Pauli Rate Equations

Pauli rate equations are classical master equations for the populations of a quantum system in a preferred basis. They appear when coherences either decay quickly, decouple by a secular approximation, or are intentionally outside the prediction target.

If $p_n(t)$ is the probability of occupying state $\lvert n\rangle$, the standard form is

$$\dot p_n = \sum_{m\ne n} \left( W_{n\leftarrow m}p_m - W_{m\leftarrow n}p_n \right), \qquad W_{n\leftarrow m}\ge0.$$

Here $W_{n\leftarrow m}$ is the transition rate from $m$ to $n$. The first term is gain into state $n$ from other states. The second term is loss out of state $n$ into other states.

The equation is quantum in origin when the rates come from matrix elements, bath spectra, tunneling amplitudes, or measurement-induced transitions. But the object being evolved is classical: a probability vector, not the full density operator.

Rate-Matrix Convention

Define a column probability vector

$$\mathbf p = (p_1,\ldots,p_N)^{\mathsf T}.$$

The rate equation can be written

$$\dot{\mathbf p} = K\mathbf p,$$

where, for $n\ne m$,

$$K_{nm} = W_{n\leftarrow m},$$

and the diagonal entries are

$$K_{nn} = - \sum_{m\ne n} W_{m\leftarrow n}.$$

With this column convention,

$$\sum_n K_{nm}=0$$

for every source state $m$. Therefore the total probability is conserved:

$$\frac{d}{dt} \sum_n p_n = \sum_{n,m}K_{nm}p_m = 0.$$

If $p_n(0)\ge0$ and $\sum_n p_n(0)=1$, then the finite-time map

$$\mathbf p(t) = e^{Kt}\mathbf p(0), \qquad t\ge0,$$

is a stochastic map under the usual finite-dimensional Markov assumptions. In this sense, Pauli rate equations are continuous-time Markov jump processes written in the language of quantum populations.

From a Lindblad Equation

A simple Lindblad embedding uses jump operators

$$L_{n\leftarrow m} = \sqrt{W_{n\leftarrow m}}\, \lvert n\rangle\langle m\rvert, \qquad n\ne m.$$

The density operator obeys

$$\dot\rho = - \frac{i}{\hbar}[H,\rho] + \sum_{n\ne m} \mathcal D[L_{n\leftarrow m}]\rho,$$

with

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

If $H$ is diagonal in the $\{\lvert n\rangle\}$ basis, the diagonal entries $p_n=\rho_{nn}$ obey the Pauli rate equation.

To see this, note that

$$L_{n\leftarrow m}\rho L_{n\leftarrow m}^\dagger = W_{n\leftarrow m} \rho_{mm} \lvert n\rangle\langle n\rvert.$$

This term adds population to $n$ when the system was in $m$. The anticommutator part removes population from the source state $m$. Taking the $nn$ matrix element gives

$$\dot\rho_{nn} = \sum_{m\ne n} W_{n\leftarrow m}\rho_{mm} - \rho_{nn} \sum_{m\ne n} W_{m\leftarrow n}.$$

That is exactly

$$\dot p_n = \sum_{m\ne n} \left( W_{n\leftarrow m}p_m - W_{m\leftarrow n}p_n \right).$$

What Happens to Coherences

The same jump embedding also predicts coherence decay. For $n\ne m$, define the total rate out of state $n$ by

$$\Gamma_n^{\mathrm{out}} = \sum_{r\ne n} W_{r\leftarrow n}.$$

For a diagonal Hamiltonian,

$$H = \sum_n E_n \lvert n\rangle\langle n\rvert,$$

the off-diagonal entry $\rho_{nm}$ obeys

$$\dot\rho_{nm} = - i\omega_{nm}\rho_{nm} - \frac12 \left( \Gamma_n^{\mathrm{out}} + \Gamma_m^{\mathrm{out}} \right) \rho_{nm},$$

where

$$\omega_{nm} = \frac{E_n-E_m}{\hbar}.$$

Additional dephasing channels can damp $\rho_{nm}$ faster. The important lesson is that a population rate equation does not uniquely specify the coherence dynamics. Many different quantum generators can produce the same equation for $\{p_n\}$ while differing on off-diagonal density-matrix elements.

Therefore a Pauli rate equation is enough only when the prediction target is population dynamics. It is not enough for Ramsey fringes, quantum beats, interference, coherent transport, entanglement, or measurement backaction.

Weak-Coupling Origin

In a weak-coupling open-system derivation, one often starts from

$$H = H_S+H_B+H_I, \qquad H_I = \sum_\alpha A_\alpha\otimes B_\alpha.$$

Let

$$H_S = \sum_n E_n \lvert n\rangle\langle n\rvert$$

for simplicity. After Born, Markov, and secular approximations, the master equation separates into Bohr-frequency blocks. Populations in the energy basis then obey a Pauli equation.

