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I'm reading the paper "Quantum secret sharing" by Hillery, Bužek and Berthiaume (Phys. Rev. A, 1999; preprint on arXiv) and I have a question about it.

Let's say Alice wants to send Bob qubit in state $|\psi_A\rangle = \alpha|0\rangle_A + \beta|1\rangle_A$, and she has one qubit from GHZ state $\frac{1}{\sqrt 2}(|0\rangle_a|00\rangle +|1\rangle_a|11\rangle)$. Then she performs a Bell basis
$$|\Psi_{\pm}\rangle_{Aa}=\tfrac{1}{\sqrt 2}(|00\rangle_{Aa}\pm|11\rangle_{Aa} \\ |\Phi_{\pm}\rangle_{Aa}=\tfrac{1}{\sqrt 2}(|01\rangle_{Aa}\pm|10\rangle_{Aa}$$ measurement on the qubits $(A, a)$.

In terms of projection operators and/or measurement operators, how can I get an expression for this 4-qubit state: $$\begin{aligned} |\Psi\rangle_4 = \tfrac{1}{2}[\, &|\Psi_{+}\rangle_{Aa}(\alpha|00\rangle +\beta|11\rangle) \\ +\ &|\Psi_{-}\rangle_{Aa}(\alpha|00\rangle -\beta|11\rangle) \\ +\ &|\Phi_{+}\rangle_{Aa}(\beta|00\rangle +\alpha|11\rangle) \\ +\ &|\Phi_{-}\rangle_{Aa}(-\beta|00\rangle +\alpha|11\rangle) \,] ?\end{aligned}$$

(I understand that if just simplify this expression I get $$(\alpha|0\rangle_A + \beta|1\rangle_A) \tfrac{1}{\sqrt 2}(|0\rangle_a|00\rangle +|1\rangle_a|11\rangle.)$$

For reference, here are the relevant sections of the paper: 1, 2

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    $\begingroup$ Welcome to Cryptography. There is also Quantum.SE and this suits more into them. $\endgroup$ – kelalaka May 5 at 17:06
  • $\begingroup$ Any sufficiently advanced technology is indistinguishable from magic (Clarke's Third Law). $\endgroup$ – fgrieu May 5 at 17:56
  • $\begingroup$ This is a cross-posting with Quantum.SE. Could you delete one copy, preferable this one. $\endgroup$ – kelalaka May 6 at 21:40

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