8
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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). $\endgroup$ Apr 5, 2020 at 15:29

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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". $\endgroup$ May 11, 2020 at 17:30
3
$\begingroup$

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 $(\mathcal{S}[B])_{nn'} = \sum_{m,m'=1}^N \mathcal{S}_{(nn')(mm')}B_{mm'}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.