For a transition $m\to n$, define

$$\omega_{mn} = \frac{E_m-E_n}{\hbar}.$$

With the convention used in Secular Approximation, positive $\omega_{mn}$ means the system loses energy. A typical weak-coupling rate has the schematic form

$$W_{n\leftarrow m} = \sum_{\alpha,\beta} \gamma_{\alpha\beta}(\omega_{mn}) \langle n\rvert A_\beta\lvert m\rangle \langle m\rvert A_\alpha\lvert n\rangle,$$

where $\gamma_{\alpha\beta}(\omega)$ is built from bath correlation spectra. The exact index order depends on Fourier-transform and coupling-operator conventions, so the reliable rule is: rates sample the bath spectrum at the transition frequency and are weighted by system matrix elements connecting the two levels.

The secular approximation is what makes the population block close. A nonsecular Redfield Equation can couple populations to coherences, especially near degeneracies or in systems with coherent transport.

Detailed Balance

If the bath is a single equilibrium reservoir and the weak-coupling thermal assumptions hold, the rates satisfy detailed balance with respect to the Gibbs distribution

$$p_n^\beta = \frac{e^{-\beta E_n}}{Z}.$$

The population detailed-balance condition is

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

Equivalently,

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

Transitions that raise the system energy are Boltzmann suppressed relative to the reverse transition. This is the population-level form of the thermal consistency discussed in Detailed Balance and Thermal Master Equations.

For nonequilibrium reservoirs, such as biased leads, driven baths, or multiple temperatures, detailed balance may be replaced by local balance relations or may fail entirely. The rate equation can still be a valid Markov model, but its steady state is then generally not a Gibbs state.

Two-State Example

Let $\lvert g\rangle$ and $\lvert e\rangle$ be ground and excited states. Write

$$\Gamma_\downarrow = W_{g\leftarrow e}, \qquad \Gamma_\uparrow = W_{e\leftarrow g}.$$

The excited-state population obeys

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

Equivalently,

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

The steady state is

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

The nonzero relaxation eigenvalue is

$$\lambda_1 = - (\Gamma_\downarrow+\Gamma_\uparrow),$$

so the population relaxation time is

$$\tau_{\mathrm{rel}} = \frac{1} {\Gamma_\downarrow+\Gamma_\uparrow}.$$

At zero temperature, $\Gamma_\uparrow=0$ and the steady state is the ground state. At finite temperature, detailed balance gives

$$\frac{\Gamma_\uparrow} {\Gamma_\downarrow} = e^{-\beta(E_e-E_g)}.$$

Multistate Structure

For many states, it is helpful to think of a directed graph:

• vertices are basis states;
• directed edges are nonzero rates $W_{n\leftarrow m}$;
• total outgoing rates set residence times;
• closed communicating classes determine long-time support;
• absorbing states are states or subspaces with no outgoing transitions.

If the graph is irreducible and finite, the rate matrix has a unique stationary distribution with strictly positive entries. If the graph has disconnected components or absorbing states, there can be multiple stationary distributions or memory of the initial component.

This graph language is often the fastest way to diagnose a rate equation before doing algebra. It also explains why a population equation can have multiple fixed points even when every listed transition rate is nonnegative.

Relation to Steady States and Liouvillian Modes

The rate matrix $K$ is the population-sector generator. Its zero eigenvectors are stationary population distributions:

$$K\mathbf p_{\mathrm{ss}}=0.$$

Nonzero eigenvalues determine relaxation modes:

$$\mathbf p(t) = \mathbf p_{\mathrm{ss}} + \sum_a c_a e^{\lambda_a t}\mathbf v_a.$$

For a stable finite Markov process, the nonzero eigenvalues have negative real parts. Complex eigenvalues can appear in irreversible cyclic networks, producing damped oscillatory population modes even though no quantum coherence is being tracked.

In a full Lindblad model, these population modes are part of the Liouvillian spectrum. The coherence modes may decay on different time scales. The general language of fixed points, gaps, and metastability is collected in Steady States and Relaxation, and numerical checks are described in Solving Lindblad Equations.

Common Uses

Pauli rate equations are useful when the basis states are long-lived alternatives and transitions are incoherent on the time scale of interest. Typical uses include:

• spontaneous emission and thermal excitation when only populations are observed;
• sequential tunneling through Coulomb-blockaded charge states;
• incoherent hopping between localized states;
• laser and pumping models where coherences have been adiabatically eliminated;
• chemical and molecular population transfer after fast dephasing;
• coarse-grained measurement records that count jumps between classical outcomes.

For mesoscopic conductors, the same structure appears with reservoir occupation factors and chemical potentials; see Mesoscopic Transport.

