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Alice wants to send Bob qubit in state $|\psi_A\rangle = \alpha|0\rangle_A + \beta|1\rangle_A$ , also she has one qubit from GHZ state $\frac{1}{\sqrt 2}(|0\rangle_a|00\rangle +|1\rangle_a|11\rangle)$.

Then she performs Bell basis
($|\Psi_{\pm}\rangle_{Aa}=\frac{1}{\sqrt 2}(|00\rangle_{Aa}\pm|11\rangle_{Aa})$, $ |\Phi_{\pm}\rangle_{Aa}=\frac{1}{\sqrt 2}(|01\rangle_{Aa}\pm|10\rangle_{Aa})$ measurement on qubits $(A, a)$.

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

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

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    $\begingroup$ $\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}$ Your two states $\ket{\Psi_4}$ and $\ket{\psi_a}\ket{GHZ}$ are indentical. The first is just a regrouped version of the second. You'll see that if you just tensor-multiply it out... $\endgroup$ – draks ... May 6 at 9:30

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