I understand that the optimal value, i.e 2/3, can be achieved if Alice and Bob are connected via a classical channel, and Alice makes a projective measurement. Note that this is just one of the ways of formalizing the experiment. The protocol is as follows: Alice measures an unknown state, say $|\psi\rangle = \cos(\theta/2)|0\rangle+\sin(\theta/2)e^{i\phi}|1\rangle$, in $\sigma_{z}$ basis, and sends the result to Bob (encoded in a classical bit). Depending on the classical information received by Bob, he either guesses $|0\rangle$ and $|1\rangle$.

In this case, we consider the closeness (or fidelity score) to be the sum of probability of getting $|0\rangle$ state multiplied by the fidelity measure, $|\langle0|\psi\rangle|^{2}$, and similarly for $|1\rangle$ state. So, if we average out the fidelity ($p(|0\rangle||\psi\rangle)*|\langle0|\psi\rangle|^{2}+p(|1\rangle||\psi\rangle)*|\langle1|\psi\rangle|^{2}$) across all possible input states in the Bloch sphere:

$$ F_{cl} = \int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}\frac{\sin\theta d\theta d\phi}{4\pi}(\cos^{2}(\theta/2)*\cos^{2}(\theta/2)+\sin^{2}(\theta/2)*\sin^{2}(\theta/2)) = \frac{2}{3} $$

But in the original paper, it is mentioned that this is just one of the experiments by which we can attain the optimal value (i.e 2/3). In fact, we need to consider more general measurements for obtaining the optimal value. I'd like to know if there is a simple proof using POVM formalism, for example, consider {$E_{i}$} to be the set of operators which are (i) positive and (ii) obey $\sum_{i} E_{i} = \texttt{I}$; can you provide a similar proof as above (using these operators)?

  • $\begingroup$ The above proof is not trvial for the general problem, i.e when N particles with k dimensions are given: paper by V. Bužek and Artur Ekert. In that paper, they show not only the existence of such a POVM (which attains the optimal value), but also give the algorithm for finding such a POVM. But, I think that there is an easier proof (via POVM) for a two-level quantum system (a qubit). $\endgroup$ – Vijeth Aradhya Mar 4 '18 at 6:41

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