I am attempting to learn a bit more about open quantum systems.

Often we derive master equations or Heisenberg-Langevin equations where we have something like

\begin{align} \dot{\rho}(t) = \mathcal{L}[\rho(t)]\\ \dot{A}(t) = \mathcal{L}^{\dagger}[A(t)] \end{align}

Here $\rho$ is the density matrix of the system and $A$ is an arbitrary system operator and $\mathcal{L}$ Lindbladian superoperator given by

$$ \mathcal{L}[\mathcal{O}] = -\frac{i}{\hbar}[H,\mathcal{O}] + \sum_i \gamma_i\left(L_i\mathcal{O}L_i^{\dagger} - \frac{1}{2}\left\{L_i^{\dagger}L_i,\mathcal{O}\right\}\right) $$

for any operator $\mathcal{O}$. See for example this post.

It is said that the Lindblad form gives the most general form to generate Markovian dynamics.

Clearly there must be more general master equations which are non-Markovian. It seems to me that one could derive the most general form in the following way. Suppose we have a system which comes from a tensor product of two systems, $A$ and $B$. We have

\begin{align} \rho(t) &= U(t)\rho(0) U^{\dagger}(t) = e^{\mathcal{L}t}[\rho(0)]\\ \dot{\rho}(t) &= \left[\dot{U}(t)U^{\dagger}(t),\rho(t) \right] = -\frac{i}{\hbar}[H(t),\rho(t)] = \mathcal{L}[\rho(t)] \end{align}

where $U$ represents the time evolution operator. We then take the partial trace to get

\begin{align} \rho_A(t) &= \text{Tr}_B[\rho(t)]=\text{Tr}_B\left[e^{\mathcal{L}t}\rho(0)\right]\\ \dot{\rho}_A(t) &= \frac{d}{dt}\text{Tr}_B\left[\rho(t)\right] = \text{Tr}_B\left[\dot{\rho}(t) \right] = \text{Tr}_B[\mathcal{L}[\rho(t)]] = \mathcal{L}_A[\rho_A(t)] \end{align}

Here $\mathcal{L}$ generates unitary dynamics for $\rho$, while tracing out the dynamics of system $B$ leads to $\mathcal{L}_A$ to generate the not-necessarily unitary dynamics for $\rho_A$.

My questions are as follows.

1) Is what I have described actually the way to derive the most general form for a quantum master equation?

2) I guess the derivation above will not lead to the Lindblad form because it does not necessarily generate Markovian dynamics? Is this correct?

3) Is it the restriction that $e^{\mathcal{L}_A t}$ is an element of a dynamical semi-group the mathematical fact which leads to $\mathcal{L}_A$ having the Lindblad form? Is the dynamical semigroup restriction the same as requiring Markovian dynamics?


1 Answer 1


I have done more reading and see an important point I missed above. In my last line for $\dot{\rho_A(t)}$ I made the step

$$ \text{Tr}_B(\mathcal{L}[\rho(t)]) = \mathcal{L}_A[\rho_A(t)] $$

This step is not justified in general. Namely, the problem is as follows. I think it is true that $\text{Tr}_B(\mathcal{L}[\rho(t)])$ can be written as a function of $\rho_A$, however, the dependence may not only include $\rho_A(t)$. In particular it may depend on $\rho_A$ at time $t$ as well as other earlier times. Physically this corresponds to the idea that information about system $A$ could leak into system $B$, then, if system $B$ has memory this information can feed back into system $A$ meaning that the current time evolution of system $A$ will depend, in general, on the prior state of system $A$ and all earlier times.

In physics we like differential equations for objects. We like it when the time evolution for an object at time $t$ is related to a function of that object also at time $t$. We can see that the above paragraph messes with this desiderata. This is why we impose the Markovity of system $B$ which ensures the master equation can be written in the form above.

I still await an answer from someone more expert than me but here are my answers to my own questions.

1) What I described was not quite the most general form because I made the above unjustified assumption. I still think that

\begin{align} \dot{\rho}_A(t) = \text{Tr}_B[\mathcal{L}[\rho(t)]] \end{align}

Is the most general form we can write down where $\mathcal{L}$ generates unitary dynamics for $\rho(t)$. Unfortunately I don't really know how to write the RHS of this equation as a function of $\rho_A$ as I would like to. Even given an explicit form for $\rho(t)$ and $\mathcal{L}$ I wouldn't know how to do it I think..

2) It is correct that this general form does not necessarily lead to a Lindblad form.

3) I am pretty certain the time evolution operator being part of a semi-group is the same as the Markov requirement.

To summarize and get to the heart of the question: if we even write down

$$ \dot{\rho}_A(t) = \mathcal{L}_A[\rho_A(t)] $$

We have ALREADY assumed Markov evolution. This means that $\mathcal{L}$ must be of the Lindblad form [1]. If we want something more general we must abandon the idea of $\dot{\rho}_A(t)$ depending only on $\rho_A(t)$ and not $\rho_A(t')$ for other times $t'$. This is not something I personally would like to abandon so I'll stick to Markov evolution and Lindblad forms for the time being!

[1] G. Lindblad, On the generators of quantum dynamical semigroups, Commun. Math. Phys 48, 119, (1976).


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