3.1.2 Born-Markov approximation

Starting with the extended Liouville-von Neumann equation (27), we perform the trace TrB over the variables of the bath (see Eq. (32)),

$$\frac{\partial}{\partial t}\rho\_s(t) - \frac{1}{\mathbf{i}\,\hbar}[H\_s,\rho\_s(t)] = \frac{1}{\mathbf{i}\,\hbar}\mathbf{T}\mathbf{r}\_\mathbf{B}[H\_{\rm int},\rho(t)]\tag{34}$$

since the remaining terms disappear and <sup>1</sup> <sup>i</sup> <sup>ℏ</sup> TrBðHBρðÞ�t ρð Þt HBÞ ¼ 0 because of cyclic invariance of the trace TrB. To obtain a closed equation for ρsð Þt , the full nonequilibrium statistical operator ρð Þt occurring on the right-hand side has to be eliminated.

For this, we calculate the time evolution of the irrelevant part of the statistical operator ΔρðÞ¼ t ρðÞ�t ρrelð Þt ,

$$\frac{\partial}{\partial t} \Delta \rho(t) = \frac{\partial}{\partial t} \rho(t) - \left[\frac{\partial}{\partial t} \rho\_s(t)\right] \rho\_\mathcal{B} \tag{35}$$

inserting the time evolution for ρð Þt (8) and ρsð Þt (34) given above:

$$\rho \left( \frac{\partial}{\partial t} + \varepsilon \right) \Delta \rho(t) = \frac{1}{\mathbf{i}\hbar} [H, \rho(t)] - \frac{1}{\mathbf{i}\hbar} [H\_{\mathrm{i}}, \rho\_{\mathrm{i}}(t)] \rho\_{\mathrm{B}} - \rho\_{\mathrm{B}} \frac{1}{\mathbf{i}\hbar} \mathrm{Tr}\_{\mathrm{B}}[H\_{\mathrm{int}}, \rho(t)]. \tag{36}$$

We eliminate ρðÞ¼ t ΔρðÞþt ρsð Þt ρ<sup>B</sup> and collect all terms with Δρð Þt on the left-hand side. We can assume that h i Hint <sup>B</sup> ¼ TrB Hintρ<sup>B</sup> ð Þ¼ 0 because the heat bath do not exert external forces on the system (if not, replace Hs by Hs þ h i Hint <sup>B</sup> and Hint by Hint � h i Hint B) so that also TrB Hint; ρ<sup>B</sup> ð Þ ½ � ρsðÞ¼ t 0 and the last term �ρ<sup>B</sup> 1 <sup>i</sup><sup>ℏ</sup> TrBf g <sup>H</sup>intρ<sup>B</sup> <sup>ρ</sup>sðÞþ<sup>t</sup> <sup>ρ</sup>Bρsð Þ<sup>t</sup> <sup>1</sup> <sup>i</sup><sup>ℏ</sup> TrBf g ρBHint vanishes. Also, the term 1 <sup>i</sup><sup>ℏ</sup> HB; ρsð Þt ρ<sup>B</sup> ½ � disappears since HB; ρ<sup>B</sup> ½ �¼ 0. We obtain

$$\left(\frac{\partial}{\partial t} + \varepsilon\right)\Delta\rho(t) - \frac{1}{\mathbf{i}\,\hbar}[(H\_\mathrm{i} + H\_{\mathrm{int}} + H\_\mathrm{B}), \Delta\rho(t)] + \rho\_\mathrm{B}\frac{\mathbf{1}}{\mathbf{i}\,\hbar}\mathrm{Tr}\_\mathrm{B}[H\_{\mathrm{int}}, \Delta\rho(t)] = \frac{1}{\mathbf{i}\,\hbar}[H\_{\mathrm{int}}, \rho\_\mathrm{i}(t)\rho\_\mathrm{B}].\tag{37}$$

The deviation Δρð Þt vanishes when Hint ! 0. In lowest order with respect to Hint, the solution is found as

$$\Delta\rho(t) = \int\_{-\infty}^{t} \mathbf{d}t' e^{-c(t-t')} e^{\frac{i}{\hbar}(t-t')(H\_t+H\_\mathbb{R})} \frac{\mathbf{1}}{\mathbf{i}\hbar} [H\_{\text{int}}, \rho\_\circ(t')\rho\_\circ] e^{-\frac{i}{\hbar}(t-t')(H\_t+H\_\mathbb{R})}.\tag{38}$$

Inserting the solution (38) into the equation of motion of ρsð Þt (34), a closed equation of evolution is obtained eliminating ρð Þt . In the lowest (second) order with respect to the interaction considered here, memory effects are neglected. We can use the unperturbed dynamics to replace ρ<sup>s</sup> t <sup>0</sup> ð Þ¼ <sup>e</sup>� <sup>1</sup> <sup>i</sup> <sup>ℏ</sup> t�t <sup>0</sup> ð ÞHs ρsð Þt e 1 <sup>i</sup> <sup>ℏ</sup> t�t <sup>0</sup> ð ÞHs and <sup>H</sup>intð Þ¼ <sup>τ</sup> <sup>e</sup>� <sup>1</sup> <sup>i</sup> <sup>ℏ</sup>τð Þ HsþH<sup>B</sup> Hinte 1 <sup>i</sup> <sup>ℏ</sup>τð Þ HsþH<sup>B</sup> so that after a shift of the integration variable

$$\frac{\partial}{\partial t}\rho\_s(t) - \frac{1}{\mathrm{i}\hbar}[H\_\mathrm{i},\rho\_s(t)] = -\frac{\mathrm{1}}{\hbar^2} \int\_{-\infty}^0 \mathrm{d}\tau \, e^{\mathrm{cr}} \mathrm{Tr}\_\mathrm{B}[H\_\mathrm{int},[H\_\mathrm{int}(\tau),\rho\_s(t)\rho\_\mathrm{B}]] = D[\rho\_s(t)]. \tag{39}$$

This result is described as quantum master equation in Born approximation. For higher orders of Hint see [4, 5].
