4
$\begingroup$

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?

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
    May 12 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
    May 12 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
    May 12 at 13:55
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.