# 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".

• Base on what - the explicit form of the Lindblad operator? – Norbert Schuch Apr 17 at 17:33

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 $$$$\frac{d\hat{\rho}}{dt}={\mathcal{L}}_{0}\hat{\rho}+{\mathcal{L}}_{J} \hat{\rho},\tag{1}$$$$ where $${\mathcal{L}}_{0}$$ and $${\mathcal{L}}_{J}$$ are linear operations on the linear space of density matrices defined by \begin{align} {\mathcal{L}}_{0}\hat{\rho} &=\frac{1}{i\hbar }\left( \hat{H}_{\rm eff}\hat{% \rho}-\hat{\rho}\hat{H}_{\rm eff}^{\dagger }\right) , \label{eqn:liouvillian_free} \\ {\mathcal{L}}_{J}\hat{\rho} &=\sum_{j}\hat{S}_{j}\hat{\rho}\hat{S}% _{j}^{\dagger }\equiv \sum_{j}{\mathcal{L}}_{j}\hat{\rho}. \label{eqn:liouvillian_jump} \end{align} We move to an interaction picture via the transformation, $$$$\tilde{\rho}\left( t\right) =e^{-{\mathcal{L}}_{0}t}\hat{\rho}\left( t\right) , \label{eqn:int_picture_transform}$$$$ in which case the master equation reduces to $$$$\frac{d\tilde{\rho}\left( t\right) }{dt}=e^{-{\mathcal{L}}_{0}t}\hat{ \mathcal{L}}_{J}e^{{\mathcal{L}}_{0}t}\tilde{\rho}\left( t\right) . \label{eqn:ME_interaction_picture}$$$$ We formally integrate this equation, yielding $$$$\tilde{\rho}\left( t\right) =\tilde{\rho}\left( t_{0}\right) +\int_{t_{0}}^{t}dt_{1}e^{-{\mathcal{L}}_{0}t_{1}}{\mathcal{L}} _{J}e^{{\mathcal{L}}_{0}t_{1}}\tilde{\rho}\left( t_{1}\right) . \label{eqn:ME_integral_form}$$$$ Iterating this integral equation once yields $$$$\tilde{\rho}\left( t\right) =\tilde{\rho}\left( t_{0}\right) +\int_{t_{0}}^{t}dt_{1}e^{-{\mathcal{L}}_{0}t_{1}}{\mathcal{L}} _{J}e^{{\mathcal{L}}_{0}t_{1}}\tilde{\rho}\left( t_{0}\right) +\int_{t_{0}}^{t}dt_{2}\int_{t_{0}}^{t_{2}}dt_{1}e^{-{\mathcal{L}} _{0}t_{2}}{\mathcal{L}}_{J}e^{{\mathcal{L}}_{0}\left( t_{2}-t_{1}\right) }{\mathcal{L}}_{J}e^{{\mathcal{L}}_{0}t_{1}} \tilde{\rho}\left( t_{1}\right) . \label{eqn:ME_twice_iterated}$$$$ 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, $$$$\hat{\rho}_{t_{1},j_{1};t_{2},j_{2};\dots ;t_{n},j_{n};t_{0}}\left( t\right) =e^{{\mathcal{L}}_{0}\left( t-t_{n}\right) }{\mathcal{L}}_{j_{n}}e^{% {\mathcal{L}}_{0}\left( t_{n}-t_{n-1}\right) }{\mathcal{L}}% _{j_{n-1}}\cdots {\mathcal{L}}_{j_{2}}e^{{\mathcal{L}}_{0}\left( t_{2}-t_{1}\right) }{\mathcal{L}}_{j_{1}}e^{{\mathcal{L}}_{0}\left( t_{1}-t_{0}\right) }\hat{\rho}\left( t_{0}\right) , \tag{2}$$$$ 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. $$$$\hat{\rho}\left( t\right) =e^{{\mathcal{L}}_{0}\left( t-t^{\prime }\right) }\hat{\rho}\left( t^{\prime }\right),$$$$ which is the solution to equation (1) with $$\hat{\mathcal{L}}_{J}$$ set to zero and initial condition, $$\hat{\rho}\left(t^{\prime }\right)$$. The other is the application of some jump operator to the density matrix, i.e. $$$${\mathcal{L}}_{j}\hat{\rho}=\hat{S}_{j}\hat{\rho}\hat{S}_{j}^{\dagger }.