# Is purity convex in time?

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.

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.

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:

• @nlupugla 1. Not sure what you mean, I specified a Lindbladian in my counterexample (through 1 Lindblad-$V$) which is, of course, what characterizes CPTP dynamics. 2. That's fair and I must admit I don't know how I missed that in your question. But yes, your argument does indeed show the (strictly stronger) statement that $L$ being normal is enough; my bad! I adjusted my answer accordingly. Commented Jun 6 at 21:27
• Neat that there are examples with such a behavior. Can this be understood as a combination of Hamiltonian evolution and damping towards a specific point on the Bloch sphere (I would imagine if the two are in-plane, the state would spiral around the equilibrium position)? Commented Jun 7 at 8:10
• @NorbertSchuch Indeed, using standard methods (1, 2) it's easy to see that the $L$ induced by $V$ from my answer is equivalent to a Lindbladian with (traceless) Lindblad-V $|0\rangle\langle 1|$ & Hamiltonian $-\sigma_y/2$. So as dissipation drives the maximally mixed state to the ball's edge the Hamiltonian induces this spiraling; but only together do $H$ and $V$ leave ((2/3,-1/3),(-1/3,1/3)) invariant (i.e. this attractive fixed point is neither a fixed point of the $H$ nor the $V$ part individually) Commented Jun 7 at 9:46
• @FrederikvomEnde Ah, thanks for the clarification regarding the Lindbladian. I misunderstood and thought your Lindbladian was just a left-multiplication by $V$. Commented Jun 7 at 13:34
• @nlupugla Ah, ok: So you care about a case where it is not monotonous, yet convex. Sounds even more tricky, I would think. Maybe it makes more sense if you ask a new question with your specific model, rather than asking a very general question (not that this is bad, on the contrary: but given Frederik's answer it seems likely there isn't a simple answer). Commented Jun 7 at 17:32