What are the possible channels preserving purity of all input states?

Question

Consider channels $\Phi$ such that $\Phi(|\psi\rangle\!\langle\psi|)$ is pure for all $|\psi\rangle$. Is there a simple way to characterise channels with this property?

Let's suppose $\Phi$ acts between input and output spaces of the same dimension. Two classes of such "purity-preserving" channels that immediately stand out are then unitary channels, $\Phi_U(\rho)=U\rho U^\dagger$ for some unitary $U$, and "replace channels" of the form $\Phi_{|u\rangle}(\rho)= \operatorname{Tr}(\rho) |u\rangle\!\langle u|$ for some $u$. These are quite "opposite", in that $\Phi_U$ acts transitively on pure states, while $\Phi_{|u\rangle}$ sends all pure states into a single point.

Is there a more general way to characterise these channels? As an alternative formulation, this should amount to asking, given a map $\Phi_f:\operatorname{Lin}(\mathcal H)\to \operatorname{Lin}(\mathcal H)$ such that there is some $f:\mathcal H\to\mathcal H$ such that $\Phi_f(\mathbb{P}_\psi)=\mathbb{P}_{f(\psi)}$ for all $\psi\in\mathcal H$, where $\mathbb{P}_u\equiv |u\rangle\!\langle u|$, what are the possible functions $f$ such that $\Phi_f$ is a channel?

Answer 1

A (partial) answer can be found in 'Transformations on tensor product spaces' of M Marcus, BN Moyls (1959), where they show that any map of matrices to itselves $f: M_n\to M_n$ that leaves the rank-1 matrices invariant must be of the form $f(X) = AXB$ or $f(X) = AX^TB$ for some invertible matrices $A$ and $B$. See the Corollary below Theorem 1 in the paper. Note that this only classifies those maps that are one-to-one for rank-1 matrices.

Now if you want your map to be a quantum channel, then it should preserve positivity, and it is then not too hard to show that we must have $B = A^*$, and if it is trace-preserving we must also have $AB=I$, so that $A$ must indeed be a unitary.

Answer 2

There are no other examples.

Consider the Kraus representation $\Phi(X)=\sum_a K_a X K_a^\dagger$.

If $\Phi_f(\mathbb{P}_\psi)=\mathbb{P}_{f(\psi)}$ is pure then all $K_a \mathbb{P}_\psi K_a^\dagger = \lambda_{a,\psi} \mathbb{P}_{f(\psi)} $ equal to the same pure state up to a factor. Thus $K_a|\psi\rangle = \mu_{a,\psi} |f(\psi)\rangle$ for all $a$.

Consider two different indices $1,2$ (if Kraus rank is $1$ then the sole $K_1$ has to be unitary).

Let $K_1v_1 = w_1$, $K_1v_2 = w_2$, where $w_1,w_2$ are non-collinear vectors. Then $K_2v_1 = \mu_1 w_1$, $K_2v_2 = \mu_2w_2$, and $K_2(v_1+v_2) = \mu_3(w_1+w_2)$. Hence $\mu_1=\mu_2=\mu_3$.

It follows that on any subspace $S \subset H$, such that $K_1 |_S$ is invertible, we must have $K_2|_S = \mu K_1|_S$.

In particular, if $H_1 = {\rm Ker}(K_1)$, then it must be $K_2 |_{H_1^\perp} = \mu K_1 |_{H_1^\perp}$.

Let $v_1 \in H_1$, $v_2 \in H_1^\perp$. Assume $K_2v_1 \neq 0$ and note that
$K_2(v_1+v_2) = cK_1(v_1+v_2)$ $\implies$ $K_2v_1+\mu K_1v_2 = cK_1v_2$ $\implies$ $K_2v_1 = (\mu-c)K_1v_2$. Similarly, $K_2v_1 = (\mu-c')K_1v_2'$ for $v_2' \in H_1^\perp$.

It follows that if ${\rm dim}H_1^\perp>1$ then $K_2(H_1)=0$ as well, thus $K_2 = \mu K_1$. We can neglect this case.

Hence we can assume that ${\rm dim}H_i^\perp=1$ (and thus ${\rm rank}K_i = 1$) for every $i$. It's then easy to show that $\Phi(X) = {\rm Tr}(X)| u\rangle\langle u|$.