Negativity of the real part of eigenvalues of Lindblad operators I'm looking for a proof of the fact that the real part of eigenvalues of Lindblad operators is always negative. So far I have only found handwavy arguments such as "things should not blow up at infinitely long-time".
 A: This may not be completely general, but I understand the positivity of the Lindbladian in terms of unravelings of the master equation in terms of quantum trajectories in which the initial density matrix evolves under an effective non-Hermitian Hamiltonian between incoherent jumps.
Main reference: Section IV.C of The quantum-jump approach to dissipative dynamics
in quantum optics, by Plenio and Knight, published in the APS journal Review of Modern Physics.
A Dyson-like series for the master equation
Consider a general master equation in the Lindblad form, given by
\begin{align}
\frac{d\hat{\rho}}{dt} &=\frac{1}{i\hbar }\left( \hat{H}_{\rm eff}\hat{\rho}-%
\hat{\rho}\hat{H}_{\rm eff}^{\dagger }\right) +\sum_{j}\hat{S}_{j}\hat{\rho}\hat{S}%
_{j}^{\dagger },
\end{align}
where
\begin{align}
\hat{H}_{\rm eff} &=\hat{H}_{0}-\frac{i\hbar }{2}\sum_{j}\hat{S}_{j}^{\dagger }%
\hat{S}_{j}\equiv \hat{H}_{0}-i\frac{\hbar }{2}\hat{D}.
\end{align}
We will derive a Dyson-type expansion for this equation. Re-write the master
equation as
\begin{equation}
\frac{d\hat{\rho}}{dt}={\mathcal{L}}_{0}\hat{\rho}+{\mathcal{L}}_{J}
\hat{\rho},\tag{1}
\end{equation}
where ${\mathcal{L}}_{0}$ and ${\mathcal{L}}_{J}$ are linear
operations on the linear space of density matrices defined by 
\begin{align}
{\mathcal{L}}_{0}\hat{\rho} &=\frac{1}{i\hbar }\left( \hat{H}_{\rm eff}\hat{%
\rho}-\hat{\rho}\hat{H}_{\rm eff}^{\dagger }\right) ,
\label{eqn:liouvillian_free} \\
{\mathcal{L}}_{J}\hat{\rho} &=\sum_{j}\hat{S}_{j}\hat{\rho}\hat{S}%
_{j}^{\dagger }\equiv \sum_{j}{\mathcal{L}}_{j}\hat{\rho}.
\label{eqn:liouvillian_jump}
\end{align}
We move to an interaction picture via the transformation,
\begin{equation}
\tilde{\rho}\left( t\right) =e^{-{\mathcal{L}}_{0}t}\hat{\rho}\left(
t\right) ,  \label{eqn:int_picture_transform}
\end{equation}
in which case the master equation reduces to
\begin{equation}
\frac{d\tilde{\rho}\left( t\right) }{dt}=e^{-{\mathcal{L}}_{0}t}\hat{
\mathcal{L}}_{J}e^{{\mathcal{L}}_{0}t}\tilde{\rho}\left( t\right) .
\label{eqn:ME_interaction_picture}
\end{equation}
We formally integrate this equation, yielding
\begin{equation}
\tilde{\rho}\left( t\right) =\tilde{\rho}\left( t_{0}\right)
+\int_{t_{0}}^{t}dt_{1}e^{-{\mathcal{L}}_{0}t_{1}}{\mathcal{L}}
_{J}e^{{\mathcal{L}}_{0}t_{1}}\tilde{\rho}\left( t_{1}\right) .
