# Why can't I derive Lindblad equation in this way(by Taylor expand)?

Assume a system with a Hamiltonian

\begin{aligned} H & =H_s+H_b+H_{sb}\\ &=\hbar\omega_0 a^{\dagger}a+\hbar\omega\alpha^{\dagger}\alpha+\hbar(\kappa\alpha^{\dagger}a+\kappa^*\alpha a^{\dagger}) \end{aligned} And we can describe the motion of density matrix of system by \begin{aligned} \rho_s(t+dt)&=tr_b\{e^{-iHdt/\hbar}\rho_s(t)\otimes|0\rangle\langle0|e^{iHdt/\hbar}\}\\ &=\sum_{k}\langle k|e^{-iHdt/\hbar}\rho_s(t)\otimes|0\rangle\langle0|e^{iHdt/\hbar}|k\rangle \end{aligned} In the Karus representation, we can introduce $$M_k(dt)=\langle k|e^{-iHdt/\hbar}|0\rangle$$ then we can rewrite $$\rho_s(t+dt)=\sum_k M_k(dt)\rho_s(t) M_k^{\dagger}(dt)$$ We can calculate $M_k$ \begin{aligned} M_0(dt) &= e^{-iH_s dt/\hbar}\langle 0|(1+(-iH_{sb}dt/\hbar)+(-iH_{sb}dt/\hbar)^2/2)+o(dt^2)|0\rangle\\ &=e^{-iH_s dt/\hbar}\langle 0|(-\hbar^2|k|^2a^{\dagger}a/2)|0\rangle\\ &=e^{-iH_s dt/\hbar}(-\hbar^2dt^2|k|^2a^{\dagger}a/2) \end{aligned} In a same way, we can get $$M_1(dt) = e^{-iH_s dt/\hbar}(-i\hbar dt^2\kappa a)$$ and for $k>=2$ $$M_k(dt) = o(dt^2)$$ then, we get the equation \begin{aligned} \rho_s(t+dt)&=\rho_s(t)-(idt/\hbar)[H_s,\rho_s(t)]-dt^2(\hbar^2|\kappa|^2/2)(a^{\dagger}a\rho_s(t)\\ &+\rho_s(t)a^{\dagger}a+2a\rho_s(t)a^{\dagger})+o(dt^2) \end{aligned} We find the $\mathcal{L}(\rho)$ is ~$dt^2$, but in the Lindblad equation, this term should be ~ $dt$.

I can't find where is wrong. But it's really different from Lindblad equation.

• Furthermore, $dt_{}^{2}$ term doesn't look like "Trace Preserving". – Sunyam Jul 9 '18 at 20:04
• I am not sure if this is the main problem, but your calculation of $M_k$ seems incorrect, since $e^H \neq e^{H_s} e^{H_b} e^{H_{sb}}$ – Noiralef Jul 10 '18 at 6:12
• $e^H \neq e^{H_s} e^{H_b+H_{sb}}$ will not change the result.This introduce a correction $e^{[H_s,H_b+H_{sb}]/2}$~ $dt^2$ and if we expand it in $M_1$ and $M_0$, will find this correction term will result in 0. – W.Eadric Jul 10 '18 at 12:09