Detailed Balance

Detailed balance is the equilibrium consistency condition that relates forward and reverse transition rates. In open quantum systems it explains why a single thermal bath tends to drive an undriven weakly coupled system toward a Gibbs state, and why upward thermal transitions are suppressed at low temperature.

For a two-level system with transition frequency $\omega_0>0$, the most familiar form is

$$\frac{\Gamma_\uparrow}{\Gamma_\downarrow} = e^{-\beta\hbar\omega_0}.$$

The downward rate $\Gamma_\downarrow$ describes relaxation from excited to ground state. The upward rate $\Gamma_\uparrow$ describes thermal excitation. Their ratio is fixed by the bath temperature if the bath is a single equilibrium reservoir and the weak-coupling Markovian assumptions apply.

Detailed balance is stronger than merely having some steady state. It is a statement of equilibrium reversibility: in equilibrium, every elementary transition is balanced by its reverse transition with the correct Gibbs weights.

Helpful background. Bohr-frequency components identify upward and downward transitions; Pauli rate equations describe their populations. Work first in finite dimension at finite temperature; zero temperature is a limit.

Classical rate balance and probability currents

For a classical Markov process with states $m,n$ and transition rates $W_{m\leftarrow n}$, detailed balance with respect to an equilibrium distribution $p_n^{\mathrm{eq}}$ means

$$W_{m\leftarrow n}p_n^{\mathrm{eq}} = W_{n\leftarrow m}p_m^{\mathrm{eq}}.$$

If

$$p_n^{\mathrm{eq}} = \frac{e^{-\beta E_n}}{Z},$$

then, for a pair with nonzero reverse rate,

$$\frac{W_{m\leftarrow n}}{W_{n\leftarrow m}} = e^{-\beta(E_m-E_n)}.$$

This equation says that transitions raising the system energy are Boltzmann suppressed relative to the reverse transitions.

Quantum detailed balance reduces to this statement for populations when the population sector closes, as in a nondegenerate secular energy basis. Degenerate sectors can retain coupled coherences. The corresponding population dynamics is treated in Pauli Rate Equations. The full quantum version also tracks coherences and operator structure.

Qubit Upward and Downward Rates

Let the ground and excited states be $\lvert g\rangle$ and $\lvert e\rangle$, with

$$E_e-E_g=\hbar\omega_0, \qquad \omega_0>0.$$

A finite-temperature two-level master equation has

$$\frac{d\rho}{dt} = \Gamma_\downarrow\mathcal D[\sigma_-]\rho + \Gamma_\uparrow\mathcal D[\sigma_+]\rho,$$

where

$$\sigma_-=\lvert g\rangle\langle e\rvert, \qquad \sigma_+=\lvert e\rangle\langle g\rvert.$$

The excited-state population obeys

$$\dot p_e = -\Gamma_\downarrow p_e + \Gamma_\uparrow p_g, \qquad p_g=1-p_e.$$

At stationarity,

$$\Gamma_\downarrow p_e^{\mathrm{ss}} = \Gamma_\uparrow p_g^{\mathrm{ss}},$$

so

$$\frac{p_e^{\mathrm{ss}}}{p_g^{\mathrm{ss}}} = \frac{\Gamma_\uparrow}{\Gamma_\downarrow}.$$

For a thermal steady state,

$$\frac{p_e^{\beta}}{p_g^{\beta}} = e^{-\beta\hbar\omega_0}.$$

Therefore detailed balance requires

$$\frac{\Gamma_\uparrow}{\Gamma_\downarrow} = e^{-\beta\hbar\omega_0}.$$

At zero temperature, $\beta\to\infty$, this gives $\Gamma_\uparrow/\Gamma_\downarrow\to0$: an ordinary passive bath cannot thermally excite the system.

Bohr-Frequency Convention

In the weak-coupling construction used in this volume, system coupling operators are decomposed as

$$A_\alpha(\omega) = \sum_{\epsilon'-\epsilon=\hbar\omega} \Pi(\epsilon)A_\alpha\Pi(\epsilon').$$

They satisfy

$$[H_S,A_\alpha(\omega)] = -\hbar\omega A_\alpha(\omega).$$

Thus positive $\omega$ labels an operator that lowers the system energy by $\hbar\omega$. It is associated with emission or downward relaxation. For Hermitian coupling operators $A_\alpha$, the reverse operator is

$$A_\alpha^\dagger(\omega) = A_\alpha(-\omega),$$

which raises the system energy.

With this convention, a thermal bath spectrum satisfies the single-operator relation

$$S(-\omega) = e^{-\beta\hbar\omega}S(\omega), \qquad \omega>0.$$

The upward rate samples the negative-frequency spectrum and is Boltzmann suppressed relative to the downward rate.

Fourier-transform signs differ across books. Before comparing formulas, check whether the source uses $e^{i\omega t}$ or $e^{-i\omega t}$ and whether positive $\omega$ labels absorption by the system or emission into the bath.

