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?


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).

| cite | improve this answer | |
  • $\begingroup$ 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 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. $\endgroup$ – Goffredo_Gretzky Oct 20 at 23:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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