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I am trying to understand the quantum key distribution and Quantum Bit Error Rate (QBER). I have question: Why is the QBER given by the average of the diagonals of the crosstalk matrix? Why the need to calculate the value using the diagonals?

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    $\begingroup$ Fixed: the B of QBER stands for "bit" (an elementary information), not "bet" (as in "bet the farm on quantum", like some want the motto of research budgets to be). The question begs for another: the diagonal of what? $\endgroup$
    – fgrieu
    Dec 13, 2020 at 11:45
  • $\begingroup$ @fgrieu: diagonals of the crosstalk matrix. $\endgroup$
    – Rasha rashed
    Dec 13, 2020 at 18:24
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    $\begingroup$ @Rasharashed : Reminds how it is computed in your text in the question itself. QBERs have been computed for decades with different ways, all equivalent, and your question assumes we all know the specific formulation you refer to. I have been working in QKD for twenty years now, and I don’t understand the question, despite knowing how to compute a QBER $\endgroup$
    – Frédéric Grosshans
    Dec 14, 2020 at 8:54
  • $\begingroup$ @FrédéricGrosshans, many thanks for your reply. I assume this cryptography site. I am very sorry if make question is not clear. I mean as you know calculate the QBER using corsstalk matrix. Take the average of the diagonals element. Why is we choose the diagonals element? $\endgroup$
    – Rasha rashed
    Dec 15, 2020 at 5:20
  • $\begingroup$ @MaartenBodewes, yes please $\endgroup$
    – Rasha rashed
    Dec 17, 2020 at 3:07

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I do not know how the crosstalk matrix works. But the way I understand QBER is the following:

QBER is the mismatch probability of the signals sent and receieved between Alice and Bob. Let Alice send 100 signals to Bob. Bob makes his own measurement choices and at the end, they compare their results. Let's say, in 60 of these 100 signals, Bob measured in the 'right' basis, i.e., his result should match Alice's preparation. Meaning, in these 60 cases, if Alice intended to send classical bit 0 (or 1), Bob should receive classical bit 0 (or 1).

However, due to some unwanted noise (Eve!), among these 60 cases, some of their results differed. Let in 5 of these 60 cases, Alice wanted to send 0, Bob should have received 0 (because he made the right measurement choice), but got 1. The other case is also possible. I.e., let 7 of the 60 cases, Alice sent a 1, but Bob received 0. The QBER is then: $$ QBER = \frac{5}{60} + \frac{7}{60}. $$

Does it help a little? I welcome the community members to comment on this answer to confirm my understanding too!

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