# Explicit calculation of $\rho_{S}(t) := \textrm{Tr}_{\textrm{oscillator}}\left(e^{-itH}\rho(0)e^{itH}\right)$

I'm trying to calculate $\rho_{S}(t) := \textrm{Tr}_{\textrm{oscillator}}\left(e^{-itH}\rho(0)e^{itH}\right)$ for the following specific case, which is an energy conserving interaction with only one oscillator, so my Hermitian Matrix is:, $$H = \omega\sigma_{z} + \omega_{o}a^{*}a + V\otimes(ga^{*} + \bar{g}a),$$ where $\omega_{o}$ is initial angular velocity, $a \in \mathbb{C}, V \in \mathbb{R}$, $\rho(0) = \rho_{S}(0)\otimes\rho_\beta^{1}\otimes\rho_{\beta}^{2}\otimes\dots$, $g$ is a form factor, and, of course, $$\sigma_{z} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}$$ To calculate $\rho_{S}(t)$ explicitly I need to first simplify the following expression: $$e^{-it[\omega\sigma_{z} + \omega_{0}a^{*}a + V\sigma_{z}\otimes(ga^{*} + \bar{g}a)]}\left(\rho_{S}\otimes\rho_{\beta}\right)e^{it[\omega\sigma_{z} + \omega_{0}a^{*}a + V\sigma_{z}\otimes(ga^{*} + \bar{g}a)]}$$ Now The person that attempted to explain this process to me suggested that I make use of the fact that $$\sigma_{z}\lvert\uparrow\rangle = \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}\begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} = \lvert\uparrow\rangle$$ and $$\sigma_{z}\lvert\downarrow\rangle = \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}\begin{pmatrix} 0 \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \\ \end{pmatrix} = -\lvert\downarrow\rangle,$$ along with the fact that $\lvert\uparrow\rangle\langle\uparrow\rvert + \lvert\downarrow\rangle\langle\downarrow\rvert = \mathbb{I}$ in the following way: $$e^{-it[\omega\sigma_{z} + \omega_{0}a^{*}a + V\sigma_{z}\otimes(ga^{*} + \bar{g}a)]}\left(\rho_{S}\otimes\rho_{\beta}\right)e^{it[\omega\sigma_{z} + \omega_{0}a^{*}a + V\sigma_{z}\otimes(ga^{*} + \bar{g}a)]} = \\ e^{-it[\omega +\omega_{0}a^{*}a + V(ga^{*} + \bar{g}a)]}\lvert\uparrow\rangle\langle\uparrow\rvert(\rho_{S}(0)\otimes\rho_{\beta})\lvert\downarrow\rangle\langle\downarrow\rvert e^{it[-\omega +\omega_{0}a^{*}a - V(ga^{*} + \bar{g}a)]}$$ Now, I understand the basic premise here, that this lets us remove some matrix terms from our exponential, but here arises my first question: I understand that this is a legal move because of $\lvert\uparrow\rangle\langle\uparrow\rvert + \lvert\downarrow\rangle\langle\downarrow\rvert = \mathbb{I}$, but I was told there are three other cases I need to consider to have a complete picture of the dynamics. It seems pretty clear, to me, that one of these cases is going to be $e^{-it[\dots]}\lvert\downarrow\rangle\langle\downarrow\rvert(\rho_{S}(0)\otimes\rho_{\beta})\lvert\uparrow\rangle\langle\uparrow\rvert e^{it[\dots]}$. But I'm not certain about the other two: $$e^{-it[\dots]}\lvert\downarrow\rangle\langle\downarrow\rvert(\rho_{S}(0)\otimes\rho_{\beta})\lvert\downarrow\rangle\langle\downarrow\rvert e^{it[\dots]}?$$ $$e^{-it[\dots]}\lvert\uparrow\rangle\langle\uparrow\rvert(\rho_{S}(0)\otimes\rho_{\beta})\lvert\uparrow\rangle\langle\uparrow\rvert e^{it[\dots]}?$$ $$e^{-it[\dots]}\lvert\uparrow\rangle\langle\downarrow\rvert(\rho_{S}(0)\otimes\rho_{\beta})\lvert\downarrow\rangle\langle\uparrow\rvert e^{it[\dots]}?$$ $$e^{-it[\dots]}\lvert\downarrow\rangle\langle\uparrow\rvert(\rho_{S}(0)\otimes\rho_{\beta})\lvert\uparrow\rangle\langle\downarrow\rvert e^{it[\dots]}?$$ How does one know which are the four cases needed?

Proceeding with the example calculation, the terms were rearranged in the following manner: $$e^{-it[\omega +\omega_{0}a^{*}a + V(ga^{*} + \bar{g}a)]}\lvert\uparrow\rangle\langle\uparrow\rvert(\rho_{S}(0)\otimes\rho_{\beta})\lvert\downarrow\rangle\langle\downarrow\rvert e^{it[-\omega +\omega_{0}a^{*}a - V(ga^{*} + \bar{g}a)]} = \\ e^{-2i\omega t}\langle\uparrow\rvert\rho_{S}(0)\rvert\downarrow\rangle\lvert\uparrow\rangle\langle\downarrow\rvert\otimes e^{-it[\omega_{0}a^{*}a + V(ga^{*} + \bar{g}a])}\rho_{\beta}e^{it[\omega_{0}a^{*}a + V(ga^{*} + \bar{g}a])}$$ I understand that some of this rearrangement of terms is allowed by properties of power series, but I am having a hard time breaking it down into smaller steps. I know that this whole expression can be simplified further via a polaron transformation, but I want to understand these basic steps first.

You are supposed to make use of the two-level identity, which means there can be only 4 terms to $\rho_S(0)$: $$\rho_S(0) = \mathbb{I}\;\rho_S(0)\; \mathbb{I} = \left(|\uparrow\rangle\langle\uparrow| + |\downarrow\rangle\langle\downarrow|\right) \rho_S(0) \left(|\uparrow\rangle\langle\uparrow| + |\downarrow\rangle\langle\downarrow|\right) = \\ = |\uparrow\rangle\langle\uparrow|\rho_S(0)|\uparrow\rangle\langle\uparrow|\; + \;|\uparrow\rangle\langle\uparrow|\rho_S(0)|\downarrow\rangle\langle\downarrow|\; + \\ +\;|\downarrow\rangle\langle\downarrow|\rho_S(0)|\uparrow\rangle\langle\uparrow| \; + \;|\downarrow\rangle\langle\downarrow|\rho_S(0)|\downarrow\rangle\langle\downarrow|\; =$$ $$\equiv \langle\uparrow|\rho_S(0)|\uparrow\rangle\;|\uparrow\rangle\langle\uparrow|\; + \; \langle\uparrow|\rho_S(0)|\downarrow\rangle\;|\uparrow\rangle\langle\downarrow|\; + \\ + \; \langle\downarrow|\rho_S(0)|\uparrow\rangle\;|\downarrow\rangle\langle\uparrow| \; + \; \langle\downarrow|\rho_S(0)|\downarrow\rangle\;|\downarrow\rangle\langle\downarrow|\;$$ Insert into $\rho_S(0)\otimes\rho_\beta$ and follow through.