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So suppose we have two 2-qubit bell states $|\Psi_{AB}\rangle$ and $|\Psi_{BC}\rangle$ defined the usual way. I want to create a three qubit pure state from qubits A,B, and C such that the entanglements of qubits AB and BC are preserved. I've seen this done with some of the specific bell states, but I wasn't sure if it can be universally done with any of the 16 combinations of 2-qubit Bell states. Can it be done in a general case?

Edit: Meant a Pure State not Bell State! Edit 2: Rephrased for clarity.

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    $\begingroup$ I don't understand, can you elaborate a bit more the question? What about $\psi_{AB}\otimes \psi_{BC}$? $\endgroup$
    – Tobias Fünke
    Commented Nov 30, 2023 at 15:29
  • $\begingroup$ I meant a combined pure state, sorry!!! $\endgroup$
    – M Rozzzz
    Commented Nov 30, 2023 at 15:33
  • $\begingroup$ Yeah sorry, I'm quite unclear about that, I'll edit momentarily! $\endgroup$
    – M Rozzzz
    Commented Nov 30, 2023 at 15:43

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A maximally entangled state between AB and BC is impossible to start with. This is due to monogamy of entanglement. The more entanglement B has with A, the less it can have with C.

See https://www.quantiki.org/wiki/monogamy-entanglement for a simplified treatment of this issue, or https://arxiv.org/abs/quant-ph/9907047 for one of the original expositions on this.

But A, B and C can be made to share some degree of entanglement. In fact, a GHZ state is such a 3 photon state. See for example https://arxiv.org/abs/quant-ph/9810035 or https://www.drchinese.com/David/Bell-MultiPhotonGHZ.pdf

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  • $\begingroup$ In that case, is it possible to at least create a pure state preserving the entanglement of A and B while somehow describing some entanglement of B and C? $\endgroup$
    – M Rozzzz
    Commented Nov 30, 2023 at 15:54
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    $\begingroup$ @MRozzzz Not if AB is in a Bell State (which is a maximally entangled state). You must start out with maximally entangled AB and CD (i.e. ABxCD). You can, by suitable distillation, convert that to maximally entangled ABC (GHZ=000+111). Note that a measurement on A giving a 1 (for example), will certainly yield a 1 for B and C as well. However, B and C are now separable and will NOT be entangled at all. On the other hand, a 3 photon W state (W=100+010+001) yields an entangled BC after measurement of A. A simplified treatment of that is: en.wikipedia.org/wiki/W_state $\endgroup$
    – DrChinese
    Commented Nov 30, 2023 at 16:59

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