# Do all Hermiticity-preserving maps generate completely positive maps?

I am confused about what kinds of maps are valid infinitesimal generators of completely positive maps. I know that any Markovian completely positive map can be written in the form $$e^{t \mathcal{L}}$$, where $$\mathcal{L}$$ is the Linbladian, a time-independent and Hermiticity-preserving map. My question is about the converse: if $$t \mathcal{L}$$ is a Hermiticity-preserving map, is it true that $$e^{t \mathcal{L}}$$ is a completely positive map? If not, are there any useful sufficient conditions on a map that ensure its exponential is completely positive, or even just positive if that's simpler?

• Your title and question disagree. Positive or completely positive? Commented May 27 at 18:25
• Thanks for pointing that out! I'll edit my title. Commented May 27 at 20:25

First a basic observation: if all Hermitian preserving $$\mathcal L$$ gave rise to completely positive dynamics $$e^{t\mathcal L}$$ for all $$t\geq 0$$, then so would $$-\mathcal L$$ (still Hermitian preserving!) meaning $$e^{t\mathcal L}$$ is completely positive for all $$t\leq 0$$, as well. But then $$e^{t\mathcal L}$$ had completely positive inverse $$e^{-t\mathcal L}$$ for all $$t\geq 0$$ meaning $$e^{t\mathcal L}$$ would have to have Kraus rank 1 for all times, thus restricting $$e^{t\mathcal L}$$ to a strict subset of all completely positive maps. In other words this basic argument shows that there has to be an extra condition which characterizes complete positivity of $$e^{t\mathcal L}$$. Indeed the following result is well known:

Given any linear map $$\mathcal L:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$$, the following statements hold.

1. $$e^{t\mathcal L}$$ is Hermitian preserving for all $$t$$ if and only if $$\mathcal L$$ is Hermitian preserving.
2. If $$\mathcal L$$ is Hermitian preserving, then the following are equivalent:
• $$\mathcal L$$ is conditionally completely positive, that is, $$({\bf1}-|\Omega\rangle\langle\Omega|)\mathsf C(\mathcal L)({\bf1}-|\Omega\rangle\langle\Omega|)\geq 0$$ where $$|\Omega\rangle:=\frac1{\sqrt n}\sum_j|j\rangle\otimes|j\rangle$$ is the maximally entangled state and $$\mathsf C(\mathcal L)=({\rm id}\otimes\mathcal L)(|\Omega\rangle\langle\Omega|)$$ is the Choi state corresponding to $$\mathcal L$$
• There exist $$\Phi$$ completely positive and $$K\in\mathbb C^{n\times n}$$ such that $$\mathcal L=\Phi+K(\cdot)+(\cdot)K^\dagger$$
• $$e^{t\mathcal L}$$ is completely positive for all $$t$$

