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 $$\Lambda[\rho]=tr_\mathcal{A}(U^\dagger (\rho\otimes \sigma)U).$$


2 Answers 2


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 $U$.

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    $\begingroup$ "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
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    $\begingroup$ 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

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.


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