What properties do Kraus operators of Markovian processes have?

Question

It is well-known that the Kraus operator can describe more kinds of processes than master equations. For example, the master equation cannot describe non-markovian processes while the Kraus operator can, and all master equations can be translated into the Kraus operator frame.

Since all the master equations can only describe the markovian process, so my question is: does this kind of Kraus operator which describes the markovian process have some property that we can classify them from the Kraus operator that describes the non-Markovian process?

Answer 1

I think this question is generally difficult because there is no standard metric for non-Markovianity, for example this paper would suggest you try to express the evolution in a time-local canonical (Lindblad) form and then look at the negativity of the rates, but other metrics may not agree for certain channels. Perhaps it is easier to answer the simpler question of when can channel be expressed in a time-independent Lindblad form (i.e. the Lindblad operators are constant). Working backwards from the master equation, as you mentioned, you can find the Kraus operators for some Markovian evolution from the Lindblad operators $$ \begin{align} K_m = \sqrt{\delta t} L_m + O(\delta t)\quad K_0 = I - \delta t \bigg(\frac{1}{2}\sum_{m>0} L_m^\dagger L_m + iH\bigg) + O(\delta t^2) \end{align} $$ which reproduces the evolution as $\delta t \rightarrow dt$. If you are trying to frame your evolution in terms of some set of Kraus operators that are parametrized in time, then expanding them for small time must yield operators that scale like $\sqrt{\delta t}$ so that the evolution can be written as a first-order differential equation. An example (section 8.4.2.A) shows that if you have a $Z$-dephasing channel \begin{align} K_0 &= \sqrt{1-p} I, \quad K_1 = \sqrt{p} Z\\ \mathcal{E}(\rho) &= (1-p) \rho + p Z \rho Z \end{align} and assume $p = \frac{1}{2} (1-\Gamma)$, then an evolution described by $\Gamma = e^{-\gamma t}$ will indeed be Markovian. If instead your environment is Gaussian with $\Gamma = e^{-(\gamma t)^2}$ then trying to reformulate the evolution into a time-independent Lindblad form won't work. So I would say that if you are approaching this more phenomenologically, then the Markovianity of the system-environment evolution depends on the functions you use to describe the parameterized channel. To actually quantify how non-Markovian something is a harder question altogether, and depends on the chosen metric.

Answer 2

I guess from the operator sum representation $\rho(t) = \sum_k K_k \rho(t_0) K_k^\dagger$ alone you won't be able to make any statement about non-Markovianity, since you are missing interesting features:

1. (Domain) Is this dynamical map valid for all initial reduced density matrices? If so, the initial state might be a product state $\rho(t_0) \otimes \rho_E$
2. (Decomposition property) Does this dynamical map have the semigroup property $\mathcal{E}(t_2, t_0) = \mathcal{E}(t_2, t_1) \mathcal{E}(t_1, t_0),$

$$\sum_k K_k(t_2,t_0) (\cdot) K_k(t_2,t_0)^\dagger =? \sum_{ij} K_i(t_2, t_1) K_j(t_1, t_0) (\cdot) K_j(t_1,t_0)^\dagger K_i(t_2,t_1)^\dagger$$

Regarding Lindblad equations: the mentioned paper examines time-local Lindblad equations over time (during which non-Markovianity and thus correlations can form) - correlations between system and environment can lead to non completely positive Lindblad equations with negative decoherence rates. Some say that when integrating over this partly Lindblad equation you will still get a completely positive map but I haven't seen a rigorous proof yet.