3
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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 $|s-t|$.

$\endgroup$
  • $\begingroup$ 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)? $\endgroup$ – Greg.Paul Jun 19 at 19:12
  • $\begingroup$ 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 $\tau$? $\endgroup$ – Greg.Paul Jun 19 at 19:18
  • 1
    $\begingroup$ 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. $\endgroup$ – Mark Mitchison Jun 21 at 8:40
  • $\begingroup$ 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? $\endgroup$ – Greg.Paul Jun 21 at 13:03
  • 1
    $\begingroup$ Yes I suspect Davies' weak-coupling limit must also be related to this although I don't know how to make the connection precise. Indeed, that is van Kampen's book, I recommend it (although it's much more general than open quantum systems and thus less detailed than, say, Breuer & Petruccione). $\endgroup$ – Mark Mitchison Jun 22 at 23:14

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.