# Born-Markov Approximation: Why is $\rho_{I}(s) \to \rho_{I}(t)$ taken, and not $\rho(s) \to \rho(t)$?

I am following along Chapter 3 of Breuer and Petruccione's book. For a Hilbert space $$\mathcal{H}_{S} \otimes \mathcal{H}_{R}$$ and Hamiltonian $$H = H_{S} \otimes \mathbb{I}_{R} + \mathbb{I}_{S} \otimes H_R + g H_{\mathrm{int}}$$ where $$H_S$$ is the Hamiltonian for the system of interest, and $$H_R$$ is the Hamiltonian for the reservoir (which will later be traced out). For simplicity, assume this Hamiltonian is time-independent. The Liouville equation for the full system's Schrodinger-picture density matrix $$\sigma(t)$$ is given by $$\frac{d \sigma(t)}{d t} = - i [ H, \sigma(t) ]$$ Switching to the interaction picture where $$\sigma_{I}(t) := e^{+iH_At} \otimes e^{+iH_Bt} \sigma(t)\ e^{-iH_At} \otimes e^{-iH_Bt}$$ and $$V(t) := e^{+iH_At} \otimes e^{+iH_Bt} H_{\mathrm{int}} \ e^{-iH_At} \otimes e^{-iH_Bt}$$ we then get the interaction-picture Liouville equation $$\frac{d \sigma_I(t)}{d t} = - i [ V(t), \sigma_I(t) ]$$

Defining $$\rho(t) := \mathrm{Tr}_{R}\left[ \sigma(t) \right]$$ to be the Schrodinger-picture reduced density matrix for the system, and then defining the interaction-picture version of this as $$\rho_I(t) := e^{+ i H_A t}\rho(t)e^{- i H_A t}$$ we then get the equation of motion (to lowest order in the coupling) $$\frac{d\rho_{I}(t)}{dt} \simeq - g^2 \int_0^t ds\ \mathrm{Tr}_{R}\left( \big[ V(t), [ V(s) , \sigma_I(s) ] \big] \right)$$ where we have also assumed $$\mathrm{Tr}_R[ V(t), \sigma(0) ]=0$$.

Furthermore, in the Born Approximation, or weak coupling approximation we assume that the reservoir is unaffected by the system in that $$\sigma(t) \simeq \rho(t) \otimes \varrho_{R}$$ for all times $$t$$ where $$\varrho_{R}$$ is some static state of the reservoir. This results in the integro-differential equation $$\frac{d\rho_{I}(t)}{dt} \simeq - g^2 \int_0^t ds\ \mathrm{Tr}_{R}\left( \big[ V(t), [ V(s) , \rho_I(s) \otimes \varrho_{R} ] \big] \right)$$ which is hard to solve. For this reason, the Markov approximation is taken, where we first replace $$\rho_{I}(s) \to \rho_{I}(t)$$ (which assumes $$\rho_{I}(s)$$ is slowly-varying) $$\frac{d\rho_{I}(t)}{dt} \simeq - g^2 \int_0^t ds\ \mathrm{Tr}_{R}\left( \big[ V(t), [ V(s) , \rho_I(t) \otimes \varrho_{R} ] \big] \right)$$ and under the assumption that the integrand disappears sufficiently fast for times $$s \gg \tau_{R}$$ (where $$\tau_R$$ is the timescale over which correlation functions for the reservoir decay), we switch the integration variable $$s \to t - s$$, and then take the upper limit to $$\infty$$ so that: $$\frac{d\rho_{I}(t)}{dt} \simeq - g^2 \int_0^\infty ds\ \mathrm{Tr}_{R}\left( \big[ V(t), [ V(t-s) , \rho_I(t) \otimes \varrho_{R} ] \big] \right) \ .$$

My Question: Why do you replace $$\rho_{I}(s) \to \rho_{I}(t)$$ and not make the replacement $$\rho(s) \to \rho(t)$$ (in the Schrodinger-picture)?

