Lindblad from infinitesimal Kraus sum representation

Physics Asked by John Doe on December 12, 2020

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 $M_{0} = I + partial
    t(-imathcal{H}+K) + mathcal{O}(partial t^2)$
    with $K$ hermitian
    while all others have the form $M_{k} = sqrt{partial t}L_{k} +
    mathcal{O}(partial t)$
    , so that we ensure $rho(t + partial t) =
    rho(t) + partial rho partial t$

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 $M_0 = I + partial
t(-imathcal{H}+K) + mathcal{O}(partial t^2)$
should be of this form?

One Answer

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: $$ dot{rho}(t)=lim_{dtrightarrow 0}frac{rho(t+dt)-rho(t)}{dt}=-i[H,rho(t)]+sum_k gamma_k left(L_krho(t)L_k^dagger-frac{1}{2}{L_k^dagger L_k,rho(t)} right). $$ 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: $$ begin{split} phi_{delta t}[rho(t)]=&left(mathbb{I}-i Hdelta t-frac{1}{2}sum_k gamma_k L_k^dagger L_k delta tright)rho(t)left(mathbb{I}+i Hdelta t-frac{1}{2}sum_k gamma_k L_k^dagger L_k delta tright)\ &+sum_kgamma_k L_krho(t)L_k^daggerdelta t+O(delta t^2). end{split} $$ 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 $M_0=mathbb{I}-delta t(i H-K)$, $M_k=sqrt{gamma_kdelta t}L_k$, 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).

Correct answer by Goffredo_Gretzky on December 12, 2020

Add your own answers!

Related Questions

Time symmetry and general relativity

2  Asked on December 30, 2020 by flyingwaffle


Can light take a (faster) detour?

2  Asked on December 30, 2020 by dennis-jaheruddin


Ask a Question

Get help from others!

© 2022 All rights reserved. Sites we Love: PCI Database, MenuIva, UKBizDB, Menu Kuliner, Sharing RPP