$$$$ If the density matrix is a pure state, $$$$\hat{\rho}\left( t\right) =\left\vert \psi \left( t\right) \right\rangle \left\langle \psi \left( t\right) \right\vert ,$$$$ then evolution under the effective Hamiltonian is equivalent to a Schr\"{o}dinger type evolution, $$$$\frac{d}{dt}\left\vert \psi \left( t\right) \right\rangle =\frac{1}{i\hbar }% \hat{H}_{\rm eff}\left\vert \psi \left( t\right) \right\rangle ,$$$$ which can be proven by using the product rule on the expression, $$\frac{d}{dt% }\left\vert \psi \left( t\right) \right\rangle \left\langle \psi \left( t\right) \right\vert$$. The formal solution of this equation is $$$$\left\vert \psi \left( t\right) \right\rangle =e^{-i\hat{H}_{\rm eff}t/\hbar }\left\vert \psi \left( 0\right) \right\rangle , \label{eqn:solution_SE_effective}$$$$ which can be written in terms of density matrices as $$$$\left\vert \psi \left( t\right) \right\rangle \left\langle \psi \left( t\right) \right\vert =\hat{\rho}\left( t\right) =e^{{\mathcal{L}}_{0}t} \hat{\rho}\left( 0\right) =e^{-i\hat{H}_{\rm eff}t/\hbar }\left\vert \psi \left( 0\right) \right\rangle \left\langle \psi \left( 0\right) \right\vert e^{i \hat{H}_{\rm eff}^{\dagger }t/\hbar }. \label{eqn:solution_SE_effective_DM}$$$$ This means that $$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.
• This is a very nice explanation of the quantum jump approach but I think something is missing here. Proving that a map is positive is not the same as showing that its generator has non-positive eigenvalues (which is what the question asks for). As a trivial counter-example, consider the "Liouvillian" $\mathcal{L} = \lambda\mathcal{I}$, proportional to the identity $\mathcal{I}$ with real eigenvalues $\lambda$. The "master equation" $\dot{\rho} = \mathcal{L}\rho$ generates a positive map for any $\lambda$ (including $\lambda>0$) but of course is not trace preserving unless $\lambda=0$. – Mark Mitchison Apr 18 at 14:01
• @MarkMitchison. I think what's missing from my "derivation" of positivity, as you say, is the conservation of probability. This can be done by noting that the probability lost when evolving under the effective Hamiltonian is gained (in some sense) by the different jump channels. There is a way to make this precise and prove this in the above formalism, but I can't remember how that goes. – march Apr 18 at 15:14
• What I am saying is that the OP is not asking for a proof of positivity of the map (which can be found in the original papers by Gorini, Kossakowski, Sudarshan and Lindblad). They are asking for a proof of the non-positivity of the generator. Obviously, those two properties are related, but the question is why! – Mark Mitchison Apr 18 at 17:26
• @MarkMitchison. I see. So what I've done is directed at proving the positivity of the map. It seemed to me that based on OP's mentioning of the "hand-waving" arguments for things not blowing up, that showing that things don't blow up would be sufficient, but you're right that that's not directly related to the generator of the map. I guess I thought that a "derivation" that doesn't require us to go into super-operator land would be nice, and the nice things about going the quantum trajectory route is that we can further talk only about pure states, but again, that misses the point of the post. – march Apr 19 at 1:32
• Yes exactly, I am not asking about the positivity of the map but rather on the negativity of the eigenvalues of the map. As Mark said, these are probably related but it's still unclear to me how. – Tony Jin Apr 19 at 17:29