The Lindblad equation

Intuition

When the environment forgets quickly — each record is carried off and replaced by fresh surroundings faster than anything the system does — the channel over a short time $dt$ depends only on the present state, and the density matrix obeys a differential equation of its own: the Lindblad equation. Its first term is the von Neumann commutator; the others, one for each jump operator $\hat{L}_{m}$ , describe jumps and the gradual loss of amplitude in not jumping. The lesson derives it from Kraus operators whose jumps have amplitudes of order $\sqrt{dt}$ , so that their probabilities grow in proportion to $dt$ . It keeps the trace and positivity; decay and dephasing are its two simplest cases.

A drunkard’s walk seen from far away is smooth diffusion: each step is random, but steps are so short and so many that only their average shows. The Lindblad equation is the smooth law left when the environment’s many quick kicks are averaged.

The Bloch vector of a decaying qubit started in $\ket{+x}$ , with the ground state at the top: $x=e^{-\gamma t/2}$ and $z=1-e^{-\gamma t}$ , so it moves along through the inside of the ball — mixed on the way — to the pure ground state.

For a Markovian environment, with jump operators $\hat{L}_{m}$ and $\{\hat{A},\hat{B}\}=\hat{A}\hat{B}+\hat{B}\hat{A}$ :

\begin{aligned}\frac{d\hat{\rho}}{dt}&=-\frac{i}{\hbar}\bigl[\hat{H},\hat{\rho}\bigr]\\ &\quad+\sum_{m}\Bigl(\hat{L}_{m}\hat{\rho}\hat{L}_{m}^{\dagger}-\tfrac12\bigl\{\hat{L}_{m}^{\dagger}\hat{L}_{m},\hat{\rho}\bigr\}\Bigr)\end{aligned}

Properties

Markovian: the environment relaxes much faster than the system changes, so the future depends only on the present $\hat{\rho}$ — the hierarchy $\tau_{\text{relax}}\ll dt\ll\tau$ .

The Lindblad equation from Kraus operators

\begin{aligned}&\hat{M}_{0}=\hat{1}-\Bigl(\frac{i}{\hbar}\hat{H}+\tfrac12\sum_{m}\hat{L}_{m}^{\dagger}\hat{L}_{m}\Bigr)dt\\ &\hat{M}_{m}=\hat{L}_{m}\sqrt{dt}\ \Longrightarrow\ \text{the Lindblad equation}\end{aligned}

Over a short step the environment either records nothing or records a jump. The no-jump operator is the identity minus terms of order $dt$ ; a jump has amplitude of order $\sqrt{dt}$ , so its probability is of order $dt$ . The requirement that the probabilities add to one fixes the anti-Hermitian part of the no-jump operator. Apply one step of the channel, keep first order in , and divide.

Proof steps

\begin{aligned}\hat{M}_{0}&=\hat{1}-\frac{i}{\hbar}\hat{H}\,dt-\tfrac12\sum_{m}\hat{L}_{m}^{\dagger}\hat{L}_{m}\,dt\\ \hat{M}_{m}&=\hat{L}_{m}\sqrt{dt}\end{aligned}

No jump, or jump $m$ , over a short step.

\hat{M}_{0}^{\dagger}\hat{M}_{0}+\sum_{m}\hat{M}_{m}^{\dagger}\hat{M}_{m}=\hat{1}+O(dt^{2})

The term $-\tfrac12\sum\hat{L}^{\dagger}\hat{L}\,dt$ is exactly what makes the probabilities add to one.

\hat{\rho}(t+dt)=\hat{M}_{0}\hat{\rho}\hat{M}_{0}^{\dagger}+\sum_{m}\hat{M}_{m}\hat{\rho}\hat{M}_{m}^{\dagger}

One step of the channel.

\begin{aligned}&\hat{\rho}(t+dt)-\hat{\rho}(t)=-\frac{i}{\hbar}[\hat{H},\hat{\rho}]\,dt\\ &\quad+\sum_{m}\Bigl(\hat{L}_{m}\hat{\rho}\hat{L}_{m}^{\dagger}-\tfrac12\{\hat{L}_{m}^{\dagger}\hat{L}_{m},\hat{\rho}\}\Bigr)dt\end{aligned}

Keep the terms of first order in $dt$ .

\begin{aligned}\frac{d\hat{\rho}}{dt}&=-\frac{i}{\hbar}[\hat{H},\hat{\rho}]\\ &\quad+\sum_{m}\Bigl(\hat{L}_{m}\hat{\rho}\hat{L}_{m}^{\dagger}-\tfrac12\{\hat{L}_{m}^{\dagger}\hat{L}_{m},\hat{\rho}\}\Bigr)\end{aligned}

Divide by $dt$ .

Applications

Practice

Two Kinds of Term

The commutator term is the system’s own evolution, as in the von Neumann equation. The other terms are the environment’s: jumps, and the loss of amplitude that goes with not jumping.

Try it

What does the first term of the Lindblad equation, $-\frac{i}{\hbar}[\hat{H},\hat{\rho}]$ , describe?