# How to efficiently check if a superoperator is Lindbladian?

Superoperators are linear maps on the vector space of linear operator. The Lindbladian superoperators are the important subset that can be expressed in the form $$\mathcal{L}[\rho] = -i (H \rho - \rho H) + \sum_i L_i\rho L_i^\dagger - \frac{1}{2}(L_i^\dagger L_i \rho + \rho L_i^\dagger L_i)$$ for some Hermitian operator $$H$$ with positive anti-Hermitian part, $$H-H^\dagger \ge 0$$, and some set of operators $$\{L^{i}\}$$. Lindbladians are important in the study of open quantum systems because they generate completely positive trace-preserving maps on density matrices, $$\exp(t \mathcal{L})$$ for $$t\ge 0$$, which describe Markovian dynamics.

What is the most "checkable" criteria for determining whether a given superoperator $$\mathcal{S}$$ is Lindbladian? (Obviously, exhaustively searching for $$H$$ and $$\{L^{i}\}$$ is not reasonable.)

For concreteness, suppose our operators $$\rho$$ act on $$N$$ dimension vectors spaces, so we can organize their matrix elements into a vector of length $$N^2$$, making the superoperators $$N^2 \times N^2$$ matrices. What algorithm can efficiently check whether a given $$N^2 \times N^2$$ matrix is Lindbladian?

• Very nice question. I have posted a partial answer. I will try to think about this a bit more (but probably someone a bit more mathematically minded on here can already do better). Apr 5, 2020 at 15:29

Depending on what you need this is only a (very) partial answer to the question, because it includes only necessary conditions for an operator to be Lindbladian, not sufficient ones (as far as I know).

Lindbladians have two important properties: they generate (1) trace-preserving and (2) completely positive evolution.

1. A necessary condition for a superoperator $$\mathcal{L}$$ to generate a trace-preserving evolution is $$\mathcal{L}^\dagger[\mathbb{1}] = 0.$$ It is easy to check since it requires only Hermitian conjugation of the matrix $$\mathcal{L}$$ and multiplication with the identity "vector" $$\mathbb{1}$$.

2. A necessary condition for $$\mathcal{L}$$ to generate a positive semi-group is that the real part of the eigenvalues of $$\mathcal{L}$$ are non-positive. This is less easy to check since one has to diagonalise an $$N\times N$$ matrix.

• With apologies, I edited my question to re-define "Lindbladian" superoperators as generating CP evolution that is not necessarily trace-preserving (TP). (In the original version of my question, I required TP as well, and your partial answer assumes this.) I did this because (1) that condition is more "natural" (has a single compact expression) and (2) I found out Lindblad already called the generators of CPTP evolution "completely dissipative". May 11, 2020 at 17:30

First, note that if we allow the Hamiltonian $$H$$ to have an anti-Hermitian part which is a positive semi-definite operator, $$H-H^\dagger \ge 0$$, then $$\mathcal{L}$$ still generates CP time evolution $$e^{t\mathcal{L}}$$; it's just not trace preserving unless $$H-H^\dagger=0$$. Let's call the not-necessarily-trace-preserving class of superoperators Lindbladian, and call the trace-preserving subset completely dissipative. (The latter name is the terminology that Lindblad originally used.)

Then with some effort one can show $$\mathcal{L}$$ is Lindbladian if and only if $$\qquad\qquad\qquad\qquad\mathcal{P} \mathcal{L}^{\mathrm{PT}} \mathcal{P} \ge 0,\qquad\qquad\qquad\qquad (1)$$ where $$\mathcal{P} \equiv \mathcal{I} - \mathcal{I}^{\mathrm{PT}}/N = \mathcal{P}^2$$ is the "superprojector" that removes an operator's trace, so that $$\mathcal{P}[B] = B - (\mathrm{Tr}[B]/N)I$$. Note that $$\mathcal{S} \ge 0$$ means that a superoperator $$\mathcal{S}$$ is a positive operator (when considered as an operator on the space of operators/matrices) in the sense of being Hermitian with positive eigenvalues or, equivalently, that $$\langle B, \mathcal{S}[B]\rangle \ge 0$$ for all operators $$B$$, where $$\langle B, C \rangle \equiv \mathrm{Tr}[B^\dagger C]$$ is the Hilbert-Schmidt inner product on the space of operators. This is a distinct condition from $$\mathcal{S}$$ being positivity preserving, i.e., $$B\ge 0 \Rightarrow \mathcal{S}[B]\ge 0$$, which is (confusingly) usually described as $$\mathcal{S}$$ being a "positive map".

An equivalent condition to Eq. (1) is $$\qquad\qquad\overline{P}_\Psi[ (\mathcal{L}\otimes \mathcal{I})(|\Psi \rangle\langle \Psi|)] \overline{P}_\Psi \ge 0,\qquad\qquad(2)$$ where $$|\Psi \rangle = N^{-1} \sum_{n=1}^N|n\rangle|n\rangle$$ is some maximally entangled state and $$\overline{P}_\Psi=I - |\Psi \rangle\langle \Psi|$$ projects onto the orthogonal subspace. (This condition is independent of the choice of basis $$\{|n\rangle\}$$ and hence the choice of maximally entangled state $$|\Psi \rangle$$.) Eq. (2) is also a condition about positivity of a linear operator, but in this case it's a condition on a tensor product of two ($$N \times N$$) density matrices rather than a condition on a single ($$N^2 \times N^2$$) superoperator as in Eq. (1).

Eq. (2) is the form the Lindbladian condition appears in some monographs like Wolf's "Quantum Channels and Operations: Guided Tour" [PDF] (see eq. (7.15)) and, I think, Tarasov's "Quantum Mechanics of Non-Hamiltonian and Dissipative Systems" (see Sec. 15.8 and 15.9). I prove Eq. (1) in a self-contained and elementary way in a blog post here.

If we want to further check whether $$\mathcal{L}$$ is completely dissipative and hence generates trace-preserving evolution (for all $$B$$, $$\mathrm{Tr}[e^{t\mathcal{L}}[B]] = \mathrm{Tr}[B]$$ or, equivalently, $$\mathrm{Tr}[ \mathcal{L}[B]]=0$$), then we just need to confirm a vanishing partial-trace condition, $$0 = \sum_{p=1}^N \mathcal{L}_{(pp)(nm)},$$ using the index convention $$(\mathcal{S}[B])_{nn'} = \sum_{m,m'=1}^N \mathcal{S}_{(nn')(mm')}B_{mm'}$$.