KMS Condition

For a bath in thermal equilibrium,

$$\rho_B = \frac{e^{-\beta H_B}}{Z_B},$$

correlation functions obey the Kubo–Martin–Schwinger condition. Assume a well-defined equilibrium state and correlators (a finite-volume Gibbs state or the appropriate reservoir limit). For Hermitian bath operators $B_\alpha(t)$ and $B_\beta(0)$, define

$$C_{\alpha\beta}(t) = \operatorname{Tr}_B \left[ B_\alpha(t)B_\beta(0)\rho_B \right].$$

With the spectrum convention

$$\Gamma_{\alpha\beta}(\omega) = \int_{-\infty}^{\infty} dt\, e^{i\omega t} C_{\alpha\beta}(t),$$

the KMS relation gives exactly

$$\Gamma_{\alpha\beta}(-\omega) = e^{-\beta\hbar\omega} \Gamma_{\beta\alpha}(\omega),$$

for the stated definitions. No extra complex conjugation is needed beyond the exchanged indices. These $\Gamma$ are spectra; if $B_\alpha$ has energy units, rate matrices include the coupling factor $\lambda^2/\hbar^2$.

For one Hermitian bath operator this becomes the common detailed-balance spectrum relation:

$$S(-\omega) = e^{-\beta\hbar\omega}S(\omega).$$

This is the microscopic origin of the qubit rate ratio.

Gibbs Stationarity

For a system Hamiltonian

$$H_S = \sum_n E_n\Pi_n,$$

the Gibbs state is

$$\rho_\beta = \frac{e^{-\beta H_S}}{\operatorname{Tr}(e^{-\beta H_S})}.$$

A thermal weak-coupling generator should satisfy

$$\mathcal L(\rho_\beta)=0$$

for the standard Davies weak-coupling generator. Its Lamb-shift Hamiltonian commutes with $H_S$, so it does not spoil the bare Gibbs stationary state. Replacing $H_S$ by $H_S+H_{\mathrm{LS}}$ in the Gibbs weights is not automatically consistent with rates evaluated at the bare gaps. Corrected equilibrium beyond that approximation requires a consistent higher-order calculation.

Detailed balance is stronger than stationarity. One common quantum definition requires the dissipative part of the Heisenberg generator to be self-adjoint in the thermal inner product

$$\langle A,B\rangle_\beta=\operatorname{Tr}(\rho_\beta A^\dagger B),$$

after separating the Hamiltonian commutator, with the Hamiltonian commuting with $\rho_\beta$. Other definitions use a symmetrically weighted inner product or include time reversal; state which definition is meant when comparing theorems. The weak-coupling thermal construction supplies more structure than merely solving $\mathcal L(\rho_\beta)=0$.

For a simple counterexample to the converse, take three degenerate levels and only clockwise jumps $\sqrt k|n+1\!\pmod3\rangle\langle n|$. The uniform Gibbs state $I/3$ is stationary because each level gains and loses $k/3$. Yet every edge has a clockwise probability current $k/3$ and zero reverse flow. A thermal stationary distribution alone therefore does not certify equilibrium dynamics.

Harmonic Oscillator Bath Example

For a harmonic oscillator coupled to a thermal reservoir, a common Lindblad equation is

$$\frac{d\rho}{dt} = - \frac{i}{\hbar}[\hbar\omega a^\dagger a,\rho] + \kappa(\bar n+1)\mathcal D[a]\rho + \kappa\bar n\,\mathcal D[a^\dagger]\rho.$$

The downward rate is proportional to $\bar n+1$: stimulated plus spontaneous emission. The upward rate is proportional to $\bar n$: thermal absorption.

For a Bose occupation

$$\bar n = \frac{1}{e^{\beta\hbar\omega}-1},$$

the ratio is

$$\frac{\kappa\bar n}{\kappa(\bar n+1)} = e^{-\beta\hbar\omega}.$$

The actual rates are $W_{n-1\leftarrow n}=\kappa(\bar n+1)n$ and $W_{n\leftarrow n-1}=\kappa\bar n n$; the same factor $n$ cancels for a reverse pair. It would be incorrect to compare downward and upward transitions leaving the same level, whose matrix elements differ.

Fluctuation–Dissipation Relation

Detailed balance is closely related to fluctuation–dissipation relations. For a stationary bath operator $B(t)$, define the symmetrized noise spectrum

$$S_{BB}^{\mathrm{sym}}(\omega) = \frac12 \int_{-\infty}^{\infty} dt\, e^{i\omega t} \langle \{B(t),B(0)\}\rangle$$

and the retarded susceptibility

$$\chi_{BB}^R(t) = \frac{i}{\hbar} \theta(t) \langle [B(t),B(0)]\rangle.$$

For a perturbation $H_{\mathrm{probe}}=-f(t)B$ and the same $e^{i\omega t}$ Fourier transform for the response, equilibrium gives

$$S_{BB}^{\mathrm{sym}}(\omega) = \hbar \coth\!\left( \frac{\beta\hbar\omega}{2} \right) \operatorname{Im}\chi_{BB}^R(\omega).$$

This relation says that equilibrium noise and linear response are not independent. The same KMS condition that fixes upward and downward transition rates also constrains the balance between fluctuations and dissipation.

