise non-vacuum inputs can change the density-matrix generator even when the operator Langevin equation keeps the same boundary form; the input correlations then carry the extra occupation or squeezing data.

Damped Harmonic Oscillator

Let $a$ be a harmonic-oscillator annihilation operator with Hamiltonian

$$H = \hbar\omega_c a^\dagger a,$$

and let the oscillator couple to one Markov port through

$$L = \sqrt{\kappa}\,a.$$

Set $X=a$ in the general Langevin equation. Since $[a,L]=0$ and

$$[L^\dagger,a] = -\sqrt{\kappa},$$

one obtains

$$\dot a = i[H/\hbar,a] - \frac{\kappa}{2}a - \sqrt{\kappa}\,b_{\mathrm{in}}(t).$$

Using $i[H/\hbar,a]=-i\omega_c a$ gives

$$\dot a = - \left( i\omega_c+\frac{\kappa}{2} \right)a - \sqrt{\kappa}\,b_{\mathrm{in}}(t).$$

The field-amplitude decay rate is $\kappa/2$, while the photon-number decay rate is $\kappa$. This is one of the most common convention traps in cavity physics.

Why the Noise Term Is Required

It is tempting to keep the damping term and drop the input noise, especially when the input is vacuum. That produces an invalid quantum equation.

Let

$$\lambda = i\omega_c+\frac{\kappa}{2}.$$

The solution of the damped-oscillator Langevin equation from time $t_0$ to $t$ is

$$a(t) = e^{-\lambda(t-t_0)}a(t_0) - \sqrt{\kappa} \int_{t_0}^{t} ds\, e^{-\lambda(t-s)} b_{\mathrm{in}}(s).$$

Assume the incoming field commutes with the initial oscillator operator. The commutator at time $t$ is then

$$\begin{aligned} [a(t),a^\dagger(t)] &= e^{-\kappa(t-t_0)} \\ &\quad + \kappa \int_{t_0}^{t} ds\, e^{-\kappa(t-s)} \\ &= 1. \end{aligned}$$

The integral contribution from the input-field commutator exactly restores the missing part. If the noise term were removed, the commutator would decay as $e^{-\kappa(t-t_0)}$, contradicting $[a,a^\dagger]=1$.

This is the operator reason vacuum noise cannot be replaced by zero in a damped quantum oscillator.

Frequency-Domain Susceptibility

The same equation displays the oscillator as a linear filter. With Fourier convention

$$f[\omega] = \int_{-\infty}^{\infty} dt\,e^{i\omega t}f(t),$$

the cavity equation gives

$$a[\omega] = -\sqrt{\kappa}\, \chi_c(\omega) b_{\mathrm{in}}[\omega],$$

where

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

The susceptibility tells which input-noise frequencies enter the oscillator. Combined with

$$b_{\mathrm{out}}(t) = b_{\mathrm{in}}(t)+\sqrt{\kappa}\,a(t),$$

it determines reflection, transmission, output spectra, and measured homodyne or heterodyne records.

Thermal Inputs and Relaxation

For a thermal input field, the operator equation has the same form, but the noise correlations change. The oscillator occupation relaxes toward the input occupation:

$$\frac{d}{dt} \langle a^\dagger a\rangle = -\kappa \left( \langle a^\dagger a\rangle-\bar n \right).$$

Thus

$$\langle a^\dagger a\rangle(t) = \bar n + \left[ \langle a^\dagger a\rangle(t_0)-\bar n \right] e^{-\kappa(t-t_0)}.$$

At zero temperature, $\bar n=0$, and the oscillator relaxes to vacuum. At finite temperature, emission and absorption terms balance at $\bar n$. The corresponding density-matrix description is the thermal oscillator master equation with both $a$ and $a^\dagger$ dissipators.

Relation to Measurement Records

The input field is not automatically a measurement record. It is part of the dynamical boundary condition. A record appears only after an output field is measured.