\label{eqn:ME_integral_form}
\end{equation}
Iterating this integral equation once yields
\begin{equation}
\tilde{\rho}\left( t\right) =\tilde{\rho}\left( t_{0}\right)
+\int_{t_{0}}^{t}dt_{1}e^{-{\mathcal{L}}_{0}t_{1}}{\mathcal{L}}
_{J}e^{{\mathcal{L}}_{0}t_{1}}\tilde{\rho}\left( t_{0}\right)
+\int_{t_{0}}^{t}dt_{2}\int_{t_{0}}^{t_{2}}dt_{1}e^{-{\mathcal{L}}
_{0}t_{2}}{\mathcal{L}}_{J}e^{{\mathcal{L}}_{0}\left(
t_{2}-t_{1}\right) }{\mathcal{L}}_{J}e^{{\mathcal{L}}_{0}t_{1}}
\tilde{\rho}\left( t_{1}\right) .  \label{eqn:ME_twice_iterated}
\end{equation}
If this iteration is carried on for an infinite number of terms, we realize
the Dyson expansion,
\begin{align}
\tilde{\rho}\left( t\right) &=
\tilde{\rho}\left( t_{0}\right) +\sum_{n=1}^{\infty
}\int_{t_{0}}^{t}dt_{n}\int_{t_{0}}^{t_{n}}dt_{n-1}\cdots
\int_{t_{0}}^{t_{3}}dt_{2}\int_{t_{0}}^{t_{2}}dt_{1}  \nonumber \\
&\quad\mbox{}\times e^{-{\mathcal{L}}_{0}t_{n}}{\mathcal{L}}_{J}e^{\hat{
\mathcal{L}}_{0}\left( t_{n}-t_{n-1}\right) }{\mathcal{L}}_{J}\cdots 
{\mathcal{L}}_{J}e^{{\mathcal{L}}_{0}\left( t_{2}-t_{1}\right) }\hat{
\mathcal{L}}_{J}e^{{\mathcal{L}}_{0}t_{1}}\tilde{\rho}\left(
t_{0}\right) . 
\end{align}
Moving out of the interaction picture and plugging in the more explicit form
of the jump Liouvillian, this becomes
\begin{align}
\hat{\rho}\left( t\right) &=e^{{\mathcal{L}}_{0}\left( t-t_{0}\right) }
\hat{\rho}\left( t_{0}\right) +\sum_{n=1}^{\infty }\sum_{j_{1},\dots
,j_{n}}\int_{t_{0}}^{t}dt_{n}\int_{t_{0}}^{t_{n}}dt_{n-1}\cdots
\int_{t_{0}}^{t_{3}}dt_{2}\int_{t_{0}}^{t_{2}}dt_{1}  \nonumber \\
&\quad\mbox{}\times e^{{\mathcal{L}}_{0}\left( t-t_{n}\right) }{\mathcal{L}}
_{j_{n}}e^{{\mathcal{L}}_{0}\left( t_{n}-t_{n-1}\right) }\hat{\mathcal{L}
}_{j_{n-1}}\cdots {\mathcal{L}}_{j_{2}}e^{{\mathcal{L}}_{0}\left(
t_{2}-t_{1}\right) }{\mathcal{L}}_{j_{1}}e^{{\mathcal{L}}_{0}\left(
t_{1}-t_{0}\right) }\hat{\rho}\left( t_{0}\right) . 
\end{align}
Quantum trajectories
The quantity inside all of the sums,
\begin{equation}
\hat{\rho}_{t_{1},j_{1};t_{2},j_{2};\dots ;t_{n},j_{n};t_{0}}\left( t\right)
=e^{{\mathcal{L}}_{0}\left( t-t_{n}\right) }{\mathcal{L}}_{j_{n}}e^{%
{\mathcal{L}}_{0}\left( t_{n}-t_{n-1}\right) }{\mathcal{L}}%
_{j_{n-1}}\cdots {\mathcal{L}}_{j_{2}}e^{{\mathcal{L}}_{0}\left(
t_{2}-t_{1}\right) }{\mathcal{L}}_{j_{1}}e^{{\mathcal{L}}_{0}\left(
t_{1}-t_{0}\right) }\hat{\rho}\left( t_{0}\right) ,
\tag{2}
\end{equation}
has a simple physical interpretation. In order to get at this interpretation,
we make a few observations.
There are two types of operations that go on in this expression. One is the 
free evolution under the effective Hamiltonian, i.e.
\begin{equation}
\hat{\rho}\left( t\right) =e^{{\mathcal{L}}_{0}\left( t-t^{\prime
}\right) }\hat{\rho}\left( t^{\prime }\right),
\end{equation}
which is the solution to equation (1) with $\hat{\mathcal{L}}_{J}$ set to zero and initial condition, $\hat{\rho}\left(t^{\prime }\right) $. The other is the application of some jump operator to the density matrix, i.e.
\begin{equation}
{\mathcal{L}}_{j}\hat{\rho}=\hat{S}_{j}\hat{\rho}\hat{S}_{j}^{\dagger }.
