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In Breuer's book, he deduces quantum master equation using following steps:

$(1). \frac{d}{dt}\rho(t)=-i[H_{I},\rho(t)]$

then the solution for equ.(1) can be written as

$(2).{\rho(t)}=\rho(0)-i\int_{0}^{t}ds[H_{I},\rho(t)]$

By plugging equ(2) to equ(1), we will have

$(3).\frac{d}{dt}\rho_{s}(t)=-\int_{0}^{t}ds\ \ tr_{B}[H_{I}(t),[H_{I}(s),\rho(s)]]$

where $H_{I},\rho$ are interaction hamiltonian and density matrix for the whole system in interaction picture. $tr_{B}$ is partial trace over reservoir.

In the book, the authors say to obtain equ(3),

$(4).tr_{B}[H_{I},\rho(0)]=0$ is assumed.

So my question is why do we assume equ.(4), is there any physics behind it? Like we take Bohn approximation, the coupling between between the system and the reservoir is small?

xiang sun
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1 Answers1

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I do not know the book and the derivation given therein, but I guess, the physical assumption behind eq. (4) is that the system and the reservoir are uncorrelated at time zero. This assumption probably makes the subsequent calculations easier.

flaudemus
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  • Thanks for your answer. Yes, it is for easier calculations. I found in a lecture note that if it's not zero, the nonzero part can be absorbed into the Hamiltonian of the system. – xiang sun Mar 06 '19 at 15:46
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    @xiangsun can you provide an answer to your own question on how the nonzero part can be absorbed into the hamiltonian? – Quantumwhisp Apr 03 '23 at 11:36