Unital dynamics are mixedness increasing

After years of not having an intuitive interpretation of the unital condition on CP maps, I recently learned a beautiful one: unitality means the dynamics never decreases the state’s mixedness, in the sense of the majorization partial order.

Consider the Lindblad dynamics generated by a set of Lindblad operators L_k, corresponding to the Lindbladian

(1)   \begin{align*} \mathcal{L}[\rho] = \sum_k\left(L_k\rho L_k^\dagger - \{L_k^\dagger L_k,\rho\}/2\right) \end{align*}

and the resulting quantum dynamical semigroup \Phi_t[\rho] = e^{t\mathcal{L}}[\rho]. Let

(2)   \begin{align*} S_\alpha[\rho] = \frac{\ln\left(\mathrm{Tr}[\rho^\alpha]\right)}{1-\alpha}, \qquad \alpha\ge 0 \end{align*}

be the Renyi entropies, with S_{\mathrm{vN}}[\rho]:=\lim_{\alpha\to 1} S_\alpha[\rho] = -\mathrm{Tr}[\rho\ln\rho] the von Neumann entropy. Finally, let \prec denote the majorization partial order on density matrices: \rho\prec\rho' exactly when \mathrm{spec}[\rho]\prec\mathrm{spec}[\rho'] exactly when \sum_{i=1}^r \lambda_i \le \sum_{i=1}^r \lambda_i^\prime for all r, where \lambda_i and \lambda_i^\prime are the respective eigenvalues in decreasing order. (In words: \rho\prec\rho' means \rho is more mixed than \rho'.) Then the following conditions are equivalent:None of this depends on the dynamics being Lindbladian. If you drop the first condition and drop the “t” subscript, so that \Phi is just some arbitrary (potentially non-divisible) CP map, the remaining conditions are all equivalent.a  

  • \mathcal{L}[I]=0
  • \Phi_t[I]=I: “\Phi_t is a unital map (for all t)”
  • \frac{\mathrm{d}}{\mathrm{d}t}S_\alpha[\Phi_t[\rho]] \ge 0 for all \rho, t, and \alpha: “All Renyi entropies are non-decreasing”
  • \Phi_t[\rho]\prec\rho for all t: “\Phi_t is mixedness non-decreasing”
  • \Phi_t[\rho] = \sum_j p_j U^{(t)}_j\rho U^{(t)\dagger}_j for all t and some unitaries U^{(t)}_j and probabilities p_j.

The non-trivial equivalences above are proved in Sec. 8.3 of Wolf, “Quantum Channels and Operations Guided Tour“.See also “On the universal constraints for relaxation rates for quantum dynamical semigroup” by Chruscinski et al [2011.10159] for further interesting discussion.b  

Note that having all Hermitian Lindblad operators (L_k = L_k^\dagger) implies, but is not implied by, the above conditions. Indeed, the condition of Lindblad operator Hermiticity (or, more generally, normality) is not preserved under the unitary gauge freedom L_k\to L_k^\prime = \sum_j u_{kj} L_j (which leaves the Lindbladian \mathcal{L} invariant for unitary u.)… [continue reading]

Lindblad operator trace is 1st-order contribution to Hamiltonian part of reduced dynamics

In many derivations of the Lindblad equation, the authors say something like “There is a gauge freedomA gauge freedom of the Lindblad equation means a transformation we can to both the Lindblad operators and (possibly) the system’s self-Hamiltonian, without changing the reduced dynamics.a   in our choice of Lindblad (“jump”) operators that we can use to make those operators traceless for convenience”. However, the nature of this freedom and convenience is often obscure to non-experts.

While reading Hayden & Sorce’s nice recent paper [arXiv:2108.08316] motivating the choice of traceless Lindblad operators, I noticed for the first time that the trace-ful parts of Lindblad operators are just the contributions to Hamiltonian part of the reduced dynamics that arise at first order in the system-environment interaction. In contrast, the so-called “Lamb shift” Hamiltonian is second order.

Consider a system-environment decomposition \mathcal{S}\otimes \mathcal{E} of Hilbert space with a global Hamiltonian H = H_S + H_{I} + H_E, where H_S = H_S \otimes I_\mathcal{E}, H_E = I_\mathcal{S}\otimes H_E, and H_I = \epsilon \sum_\alpha A_\alpha \otimes B_\alpha are the system’s self Hamiltonian, the environment’s self-Hamiltonian, and the interaction, respectively. Here, we have (without loss of generality) decomposed the interaction Hamiltonian into a tensor product of Hilbert-Schmidt-orthogonal sets of operators \{A_\alpha\} and \{B_\alpha\}, with \epsilon a real parameter that control the strength of the interaction.

