Lindbladian superoperator decomposition

Question

this question is about the equation that describes the open quantum system.

"The general stochastic evolution of an open quantum system can be modelled by \begin{equation} \frac{d\rho}{dt} = \mathcal{L}(\rho), \end{equation} where the superoperator $\mathcal{L}(\rho)$ can be decomposed as \begin{equation} \mathcal{L}(\rho) = \sum_i g_i S_i \rho T_i^{\dagger}, \end{equation} with unitary operator $S_i \text{ and } T_i$, and coefficients $g_i$."

This is part of the paper titled "Theory of variational quantum simulation".

The paper does not say what the superporator is, but I believe the superoperator $\mathcal{L}(\rho)$ should be expressed as $$\mathcal{L}(\rho) = -\frac{i}{\hbar}[H,\rho] + \sum_j \gamma_j \left(L_j \rho L_j^{\dagger}-\frac{1}{2}\lbrace L_j^{\dagger} L_j,\rho \rbrace \right)$$ which is the RHS of the Lindblad master equation.

My question is, can we always decompose the superoperator $\mathcal{L}(\rho)$ as $\mathcal{L}(\rho) = \sum_i g_i S_i \rho T_i^{\dagger}$? I think that $g_i, S_i \text{, and } T_i$ might not always exist.

Answer 1

Let me extend the existing answers addressing the unitarity of $S_i$ and $T_i$ as asked for in OP's question. From @E. Anikin's answer, we have $\mathcal{L}(O) = \sum_i g_i S_i O T_i^\dagger$. You can always find a basis of the space of operators with unitaries as basis states. Since the paper in OP's question is on quantum simulation, you could for instance use a decomposition into Pauli operators. This way, the sum in $\mathcal{L}$ eventually blows up (in this sense, it is not a minimal decomposition). but we can replace $S_i$ and $T_i$ by their Pauli decompositions. This means you can always find a set of unitaries $S_i$ and $T_i$ to write the Lindbladian in the desired form.

Answer 2

A superoperator is a linear operator acting on the space of operators (let us denote it $V$). Therefore, any superoperator can be represented as a matrix $\mathcal{L}_{\alpha\beta,\gamma\delta}$, where it acts on operator with a matrix $O_{\alpha\beta}$ as $$ (\mathcal{L}(O))_{\alpha\beta} = \sum_{kl}\mathcal{L}_{\alpha\beta,\gamma\delta} O_{\gamma\delta}. $$ It is easy to see that the space of $\mathcal{L}_{ij,kl}$ is isomorphic to a tensor product $V\otimes V$ (sorry, I don't care here about the difference between $V$ and its conjugate space). Any element of a $V\otimes V$ can be expressed as a sum over tensor products of elements of $V$, so $$ \mathcal{L}_{\alpha\beta,\gamma\delta} = \sum_i S^i_{\alpha\gamma} (T^i_{\delta\beta})^*. $$ Combining two above equations, we get just $$ (\mathcal{L}(O)) = \sum_i S_i O T_i^\dagger. $$ We can also change the normalization of $O$ and $T$ arbitrarily to get some constants $g_i$, so finally $$ \mathcal{L}(O)) = \sum_i g_j S_i O T_i^\dagger. $$

Answer 3

Yes, even more, $S_i$ and $T_i$ can be chosen the same. This mapping is known as a Kraus decomposition: it is the generalisation of a unitary transformation that includes decoherence. The Kraus decomposition is also known to be a completely positive trace-preserving map, meaning it maps density matrices to density matrices.

The Lindblad equation is the continuous version of this trace-preserving map; for an infinitesimal timestep you could have for example $S_i=\sqrt{dt}L_i$ and $S_0=1-dt\sum_j L_i^\dagger L_i$.

After a finite evolution time, the result is still a completely positive trace-preserving map, so the decomposition should exist in principle, although I don't have an immediate clean expression for it from the top of my head.

CAVEAT: I have assumed open quantum systems means a Markovian one, where the Lindblad equation is actually valid. More general cases of open quantum systems, with an environment that is not memory-free, exist too and their situation is more complicated.

EDIT: I missed that $S_i$ and $T_i$ have to be unitary. This is not necessarily the case for the Kraus decomposition, so choosing them the same may not be allowed in general.