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 freedom^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 always^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 forgiven^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 $H$ . (Here, $H_I(t)$ is the interaction-picture version of the interaction Hamiltonian and $\rho(0)$ is the global initial state.) However, a careful reading will show that this is in fact only necessary if you want to satisfy the stronger condition that $\mathrm{Tr}_\mathcal{E}[H_I(t) \rho(0)]=0$ (without the commutator comma), i.e., that $\mathrm{Tr}_\mathcal{E}[B_\alpha \rho_\mathcal{E}(0)]=0$ for all $\alpha$ . Several authors, including Breuer & Petruccione, suggest that we might as well do so since we can always choose to satisfy this stronger condition by making the above transformation with the choice $\Delta H_S = \epsilon \sum_\alpha A_\alpha \mathrm{Tr}_\mathcal{E}[B_\alpha \rho_{\mathcal{E}}(0)] = \epsilon \mathrm{Tr}_\mathcal{E}[H_I \rho_{\mathcal{E}}(0)]$ .

However, this choice can have the consequence of introducing (or eliminating or modifying) the trace of the Lindblad operators. Obviously, at the end you will get the same reduced dynamics, but you need to be mindful that this is happening since the Hamiltonian part of the dynamics you might want to understand can be “hiding” inside the Lindblad operators. This choice can also have a big impact on the feasibility of analytically deriving the Lindblad operators since it generically changes the energy eigenbasis of $H_S$ , which plays a very important role in the derivation.

So how do we interpret the trace-ful part of the Lindblad operators? One can check that they merely make a Hamiltonian contribution to the dynamics. Let’s uniquely expand the Lindblad operators as $L_i = L^0_i + \alpha_i I$ , where $L^0_i$ is the traceless part of $L_i$ and $\alpha_i := \mathrm{Tr}[L_i]/\mathrm{dim} [\mathcal{E}]$ is a complex number proportional to the trace. Then using the traditional definition of the Hamiltonian superoperator $\mathcal{H}_{H} [\rho]:= -i[H,\rho]$ and the dissipator superoperator $\mathcal{D}_{\{M_i\}}[\rho]:= \sum_i[M_i \rho M_i^\dagger - (M_i^\dagger M_i \rho + \rho M_i^\dagger M_i)/2]$ , we see

(2) $\begin{align*} \mathcal{D}_{\{L_i\}} [\rho_{\mathcal{S}}] &= \sum_i\left[L_i \rho_{\mathcal{S}} L_i^\dagger - (L_i^\dagger L_i \rho_{\mathcal{S}} + \rho_{\mathcal{S}} L_i^\dagger L_i)/2\right] \\ &= \sum_i\left[L^0_i \rho_{\mathcal{S}} L_i^{0 \dagger} - (L_i^{0 \dagger} L_i^0 \rho_{\mathcal{S}} + \rho_{\mathcal{S}} L_i^{0 \dagger} L_i^0)/2 \right] + \left[\alpha_i^* (L_i^0 \rho_{\mathcal{S}} - \rho_{\mathcal{S}} L_i^0 ) + \alpha_i (\rho_{\mathcal{S}} L_i^{0 \dagger} - L_i^{0 \dagger} \rho_{\mathcal{S}})\right]/2 \\ &=\mathcal{D}_{\{L_i^0\}}[\rho_{\mathcal{S}}] + \mathcal{H}_{H_{\mathcal{S}}^{(1)}}[\rho_{\mathcal{S}}] \end{align*}$

with^d the new contribution $H_{\mathcal{S}}^{(1)}:=\mathrm{Im}[\sum_i \alpha_i L_i^{0 \dagger}] = \sum_i(\alpha_i L_i^{0 \dagger} -\alpha_i^* L_i^0)/2i$ to the Hamiltonian part $\tilde{H}_{\mathcal{S}}$ of the reduced dynamics. Note that all the terms above that are quadratic in $\alpha_i$ have canceled.

Importantly, this Hamiltonian contribution from the trace-ful part of the Lindblad operators is first order in the coupling constant $\epsilon$ (hence the “ $(1)$ ” superscript) and therefore is distinct from the strictly second-order Hamiltonian contribution $H_{\mathcal{S}}^{(2)}$ , also arising during this microscopic derivation, that is commonly known as the (generalized) Lamb shift. To see this, one merely notes in the derivation that $L_i \propto \epsilon A^{(i)}$ and $H_{\mathcal{S}}^{(2)} \propto \epsilon^2 \sum_i A^{(i)\dagger} A^{(i)}$ , where the $A^{(i)}$ are operators on $\mathcal{S}$ constructed linearly during the derivation from the operators $A_\alpha$ in the interaction Hamiltonian $H_I = \epsilon \sum_\alpha A_\alpha \otimes B_\alpha$ . We end up with a Lindblad equation, (1), but now with the replacements $L_i\to L_i^0$ and $\tilde{H}_{\mathcal{S}} = H_{\mathcal{S}} + H_{\mathcal{S}}^{(2)} \to H_{\mathcal{S}} + H_{\mathcal{S}}^{(1)} + H_{\mathcal{S}}^{(2)}$ .

Hayden & Sorce, generalizing^e the work of Gorini, Kossakowski, & Sudarshan^f, show that subtracting off the Lindblad-operator traces in this way minimizes a certain norm of the dissipative superoperator associated with the Lindblad operators, and so in this sense this choice is preferred.

[I thank Dan Ranard for discussion and for bringing Hayden & Sorce to my attention.]

Footnotes

(↵ returns to text)

A 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.↵
There is also a similar freedom with the environment in the sense that we can send $H_E \to H_E + \Delta H_E$ and $\epsilon H_I \to \epsilon H_I - \Delta H_E$ .↵
Specifically, I have forgiven myself for doing this…↵
$\mathrm{Im}[M] = (M-M^\dagger)/(2i)$ just means the skew-Hermitian part of the matrix $M$ , which reduces to the imaginary part of a scalar when the matrix is $1\times 1$ .↵
Gorini, Kossakowski, & Sudarshan proved this for Markovian dynamics while Hayden & Sorce generalize to the non-Markovian case.↵
V. Gorini, A. Kossakowski, and E. C. G. Sudarshan, Completely positive dynamical semigroups of n-level systems, Journal of Mathematical Physics 17 (1976), no. 5 821–825.↵

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

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