For a broader discussion of ordered versus symmetrized bath spectra and their use in rates, see Noise Spectra. The Fluctuation–Dissipation Theorem derives the equilibrium response relation from KMS and the Lehmann representation; Fluctuation–Dissipation Relation applies it to bath noise and damping.

When Detailed Balance Fails

Detailed balance is not expected in every Markovian master equation.

It can fail for:

• multiple baths at different temperatures;
• driven systems in rotating frames;
• feedback-controlled dynamics;
• measurement-conditioned dynamics;
• nonthermal reservoirs;
• inverted media;
• approximate generators before secularization;
• strong coupling where the steady state is not the bare Gibbs state.

Failure of detailed balance does not automatically mean the equation is mathematically invalid. It means the model is not an equilibrium single-bath thermal relaxation model.

Exercises

Two-level steady state

For

$$\dot p_e = -\Gamma_\downarrow p_e + \Gamma_\uparrow(1-p_e),$$

find $p_e^{\mathrm{ss}}$ and the condition for a Gibbs steady state.

Solution

Set $\dot p_e=0$:

$$\Gamma_\downarrow p_e^{\mathrm{ss}} = \Gamma_\uparrow(1-p_e^{\mathrm{ss}}).$$

Thus

$$p_e^{\mathrm{ss}} = \frac{\Gamma_\uparrow} {\Gamma_\uparrow+\Gamma_\downarrow}.$$

The ratio of excited to ground populations is

$$\frac{p_e^{\mathrm{ss}}}{p_g^{\mathrm{ss}}} = \frac{\Gamma_\uparrow}{\Gamma_\downarrow}.$$

For a Gibbs state with spacing $\hbar\omega_0$,

$$\frac{p_e^\beta}{p_g^\beta} = e^{-\beta\hbar\omega_0},$$

so detailed balance requires

$$\frac{\Gamma_\uparrow}{\Gamma_\downarrow} = e^{-\beta\hbar\omega_0}.$$

Oscillator rate ratio

Show that for

$$\bar n=\frac{1}{e^{\beta\hbar\omega}-1},$$

one has

$$\frac{\bar n}{\bar n+1} = e^{-\beta\hbar\omega}.$$

Solution

Compute

$$\bar n+1 = \frac{1}{e^{\beta\hbar\omega}-1}+1 = \frac{e^{\beta\hbar\omega}} {e^{\beta\hbar\omega}-1}.$$

Therefore

$$\frac{\bar n}{\bar n+1} = \frac{1} {e^{\beta\hbar\omega}} = e^{-\beta\hbar\omega}.$$

KMS to upward suppression

Assume a single bath operator has spectrum satisfying

$$S(-\omega)=e^{-\beta\hbar\omega}S(\omega), \qquad \omega>0.$$

If $S(\omega_0)$ gives the downward rate and $S(-\omega_0)$ gives the upward rate, find $\Gamma_\uparrow/\Gamma_\downarrow$.

Solution

With the stated convention,

$$\Gamma_\downarrow\propto S(\omega_0), \qquad \Gamma_\uparrow\propto S(-\omega_0).$$

Therefore

$$\frac{\Gamma_\uparrow}{\Gamma_\downarrow} = \frac{S(-\omega_0)}{S(\omega_0)} = e^{-\beta\hbar\omega_0}.$$

Zero-temperature limit

What happens to $\Gamma_\uparrow/\Gamma_\downarrow$ as $T\to0$ for an ordinary thermal bath?

Solution

As $T\to0$, $\beta=1/(k_BT)\to\infty$. For $\omega_0>0$,

$$e^{-\beta\hbar\omega_0}\to0.$$

Thus

$$\frac{\Gamma_\uparrow}{\Gamma_\downarrow}\to0.$$

An ordinary zero-temperature bath can absorb energy but cannot thermally excite the system.

References

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• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002). Publisher.
• E. B. Davies, Quantum Theory of Open Systems, Academic Press (1976).
• C. W. Gardiner and P. Zoller, Quantum Noise, 3rd ed., Springer (2004). Publisher.
• R. Kubo, “Statistical-Mechanical Theory of Irreversible Processes. I. General Theory and Simple Applications to Magnetic and Conduction Problems,” Journal of the Physical Society of Japan 12, 570–586 (1957). DOI.
• P. C. Martin and J. Schwinger, “Theory of Many-Particle Systems. I,” Physical Review 115, 1342–1373 (1959). DOI.
• Á. Rivas and S. F. Huelga, Open Quantum Systems: An Introduction, Springer (2012). Author manuscript.