Is purity convex in time?

Question

Is the purity $\mathrm{Tr}[\rho(t)^2]$ of a quantum state $\rho(t)$ that evolves according to a time-independent Lindbladian $\partial_t \rho = \mathcal{L}[\rho]$ convex in time $t$?

I suspect the the answer is "no" in general, but I wonder whether there are any useful characterizations of the kinds of Lindbladians which lead to purities that are convex in time. The furthest I have gotten on my own is a sufficient condition which I suspect is not a necessary condition: if the map $\mathcal{L}^\dagger \mathcal{L} - \mathcal{L}\mathcal{L}^\dagger$ is positive ($\mathcal{L}^\dagger$ is the adjoint of $\mathcal{L}$), then the purity is convex in time. I arrived at this condition by expressing the second time derivative of purity as $\partial_t^2 \mathrm{Tr}[\rho(t)^2] = \mathrm{Tr}[\rho(t) ((\mathcal{L} + \mathcal{L}^\dagger)^2 + \mathcal{L}^\dagger \mathcal{L} - \mathcal{L} \mathcal{L}^\dagger)[\rho(t)]]$.

I'd also be interested in similar characterizations for any entanglement monotones of the form $\mathrm{Tr}[f(\rho(t))]$, where $f$ is a convex function.

Thanks!

Edit: this is a cross post from https://math.stackexchange.com/questions/4927440/are-there-any-useful-convexity-properties-of-quantum-dynamical-semigroups.

Answer 1

The following is only a partial answer: While your argument rightly shows that normal $\mathcal L$ (which means $e^{t\mathcal L}$ is unital) is sufficient for convex purity, numerics suggest this even holds for all time-independent Markovian unital processes.

Note that time independence is essential here: while unitality is equivalent to decreasing purity this is logically independent from convexity. In fact it is easy to write down an example of a time-dependent Markovian process with decreasing, but non-convex purity: consider the Lindbladian $\mathcal L(t)$ generated only by $V(t)=\sin(t)\sigma_x$. Because $V(t)$ is Hermitian $L(t)$ generates a unital process. Moreover, $\rho_0={\rm diag}(2/3,1/3)$ evolves like $$ \rho(t)=e^{\int_0^tL(s)\,ds}\rho_0=\begin{pmatrix} \frac12+\frac{1}{6} e^{\sin (t) \cos (t)-t} & 0 \\ 0 & \frac{1}{2}-\frac{1}{6} e^{\sin (t) \cos (t)-t} \end{pmatrix} $$ so the purity

—while of course decreasing—is not convex. The underlying mechanism here is that the factor $\sin(t)$ introduces varying speed of the evolution of $\rho_0$ towards the maximally mixed states which interferes with convexity. NB: the reason your proof does not work in the time-dependent realm anymore is that you get extra derivatives $\frac d{dt}L(t)$ which mess up the structure.

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The other point of my post is to confirm your suspicion that there exist time-independent Markovian processes where the purity is not convex in $t$. Numerical search yielded the following counterexample: $\rho_0={\bf1}/2$ and our Lindbladian $\mathcal L$ is generated by one operator (in the usual way): $$ V=\begin{pmatrix}1&1\\0&1\end{pmatrix} $$ Then the eigenvalues of $e^{tL}(\rho_0)$ evolve as

which leads to the following (non-convex!) purity

This behaviour also shows when looking at the trajectory of $\rho(t)$ in the Bloch ball:

(dynamics start in the middle and end up in the fixed point ((2/3,-1/3),(-1/3,1/3)). The reason for the oscillation in the purity is the little swirl the dynamics do in the end: