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Given an arbitrary state $\rho_{AB}$, is it always possible to construct an extension $\rho_{ABC}$ such that

$$Tr_B(\rho_{ABC}) := \rho_{AC} = \rho_{AB} := Tr_C(\rho_{ABC})$$

If yes, does there exist a quantum channel i.e. $\mathcal{E}: \mathcal{H}_{AB} \rightarrow \mathcal{H}_{ABC}$ that can achieve this?

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    $\begingroup$ There's quite some work on that (also/especially on the channel analogue, due to the connection to cloning), the keyword is "symmetrically extendible", I think. $\endgroup$ Jan 9 at 18:28

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No, assume $\rho_{AB}$ is pure, so that $\rho_{AB} = |u\rangle\langle u|_{{AB}}$. Since it's pure $\rho_{ABC}$ must have the form $\rho_{ABC} = \rho_{AB} \otimes \rho_{C}$. It follows that $\rho_{AC} = \rho_{A} \otimes \rho_{C}$. But if $|u\rangle_{{AB}}$ is entangled then $\rho_{AB} \neq \rho_{A} \otimes \rho_{B}$.

This is related to Monogamy of entanglement.

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  • $\begingroup$ Thanks. It seems your construction shows that such a channel cannot exist, even if $\rho_{AB}$ is separable. For example, let $\rho_{AB} = \vert u\rangle\langle u\vert_A\otimes \vert v\rangle\langle v\vert_B$. The extension in the $C$ register is then forced to be $\vert v\rangle\langle v\vert_C$, a clone of the $B$ register - does that argument make sense to show there cannot be such a channel as I hoped for, even for separable states? $\endgroup$ Jan 9 at 16:55
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    $\begingroup$ That's right, it would be a cloning, and so such a channel can't exist (despite existence of an extension). $\endgroup$
    – Danylo Y
    Jan 9 at 19:43
  • $\begingroup$ @user1936752 It is not clear what channel you do expect to exist. If it should just output rho_ABC for the specific setting, it always does: Just discard the input and discard rho_ABC. $\endgroup$ Jan 17 at 7:32
  • $\begingroup$ @NorbertSchuch yes I see what you mean. I was thinking about a channel that constructed a symmetric extension for an arbitrary input $\rho_{AB}$ - this is prevented by no-cloning in the example discussed above. $\endgroup$ Jan 17 at 10:31
  • $\begingroup$ @user1936752 Now the right question seems: In the case the existence of the extension is not ruled, is there a map which implements it? $\endgroup$ Jan 17 at 17:09
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As noted in the comment, the state part of the question is precisely the question of symmetric extensions; see for instance the thesis https://arxiv.org/abs/1103.0766 and references therein. Apart from cloning and monogamy of entanglement, it is relevant to quantum key distribution, and also separability criteria via a neat de Finetti argument (see https://arxiv.org/abs/quant-ph/0308032v3 Sec. II), though those topics are also all connected to some degree.

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