Proof: 1. It is well known that arbitrary $$A\in\mathbb C^{n\times n}$$ is Hermitian if and only if $${\rm tr}(BA)\in\mathbb R$$ for all $$B$$ Hermitian (choose $$B=|x\rangle\langle x|$$). Thus a linear map $$\Phi$$ is Hermitian preserving if and only if $${\rm tr}(B\Phi(A))\in\mathbb R$$ for all $$A,B$$ Hermitian. This readily implies the desired statement: If $$\mathcal L$$ is Hermitian preserving, then $${\rm tr}(B e^{t\mathcal L}(A))=\sum_j\frac{t^j}{j!}{\rm tr}(B\mathcal LL^j(A))\in\mathbb R$$ for all $$t\in\mathbb R$$ and all $$A,B$$ Hermitian, and if—conversely—$$e^{t\mathcal L}$$ is Hermitian preserving for all $$t$$, then $${\rm tr}(B\mathcal L(A))=\frac{d}{dt}\underbrace{{\rm tr}(B e^{t\mathcal L}(A))}_{\in\mathbb R}\Big|_{t=0}\in\mathbb R$$ as desired. 2. First, as proven in the phys.SE post linked above conditionally completely positive implies the form $$\mathcal L=\Phi+K(\cdot)+(\cdot)K^\dagger$$. Next, we want to see that such $$\mathcal L$$ gives rise to completely positive dynamics: $$e^\Phi=\sum_j\frac{\Phi^j}{j!}$$ is completely positive (sums, products, and limits of CP maps are CP) and from the exponential formula $$e^{A(\cdot)+(\cdot)B}=e^A(\cdot)e^B$$ it follows that $$e^{K(\cdot)+(\cdot)K^\dagger}=e^K(\cdot)(e^K)^\dagger$$. Together with the Trotter product formula this implies $$e^{t\mathcal L}=\lim_{n\to\infty}\big( e^{t\Phi/n}e^{t(K(\cdot)+(\cdot)K^\dagger)/n} \big)^n=\lim_{n\to\infty}\big( e^{t\Phi/n}\circ \big(e^{tK/n}(\cdot)(e^{tK/n})^\dagger\big) \big)^n$$ meaning—as before—$$e^{t\mathcal L}$$ is completely positive as a limit of (products of) completely positive maps. Finally, if $$e^{t\mathcal L}$$ is completely positive, i.e. $$\mathsf C(e^{t\mathcal L})\geq 0$$ for all $$t$$, then by linearity of the Choi formalism \begin{align*} ({\bf1}-|\Omega\rangle\langle\Omega|)\mathsf C(\mathcal L)({\bf1}-|\Omega\rangle\langle\Omega|)&= \frac{d}{dt} ({\bf1}-|\Omega\rangle\langle\Omega|)\mathsf C(e^{t\mathcal L})({\bf1}-|\Omega\rangle\langle\Omega|)\Big|_{t=0}\\ &=\lim_{t\to 0^+}({\bf1}-|\Omega\rangle\langle\Omega|)\mathsf C\Big(\frac{e^{t\mathcal L}-{\rm id}}{t}\Big)({\bf1}-|\Omega\rangle\langle\Omega|)\\ &=\lim_{t\to 0^+}\frac1t\Big(({\bf1}-|\Omega\rangle\langle\Omega|)\mathsf C(e^{t\mathcal L})({\bf1}-|\Omega\rangle\langle\Omega|)\\ &\qquad\qquad\qquad-({\bf1}-|\Omega\rangle\langle\Omega|)\mathsf C({\rm id})({\bf1}-|\Omega\rangle\langle\Omega|)\Big)\,. \end{align*} But $$\mathsf C({\rm id})=|\Omega\rangle\langle\Omega|$$ so the second term vanishes. This leaves \begin{align*} ({\bf1}-|\Omega\rangle\langle\Omega|)\mathsf C(\mathcal L)({\bf1}-|\Omega\rangle\langle\Omega|)&=\lim_{t\to 0^+}\frac1t({\bf1}-|\Omega\rangle\langle\Omega|)\mathsf C(e^{t\mathcal L})({\bf1}-|\Omega\rangle\langle\Omega|) \end{align*} which is $$\geq 0$$ because all $$\mathsf C(e^{t\mathcal L})$$ are, so the same is true for $$({\bf1}-|\Omega\rangle\langle\Omega|)\mathsf C(e^{t\mathcal L})({\bf1}-|\Omega\rangle\langle\Omega|)$$ and for its (scaled) limit. $$\square$$

• Thanks you! I wasn't aware of the term "conditionally completely positive." Commented May 27 at 21:22
• Fair point, I updated my answer to now include a full proof of the equivalence! Commented May 27 at 23:00
• Beautiful proof, thank you so much! For me, the key parts I was missing were a) you can use Trotter to circumvent the issue of $\phi$ and $\kappa$ potentially not commuting (thus making it hard to split up the exponential) and b) the role of "conditional complete positivity" is to express the generator of the map as the limit of manifestly completely positive maps. Personally, the insights would be more clear to me if $1 - |\Omega\rangle\langle\Omega|$ were written something like $\mathrm{Ker}(\mathrm{id})$. That said, your post is great and your notation is probably more familiar to most. Commented May 28 at 13:43
• But I do show the implication in question at the very end of my post? Starting from "Finally, if $e^{t\mathcal L}$ is completely positive, [...]" I show that CP dynamics imply that $({\bf1}-|\Omega\rangle\langle\Omega|)\mathsf C(\mathcal L)({\bf1}-|\Omega\rangle\langle\Omega|)$ has to be positive semi-definite (=definition of conditionally CP) because it is the limit of the positive semi-definite operators $\frac1t({\bf1}-|\Omega\rangle\langle\Omega|)\mathsf C(e^{t\mathcal L})({\bf1}-|\Omega\rangle\langle\Omega|)$. Or did I misunderstand your comment? Commented May 28 at 14:42
• Ah, you are correct, thanks. To summarize, the logic is like this. There are three logically equivalent statements: 1. $e^{t \mathcal{L}}$ is CP for $t \geq 0$ 2. $\mathcal{L}(\rho) = \phi(\rho) + K \rho + \rho K^\dagger$ 3. $\mathcal{L}$ is conditionally completely positive. The post you linked shows 2 iff 3. You prove 1 implies 3. and 2 implies 1, completing the equivalence. Commented May 28 at 14:55