6
$\begingroup$

The Stinespring dilation theorem states that any CPTP map $\Lambda$ on a system with Hilbert space $\mathcal{H}$ can be represented as $$\Lambda[\rho]=tr_\mathcal{A}(U^\dagger (\rho\otimes |\phi\rangle\langle \phi|)U)$$ where $\mathcal{A}$ is an ancilla, $|\phi\rangle\in\mathcal{A}$ is an arbitrary pure state, and $U$ is a unitary on the joint system $\mathcal{H}\otimes\mathcal{A}$. Importantly, this works for any choice of the state $|\phi\rangle\in\mathcal{A}$ -- that is, neither does $|\phi\rangle\in\mathcal{A}$ have to depend on $\Lambda$, nor is a specific choice of $|\phi\rangle\in\mathcal{A}$ required.

Would this representation still be valid if, instead of attaching an ancilla in the pure state $|\phi\rangle\langle\phi|$, we instead attached an ancilla in some arbitrary mixed state $\sigma$? That is, given a state $\sigma$ of $\mathcal{A}$, I would like to know whether for every CPTP channel $\Lambda$, there exists a unitary $U$ on $\mathcal{H}\otimes\mathcal{A}$ such that $$\Lambda[\rho]=tr_\mathcal{A}(U^\dagger (\rho\otimes \sigma)U).$$

$\endgroup$
0

2 Answers 2

7
$\begingroup$

No, that doesn't work. It's fine to use an arbitrary pure state because the unitary $U$ can always be used to take it to any pure state you want. This argument doesn't work for a mixed state, as unitaries cannot take mixed states to pure states.

As a concrete example, consider the CPTP map $$ \Lambda(\rho) = |0\rangle\langle 0| \operatorname{tr}(\rho), $$ and let $\sigma = I/d_{\mathcal A}$, the maximally mixed state on $\mathcal A$. If you now apply this CPTP map to $I/d_{\mathcal H}$, the maximally mixed state on $\mathcal H$, you should get $|0\rangle \langle 0|$, but $$ \operatorname{tr}_\mathcal{A}(U^\dagger (\rho\otimes \sigma)U) = \operatorname{tr}_\mathcal{A}(U^\dagger (I/d_{\mathcal H}\otimes I/d_{\mathcal A})U) = I/d_{\mathcal H}$$ for any unitary $U$.

$\endgroup$
6
  • 1
    $\begingroup$ "as unitaries cannot take mixed states to pure states" -- that's no reason. $\endgroup$ Mar 1, 2021 at 21:28
  • $\begingroup$ Why not? If you could take $\sigma$ to a pure state, you would be back at the usual Stinespring dilation with an arbitrary pure ancilla. $\endgroup$ Mar 2, 2021 at 9:41
  • $\begingroup$ True, but you use it as an argument for the converse direction. $\endgroup$ Mar 2, 2021 at 10:22
  • 2
    $\begingroup$ I don't. I'm just saying that the argument why you can get away with an arbitrary pure state doesn't hold for an arbitrary mixed state. You're just being pedantic. $\endgroup$ Mar 2, 2021 at 10:46
  • $\begingroup$ I see. It seems the "This doesn't hold" was ambiguous. $\endgroup$ Mar 2, 2021 at 11:06
4
$\begingroup$

No. If you take $\sigma=\tfrac1{d_A}\mathbb I$, you will have that for $\rho=\tfrac1{d_S}\mathbb I$, $$ \mathrm{tr}_A(U^\dagger(\rho\otimes\sigma)U)=\tfrac1{d_S}\mathbb I\ , $$ and thus, it will not be possible to implement any channel for which $$ \Lambda(\tfrac1{d_S}\mathbb I)\ne \tfrac1{d_S}\mathbb I\ , $$ such as any channel mapping all inputs to a constant output other than the identity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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