Dyson expansion for the density matrix

Question

I am following these notes and I am stuck on going from equation (37) to (38). In a nutshell, given $$ \frac{d \tilde{\rho}(t)}{dt} =-i\alpha[\tilde{H}_I(t),\tilde{\rho}(t)], \quad (*) $$ where $\tilde{A}$ is an operator in the interaction picture with $H_T=H_0+\alpha H_I$. This equation has the standard solution $$ \tilde{\rho}(t) = \tilde{\rho}_0-i\alpha\int_0^tds [\tilde{H}_I(s),\tilde{\rho}(s)]. $$ We can iterate $(*)$ with the above definition to yield $$ \frac{d \tilde{\rho}(t)}{dt} =-i\alpha[\tilde{H}_I(t),\tilde{\rho}_0]-\alpha^2 \int_0^tds [\tilde{H}_I(s),[\tilde{H}_I(s),\tilde{\rho}(s)]]. $$ One wishes to eliminate the dependance of $\rho$ on all previous times so we take advantage that $\alpha$ is assumed small to iterate infinitely, obtaining $$ \frac{d \tilde{\rho}(t)}{dt} =-i\alpha[\tilde{H}_I(t),\tilde{\rho}_0]-\alpha^2 \int_0^tds [\tilde{H}_I(s),[\tilde{H}_I(s),{\tilde{\rho}(t)}]] +\mathcal{O}(\alpha^3), \quad (**) $$ where now the integral doesnt integrate $\rho$ for all times.

That is what I don’t understand, I expected $(**)$ to be instead $$ \frac{d \tilde{\rho}(t)}{dt} =-i\alpha[\tilde{H}_I(t),\tilde{\rho}_0]-\alpha^2 \int_0^tds [\tilde{H}_I(s),[\tilde{H}_I(s),{\tilde{\rho}_0}]] +\mathcal{O}(\alpha^3), $$ why isn’t this the case?

Answer 1

Equation (*) says that the difference $\bar \rho(t+\delta t)-\bar \rho(t)$ is $\mathcal O(\alpha)$ , which is why the substitution from your third formula to (**) is valid. Of course you could argue the same for $\bar \rho_0$ and you would obtain another valid formula as far as i can tell; the difference between the two results you mention is all encapsulated in the $\mathcal O(\alpha^3)$ term.
For a wiser insight on why (**) is the useful one, I guess you will have to follow your notes!

Answer 2

I am simply explaining the fine and sound other answer by @tbt.

$$ \tilde{\rho}(t) = \tilde{\rho}_0-i\alpha\int_0^tds [\tilde{H}_I(s),\tilde{\rho}(s)] $$ is equivalent to $$ \tilde{\rho}(t) = \tilde{\rho}(s)-i\alpha\int_s^td\tau [\tilde{H}_I(\tau),\tilde{\rho}(\tau)] = \tilde{\rho}(s) +O(\alpha). $$ This is what you insert in
$$ \frac{d \tilde{\rho}(t)}{dt} =-i\alpha[\tilde{H}_I(t),\tilde{\rho}_0]-\alpha^2 \int_0^tds [\tilde{H}_I(t),[\tilde{H}_I(s),\tilde{\rho}(s)]], \tag{38} $$ so the second term becomes $$ -\alpha^2 \int_0^tds [\tilde{H}_I(t),[\tilde{H}_I(s),\tilde{\rho}(t)]] +O(\alpha^3), \tag{39} $$ completely equivalent to the expression at 0 you expected, provided the first Hamiltonian is at t, not s; you made a transcription mistake in your (38), pre-(**).

Answer 3

According to “The Theory of Quantum Open Systems” by Breuer and Petruccione.

[…] $$ \frac{d \tilde{\rho}(t)}{dt} =-i[\tilde{H}_I(t),\tilde{\rho}_0]-\int_0^tds [\tilde{H}_I(t),[\tilde{H}_I(s),\tilde{\rho}(s)]]. $$ In order to simplify the above equation further we perform the Markovian approximation, in which the integrand $\tilde{\rho}(s)$ is replaced by $\tilde{\rho}(t)$. In this way we obtain an equation of motion in which the time development of the state of the system at time $t$ only depends on the present state $\tilde{\rho}(t)$, $$ \frac{d \tilde{\rho}(t)}{dt} =-i[\tilde{H}_I(t),\tilde{\rho}_0]-\int_0^tds [\tilde{H}_I(t),[\tilde{H}_I(s),\tilde{\rho}(t)]]. $$ This equation is called the Redfield equation […]

So it is just a Markovian approximation.