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I am interested in solving the dual (adjoint) Lindblad master equation for a time-dependent operator $O(t)$ as follows \begin{equation} \dot{O}(t) = i[H, O(t)]+\sum_{\alpha\in I} L_\alpha ^\dagger O(t) L_\alpha -\frac{1}{2}\left\{L^\dagger_\alpha L_\alpha, O(t)\right\} \end{equation}

where the Hamiltonian $H$ is quadratic in fermionic operators, the jump operators are linear and have the form $L_\alpha = \sqrt{\gamma}c_\alpha$, $c_\alpha$ is the fermionic field, $\gamma$ is the decay rate and $I$ is a set of indices.

In particular, I am interested in the dynamics of $O(t)=c^\dagger_j(t)$ for some index $j\in I$, because I am actually interested in computing the two-time correlator $<c^\dagger(t)_j c_i(0)>$.

I know that the Lindbladian is linear in the reduced density matrix of the open system, so I expect the dual (adjoint) Lindbladian to be linear in the operator $O(t)$ too.

My question is, owing to this supposed linearity - which I expect leads to a linear differential dual Lindblad equation - whether it is possible, in principle, to write the solution of this equation as a linear combination of the fermionic operators at time $0$, i.e., \begin{equation} c^\dagger_j(t)= \sum_{\alpha\in I} V_{j,\alpha}(t)c^\dagger_\alpha(0) \end{equation}

If this was true, the two-time correlator might be easily written as \begin{equation} <c^\dagger_j(t)c_i(0)> = \sum_ {\alpha\in I} V_{j,\alpha}(t)<c^\dagger_\alpha(0)c_i(0)> \end{equation} and I could be able to retrieve the form of $V(t)$ from the equation of motion of the this two-time correlator.

I am not sure about this possibility since I am not able to retrieve explicitly the form of the matrix $V$ owing to the presence of the anticommutator in the Lindbladian, which prevents me from simplifying the expression of the dual equation as it can be done in the Heisenberg equation of closed systems. Furthermore, what confuses me is the fact that the operator $c^\dagger_j$ is also involved in the sum of the Lindbladian, even if in a time-independent way.

In short, is the dual Lindblad equation actually linear in the sense I am writing or am I lacking some theoretical information concerning linearity? I would be glad if you could provide some references as well.

GSLAM
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2 Answers2

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The "Heisenberg picture" evolution for Lindblad equations does not play nicely with the operator product: if $C=AB$ and you know the solution for the two operators $A(t)$ and $B(t)$, unfortunately $C(t) \neq A(t) B(t)$ in general, unless the time evolution is unitary. The only thing that $A(t)$ tells you how to calculate is then $\langle A(t)\rangle$. So, in a fermionic system, the solution for $c_j(t)$ itself is of very little interest because $\langle c_j(t)\rangle$ is pretty much meaningless.

However, if your real interest lies in the two-point function $G_{ij}(t) = \langle c^\dagger_i(t)c_j(0)\rangle$, then you can exploit the quantum regression theorem, which states that $\langle A(t) B(0)\rangle_\rho = {\rm tr}\left[A e^{\mathcal{L}t}(B \rho)\right]$, where $\mathcal{L}$ is the (Schrödinger-picture) Liouvillian and $\rho$ is the quantum state. In fermionic systems, this requires modification to account for exchange phases, as described in this paper by Schwarz et al.

As you should carefully verify yourself (because I haven't!), the result is a closed equation of motion for the two-point function, which in a compact matrix notation reads $$i\partial_t G(t) = (h-i\gamma\mathbb{1})\cdot G(t),$$ where $H = \sum_{i,j} h_{ij} c_j^\dagger c_i$, $\gamma$ is the decay rate as defined in the OP, and $\mathbb{1}$ is the identity matrix. See Eqs. (15) and (18) in Schwarz et al. for details.

Mark Mitchison
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Quadratic Lindbladian operators have been investigated extensively for the first time by Tomaž Prosen and his collaborators. He calls his approach third quantization because it amounts to write the Lindbladian in the space of superoperators as a sum of operators which resemble occupation number operators. Using his approach the full spectrum of the Lindbladian can be obtain with a complexity $\mathrm{poly}(N)$ where $N$ is the number of sites.

However to find the time evolution perhaps a simple strategy is the following. One first note that such Lindbladians map Gaussian states (i.e. states which are uniquely characterized by the covariance, two-point, matrix) to Gaussian states. The evolution of such states is then equivalent to a certain matrix ODE for the covariance matrix (similar to Sylvester equation). This approach has been pursued for the first time by J. Ignacio Cirac, G. Giedke and collaborators.

  1. Prosen, T. Third quantization: a general method to solve master equations for quadratic open Fermi systems. New J. Phys. 10, 043026 (2008).
  2. Prosen, T. & Žunkovič, B. Exact solution of Markovian master equations for quadratic Fermi systems: thermal baths, open XY spin chains and non-equilibrium phase transition. New J. Phys. 12, 025016 (2010).
  3. Horstmann, B., Cirac, J. I. & Giedke, G. Noise-driven dynamics and phase transitions in fermionic systems. Phys. Rev. A 87, 012108 (2013).
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