This Hamiltonian decomposition is not unique in the sense that we can alwaysThere is also a similar freedom with the environment in the sense that we can send H_E \to H_E + \Delta H_E and \epsilon H_I \to \epsilon H_I - \Delta H_E.b   send H_S \to H_S + \Delta H_S and H_I \to H_I - \Delta H_S, where \Delta H_S = \Delta H_S \otimes I_\mathcal{E} is any Hermitian operator acting only on the system. When reading popular derivations of the Lindblad equation

(1)   \begin{align*} \partial_t \rho_{\mathcal{S}} = -i[\tilde{H}_{\mathcal{S}}, \rho_{\mathcal{S}}] + \sum_i\left[L_i \rho_{\mathcal{S}} L_i^\dagger - (L_i^\dagger L_i \rho_{\mathcal{S}} + \rho_{\mathcal{S}} L_i^\dagger L_i)/2\right] \end{align*}

like in the textbook by Breuer & Petruccione, one could be forgivenSpecifically, I have forgiven myself for doing this…c   for thinking that this freedom is eliminated by the necessity of satisfying the assumption that \mathrm{Tr}_\mathcal{E}[H_I(t),\rho(0)]=0, which is crucially deployed in the “microscopic” derivation of the Lindblad equation operators \tilde{H}_{\mathcal{S}} and \{L_i\} from the global dynamics generated by H.… [continue reading]

A checkable Lindbladian condition

Summary

Physicists often define a Lindbladian superoperator as one whose action on an operator B can be written as

(1)   \begin{align*} \mathcal{L}[B] = -i [H,B] + \sum_i \left[ L_i B L_i^\dagger - \frac{1}{2}\left(L_i^\dagger L_i B + B L_i^\dagger L_i\right)\right], \end{align*}

for some operator H with positive anti-Hermitian part, H-H^\dagger \ge 0, and some set of operators \{L^{(i)}\}. But how does one efficiently check if a given superoperator is Lindbladian? In this post I give an “elementary” proof of a less well-known characterization of Lindbladians:

A superoperator \mathcal{L} generates completely positive dynamics e^{t\mathcal{L}}, and hence is Lindbladian, if and only if \mathcal{P} \mathcal{L}^{\mathrm{PT}} \mathcal{P} \ge 0, i.e.,

    \[\mathrm{Tr}\left[B^\dagger (\mathcal{P} \mathcal{L}^{\mathrm{PT}} \mathcal{P})[B]\right] \ge 0\]

for all B. Here “\mathrm{PT}” denotes a partial transpose, \mathcal{P} = \mathcal{I} - \mathcal{I}^{\mathrm{PT}}/N = \mathcal{P}^2 is the “superprojector” that removes an operator’s trace, \mathcal{I} is the identity superoperator, and N is the dimension of the space upon which the operators act.

Thus, we can efficiently check if an arbitrary superoperator \mathcal{L} is Lindbladian by diagonalizing \mathcal{P}\mathcal{L}^{\mathrm{PT}} \mathcal{P} and seeing if all the eigenvalues are positive.

A quick note on terminology

The terms superoperator, completely positive (CP), trace preserving (TP), and Lindbladian are defined below in Appendix A in case you aren’t already familiar with them.

Confusingly, the standard practice is to say a superoperator \mathcal{S} is “positive” when it is positivity preserving: B \ge 0 \Rightarrow \mathcal{S}[B]\ge 0. This condition is logically independent from the property of a superoperator being “positive” in the traditional sense of being a positive operator, i.e., \langle B,\mathcal{S}[B]\rangle  \ge 0 for all operators (matrices) B, where

    \[\langle B,C\rangle  \equiv \mathrm{Tr}[B^\dagger C]  = \sum_{n=1}^N \sum_{n'=1}^N   B^\dagger_{nn'} C_{n'n}\]

is the Hilbert-Schmidt inner product on the space of N\times N matrices. We will refer frequently to this latter condition, so for clarity we call it op-positivity, and denote it with the traditional notation \mathcal{S}\ge 0.

Intro

It is reasonably well known by physicists that Lindbladian superoperators, Eq.… [continue reading]

Lindblad Equation is differential form of CP map

The Master equation in Lindblad form (aka the Lindblad equation) is the most general possible evolution of an open quantum system that is Markovian and time-homogeneous. Markovian means that the way in which the density matrix evolves is determined completely by the current density matrix. This is the assumption that there are no memory effects, i.e. that the environment does not store information about earlier state of the system that can influence the system in the future.Here’s an example of a memory effect: An atom immersed in an electromagnetic field can be in one of two states, excited or ground. If it is in an excited state then, during a time interval, it has a certain probability of decaying to the ground state by emitting a photon. If it is in the ground state then it also has a chance of becoming excited by the ambient field. The situation where the atom is in a space of essentially infinite size would be Markovian, because the emitted photon (which embodies a record of the atom’s previous state of excitement) would travel away from the atom never to interact with it again. It might still become excited because of the ambient field, but its chance of doing so isn’t influenced by its previous state. But if the atom is in a container with reflecting walls, then the photon might be reflected back towards the atom, changing the probability that it becomes excited during a later period.a   Time-homogeneous just means that the rule for stochastically evolving the system from one time to the next is the same for all times.

Given an arbitrary orthonormal basis L_n of the space of operators on the N-dimensional Hilbert space of the system (according to the Hilbert-Schmidt inner product \langle A,B \rangle = \mathrm{Tr}[A^\dagger B]), the Lindblad equation takes the following form:

(1)   \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} \rho=- i[H,\rho]+\sum_{n,m = 1}^{N^2-1} h_{n,m}\left(L_n\rho L_m^\dagger-\frac{1}{2}\left(\rho L_m^\dagger L_n + L_m^\dagger L_n\rho\right)\right) , \end{align*}

with \hbar=1.… [continue reading]