6.1: The Lindblad Equation

Next, we will derive the Lindblad equation, which is the direct extension of the Heisenberg equation for the density operator, i.e., the mixed state of a system. We have seen in Eq. (4.30) that formally, we can write the evolution of a density operator as a mathematical map $\mathscr{E}$, such that the density operator $\rho$ transforms into

$$\rho \rightarrow \rho^{\prime}=\mathscr{E}(\rho) \equiv \sum_{k} A_{k} \rho A_{k}^{\dagger},\tag{6.1}$$

where the $A_{k}$ are the Kraus operators. Requiring that $\rho^{\prime}$ is again a density operator $\left(\operatorname{Tr}\left(\rho^{\prime}\right)=1\right)$ leads to the restriction $\sum_{k} A_{k}^{\dagger} A_{k}=\mathbb{I}$.

We want to describe an infinitesimal evolution of $\rho$, in order to give the continuum evolution later on. We therefore have that

$$\rho^{\prime}=\rho+\delta \rho=\sum_{k} A_{k} \rho A_{k}^{\dagger}\tag{6.2}$$

Since $\delta \rho$ is very small, one of the Kraus operators must be close to the identity. Without loss of generality we choose this to be $A_{0}$, and then we can write

$$A_{0}=\mathbb{I}+\left(L_{0}-i K\right) \delta t \quad \text { and } \quad A_{k}=L_{k} \sqrt{\delta t},\tag{6.3}$$

where we introduced the Hermitian operators $L_{0}$ and $K$, and the remaining $L_{k}$ are not necessarily Hermitian. We could have written $A_{0}=\mathbb{I}+L_{0} \delta t$ and keep $L_{0}$ general (non-Hermitian as well), but it will be useful later on to explicitly decompose it into Hermitian parts. We can now write

$$\begin{aligned} &A_{0} \rho A_{0}^{\dagger}=\rho+\left[\left(L_{0}-i K\right) \rho+\rho\left(L_{0}+i K\right)\right] \delta t+O\left(\delta t^{2}\right) \\ &A_{k} \rho A_{k}^{\dagger}=L_{k} \rho L_{k}^{\dagger} \delta t \end{aligned}.\tag{6.4}$$

We can substitute this into Eq. (6.2), to obtain up to first order in $\delta t$

$$\delta \rho=\left[\left(L_{0} \rho+\rho L_{0}\right)-i(K \rho-\rho K)+\sum_{k \neq 0} L_{k} \rho L_{k}^{\dagger}\right] \delta t.\tag{6.5}$$

We now give the continuum evolution by dividing by $\delta t$ and taking the limit $\delta t \rightarrow d t$:

$$\frac{d \rho}{d t}=-i[K, \rho]+\left\{L_{0}, \rho\right\}+\sum_{k \neq 0} L_{k} \rho L_{k}^{\dagger},\tag{6.6}$$

where $\{A, B\}=A B+B A$ is the anti-commutator of $A$ and $B$. We are almost there, but we must determine what the different terms mean. Suppose we consider the free evolution of the system.

Eq. (6.6) must then reduce to the Heisenberg equation for the density operator $\rho$ in Eq. (4.7), and we see that all $L_{k}$ including $L_{0}$ are zero, and $K$ is proportional to the Hamiltonian $K=H / \hbar$. Again from the general property that $\operatorname{Tr}(\rho)=1$ we have

$$\operatorname{Tr}\left(\frac{d \rho}{d t}\right)=0 \rightarrow L_{0}=-\frac{1}{2} \sum_{k \neq 0} L_{k}^{\dagger} L_{k}.\tag{6.7}$$

This finally leads to the Lindblad equation

$$\frac{d \rho}{d t}=\frac{1}{i \hbar}[H, \rho]+\frac{1}{2} \sum_{k}\left(2 L_{k} \rho L_{k}^{\dagger}-\left\{L_{k}^{\dagger} L_{k}, \rho\right\}\right).\tag{6.8}$$

The operators $L_{k}$ are chosen such that they model the relevant physical processes. This may sound vague, but in practice it will be quite clear. For example, modelling a transition $|1\rangle \rightarrow|0\rangle$ without keeping track of where the energy is going or coming from will require a single Lindblad operator

$$L=\gamma|0\rangle\langle 1|,\tag{6.9}$$

where $\gamma$ is a real parameter indicating the strength of the transition. This can model both decay and excitations.