Hermitian and non-Hermitian jump operators in Lindblad master equation

Question

Is there a way of rotating non-Hermitian jump operators for a Lindblad master equation (LME) to a basis where they are Hermitian? In other words, I have a (diagonal) LME:

$$ \dot{\rho} = -i [\mathcal{H}, \rho] + \sum_{\alpha} \gamma_{\alpha} \left[ L_{\alpha} \rho L_{\alpha}^{\dagger} - \frac{1}{2}\left\{ L_{\alpha}^{\dagger} L_{\alpha}, \rho \right\} \right] $$

where the jump operators are non-Hermitian. The simplest case is for just a single fermionic site with $\mathcal{H} = \epsilon f^{\dagger} f$ and two jump operators $L_1 = f^{\dagger}, L_2 = f, \gamma_1 = \sqrt{\Gamma_{\mathrm{in}}}, \gamma_2 = \sqrt{\Gamma_{\mathrm{out}}}$. I would like to rewrite the master equation in a non-diagonal form:

$$ \dot{\rho} = -i [\mathcal{H}, \rho] + \sum_{\mu, \nu} \gamma_{\mu \nu} \left[ A_{\mu} \rho A_{\nu}^{\dagger} - \frac{1}{2} \left\{ A_{\nu}^{\dagger} A_{\mu}, \rho \right\} \right] $$

but with the additional property that the operators $A$ are Hermitian, ie $A_{\mu} = A_{\mu}^{\dagger}$. Is it possible to do so? In principle, the $L_{\alpha}$'s and $A_{\mu}$'s are related by unitary transform $U$, which also rotates the coefficient matrix $(\gamma_{\mu \nu})$ to its diagonal form:

$$ U \gamma U^{\dagger} = \mathrm{diag}(\gamma_1, \gamma_2, \cdots) $$ $$ A_{\mu} = \sum_{\alpha} U_{\alpha \mu} L_{\alpha} $$

I have tried to determine the form of $U$ and $(\gamma_{\mu \nu})$, but I encounter inconsistencies (eg $U$ turns out to be non-unitary), so I have a hunch that this is not doable.

The reason why I'm trying to rewrite the jump operators in a Hermitian way is that I want to compare the results obtained with the LME to the more general Redfield master equation (RME):

$$ \dot{\rho} = -i \left[ \mathcal{H}, \rho \right] + \sum_{\mu, \nu} \int_0^{\infty} \mathrm{d} \tau \: \left\{ \Gamma_{\nu \mu}^{\beta}(\tau) \left[ e^{-i \tau \mathcal{H}} A_{\mu} e^{i \tau \mathcal{H}} \rho, A_{\nu} \right] + h.c. \right\} $$

In the standard derivation of the RME the interaction operators $A_{\mu}$ are always assumed to be Hermitian. In principle, from the RME one could obtain the LME by using bath correlation function of the form $\Gamma_{\nu \mu}^{\beta}(\tau) = \Gamma_{\nu \mu} \: \delta(\tau + 0^+)$, such that the time integration disappears, see for instance T. Prosen and B. Zunkovic, New J. Phys. 12, 025016 (2010).

Answer 1

Writing the Lindblad equation in non-diagonal form with Hermitian operators is always possible. This follows from the simple observation that an arbitrary operator can be expanded in a Hermitian operator basis $$L_\alpha = \sum_\mu c_{\alpha\mu} A_\mu,$$ where $A_\mu = A_\mu^\dagger$. For example, one can choose the basis \begin{align} s_{mn} &= \frac{1}{2}|m\rangle\langle n| + {\rm h.c.}, \\ a_{mn} & = \frac{{\rm i}}{2}|m\rangle\langle n| + {\rm h.c.}. \end{align} Clearly, an arbitrary operator $B = \sum_{mn} B_{mn}|m\rangle\langle n| = \sum_{mn} B_{mn}(s_{mn} - {\rm i}a_{mn})$ can be expanded in this basis. Plugging in the expansion above, one obtains the non-diagonal form with the coefficient matrix $\gamma_{\mu \nu} = \sum_\alpha c_{\nu \alpha}^* \gamma_\alpha c_{\alpha \mu}$. In your specific example, one could choose the basis $$ \begin{pmatrix} A_1 \\ A_2 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ {\rm i} & {\rm -i}\end{pmatrix}\begin{pmatrix} f \\ f^\dagger\end{pmatrix},$$ and here the coefficient matrix is manifestly unitary.

However, it is worth emphasising that the Lindblad operators in the diagonal form cannot be made Hermitian in general. To see this, note that if $L_\alpha = L_\alpha^\dagger$ then the maximally mixed state $\rho(t) = \mathbb{1}/d$ is a solution for all $t$ ($d=$ Hilbert space dimension), i.e. it satisfies $\dot{\rho} = \mathcal{L}\rho = 0$, where $\mathcal{L}$ is the Liouvillian superoperator describing the right-hand side of your master equation. As a result, the master equation generates a unital channel: $${\rm e}^{\mathcal{L}t} \mathbb{1} = \mathbb{1}, $$ where ${\rm e}^{\mathcal{L}t}$ is the time evolution operator, i.e. $\rho(t) = {\rm e}^{\mathcal{L}t} \rho(0)$. Assuming that the master equation has a unique steady state, the unital condition implies that $$\lim_{t\to\infty}{\rm e}^{\mathcal{L}t} \rho(0) = \mathbb{1},$$ for any initial condition $\rho(0)$. Most dissipative evolutions are not unital, e.g. any kind of thermalisation process is obviously non-unital because its stationary long-time state is thermal, not maximally mixed (except at infinite temperature where the thermal state is maximally mixed).

Your specific master equation generates a unital evolution only when $\Gamma_{\rm in}= \Gamma_{\rm out}$, so that the gain and decay rates are balanced. This is what one would expect for a thermal bath at infinite temperature, and it is entirely consistent with a microscopic derivation from the Redfield equation with Hermitian interaction operators $A_\mu$. The assumption of a delta-correlated bath $\Gamma_{\mu\nu}(\tau) \propto \delta(\tau)$ implies that in Fourier space $$\Gamma_{\mu\nu}(\omega) = \int {\rm d}\tau \,{\rm e}^{{\rm i}\omega \tau} \Gamma_{\mu \nu}(\tau) = {\rm const.}$$ This is clearly incompatible with the detailed balance (Kubo-Martin-Schwinger) condition $$ \Gamma_{\mu\nu}(\omega) = {\rm e}^{\beta\omega}\Gamma_{\nu\mu}(-\omega)$$ except in the infinite-temperature limit where $\beta=1/k_BT = 0$.