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I'm trying to understand how to get the second-order approximation of a density function in the interaction picture given that $$\frac{d \tilde{\rho}}{dt} = -i[\tilde{V}, \tilde{\rho}]$$ $$\tilde{\rho(t)} = \tilde{\rho_{S}(t)} \otimes \rho_E$$ where $\tilde{\rho_{S}(t)}$ describes the system and $\rho_E = \Pi_j |0_i \rangle \langle0_j|$ is the environment.

The book states the following: $$ \begin{aligned} \tilde{\rho}(t+\Delta t)-\tilde{\rho}(t)=&-i \int_{t}^{t+\Delta t} \mathrm{dt'} \quad\left[\tilde{V}\left(t^{\prime}\right), \tilde{\rho}(t)\right] \\ &+(-i)^{2} \int_{t}^{t+\Delta t} d t^{\prime} \int_{t}^{t^{\prime}} d t^{\prime \prime}\left[\tilde{V}\left(t^{\prime}\right),\left[\tilde{V}\left(t^{\prime \prime}\right), \tilde{\rho}(t)\right]\right] \end{aligned} $$

And I don't quite follow it for a couple reasons. It mostly makes sense how $$\frac{d \tilde{\rho}}{dt} = -i[\tilde{V}, \tilde{\rho}] $$ $$\implies \tilde{\rho}(t+\Delta t)-\tilde{\rho}(t)= -i \int_{t}^{t+\Delta t} \mathrm{dt'}\left[\tilde{V}\left(t^{\prime}\right), \tilde{\rho}(t)\right]$$ By just integrating from $t \rightarrow t + \Delta t$. My question about this portion is why is there t' in the argument of $V$, but just t in he argument of $\rho$?

Next, I assume you get the second term in the approximation by finding $\tilde{\rho}$ to sub into the second term of the commutator, but I'm not understanding where the $t \rightarrow t'$ bounds come from in the integral. It seems like it is saying you sub in $\tilde{\rho} = -i\int_{t}^{t'}dt'' [\tilde{V(t'')}, \tilde{\rho(t)}]$

In the end, I should be able to take the partial trace $Tr_E$ and the $-i \int_{t}^{t+\Delta t} \mathrm{dt'}\left[\tilde{V}\left(t^{\prime}\right), \tilde{\rho}(t)\right]$ portion should be equal to $0$. I don't see this, but maybe understanding the above will help.

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    $\begingroup$ Without explicitly making the calculation, I think all terms appearing to the right are just same same as the exponential operator expansion for unitary time evolution, since the equation is exactly equivalent for both $U(t,t_0)$ and $\rho(t)$ (just with a minus sign difference I think). So in a nutshell, I would look at the Dyson series for the evolution operator: en.wikipedia.org/wiki/…., and translate the problem for a time dependent $\rho$. $\endgroup$ Jun 19 at 21:30

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It is just a matter of repeated back substitution. Starting with $$ \rho(t+\Delta t) = \rho(t) - i\int_t^{\Delta t} [V(t_1),\rho(t_1)]\ \text{d}t_1 , $$ and performing a back substitution, we get $$ \rho(t+\Delta t) = \rho(t) -i \int_t^{\Delta t} [V(t_1),\rho(t)]\ \text{d}t_1 - \int_t^{\Delta t} \int_t^{t_1} [V(t_1),[V(t_2),\rho(t_2)]]\ \text{d}t_2\ \text{d}t_1 . $$ Note that the first order term now contains $\rho(t)$ in the commutator. After the next back substitution there will be a $\rho(t)$ in the double commutator as well, and so forth. Obvious, the integration variable needs to be different for each back substitution and the integration boundaries also need to change to represent that which is being substituted.

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