With the identity $$\dot{\rho}(t) = - i [H_A,\rho(t)] + e^{-iH_A t} \dot{\rho}_I(t) e^{+i H_A t}$$, you can easily write the earlier integro-differential equation in the form $$\frac{d\rho(t)}{dt} \simeq - i [H_A,\rho(t)] - g^2 \int_0^t ds\ e^{-i H_A t} \mathrm{Tr}_{R}\left( \big[ V(t), [ V(s) , e^{+i H_A s}\rho(s)e^{-i H_A s} \otimes \varrho_{R} ] \big] \right)e^{+i H_A t} \ ,$$ and then it would seem reasonable to replace $$\rho(s) \to \rho(t)$$ (assuming $$\rho(s)$$ is slowly-varying).

It seems to me like it would make more sense that $$\rho(s)$$ is slowly varying rather than $$\rho_I(s) = e^{+i H_A s}\rho(s)e^{-i H_A s}$$ (which usually includes oscillatory factors $$e^{i \Delta E t}$$ in terms of energy gaps of $$H_A$$).

What is the reason?

Roughly speaking, it is the Schrödinger-picture density operator which has rapidly oscillating phase factors. Transforming to the interaction picture removes these phase factors. The residual time dependence of $$\rho_I(t)$$ is generated only by coupling to the reservoir and is therefore slow, assuming weak coupling.
Indeed, you have already shown that $${\rm d}\rho_I/{\rm d}t = O(g^2)$$. Therefore, on the right-hand side of your integro-differential equation, you can Taylor expand the integrand as $$g^2 \rho_I(s) = g^2[\rho_I(t) + (s-t){\rm d}\rho_I/{\rm d}t + \cdots] = g^2\rho_I(t) + O(g^4),$$ and therefore you may replace $$\rho_I(s)\to\rho_I(t)$$ at second order in $$g$$. Note that this low-order expansion makes sense only if $$|s-t|$$ does not become too large. This is ensured by the fact that the reservoir correlation functions are rapidly decaying functions of $$|s-t|$$.
• Hi Mark, thanks for your great answer. About your last sentence: the reason that $|s-t|$ doesn't get too large is assuming that correlation functions decay rapidly. Is the following the correct interpretation of this statement? When we replace $\varrho_{I}(s) \simeq \varrho_{I}(t) + (s-t) \dot{\varrho}_{I}(t)$ in the above Born equation, we can already assume that $\dot{\varrho}_{I}(t)$ is small there right? The rapidly-decaying property of the correlators $C(s)$ is needed only because there will be at $\mathcal{O}(g^4)$ integrals over $s C(s)$ (and probably matrices that depend on t)? – Greg.Paul Jun 19 '19 at 19:12
• Another way maybe to get at the answer to my question: Suppose that $C(s)$ was not rapidly decaying enough (maybe it falls off like $1/s$). Is the problem with having a correlator like this is because at order $\mathcal{O}(g^4)$ in the Born Approximation, we'll have integrals over factors $s C(s) \sim s/s = 1$ aka constants being integrated from 0 to $t$, and so will be divergent for $t$ getting large? Is this why we usually want decaying correlators like $C(s) \sim e^{s/\tau}$ for some timescale $\tau$? – Greg.Paul Jun 19 '19 at 19:18
• Yes, I think that is essentially correct. The way that I think about this is the following. The small parameter $g$ can be interpreted as the characteristic energy scale of the interaction Hamiltonian. But in this case, it makes no sense to talk about $g$ being "small"; small compared to what? The answer, as explained by van Kampen in his book, is that $g$ is small compared to $\tau^{-1}$, where $\tau$ is the bath correlation time, i.e. $C(t)\approx 0$ for $t>\tau$. So then my expansion above is in terms of a small dimensionless parameter, $g\tau\ll 1$, and thus should be a good approximation. – Mark Mitchison Jun 21 '19 at 8:40
• Very interesting! Is your argument about $g$ being small compared to $\tau^{-1}$ related in any way to the weak-coupling argument of Davies (ie. his 1974 paper 'Markovian Master Equations'). There the time $t$ is scaled by $g^2$ defining a new time variable $T = g^2 t$, where you take $g \to 0$ and $t \to \infty$ in a way which $T$ stays constant. Also, what is the book by Van Kampen called? Is it Stochastic Processes in Physics and Chemistry? – Greg.Paul Jun 21 '19 at 13:03