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

Question

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.

Answer 1

Your derivation is almost correct, and it highlights in what precise sense a dilated evolution (i.e. evolution of the larger closed system under $H=H_s+H_b+H_{sb}$) is at odds with the Lindblad equation if certain base assumptions are not met.

More precisely one can prove—as done in this paper (arXiv)—that an evolution of the form $\rho(t)={\rm tr}_E(e^{iHt}(\rho_0\otimes\rho_E)e^{-iHt})$ (assuming bounded $H$, this will become important later) at $t=0$ has first derivative${}^1$ $i[{\rm tr}_\omega(H),\cdot]$ and second derivative $2L$ where $$L:={\rm tr}_E(H((\cdot)\otimes \omega)H)-\frac12{\rm tr}_\omega(H^2)(\cdot)-\frac12(\cdot){\rm tr}_\omega(H^2)$$ which is precisely of Lindblad form as ${\rm tr}_E(H((\cdot)\otimes \omega)H)$ is the composition of completely positive maps, hence completely positive. In other words, Taylor expanding around $t=0$ yields $$ \rho(t)=\rho_0+it[{\rm tr}_\omega(H),\rho_0]+t^2 L(\rho_0) +o(t^2) $$ which is of the same form as what you found.

Thus the only mathematical way (so none of the usual Lindblad-justifying physical approximations) to resolve this incompatibility is to require that the system-environment Hamiltonian is unbounded. In such an unbounded scenario taking derivatives becomes a rather delicate matter—e.g., $\frac d{dt}e^{itH}$ is understood in a weaker sense and has to be evaluated more carefully—which is why the usual derivative manipulations produce "wrong" results, i.e. the above formulas for the first and second derivative will in general be wrong if $H$ is unbounded.

To highlight the importance of unboundedness: one can show that a (finite-dimensional) Lindblad evolution $e^{t\mathcal L}(\rho_0)$ can be cast in the form ${\rm tr}_E(e^{iHt}(\rho_0\otimes\rho_E)e^{-iHt})$ for some unbounded $H$ (so, in particular, the environment has to be infinite dimensional), cf. Ch. 9.4 in Davies' book "Quantum Theory of Open Systems", 1976. In other words an unbounded system-environment Hamiltonian allows to—and is actually necessary to—move dissipative effects from $O(t^2)$ to $O(t)$.

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_{1: Here ${\rm tr}_\omega(X)$ is the partial trace with respect to $\omega$ which is the partial trace in the Heisenberg picture. It is defined as the adjoint channel of the extension map $X\mapsto X\otimes\omega$}