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I am familiar with the quantum teleportation protocol to some extent. However, I do not understand what it means when we say that "the teleportation relies on the projection of two qubits onto maximally entangled Bell states."

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    $\begingroup$ Another way to say that might be "teleportation relies on having the ability to perform a projective measurement in the Bell basis", i.e. to realize the PVM $\{|\Phi_{ij}\rangle \langle \Phi_{ij}| \}$ where $|\Phi_{ij}\rangle$ for $i,j \in \{0,1\}$ are Bell states. $\endgroup$
    – forky40
    Sep 16 at 23:13
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    $\begingroup$ @forky40 This should be an answer, not a comment :-) $\endgroup$ Sep 16 at 23:29
  • $\begingroup$ could you be more specific as to what exactly you find unclear about that sentence? Also, this answer might be relevant: quantumcomputing.stackexchange.com/a/21233/55, as well as this one on physics.SE: physics.stackexchange.com/a/274739/58382 $\endgroup$
    – glS
    Sep 17 at 7:19
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If Alice and Bob share a maximally entangled state wherein each controls one qubit: $$|\Phi^{+}\rangle=\frac{|0\rangle^{A}\otimes|0\rangle^{B}+|1\rangle^{A}\otimes|1\rangle^{B}}{\sqrt{2}}$$ and Alice wants to teleport one qubit $|\psi\rangle^{S}=a|0\rangle^{s}+b|1 \rangle^{s}$ Now, as you can see, these two states are product with one another, so the total state is $$|\psi\rangle^{S}\otimes|\Phi^{+}\rangle=a|0\rangle^{s}+b|1 \rangle^{s}\otimes\frac{|0\rangle^{A}\otimes|0\rangle^{B}+|1\rangle^{A}\otimes|1\rangle^{B}}{\sqrt{2}}$$ Now we can expand this and the group some of the terms, and as a result re-express this as a linear combination of entangled states of SA product with some state for Bob, b $$|\psi\rangle^{S}\otimes|\Phi^{+}\rangle=\frac{1}{\sqrt{2}}(a|0\rangle^{s}\otimes|0\rangle^{A}\otimes|0\rangle^{B}+a|0\rangle^{s}\otimes|1\rangle^{A}\otimes|0\rangle^{1}+b|1\rangle^{s}\otimes|0\rangle^{A}\otimes|0\rangle^{B}+b|1\rangle^{s}\otimes|1\rangle^{A}\otimes|1\rangle^{B})$$ Next, we remember that we can rewrite the product states of SA as the sum of maximally entangled bell states, ie $|00\rangle=\frac{1}{\sqrt{2}}(\frac{|00\rangle+|11\rangle}{\sqrt{2}}-\frac{|00\rangle-|11\rangle}{\sqrt{2}})$. Now, after re-writing the product terms for SA as the difference of bell states and cancelling, we are left with $$\frac{1}{2}(|\Phi^{+}\rangle^{SA}\otimes(a|0\rangle+b|1\rangle)+|\Phi^{-}\rangle^{SA}\otimes(a|0\rangle -b|1\rangle)+|\Psi^{+}\rangle^{SA}\otimes(a|1\rangle+b|0\rangle)+|\Psi^{-}\rangle^{SA}\otimes(a|1\rangle-b|0\rangle))$$ At this point, it is clear to see what they mean by projection onto a maximally entangle bell pair, as said projection on the subsystem of SA will then leave you one of the 4 states for Bob, which is what we wanted in the first place.

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