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?


3 Answers 3


TL;DR: The unitary and reset channels are the only ones that return pure output for every pure input. That's because under Stinespring dilation the requirement that $\Phi$ return pure output for every pure input translates into the requirement that the unitary acting on the larger Hilbert space of the system $S$ and its environment $E$ be non-entangling. There are only two types of non-entangling unitaries: $U(d_S)\otimes U(d_E)$ and $U(d_S)\otimes U(d_E)\circ\text{SWAP}$.

We can represent $\Phi$ as $$ \Phi(|\psi_S\rangle\langle\psi_S|)=\mathrm{tr}_E(U_{SE}\circ|\psi_S\rangle\langle\psi_S|\otimes|0_E\rangle\langle 0_E|\circ U_{SE}^\dagger).\tag1 $$ If $\Phi$ returns pure state for every pure input, then we can choose$^1$ $U_{SE}$ which does not entangle the system $S$ with its environment $E$. Such unitaries fall into two categories: product unitaries $U(d_S)\otimes U(d_E)$ and exchange unitaries $U(d_S)\otimes U(d_E)\circ\text{SWAP}$ where $\text{SWAP}$ acts on the $d_S$-dimensional Hilbert space of $S$ and some $d_S$-dimensional subspace of the Hilbert space of $E$. Any other unitary produces entanglement between $S$ and $E$ for some inputs $|\psi_S\rangle$ and this results in a mixed state after tracing out the environment.

If $\Phi$ can be described using $U_{SE}\in U(d_S)\otimes U(d_E)$, then $\Phi$ is a unitary channel. If $\Phi$ can be described using $U_{SE}\in U(d_S)\otimes U(d_E)\circ\text{SWAP}$, then $\Phi$ is a reset channel.

$^1$ Note the subtlety: $U_{SE}$ is not uniquely determined. More precisely, its action when environment is in a state $|k_E\rangle$ with $k\ne 0$ is not uniquely determined. Still, we know it does not entangle $S$ and $E$ when environment is in state $|0_E\rangle$ and we can define it by extending to a non-entangling unitary. Thus, among the unitaries that satisfy $(1)$ there exist unitaries that do not entangle $S$ and $E$. That's all we need.

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.

  • $\begingroup$ interesting, thanks. I'd say this doesn't characterise all "maps that are one-to-one for rank-1 matrices" though. If I'm reading this right, it seems it only classifies those who act like the identity on rank-1 matrices. For example, (nontrivial) unitary channels do not satisfy this $\endgroup$
    – glS
    Commented May 12, 2022 at 13:48
  • $\begingroup$ I think you might be misunderstanding. Here $X$ could for instance be your rank-1 matrix $X=|\psi\rangle\!\langle \psi|$. So these are definitely not the identity on rank-1 matrices. Just taking $A=U$ and $B=U^*$ includes the examples you gave. $\endgroup$
    – John
    Commented May 12, 2022 at 13:51
  • 1
    $\begingroup$ oh ok. So by "leaves the rank-1 matrices invariant" you mean the whole set not the individual elements, makes sense. I suppose another way to get a similar conclusion is via the result that the only invertible channels are unitary channels, though this is more general. This also means that any channel preserving pure states with Choi rank greater than one contracts the set of pure states, interesting. $\endgroup$
    – glS
    Commented May 12, 2022 at 13:55

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 $\Phi(X) = {\rm Tr}(X)| u\rangle\langle u|$.


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