Lindblad equation

Remarks

Example: let $H = \C^2$ with basis $|0\rangle, |1\rangle$, let $K = 0$, $m = 1$, $L_1 = |0\rangle\langle 1|$ and $\gamma_1 = \gamma > 0$. Then $L_1 \rho L_1^* = \rho_{11} |0\rangle\langle 0|$ and $L_1^* L_1 = |1\rangle\langle 1|$, so the $(1,1)$ entry of $\mathcal{L}(\rho)$ is $-\gamma \rho_{11}$. Hence $\rho_{11}(t) = e^{-\gamma t} \rho_{11}(0)$: the excited level decays at rate $\gamma$.

Non-example: the equation $\frac{d\rho}{dt} = -\gamma \rho$ with $\gamma > 0$ is not a Lindblad equation. Every Lindbladian satisfies $\tr \mathcal{L}(\rho) = 0$, by cyclicity of the trace, so the Lindblad equation preserves the trace. The solution $e^{-\gamma t} \rho$ loses trace, so no choice of $K$, $L_k$ and $\gamma_k$ gives this map.

The rates can be absorbed into the jump operators by replacing $L_k$ by $\sqrt{\gamma_k} L_k$. The equation is also called the GKSL equation, after Gorini, Kossakowski, Sudarshan and Lindblad, who found it independently in 1976. Their theorem says that a linear map $\mathcal{L}$ generates a semigroup $e^{t\mathcal{L}}$, $t \ge 0$, of completely positive trace-preserving maps if and only if $\mathcal{L}$ has the form above.