# Relationship between the Lindblad Equation and Redfield Equation

Both the Lindblad and Redfield Equation both model the open quantum system dynamics given a Hamiltonian and some operators. What is the relationship between the two equations? How can they transformed from Lindblad to Redfield, given the Hamiltonian, and Lindblad operators? Is this not possible?

• Actually the transformation is always from Redfield to Lindblad. The approximation imposed on the environment coupling terms of the Redfield eq is a so called long-time one that cannot be (properly) reversed from the resulting Lindblad dissipators. – udrv Jan 13 '16 at 21:34
• @udrv could you perhaps expand into an answer? – TanMath Jan 13 '16 at 22:04
• On a 2nd thought, it may still be possible to reconstruct a Redfield from a Lindblad, regardless of the long-time approx. See my answer below. – udrv Jan 14 '16 at 12:39

Usually it is the Redfield equation that is converted to Lindblad form. Doing the reverse depends strongly on the Lindblad form and the available interpretation of the Lindblad operators.

The conversion from Redfield to Lindblad goes like this:

Given a system $S$ with Hamiltonian $H_S$, interacting with an environment $E$ with Hamiltonian $H_E$, let the total hamiltonian read $$H = H_S + H_E + V$$ where $V = \sum_k{S_k\otimes E_k}$ is the system-environment interaction. Here operators $S_k$ act on the system, the $E_k$-s act on the environment, and we can always arrange things so that all $S_k$-s and $E_k$-s are self-adjoint, $S_k = S^\dagger_k$, $E_k = E^\dagger_k$. Assuming weak coupling and a large environment in thermal equilibrium, so that the total density matrix can be approximated as $$\rho(t) \approx \rho_S(t) \otimes \rho_E = \rho_S(t) \otimes \frac{e^{-\beta H_E}}{Tr_E(e^{-\beta H_E})}$$ the Redfield equation for the system's reduced density matrix in interaction picture is $$\frac{d\rho_S}{dt} = \frac{1}{(i\hbar)^2}\int_0^t{d\tau\;Tr_E\left(\;\left[V_I(t), \left[V_I(\tau) , \rho_S(t)\otimes \rho_E\right]\right]\;\right) }$$ (See for instance these MIT notes on "Open Quantum Systems").

In preparation for the desired transformation to Lindblad, let us take into account the separable form of $V$ and bring the integrand to the form $$Tr_E\left(\;\left[V_I(t), \left[V_I(\tau) , \rho_S(t)\otimes \rho_E\right]\right]\;\right) = \\ Tr_E\left(\;\left[\sum_j{S_j(t)E_j(t)}, \left[\sum_k{S_k(\tau) E_k(\tau)},\rho_S(t)\otimes \rho_E\right]\right]\;\right) = \\ = \sum_{j,k}{Tr_E\left(\;S_j(t)E_j(t)S_k(\tau)E_k(\tau)\rho_S(t)\otimes \rho_E - S_j(t)E_j(t)\rho_S(t)\otimes \rho_E S_k(\tau)E_k(\tau) - \\ -S_k(\tau)E_k(\tau)\rho_S(t)\otimes \rho_E S_j(t)E_j(t) + \rho_S(t)\otimes \rho_E S_k(\tau)E_k(\tau) S_j(t)E_j(t)\;\right)} = \\ = \sum_{j,k}{\left(\langle E_j(t)E_k(\tau)\rangle S_j(t)S_k(\tau)\rho_S(t) - \langle E_k(\tau)E_j(t)\rangle S_j(t)\rho_S(t)S_k(\tau) - \\ -\langle E_j(t)E_k(\tau)\rangle S_k(\tau)\rho_S(t)S_j(t) + \langle E_k(\tau)E_j(t)\rangle \rho_S(t)S_k(\tau)S_j(t)\right)} = \\ = \sum_{j,k}{\left(\langle E_j(t)E_k(\tau)\rangle \left[S_j(t),S_k(\tau)\rho_S(t)\right] - \langle E_k(\tau)E_j(t)\rangle \left[S_j(t),\rho_S(t)S_k(\tau)\right] \right)}$$ Also, the thermal state of the environment means that the correlation functions are of the form $$\langle E_j(t)E_k(\tau)\rangle = G_{jk}(t - \tau)$$ and have the property that $$G_{jk}(t) = G^*_{kj}(-t)$$ With this, the Redfield equation becomes $$\frac{d\rho_S}{dt} = \frac{1}{(i\hbar)^2}\int_0^t{d\tau\; \sum_{j,k}{\left(G_{jk}(t - \tau) \left[S_j(t),S_k(\tau)\rho_S(t)\right] - G^*_{jk}(t - \tau) \left[S_j(t),\rho_S(t)S_k(\tau)\right] \right)}}$$

