Unital dynamics are mixedness increasing

[EDIT 2025-6-25: The original version of this post incorrectly included mixed-unitary and Renyi-entropy-increasing in the equivalent conditions.]

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 \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 two conditions and drop the “t” subscript, so that \Phi is just some arbitrary (potentially non-divisible) CP map, the remaining three conditions are all equivalent.a  

  • \sum_k [L_k, L_k^\dagger]=0
  • \mathcal{L}[I]=0
  • \Phi_t[I]=I: “\Phi_t is a unital map (for all t\ge 0)”
  • \Phi_t[\rho]\prec\rho for all t\ge 0: “\Phi_t is mixedness non-decreasing”
  • \frac{\mathrm{d}}{\mathrm{d}t}\mathrm{Tr}[f(\Phi_t[\rho])] \ge 0 for all \rho, t\ge 0, and concave functions f:\mathbb{R}\to\mathbb{R}

The last condition implies that all Renyi entropies, including the Shannon entropy, are non-decreasing: \frac{\mathrm{d}}{\mathrm{d}t}S_\alpha[\Phi_t[\rho]] \ge 0 for all \rho, t\ge 0, and \alpha.

The non-trivial equivalences above are proved by Thm 8.8 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. [ADDED:] Indeed, one can see there exist unital Lindblad dynamics that cannot be written using Hermitian Lindblad operators. It relies on breaking detailed balance, and so the smallest counterexample is in 3 dimensions where non-trivial cycles exist. Let \mathcal{L}^{(1)} be the Lindbladian with H=0 and three non-Hermitian Lindblad operators \{L_k^{(1)}\} = \{|1\rangle\langle 0|,|2\rangle\langle 1|,|0\rangle\langle 2|\}, and let \mathcal{L}^{(2)} be a Lindbladian with any H and any Hermitian Lindblad operators L_\ell^{(2)} = L_\ell^{(2)\dagger}. Using P_k := |k\rangle\langle k| and G_{k,k'}[\mathcal{L}] := \mathrm{Tr}[P_k \mathcal{L}[P_{k'}]] = \langle k|\mathcal{L}[|k'\rangle\langle k'|]|k\rangle, it’s easy to check that G[\mathcal{L}^{(2)}] is a symmetric matrix but G[\mathcal{L}^{(1)}] is not, so no choice of Hermitian Lindblad operators can produce a Lindbladian \mathcal{L}^{(2)} equal to \mathcal{L}^{(1)}.

Footnotes

(↵ returns to text)

  1. None of this depends on the dynamics being Lindbladian. If you drop the first two conditions and drop the “t” subscript, so that \Phi is just some arbitrary (potentially non-divisible) CP map, the remaining three conditions are all equivalent.↵
  2. See also “On the universal constraints for relaxation rates for quantum dynamical semigroup” by Chruscinski et al [2011.10159] for further interesting discussion.↵
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One Comment

  1. Cool! I did not know this condition either. This seems to answer an old loophole in a famous paper on BH information loss by Banks, Peskin, and Susskind: https://inspirehep.net/literature/194105. After Eq. 9 they say (wrt a general Lindblad equation) “One might also insist that the entropy defined by \rho not decrease with time. We do not know what conditions are necessary to insure these properties, but we can state some simple sufficient conditions. “

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