The other day our professor was talking about Kraus representation of density operator and the derivation of Lindblad equation. He told that this was related to the Feynman path integrals and that we should have noticed the analogy. I am still wondering what he meant. I tried to google for any connection between path integrals and Kraus operators or between path integrals and the Lindblad equation, with zero success...

Does anyone know what this analogy could have been?

Briefly, his derivation was the following.

Consider a closed system with two subsystems $A$ and $B$, which were independent at $t=0$, thus $$\rho(t=0)=\rho^A_0\otimes\rho^B_0$$

Let also $B$ be in a pure sate $\rho^B_0=\left|in^B\right\rangle\left\langle in^B\right|$. The evolution is defined by unitary operator:

$$\rho(t)=\hat{U}(t) \left|in^B\right\rangle\rho^A_0\left\langle in^B\right|\hat{U}^\dagger(t)$$ and $$\rho^A(t)=Tr^B\left\{\hat{U}(t) \left|in^B\right\rangle\rho^A_0\left\langle in^B\right|\hat{U}^\dagger(t)\right\}$$

Taking some basis $\left|f_k^~\right\rangle$ in $B$ one can rewrite this as $$\rho^A(t)=\sum_k\left\langle f_k^~\right|\hat{U}(t) \left|in^B\right\rangle\rho^A_0\left\langle in^B\right|\hat{U}^\dagger(t)\left|f_k^~\right\rangle$$

Introducing $\hat{M}_k(t)=\left\langle f_k^~\right|\hat{U}(t) \left|in^B\right\rangle$, the last equation may be rewritten simply as: $$\rho^A(t)=\sum_k\hat{M}_k(t)\rho^A_0\hat{M}^\dagger_k(t)$$

Then: $$\rho^A(t)+\Delta\rho^A(t)=\rho^A(t+\Delta t) = \sum_k\hat{M}_k(\Delta t)\rho^A(t)\hat{M}^\dagger_k(\Delta t)$$

Choosing the basis $\left|f_k^~\right\rangle$ such that $\hat{M}_0$ is close to $\hat{1}$ and taking into account $\sum_k\hat{M}_k\hat{M}^\dagger_k=\hat{1}$, we obtain (considering only terms not smaller than $\Delta t$): $$\hat{M}_0=\hat{1}+ \left(\hat{L}_0-\frac{i\hat{H}^A}{\hbar}\right)\cdot\Delta t$$ $$\hat{M}_k=\hat{L}_k\sqrt{\Delta t},\hspace{1cm}k\neq 0,$$

Where $\hat{L}_0$ and $\hat{H}^A$ are arbitrary hermitian operators, $\hat{L}_k(k\neq0)$ are arbitrary operators. Substituting these into previous expression, taking into account that $\hat{L}_0=-\frac{1}{2}\sum_{k\neq0}\hat{L}_k^\dagger\hat{L}_k$, dividing everything by $\Delta t$ and taking the limit $\Delta t \rightarrow 0$ we get the Lindblad equation: $$\frac{d\rho^A(t)}{dt} = \frac{1}{i\hbar}\left[\hat{H}^A,\rho^A(t)\right] + \sum_{k\neq0}\left(\hat{L}_k\rho^A(t)\hat{L}_k^\dagger - \frac{1}{2} \left\{ \hat{L}_k^\dagger\hat{L}_k,\rho^A(t) \right\} \right)$$

  • $\begingroup$ Have you asked your professor what he meant? $\endgroup$
    – rob
    May 12, 2015 at 21:00

1 Answer 1


I guess your prof might have been referring to the "spirit" of the derivation, which proceeds by splitting up time into discrete chunks and considering a stroboscopic evolution over short time intervals $\Delta t$. Then by taking the limit $\Delta t \to 0$, the desired expression is obtained. A similar approach is used in the derivation of the Feynman path integral, although the similarity ends there. For example, here we obtain an expression for the time derivative of the density matrix, whereas the path integral is a prescription for calculating the density matrix itself.

Note that the derivation you have given is extremely dodgy, and incorrect in essentially all non-trivial cases. It is simply not possible to derive a Lindblad equation from the majority of Kraus operators that one could write down, and certainly not when starting from a Hamiltonian interaction between $A$ and $B$ (without invoking further approximations such as time coarse-graining and Markovianity).

You state, correctly, that the Kraus map is given by $$ \rho^A(t) = \sum_k M_k(t) \rho^A(0) M_k^\dagger(t),$$ with the Kraus operators as defined in the OP. This implies that $$\rho^A(t+\Delta t ) = \sum_k M_k(t+\Delta t ) \rho^A(0) M_k^\dagger(t+\Delta t ).$$ However, this is not the same as $$\rho^A(t+\Delta t ) = \sum_k M_k(\Delta t ) \rho^A(t) M_k^\dagger(\Delta t ). \qquad \mathrm{(wrong!)}$$ Tracing back through the derivation, you will find that the above equation is only true if the global state is given at all times by $$\rho(t) = \rho^A(t)\otimes \rho^B.$$ But clearly this will not be the case in general, and especially not if there is a non-trivial interaction between the subsystems $A$ and $B$.

  • $\begingroup$ Thanks for your reply! I also noticed the error while copying the formulas from my prof's slides, just didn't want to pay much attention to that since that is not the point of my question. By the way, I actually changed it a bit: the original formula was $\rho^A(t+\Delta t)=\sum M_k(t+\Delta t)\rho^A(t) M_k^\dagger(t+\Delta t)$, which is even more wrong IMO. $\endgroup$
    – SiLiKhon
    May 6, 2015 at 19:14
  • $\begingroup$ OK, but either way, if you professor is trying to demonstrate that you can derive a Lindblad equation from a Kraus map, he/she is misleading you. All you can show is that if the evolution of the density matrix over a short time step can be written in the Kraus form $\rho(t+\Delta t) = \sum_k M_k(\Delta t) \rho(t) M_k^\dagger(\Delta t)$, which is a hugely restrictive assumption, then $\rho(t)$ satisfies a Lindblad equation. $\endgroup$ May 6, 2015 at 19:19
  • $\begingroup$ But isn't Lindblad equation includes Born approximation, which is stated in your last line? Hence it is a special type of Kraus map in this sense. $\endgroup$
    – donnydm
    Nov 22, 2017 at 12:46
  • 2
    $\begingroup$ @donnydm Every Lindblad equation defines a Kraus map, but the point of this answer is that the converse is not true. $\endgroup$ Nov 30, 2017 at 18:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.