To retrieve a Lindblad form, we now assume that the correlation time of the environment is short, such that all $G_{jk}(t - \tau)$ are significantly different from zero only for $t \sim \tau$ and the integral can be extended to infinity. If we introduce Fourier transforms for the system couplings, $$S_k(t) = \int{d\omega\;\tilde S_k(\omega)e^{-i\omega t}}$$ we can rewrite the integral as $$\int_0^t{d\tau\; \sum_{j,k}{\left(G_{jk}(t - \tau) \left[S_j(t),S_k(\tau)\rho_S(t)\right] - G^*_{jk}(t - \tau) \left[S_j(t),\rho_S(t)S_k(\tau)\right] \right)}} = \\ \int{d\omega}\int{d\omega'}\sum_{j,k}{\int_0^\infty{d\tau\;\left(G_{jk}(t - \tau)e^{i\omega'(t-\tau)}e^{-i(\omega + \omega')t}\left[\tilde S_j(\omega), \tilde S_k(\omega') \rho_S(t)\right] - \\ - G^*_{jk}(t - \tau)e^{i\omega'(t-\tau)}e^{-i(\omega + \omega')t}\left[\tilde S_j(\omega), \rho_S(t)\tilde S_k(\omega') \right]\right)}} = \\ \int{d\omega}\int{d\omega'}\sum_{j,k}{e^{-i(\omega + \omega')t}\left(\tilde G_{jk}(-\omega') \left[\tilde S_j(\omega), \tilde S_k(\omega') \rho_S(t)\right] - \tilde G^*_{jk}(\omega') \left[\tilde S_j(\omega), \rho_S(t)\tilde S_k(\omega') \right]\right) }$$ where $G_{jk}(\omega) = \int_0^\infty{d\tau\;G_{jk}(\tau)e^{-i\omega\tau}}$. Since at long times $t$ the exponential factor $e^{-i(\omega + \omega')t}$ oscillates rapidly, the main contributions to the integral come from terms with $\omega + \omega' = 0$, so we are left with
$$\int{d\omega}\int{d\omega'}\sum_{j,k}{e^{-i(\omega + \omega')t}\left(\tilde G_{jk}(-\omega') \left[\tilde S_j(\omega), \tilde S_k(\omega') \rho_S(t)\right] - \tilde G^*_{jk}(\omega') \left[\tilde S_j(\omega), \rho_S(t)\tilde S_k(\omega') \right]\right) } \approx \\ \int{d\omega}\sum_{j,k}{\left(\tilde G_{jk}(\omega) \left[\tilde S_j(\omega), \tilde S_k(-\omega) \rho_S(t)\right] - \tilde G^*_{jk}(-\omega) \left[\tilde S_j(\omega), \rho_S(t)\tilde S_k(-\omega) \right]\right) }$$ We can simplify more if we notice that $G_{jk}(t) = G^*_{kj}(-t)$ implies $$\tilde G_{jk}(\omega) = \tilde G^*_{kj}(\omega)$$
Now change $\omega \rightarrow -\omega$ and $j \leftrightarrow k$ in the second term of the integrand to obtain $$\int{d\omega}\sum_{j,k}{\left(\tilde G_{jk}(\omega) \left[\tilde S_j(\omega), \tilde S_k(-\omega) \rho_S(t)\right] - \tilde G^*_{jk}(-\omega) \left[\tilde S_j(\omega), \rho_S(t)\tilde S_k(-\omega) \right]\right) } = \\ = \int{d\omega}\sum_{j,k}{\left(\tilde G_{jk}(\omega) \left[\tilde S_j(\omega), \tilde S_k(-\omega) \rho_S(t)\right] - \tilde G^*_{kj}(\omega) \left[\tilde S_k(-\omega), \rho_S(t)\tilde S_j(\omega) \right]\right) } = \\ = \int{d\omega}\sum_{j,k}{\tilde G_{jk}(\omega) \left( \left[\tilde S_j(\omega), \tilde S_k(-\omega) \rho_S(t)\right] - \left[\tilde S_k(-\omega), \rho_S(t)\tilde S_j(\omega) \right]\right) }$$ Expand the commutators and find that the evolution equation is now in Lindblad form: $$\frac{d\rho_S}{dt} = \\ = \frac{1}{(i\hbar)^2}\int{d\omega}\sum_{j,k}{\tilde G_{jk}(\omega) \left[ \tilde S_j(\omega)\tilde S_k(-\omega) \rho_S(t) + \rho_S(t)\tilde S_j(\omega)\tilde S_k(-\omega) - 2 \tilde S_k(-\omega) \rho_S(t)\tilde S_j(\omega)\right]}$$ We can still tidy up by taking advantage of the fact that for given $\omega$ the factors $\{G_{jk}(\omega)\}_{j,k}$ form a hermitian matrix. So we can write $$G_{jk}(\omega) = \sum_n{c_{jn}(\omega)\lambda_n(\omega) c^*_{kn}(\omega)}$$ with $\lambda_n(\omega) = \lambda^*_n(\omega)$ and $\sum_n{c_{jn}(\omega)c^*_{kn}(\omega) = \delta_{jk}}$. Note also that $S_k(t) = S^\dagger_k(t)$ means $\tilde S_k(-\omega) = \tilde S^\dagger(\omega)$. Substitute everything in the integrand, and redistribute the sums to obtain the final Lindblad form as $$\frac{d\rho_S}{dt} = \\ = \frac{1}{(i\hbar)^2}\int{d\omega}\sum_n{\lambda_n(\omega) \left[ L_n(\omega) L^\dagger_n(\omega) \rho_S(t) + \rho_S(t) L_n(\omega)L^\dagger_n(\omega) - 2 L^\dagger_n(\omega) \rho_S(t) L_n(\omega)\right]}$$ where $L_n(\omega) = \sum_j{c_{jn}}\tilde S_j(\omega)$.

