Let us consider a initial total quantum system as $\rho(0) = \rho_S(0)\otimes\rho_B$ where $\rho_S(0)$ is initial open system and $\rho_B$ is density matrix for environment.

We can use partial trace to get open system after some time, $\rho_S(t) = \mathrm{tr}_B \{ U(t, 0) [\rho_S(0) \otimes \rho_B] U^{\dagger}(t, 0) \}$

This will be a dynamical map $V(t): S(\mathcal{H}_S) \mapsto S(\mathcal{H}_S)$ where $\rho_S(t) =V(t)\rho_S(0)$. And given $\rho_B=\sum_\alpha \lambda_\alpha | \phi_\alpha > <\phi_\alpha|$ .

The question is, I am not sure how in detail get the following

$V(t)\rho_S = \sum_{\alpha, \beta} W_{\alpha, \beta}(t) \rho_S W_{\alpha, \beta}^{\dagger}(t)^{}$ where $W_{\alpha, \beta}(t) = \sqrt{\lambda_\beta} <\phi_\alpha|U(t, 0)|\phi_\beta>$.

I tried to put $\rho_B$ into the $\rho_S(t)$ as

$\rho_S(t) = \mathrm{tr}_B \{ U(t, 0) [\rho_S(0) \otimes (\sum_\alpha \lambda_\alpha | \phi_\alpha > <\phi_\alpha|)] U^{\dagger}(t, 0) \}$

But how to proceed from here ?


You are on the right track. Just use the definition of the partial trace, which is $$\mathrm{tr}_B(A) = \sum_{\alpha}\langle\phi_{\alpha}\rvert A \lvert \phi_{\alpha}\rangle$$ for any operator $A$, where the $\lvert \phi_{\alpha}\rangle$ constitute a complete orthonormal basis for the environment Hilbert space. Using this in your final expression leads directly to the required result.

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  • $\begingroup$ just tried to do this question now, it is a simple exercise if you can write the initial state as just the product of the density matrices, assuming no correlation in the beginning, which I believe is the normal assumption $\endgroup$ – Jordan Simba Jul 13 '16 at 2:59

Usually when investigating the time evolution of a subsystem one finds that a memory term enters the formalism. I do not see such a term in your approach.

A standard procedure is to use the Zwanzig projector formalism

\begin{eqnarray*} \rho (0) &=&\rho _{S}(0)\otimes \rho _{B} \\ \rho (t) &=&U(t,0)\rho (0)U^{\dagger }(t,0) \\ \partial _{t}\rho (t) &=&-i[H,\rho (t)]=-iL\rho (t) \end{eqnarray*} Projector \begin{eqnarray*} Pf &=&\rho _{B}\mathrm{tr}_{B}f\Rightarrow \;P^{2}f=Pf,\;Q=1-P \\ P\rho (0) &=&\rho _{B}\mathrm{tr}_{B}\rho _{S}(0)\otimes \rho _{B}=\rho (0),\;Q\rho (0)=0 \end{eqnarray*} Zwanzig (http://en.wikipedia.org/wiki/Zwanzig_projection_operator) \begin{eqnarray*} \partial _{t}P\rho (t) &=&-iPLP\rho (t)-iPLQ\rho (t) \\ \partial _{t}Q\rho (t) &=&-iQLQ\rho (t)-iQLP\rho (t) \\ Q\rho (t) &=&\exp [-iQLQt]Q\rho (0)-i\int_{0}^{t}ds\exp [-iQLQ(t-s)]QLP\rho (s) \\ \partial _{t}P\rho (t) &=&-iPLP\rho (t)-\int_{0}^{t}dsPLQ\exp [-iQLQ(t-s)]QLP\rho (s) \end{eqnarray*} After Laplace transforming \begin{eqnarray*} \hat{f}(z) &=&\int_{0}^{\infty }dt\exp [izt]f(t),\;{Im}z>0, \\ \int_{0}^{\infty }dt\exp [izt]\partial _{t}f(t) &=&-f(0)-iz\hat{f}(z) \end{eqnarray*} $$ -P\rho (0)-izP\hat{\rho}(z)=-iPLP\hat{\rho}(z)-PLQ[z-QLQ]^{-1}QLP\hat{\rho}% (z) $$ Another way to obtain this is first to take the Laplace transform of $\rho (t)=\exp [-iLt]\rho (0)$ and then use the Feshbach projection formula.

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  • 1
    $\begingroup$ There is nothing wrong with what the OP has written. The memory terms you are referring to appear when one tries to write a differential equation for the reduced density matrix, which is not what the question is about in this case. Rather, the OP is trying to derive a specific form of the exact (formal) solution of the dynamical equations you have described here. $\endgroup$ – Mark Mitchison Jun 2 '14 at 22:45

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