For a monitored port,

$$b_{\mathrm{out}}(t) = b_{\mathrm{in}}(t)+L(t).$$

Photon counting samples output intensity. Homodyne detection samples an output quadrature. Averaging over those records recovers the unconditional master equation, while conditioning on them gives quantum trajectories and filtering equations.

This is why the same coupling operator $L$ can appear in:

• a quantum Langevin equation for system operators;
• a Lindblad dissipator $\mathcal D[L]$ for the unconditional state;
• an input–output relation for traveling fields;
• a stochastic master equation for a measured record.

For the conditioned-state side, see Stochastic Master Equations and Quantum Filtering.

Non-Markovian Langevin Equations

The word “Langevin” is broader than the Markov input-field equation above. If a bath has memory, eliminating it can produce a retarded equation with a memory kernel and a colored noise force. For a coordinate-coupled oscillator bath, the schematic structure is

$$M\ddot q(t) + V'(q(t)) + \int_{t_0}^{t} ds\, \eta(t-s)\dot q(s) = \xi(t).$$

Here $\eta(t-s)$ is a damping kernel and $\xi(t)$ is a bath force operator. In thermal equilibrium, their correlations and commutators are constrained by the fluctuation–dissipation relation. In a Markov limit, the kernel becomes sharply localized and the noise becomes effectively white over the relevant bandwidth.

The canonical oscillator-bath model and its kernels are treated in Caldeira–Leggett Model. The purpose of the present page is the Markovian Heisenberg-noise structure and its simplest oscillator example.

Assumptions

The compact Markov quantum Langevin equation assumes:

• a continuum input channel with short correlation time;
• weak, frequency-smooth coupling over the relevant system bandwidth;
• a rotating-wave coupling for the displayed bosonic input form;
• initially specified input-field states;
• no unresolved propagation delay or coherent feedback loop;
• independent channels when multiple inputs are used;
• consistent sign and normalization conventions for $L$, $b_{\mathrm{in}}$, and $b_{\mathrm{out}}$.

When these assumptions fail, a Langevin description may still exist, but it can require colored noise, memory kernels, explicit bath modes, delay equations, or non-Markovian numerical methods.

Common Mistakes

• Dropping the input noise while keeping damping.
• Treating $b_{\mathrm{in}}(t)$ as an ordinary function rather than an operator-valued distribution.
• Confusing the system operator $a(t)$ with the traveling field $b_{\mathrm{out}}(t)$.
• Mixing sign conventions for $b_{\mathrm{out}}=b_{\mathrm{in}}\pm L$.
• Forgetting that $\kappa$ is usually an energy-decay rate, while the field amplitude decays at $\kappa/2$.
• Using symmetrized noise spectra when a normally ordered output quantity is being computed.
• Assuming a Markov Langevin equation remains valid for structured reservoirs, long delay lines, or strong coupling.
• Treating a single Langevin equation as a single classical trajectory rather than an operator equation whose measured versions require a detector model.

Cross-Links

• Quantum Noise for vacuum, thermal, classical, and technical noise.
• Correlation Functions for time-domain bath correlations.
• Noise Spectra for the frequency-domain version of the same noise.
• Fluctuation–Dissipation Relation for equilibrium constraints between noise and damping.
• Thermal and Vacuum Noise for input occupation factors and vacuum limits.
• Input–Output Theory for the boundary relation between incident and emitted fields.
• Caldeira–Leggett Model for oscillator-bath memory kernels and Brownian limits.
• Quantum Brownian Motion for damping, diffusion, decoherence, and phase-space forms.
• Quantum Optical Master Equation for the density-matrix counterpart in optical systems.
• Amplitude Damping Master Equation for the two-level loss-channel version.
• Homodyne Detection and Photon Counting for measured output fields.
• Approximation Checklist for Markov, thermal, secular, and positivity checks.

