Lindblad equation using Kraus operators and a CPTP map

This is a derivation found in [1] but with details from [2]. I am only emphasizing the details that I found interesting, because it is a very short derivation originally and the way it hides some mysteries is very enticing. Most consider a postulate of quantum mechanics that a completly positive trace preserving (CPTP) map, famous for the propriety of mapping physical systems to other physical systems, can be described by Kraus operators, specifically a map from one density matrix to another. Considering said map (we make implicit the Choi-Kraus’ theorem), then $$\hat{\rho}^{\prime}=\sum_{\mu}\hat{K}_{\mu}\hat{\rho}\hat{K}_{\mu}^{\dagger}\qquad\text{with }\quad\sum_{\mu=0}K_{\mu}^{\dagger}K_{\mu}=\mathbb{I}.$$ Lets now check that if $\hat{\rho}$ obeys the proprieties of a density matrix, then so will $\hat{\rho}^{\prime}$. That is, $\hat{\rho}\prime$ is hermitian, positive semi-definite, and unit trace. The unit trace comes from the fact that it is a CPTP map, but to show this notice $$ \text{Tr}\left(\hat{\rho}^{\prime}\right)=\sum_{\mu}\text{Tr}\left(K_{\mu}\hat{\rho}K_{\mu}^{\dagger}\right)=\text{Tr}\left(\hat{\rho}\sum_{\mu}K_{\mu}^{\dagger}K_{\mu}\right)=\text{Tr}\left(\hat{\rho}\right), $$ where we used the cyclic propriety of traces and the unit sum of the Kraus operators. Since $\hat{\rho}$ is hermitian, $\hat{\rho}^{\prime\dagger}=\sum_{\mu}\left(K_{\mu}\hat{\rho}K_{\mu}^{\dagger}\right)^{\dagger}=\sum_{\mu}K_{\mu}\hat{\rho}K_{\mu}^{\dagger}=\hat{\rho}^{\prime}$. And finally, since $\hat{\rho}$ is positive semi-definite, $\langle\psi|\hat{\rho}|\psi\rangle\geq0,\quad\forall|\psi\rangle$, then in particular $\langle\psi|K_{\mu}\hat{\rho}K_{\mu}^{\dagger}|\psi\rangle\geq0,\quad\forall K_{\mu}^{\dagger}|\psi\rangle$, then given $|\psi\rangle$ we have $\langle\psi|\hat{\rho}^{\prime}|\psi\rangle=\sum_{\mu}\langle\psi|K_{\mu}\hat{\rho}K_{\mu}^{\dagger}|\psi\rangle\geq0$.

