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

Norbert Schuch
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Jacob Drori
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2 Answers2

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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 $U$.

Norbert Schuch
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Mateus Araújo
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    "as unitaries cannot take mixed states to pure states" -- that's no reason. – Norbert Schuch Mar 01 '21 at 21:28
  • 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. – Mateus Araújo Mar 02 '21 at 09:41
  • True, but you use it as an argument for the converse direction. – Norbert Schuch Mar 02 '21 at 10:22
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    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. – Mateus Araújo Mar 02 '21 at 10:46
  • I see. It seems the "This doesn't hold" was ambiguous. – Norbert Schuch Mar 02 '21 at 11:06
  • What seems a potentially interesting question is "given a target channel, which $\sigma$ are admissible to realize the channel". It seems that there must be enough "purity" in the state to allow for all outputs, but I'm not sure how to quantify this, or whether this is the only constraint. – Norbert Schuch Mar 02 '21 at 12:17
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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.

Norbert Schuch
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