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I know that noise is a big challenge for running this protocol because it is hard to understand whether the error is produced by noise or eavesdropper. To make the success rate higher you assume that the noise is also caused due to eavesdropping. As you know this protocol heavily relies on probability calculations. So, while calculating the probability for ...


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The Bures metric is the limit of the Bures distance for two infinitesimally close density matrices $\rho$ and $\rho+d\rho$. The Bures distance however, is not unique. It depends on the space on which we integrate the Bures metric. The Bures distance for positive matrices differs from that of density matrices (with a unit trace), however they have the same ...


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The standard noisy approach is not to try to determine the presence of an eavesdropper as such, but to create a final key where, even if there is an eavesdropper, you can still be confident that the eavesdropper has negligible information about the key. So you aren't trying to distinguish between noise and eavesdropping, but pessimistically assuming that ...


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If there is an evasdroping, an error rate is higher than some long-term average of a quantum communication channel used for a quantum key distribution. So, when the error rate is high, the key is deleted and new one is distribuited again until noise level is at acceptable level. A natural noise can be reduced with classical error correction, something like ...


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Background If $v_1, v_2, \dots, v_n$ is an orthonormal basis in the inner product space $V$, then any vector $u\in V$ can be expressed as a linear combination $$ u = \alpha_1 v_1 + \alpha_2 v_2 + \dots + \alpha_n v_n.\tag1 $$ Moreover, the coefficients can be computed using $\alpha_k=\langle v_k, u\rangle$, as can be seen by applying $\langle v_k, .\rangle$ ...


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Let $\mathcal{G}_n$ denote the Pauli group on $n$ qubits. An $n$-qubit state $|\psi\rangle$ is called a stabilizer state if there exists a subgroup $S \subset \mathcal{G}_n$ such that $|S|=2^n$ and $A|\psi\rangle = |\psi\rangle$ for every $A\in S$. For example, $(|00\rangle+|11\rangle)/\sqrt2$ is a stabilizer state, because it is a $+1$ eigenstate of the ...


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