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To derive quantum master equation using perturbative and born approximation we get the equation for state evolution as (from Liouville-von Neumann)

$$ \frac{d}{dt}\rho(t)=-i[H_I(t),\rho(0)]-\int_0^t ds[H_I(t),[H_I(s),\rho(0)]]+i\int_0^t \int_0^sdsdt'[H_I(t),[H_I(s),[H_I(t'),\rho(0)]]]+... $$

Using Born and second order approximation and also tracing the entire equation for all bath degree of freedom

$$ \frac{d}{dt}\rho_s(t)=-(i)tr_E([H_I(t),\rho(0)])-\int_0^t ds tr_E([H_I(t),[H_I(s),\rho(0)]]) $$

In Breuer and many other reference book they "assume" that $tr_E([H_I(t),\rho(0)])=0$. I am trying to understand this assumption.

If we breakdown interaction Hamiltonian as tensor product of observable in system $S$ and in bath $E$ then

$$ H_I(t)=S(t)\otimes E(t)$$ $$tr_E([S(t)\otimes E(t), \rho_S(0)\otimes \rho_E(0)])=S(t)\rho_S(0)\otimes tr_E(E(t)\rho_E(0))-\rho_S(0)S(t)\otimes tr_E(\rho_E(0)E(t))$$

One way to make this assumption true if one have $$ tr_E(E(t)\rho_E(0))=tr_E(\rho_E(0)E(t))= <E(t)>=0 $$

How can this be true ? I read some Weiss and somehow they connect it with noise term on Langevin equation which have zero expectation value but I don't understand how they make the "jump" to connect quantum bath interaction to noise in classical mechanics. Are they analog for one another or we use the noise in classical mechanics to work the problem of open quantum system because the field is more well established ?

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  • $\begingroup$ Not sure about applicability in your context, but often interaction hamiltonian is 2x2 matrix with zero diagonal and $\rho(0)$ is 2x2 matrix with single 1 for the ground state and all other entries are zero. Then you can verify the relation $tr_E([H_I,\rho(0)]=0$ explicitly, the commutator has zeros on its diagonal. $\endgroup$ Mar 16 at 16:01
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The question is how to justify assuming that this quantity is zero: $$ \mathrm{trace}_E\Big(\big[H_I(t),\rho(0)\big]\Big) $$ Short answer: Even if it isn't zero, we can absorb it into other terms.

Details: As suggested in the question, let's suppose $$ H_I(t)= S(t)\otimes E(t) $$ and $$ \rho(0) = \rho_S(0)\otimes \rho_E(0) $$ so that $$ \mathrm{trace}_E\Big(\big[H_I(t),\rho(0)\big]\Big) = \big[S(t),\rho_S(0)\big] \,\big\langle E(t)\big\rangle \tag{1} $$ with $$ \big\langle E(t)\big\rangle \equiv \mathrm{trace}\big(E(t)\rho_E(0)\big). $$ We can rewrite $H_I$ like this: \begin{align*} H_I(t) &= S(t)\otimes E(t) \\ &= \big(S(t) \otimes 1\big) \big\langle E(t)\big\rangle + S(t)\otimes \Big(E(t)-\big\langle E(t)\big\rangle\Big). \end{align*} In the last expression, the first term can be considered part of the $S$-only Hamiltonian, and the second term is a modified version of the interaction Hamiltonian. If we re-do the previous calculation that led to equation (1) using this modified interaction Hamiltonian, we get the same equation again except with $E(t)$ replaced by $E(t)-\langle E(t)\rangle$, whose expectation value (trace over the environment) is zero.

Reference: Pages 72-73 in section 8.4.3 of the MIT lecture notes for "Quantum Theory of Radiation Interactions" (https://ocw.mit.edu/courses/nuclear-engineering/22-51-quantum-theory-of-radiation-interactions-fall-2012/lecture-notes/MIT22_51F12_Ch8.pdf)

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