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Define a matrix differential equation $$\dot{X}=A(t)X(t),$$ where $X=[x1,x2,...]^T$ is a 1D vector and $A(t)$ is a complex-valued time-dependent matrix.

This system can be solved by $$X(t)= \mathcal{T}\exp\left[\int_0^t A(t')dt'\right] X(0),$$ where we introduce a time-ordered exponential. In general, this is an arduous task. For my specific case:

$\bullet$ My time-dependent matrix has the form of $A(t)=A_0+\alpha(t)A_1$, where $\alpha(t)$ is an Orstein-Uhlenbeck process, or in other words, colored noise. This is not relevant (Itô calculus is not necessary here because of the finite correlation time of the process).

$\textbf{If}$ the noise is slow enough such that $\alpha(t)\rightarrow \alpha$ (let's call it the static limit), then one can solve the matrix exponential without the time-ordering and everything is fine. I can solve for $X(t)$ and then, I can average over $\alpha$.

My goal is now to go to a quasi-static limit, i.e. we keep the time-dependence on $\alpha(t)$ assuming this is still to be slowly varying. Now I do need to compute the time-ordered exponential.

My question is: Is there any approximation I could use? Any techniques to solve it?

Note: Time-ordered exponentials can be simplified if the matrix commutes at different times $[A(t_1),A(t_2)]=0$. However, this is not the case for me.

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  • $\begingroup$ Essentially what you are asking is an adiabatic approximation. There are results when the generator $A(t)$ Is antihermitian and (in the more general case) when it is a generator of a contraction semigroup. The best reference for the antihermitian case is still the paper by Kato. $\endgroup$
    – lcv
    Commented Nov 7, 2023 at 11:17
  • $\begingroup$ To be precise, the adiabatic approximation is an expansion into derivatives of $\alpha(t)$, so the first term is indeed what you would have if $\alpha$ was constant (which is your limiting case). If you are interested I can provide more details. $\endgroup$
    – lcv
    Commented Nov 7, 2023 at 11:21
  • $\begingroup$ You are right, I am performing an adiabatic approximation of the noise. To the first order (when it's constant), I know the solution and I am now interested in the next orders which depend on the derivative of $\alpha$. I am unaware of the work by Kato. Could you please share some references, please? $\endgroup$
    – J.Agusti
    Commented Nov 7, 2023 at 11:23

1 Answer 1

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The answer to your question is given by the adiabatic approximation (or adiabatic series). The first proof for the case when $A(t)$ is anti-hermitian has been given by Born and Fock, later elegantly extended by Kato [1]. Note that in this case the norm of $X$ is preserved by the dynamics which avoids dangerous blow-up. A generalization of that result can be given when $A(t)$ generates a contraction semi-group. Essentially this means that the generator of $A(t)$ (i.e. that operator $G(t)$ that satisfies $A(t)= \exp{G(t)}$) has spectrum that satisfies $\mathrm{Re}(\sigma(t)) \le 0$ for all $t$. This condition also avoids that the solution blows up. A physics reference is given in [2] where the whole series is presented.

Note that the first step, often overlooked in the physics literature, is that of making the problem well defined mathematically. Essentially this means that we have to decide how to approach the limit $\alpha(t) \to$ constant. There are infinite ways in which this limit can be approached and it is unlikely (and indeed false) that we can find a solution for all these cases. For example $\alpha(t)$ could oscillate with some resonant frequency of $A_0$ thus amplifying its response as in parametric amplification. A meaningful approach is to consider $A(t) \to A(t/T)$ and then consider the limit $T\to \infty$. One is then led to consider an expansion in $\epsilon = 1/T$.

Added

Given that your generator is in Lindblad form for any $t$, for the adiabatic expansion of the solution, what you call $X(t)$, (that is, and expansion in series of $\epsilon$) you can look at Eq. (4a) of [3]. This is a simplification of Theorem 6 of [4].

References

[1] T. Kato, J. Phys. Soc. Jpn. 5, 435 (1950)

[2] L. Campos Venuti, T. Albash, D. A. Lidar, and P. Zanardi, Adiabaticity in open quantum systems, Phys. Rev. A 93, 032118 – Published 14 March 2016. arXiv version

[3] L. Campos Venuti, D. A. Lidar, Phys. Rev. A 98, 022315 (2018) arXiv version

[4] J. E. Avron, M. Fraas, G. M. Graf, and P. Grech, Commun. Math. Phys. 314, 163 (2012)

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  • $\begingroup$ Thanks for the detailed answer and the references! I have a couple of questions: My matrix $A(t)$ is not anti-hermitian. However, the origin of my equation is a master equation, so the $A(t)$ is the Lindbladian operator. Is it enough to ensure that it generated a contraction semi-group? $\endgroup$
    – J.Agusti
    Commented Nov 7, 2023 at 12:50
  • $\begingroup$ About how to define the noise, maybe I am being naïve here but I just do the substitution $\alpha(t)\rightarrow \alpha$, and as I know the distribution of $P(\alpha)$, it allows me to then take some statistical average. $\endgroup$
    – J.Agusti
    Commented Nov 7, 2023 at 12:51
  • $\begingroup$ Re your first question: yes, a Lindbladian generates a contraction semigroup. This is easy to show and I believe it is also shown in [2]. $\endgroup$
    – lcv
    Commented Nov 7, 2023 at 14:00
  • $\begingroup$ Regarding your second question, I'm afraid I don't understand it :) $\endgroup$
    – lcv
    Commented Nov 7, 2023 at 14:01
  • $\begingroup$ After checking the paper, I realized this boils down to the adiabatic theorem in open systems. I have already tried doing this using the work of Sarandy (journals.aps.org/pra/abstract/10.1103/PhysRevA.71.012331). I couldn't go on because my matrix is 9x9, so any Jordan decomposition/diagonalization was limited. $\endgroup$
    – J.Agusti
    Commented Nov 7, 2023 at 15:32

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