References

• C. W. Gardiner and M. J. Collett, “Input and output in damped quantum systems: Quantum stochastic differential equations and the master equation,” Physical Review A 31, 3761–3774 (1985).
• C. W. Gardiner and P. Zoller, Quantum Noise, 3rd ed., Springer (2004).
• D. F. Walls and G. J. Milburn, Quantum Optics, 2nd ed., Springer (2008).
• H. J. Carmichael, An Open Systems Approach to Quantum Optics, Springer (1993).
• H. M. Wiseman and G. J. Milburn, Quantum Measurement and Control, Cambridge University Press (2010).
• A. A. Clerk, M. H. Devoret, S. M. Girvin, F. Marquardt, and R. J. Schoelkopf, “Introduction to quantum noise, measurement, and amplification,” Reviews of Modern Physics 82, 1155–1208 (2010).
• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002).

Exercises

Recover the oscillator equation

Starting from the general one-channel Langevin equation, set $L=\sqrt{\kappa}a$ and $X=a$. Derive the damped oscillator equation.

Solution

The relevant commutators are

$$[a,L]=[a,\sqrt{\kappa}a]=0,$$

and

$$[L^\dagger,a] = [\sqrt{\kappa}a^\dagger,a] = -\sqrt{\kappa}.$$

The dissipative drift becomes

$$\frac12 L^\dagger[a,L] + \frac12[L^\dagger,a]L = 0 - \frac{\kappa}{2}a.$$

The noise terms give

$$b_{\mathrm{in}}^\dagger(t)[a,L] + [L^\dagger,a]b_{\mathrm{in}}(t) = -\sqrt{\kappa}\,b_{\mathrm{in}}(t).$$

Including the Hamiltonian part,

$$\dot a = i[H/\hbar,a] - \frac{\kappa}{2}a - \sqrt{\kappa}\,b_{\mathrm{in}}(t).$$

Preserve the commutator

Use the formal solution of the damped oscillator to show that $[a(t),a^\dagger(t)]=1$ for vacuum or thermal Markov input.

Solution

The preservation uses the input-field commutator, not the input-field state. With

$$a(t) = e^{-\lambda T}a(t_0) - \sqrt{\kappa} \int_{t_0}^{t} ds\, e^{-\lambda(t-s)} b_{\mathrm{in}}(s),$$

where $T=t-t_0$ and $\lambda=i\omega_c+\kappa/2$, the initial oscillator contributes $e^{-\kappa T}$. The input field contributes

$$\kappa \int_{t_0}^{t} ds \int_{t_0}^{t} ds'\, e^{-\lambda(t-s)} e^{-\lambda^*(t-s')} \delta(s-s') = 1-e^{-\kappa T}.$$

Adding both pieces gives $[a(t),a^\dagger(t)]=1$.

Thermal steady occupation

Given

$$\frac{d}{dt} \langle a^\dagger a\rangle = -\kappa \left( \langle a^\dagger a\rangle-\bar n \right),$$

find the steady-state occupation and relaxation time.

Solution

Set the time derivative to zero:

$$0 = -\kappa \left( \langle a^\dagger a\rangle_{\mathrm{ss}}-\bar n \right).$$

Thus

$$\langle a^\dagger a\rangle_{\mathrm{ss}} = \bar n.$$

The deviation from steady state obeys

$$\frac{d}{dt} \left( \langle a^\dagger a\rangle-\bar n \right) = -\kappa \left( \langle a^\dagger a\rangle-\bar n \right),$$

so the occupation relaxes on the timescale $1/\kappa$.

Why damping alone fails

Suppose one writes $\dot a=-(i\omega_c+\kappa/2)a$ without the input noise. What happens to $[a(t),a^\dagger(t)]$?

Solution

The solution is

$$a(t) = e^{-(i\omega_c+\kappa/2)(t-t_0)} a(t_0).$$

Therefore

$$[a(t),a^\dagger(t)] = e^{-\kappa(t-t_0)} [a(t_0),a^\dagger(t_0)] = e^{-\kappa(t-t_0)}.$$

The canonical commutator would decay, which is impossible for a valid oscillator operator. The missing input noise supplies the commutator needed to keep $[a,a^\dagger]=1$.