# Dyson expansion for the density matrix

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?

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!

• What do you mean? To me, $(*)$ is simply a restatement of Schrödinger's equation, I am not aware of any first order approximation in $(*)$. Furthermore, the $\mathcal{O}(\alpha^3)$ term in (**) bothers me as there is an $\alpha^3$ inside the integral via the $\rho(t)$... Feb 12, 2021 at 12:01
• I did not say that (*) is an approximation; its solutions tells you that the difference between density matrices at different times is of order $\alpha$. Remember the definition of big O notation. The fact that there is an $\mathcal O(\alpha^3)$ doesn't preclude the presence of higher orders in the integral.
– tbt
Feb 12, 2021 at 21:04

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-(**).

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.