I have a few basic queries regarding a proof in the set of notes MIT: Open Quantum Systems, the following is stated:

We can derive the Lindblad equation from an infinitesimal evolution described by the Kraus sum representation with the following steps:

  1. From the Kraus sum we can write the evolution of $\rho$ to $t + \partial t$ as: $\rho(t+\partial t) = \sum_{k}M_{k}(\partial t) \rho(t) M_{k}^{\dagger}(\partial t)$.

  2. We now take the limit of the infinitesimal time, $\partial t \to 0$. We only keep terms up to first order in $\partial t, \rho(t + \partial t) = \rho(t) + \partial t \partial \rho$. This implies that the Kraus operator should be expanded as $M_{k} = M_{k}^{(0)} + \sqrt{\partial t}M_{k}^{(1)} + \partial t M_{k}^{(2)}+ ...$. Then there is one Kraus operator such that $M_{0} = I + \partial t(-i\mathcal{H}+K) + \mathcal{O}(\partial t^2)$ with $K$ hermitian while all others have the form $M_{k} = \sqrt{\partial t}L_{k} + \mathcal{O}(\partial t)$, so that we ensure $\rho(t + \partial t) = \rho(t) + \partial \rho \partial t$.

Question: Why does keeping first order terms imply that the Kraus operators should and can be expanded as a power series as stated? Also, why does it follow that Kraus operator $M_0 = I + \partial t(-i\mathcal{H}+K) + \mathcal{O}(\partial t^2)$ should be of this form?


1 Answer 1


I think that your notes want to show that any (time-independent) Markovian master equation is written in the Gorini-Kossakowski-Sudarshan-Lindblad (GKLS) form. My feeling is that they are ignoring some mathematical details, but intuitively their procedure is sound. The rigorous proof of the equivalence Markovianity-GKLS form is usually a bit more elaborate, and, for instance, you can find it in the original papers [1,2] or in the standard textbook by Breuer and Petruccione [3].

In my opinion, trying to follow your notes to get to the desired equivalence may be quite confusing. I just would like to point out that the appearance of the time-dependent Kraus operators $M_k(\delta t)$, expanded as you have written for small $\delta t$, is an ansatz, i.e. a priori is not due to any mathematical constraint, but we introduce it for our convenience. Anyway, I suggest you checking the rigorous proof [3] and trying to compare each step with the discussion in your notes. You can see that, ultimately, they follow the same lines.

I have to say, however, that the approach of your notes is very useful to obtain the Kraus decomposition of the quantum map associated to a given master equation. Let us start from the GKLS form of a Markovian dynamics: $$ \dot{\rho}(t)=\lim_{dt\rightarrow 0}\frac{\rho(t+dt)-\rho(t)}{dt}=-i[H,\rho(t)]+\sum_k \gamma_k \left(L_k\rho(t)L_k^\dagger-\frac{1}{2}\{L_k^\dagger L_k,\rho(t)\} \right). $$ We want to find the Kraus decomposition of the quantum map $\phi_{\delta t}$ such that $\phi_{\delta t}[\rho(t)]=\rho(t+\delta t)$, for a small but finite $\delta t$. We have $\phi_{\delta t}[\rho(t)]=\rho(t)+\mathcal{L}[\rho(t)]\delta t+O(\delta t^2)$, that can be rewritten as: $$ \begin{split} \phi_{\delta t}[\rho(t)]=&\left(\mathbb{I}-i H\delta t-\frac{1}{2}\sum_k \gamma_k L_k^\dagger L_k \delta t\right)\rho(t)\left(\mathbb{I}+i H\delta t-\frac{1}{2}\sum_k \gamma_k L_k^\dagger L_k \delta t\right)\\ &+\sum_k\gamma_k L_k\rho(t)L_k^\dagger\delta t+O(\delta t^2). \end{split} $$ In conclusion, by setting $K=-\frac{1}{2}\sum_k \gamma_k L_k^\dagger L_k$, $\phi_{\delta t}$ can be decomposed through the Kraus operators $M_0=\mathbb{I}-\delta t(i H-K)$, $M_k=\sqrt{\gamma_k\delta t}L_k$, up to a precision of the order of $O(\delta t^2)$. Note that this does not tell us how to decompose the general quantum map $\phi_\tau[\rho(t)]=\sum_k \tilde{M}_k(\tau)\rho(t)\tilde{M}_k^\dagger(\tau)$ which drives the evolution for any large time $\tau$, and, as far as I know, such a decomposition is in general not easy to find (one has to solve the master equation, find the Choi matrix, etc...). However, it provides us with a great method to reconstruct the dynamics generated by the master equation via repeated applications of the map $\phi_{\delta t}$, within a certain precision bounded by $O(\delta t^2)$. As you can guess, this is very important for the quantum simulation of open systems: the Kraus operators $M_0$ and $M_k$ may be obtained as the first-order expansion of some unitary operators (quantum gates) $U(\delta t)$.

[1] G. Lindblad, Comm. Math. Phys. 48, 119 (1976).

[2] V. Gorini, A. Kossakowski, and E. C. G. Sudarshan, J. Math. Phys. 17, 821 (1976).

[3] H.-P. Breuer and F. Petruccione, The theory of open quantum systems (Oxford University Press, 2002).

  • $\begingroup$ Thanks for the response. So you are instead starting from the Lindblad and finding the Kraus decomposition (does seem clearer from that direction). I will check out the references as well for the rigorous proof. $\endgroup$
    – John Doe
    Oct 20, 2020 at 17:09
  • 2
    $\begingroup$ Exactly. The converse is true as well, but the proof in your notes skips many mathematical steps and is kind of sloppy on certain subtleties. Have a look at the nice proof in Breuer&Petruccione and you'll see how the Kraus decomposition+Markovianity imply the Lindblad form. $\endgroup$ Oct 20, 2020 at 23:41

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