Common Mistakes

Treating population dynamics as the full quantum state

A rate equation does not predict phases or coherences. If an observable has off-diagonal matrix elements in the chosen basis, the rate equation is not enough.

Using rates without a basis

The variables $p_n$ are populations in a declared basis. Energy eigenstates, localized sites, charge states, dressed states, and measurement pointer states can give different rate equations.

Secularizing through near degeneracies

When two Bohr frequencies are separated by a scale comparable to the relaxation rate, coherences may affect populations. Blindly dropping them can remove quantum beats, dark states, or interference-assisted transport.

Calling every steady state thermal

A rate equation can have a stationary distribution without satisfying detailed balance. Equilibrium thermal interpretation requires the appropriate Gibbs or grand-canonical balance relations.

Forgetting that rates may be time dependent

The matrix $K$ above is time independent. Driven systems, quenches, feedback, or aging reservoirs may require $K(t)$, and then the simple semigroup formula $e^{Kt}$ no longer applies.

Inferring a unique Lindblad equation from populations alone

The same $W_{n\leftarrow m}$ can be embedded in different quantum generators with different dephasing, Hamiltonian, and unraveling structure. Population data alone usually do not identify the full open-system model.

Exercises

1. Trace conservation. Starting from

$$\dot p_n = \sum_{m\ne n} \left( W_{n\leftarrow m}p_m - W_{m\leftarrow n}p_n \right),$$

show that $\sum_n p_n$ is constant.

Solution

Sum over $n$:

$$\sum_n\dot p_n = \sum_{n,m:n\ne m} W_{n\leftarrow m}p_m - \sum_{n,m:n\ne m} W_{m\leftarrow n}p_n.$$

In the second double sum, exchange the dummy labels $m$ and $n$. It becomes

$$\sum_{m,n:m\ne n} W_{n\leftarrow m}p_m,$$

which is identical to the first double sum. The difference is zero, so total probability is conserved.

2. Two-state solution. Solve

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

for $p_e(t)$.

Solution

The steady state is

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

Subtract it:

$$\frac{d}{dt} \left( p_e-p_e^{\mathrm{ss}} \right) = - (\Gamma_\downarrow+\Gamma_\uparrow) \left( p_e-p_e^{\mathrm{ss}} \right).$$

Thus

$$p_e(t) = p_e^{\mathrm{ss}} + \left[ p_e(0)-p_e^{\mathrm{ss}} \right] e^{-(\Gamma_\downarrow+\Gamma_\uparrow)t}.$$

3. Gibbs detailed balance. Suppose

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

for every connected pair. Show that $p_n^\beta\propto e^{-\beta E_n}$ is stationary.

Solution

For the Gibbs distribution,

$$\frac{p_n^\beta}{p_m^\beta} = e^{-\beta(E_n-E_m)}.$$

The assumed rate relation gives

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

Therefore each gain term is paired with an equal loss term. In the rate equation for each $n$, all terms cancel pairwise, so $\dot p_n=0$.

4. Lindblad embedding. For

$$L_{n\leftarrow m} = \sqrt{W_{n\leftarrow m}}\, \lvert n\rangle\langle m\rvert,$$

show that the jump term contributes $W_{n\leftarrow m}\rho_{mm}$ to $\dot\rho_{nn}$.

Solution

Compute

$$L_{n\leftarrow m}\rho L_{n\leftarrow m}^\dagger = W_{n\leftarrow m} \lvert n\rangle\langle m\rvert \rho \lvert m\rangle\langle n\rvert.$$

Since

$$\langle m\rvert\rho\lvert m\rangle = \rho_{mm},$$

this equals

$$W_{n\leftarrow m}\rho_{mm} \lvert n\rangle\langle n\rvert.$$

Taking the $nn$ matrix element gives $W_{n\leftarrow m}\rho_{mm}$.

5. Same populations, different coherences. Add a pure-dephasing term

$$\gamma_\phi\mathcal D[ \lvert n\rangle\langle n\rvert ]\rho$$

to the Lindblad embedding for one state $n$. Does it change the Pauli rate equation for the populations?

Solution

No. A diagonal projector dephasing term leaves all diagonal density-matrix entries unchanged, so it does not alter any $p_k=\rho_{kk}$ equation.

It does, however, change coherences involving $\lvert n\rangle$. This illustrates why the population rate equation does not uniquely determine the full quantum master equation.

References

• N. G. van Kampen, Stochastic Processes in Physics and Chemistry, 3rd ed., North-Holland (2007).
• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002).
• C. W. Gardiner and P. Zoller, Quantum Noise, 3rd ed., Springer (2004).
• R. Alicki and K. Lendi, Quantum Dynamical Semigroups and Applications, Springer (2007).
• C. W. Gardiner, Handbook of Stochastic Methods, 3rd ed., Springer (2004).