Lindblad's master equation

To derive Lindblad's equation, consider $\hat{\rho}^{\prime}$ is very close to $\hat{\rho}$, that is, an infinitesimal transformation from the state at time $t$ to time $t+dt$. This hides an implicit Markovianity assumption, because the system can be described enterily by the next step in time with no memory effects, that is, bigger time steps don't need to be considered. Due to the trace preserving condition $\sum_{\mu=0}K_{\mu}^{\dagger}K_{\mu}=\mathbb{I}$, $$K_{0}^{\dagger}K_{0}\approx\mathbb{I}+\left(A+A^{\dagger}\right)dt$$ $$\sum_{\mu\neq0}K_{\mu}^{\dagger}K_{\mu}=\sum_{\mu\neq0}L_{\mu}^{\dagger}L_{\mu}dt,$$ where the operators $L_{\mu}$ are the Lindblad operators representing quantum jumps between states. Notice this is where we introduce our assumption of an open system, because we arbitrarily separated a part of the evolution of the system in one $K_0$ and the others account for dissipative terms that are time dependent. In particular, this separation is only possible in weak coupling baths, since we considered only up to first order terms in time, making most of the dynamics unitary and a small portion of quantum jumps that scale with $\sqrt{dt}$. We have then $$\frac{A+A^{\dagger}}{2}=-\frac{1}{2}\sum_{\mu\neq0}L_{\mu}^{\dagger}L_{\mu},$$ making this the hermitian part of $A$. The anti-hermitian can be chosen as $iH$, since for an infinitesimal evolution of $K_{0}$ by $e^{iHdt}$ it gives an added anti-hermitian term $iH$. So $A=-\frac{1}{2}\sum_{\mu\neq0}L_{\mu}^{\dagger}L_{\mu}+iH$ and $$K_{0}=\mathbb{I}+\left(-\frac{1}{2}\sum_{\mu\neq0}L_{\mu}^{\dagger}L_{\mu}+iH\right)dt$$ $$K_{\mu}=L_{\mu}\sqrt{dt},\quad\mu\neq0.$$ Using this expression on Eq. ([eq:5]), $$ \hat{\rho}\left(t+dt\right) =\sum_{\mu}\hat{K}_{\mu}\hat{\rho}\hat{K}_{\mu}^{\dagger} =\hat{K}_{0}\hat{\rho}\hat{K}_{0}^{\dagger}+\sum_{\mu\neq0}\hat{K}_{\mu}\hat{\rho}\hat{K}_{\mu}^{\dagger} $$ $$ =\hat{\rho}-\hat{\rho}(\frac{1}{2}\sum_{\mu\neq0}L_{\mu}^{\dagger}L_{\mu}+iH)dt +(-\frac{1}{2}\sum_{\mu\neq 0}L_{\mu}^{\dagger}L_{\mu}\hat{\rho}(\mathbb{I}-(\frac{1}{2}\sum_{\mu\neq0}L_{\mu}^{\dagger}L_{\mu}+iH)dt) $$ $$+iH\hat{\rho}(\mathbb{I}-(\frac{1}{2}\sum_{\mu\neq0}L_{\mu}^{\dagger}L_{\mu} +iH)dt))dt+\sum_{\mu\neq0}\hat{L}_{\mu}\hat{\rho}\hat{L}_{\mu}^{\dagger}dt $$ Now discarding terms of higher than second order on $dt$ and playing with the differentials, we obtain $$\frac{\hat{\rho}\left(t+dt\right)-\hat{\rho}}{dt} =-\frac{1}{2}\hat{\rho}\sum_{\mu\neq0}L_{\mu}^{\dagger}L_{\mu}+i\hat{\rho}H-\frac{1}{2}\sum_{\mu\neq0}L_{\mu}^{\dagger}L_{\mu}\hat{\rho}-iH\hat{\rho}+\sum_{\mu\neq0}\hat{L}_{\mu}\hat{\rho}\hat{L}_{\mu}^{\dagger}\dot{\hat{\rho}} $$ $$ =-\frac{1}{2}\left\{ \hat{\rho},\sum_{\mu\neq0}L_{\mu}^{\dagger}L_{\mu}\right\} -i\left[H,\hat{\rho}\right]+\sum_{\mu\neq0}\hat{L}_{\mu}\hat{\rho}\hat{L}_{\mu}^{\dagger}\dot{\hat{\rho}}=-i\left[H,\hat{\rho}\right]+\sum_{\mu\neq0}\left(\hat{L}_{\mu}\hat{\rho}\hat{L}_{\mu}^{\dagger}-\frac{1}{2}\left\{ \hat{\rho},L_{\mu}^{\dagger}L_{\mu}\right\} \right) $$

To account for the rate of transitions, simply rescale the definition of the Lindblad operator, $\hat{L}_{\mu}\to\sqrt{\gamma_{\mu}}\hat{L}_{\mu}$, where $\gamma_{\mu}$ is the probability of a jump $L_{\mu}$, $$ \dot{\hat{\rho}}=-i\left[H,\hat{\rho}\right]+\sum_{\mu\neq0}\gamma_{\mu}\left(\hat{L}_{\mu}\hat{\rho}\hat{L}_{\mu}^{\dagger}-\frac{1}{2}\left\{ \hat{\rho},L_{\mu}^{\dagger}L_{\mu}\right\} \right). $$ This gives back our closed system result when all $\gamma_\mu$'s are zero. I will give more comments on this derivation later as an update.

References:

[1] "Many Body Open Quantum Systems", R. Fazio et al. (2025)

[2] "A short introduction to the Lindblad Master Equation", D. Manzano (2020).

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