\end{equation}
If the density matrix is a pure state, 
\begin{equation}
\hat{\rho}\left( t\right) =\left\vert \psi \left( t\right) \right\rangle
\left\langle \psi \left( t\right) \right\vert ,  
\end{equation}
then evolution under the effective Hamiltonian is equivalent to a Schr\"{o}dinger type evolution,
\begin{equation}
\frac{d}{dt}\left\vert \psi \left( t\right) \right\rangle =\frac{1}{i\hbar }%
\hat{H}_{\rm eff}\left\vert \psi \left( t\right) \right\rangle ,
\end{equation}
which can be proven by using the product rule on the expression, $\frac{d}{dt%
}\left\vert \psi \left( t\right) \right\rangle \left\langle \psi \left(
t\right) \right\vert $. The formal solution of this equation is
\begin{equation}
\left\vert \psi \left( t\right) \right\rangle =e^{-i\hat{H}_{\rm eff}t/\hbar
}\left\vert \psi \left( 0\right) \right\rangle ,
\label{eqn:solution_SE_effective}
\end{equation}
which can be written in terms of density matrices as
\begin{equation}
\left\vert \psi \left( t\right) \right\rangle \left\langle \psi \left(
t\right) \right\vert =\hat{\rho}\left( t\right) =e^{{\mathcal{L}}_{0}t}
\hat{\rho}\left( 0\right) =e^{-i\hat{H}_{\rm eff}t/\hbar }\left\vert \psi \left(
0\right) \right\rangle \left\langle \psi \left( 0\right) \right\vert e^{i
\hat{H}_{\rm eff}^{\dagger }t/\hbar }.  \label{eqn:solution_SE_effective_DM}
\end{equation}
This means that $e^{{\mathcal{L}}_{0}t}$ has the effect of evolving a
pure state into another pure state according to the non-Hermitian
Hamiltonian, $\hat{H}_{\rm eff}$. The action of ${\mathcal{L}}_{j}$ also
preserves the purity of a state, as can be seen by noting that $\hat{
\mathcal{L}}_{j}\left( \left\vert \psi \right\rangle \left\langle \psi
\right\vert \right) $ is equivalent to the expression, $\hat{S}
_{j}\left\vert \psi \right\rangle \left\langle \psi \right\vert \hat{S}
_{j}^{\dagger }$. Thus, if the initial state is pure, then the quantity defined
in equation (2) defined above is at all times a pure state.
Positivity
As I understand it, positivity is guaranteed by two things:


*

*First, the effective Hamiltonian is the sum of a Hermitian matrix and $-i$ times a non-negative matrix, since $\hat{D}$, defined in the very second equation above, is manifestly non-negative. The (right-)eigenvalues of the effective Hamiltonian must therefore have negative imaginary parts. Finally, when multiplying again by $-i$ (in defining the corresponding Lindblad super-operator; see the first equation above), the real part is necessarily negative. Alternatively, one can see that under the action of the effective Hamiltonian, the norm of the state decreases but never goes negative, which is also implies positivity.

*Second, the application of the jump operator is also manifestly positive, as it transforms a pure state into another pure state.

A: It is enough to prove that $Re( \mathcal{L}) \leq 0$. Since the Hamiltonian part is skew it does not matter for the proof of dissipativity.
We can look at the single Lindblad operators individually.
\begin{align*}
\mathcal{D}( \rho) = L \rho L^* -  \frac{1}{2} \{ L^* L , \rho\}
\end{align*} 
Then
\begin{align*}
\mathcal{D}^*( \rho) = L^* \rho L -  \frac{1}{2} \{ L^* L , \rho\}
\end{align*} 
and thus
\begin{align*}
2Re( \mathcal{D}) (\rho)  = L \rho L^* +  L^* \rho L  -  \{ L^* L , \rho\} = L \rho L^* -  L^* \rho L  +  L^* L \rho + \rho  L^* L
\end{align*} 
Proving dissipativity of  $\mathcal{D}$ amounts to proving $Re(\langle \rho, \mathcal{D} \rho \rangle) = \langle \rho, Re(\mathcal{D}) \rho \rangle \leq 0 $.
Thus,
\begin{align*}
 \langle \rho, Re(\mathcal{D}) \rho \rangle = Tr( \rho^* (L \rho L^* +  L^* \rho L  -  L^* L \rho - \rho  L^* L) ) = Tr( - ( \rho L - L\rho) ( \rho L - L\rho)^* )  \leq 0 
\end{align*} 
where we used that any operator of the form $A A^* $ is positive.
This is described in https://arxiv.org/abs/2206.09879.