On the conversion from Lindblad to Redfield:

The last Lindblad form above by itself wouldn't be sufficient to reverse to the Redfield equation, as the information on the particular form of the system couplings $S_j$ and correlation functions $G_{jk}$ is lost. The form that retains the Fourier transforms $\tilde G_{jk}(\omega)$ and $\tilde S_j(\omega)$ is much more useful for this purpose. The information on the specific environment couplings is still lost. But it should be possible to retrieve the $S_j$-s and $G_{jk}$-s, which is sufficient to reconstruct the Redfield equation, regardless of the additional approximations involved in deriving the Lindblad equation.

• isn't the 'conjugate equality' property of the correlation functions dependent on the operators Ej being Hermitian? When the Ej are not Hermitian, how should one proceed? – transistor Dec 24 '16 at 18:47
• The reason why "we can safely assume that all $S_k$-s and $E_k$-s are self-adjoint" is that any operator $A$ can be written as a sum of 2 self-adjoint terms (actually hermitian, but with the right domain and range self-adjoint), as in $A = B + iC$, $B = B^\dagger = (A+A^\dagger)/2$, $C=C^\dagger = i(A^\dagger-A)/2$. If some $S_k$ and/or $E_k$ are not self-adjoint, simply use this decomposition and take into account that $V$ itself must be self-adjoint, i.e. it must contain a term in $S_k^\dagger$ and $E_k^\dagger$ too. The surviving terms will be products of self-adjoint operators. – udrv Dec 24 '16 at 23:29