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15
answered Entangled electron-positron pair
Oct
8
comment Extending the idea of superdense coding
If you could produce $2^{n}$ orthogonal states this would allow you to signal because you would have to change the reduced density matrix of the $n-1$ qubits to obtain that many orthogonal states (since you will have as many states as it is possible to obtain on $n$ qubits)
Oct
8
comment Extending the idea of superdense coding
@braindead $2^{n}$ states, yes; $2^{n}$ orthogonal states, no. You can see this by decomposing our bipartite state using the Schmidt decomposition $|\psi_{tot}\rangle=\Sigma\lambda_{i}|\psi_{i}\rangle|e_{i}\rangle$. Restricting ourselves to unitaries on the second party (single qubit) we can only obtain $4$ mutually orthogonal states because the only things we can change are the Schmidt basis "$e$" and the phase relationship between the two terms.
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answered What is postselection?
Oct
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answered Extending the idea of superdense coding