# Quantum bit error rate

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?

• 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? Dec 13 '20 at 11:45
• @fgrieu: diagonals of the crosstalk matrix.
– Rasha rashed
Dec 13 '20 at 18:24
• @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 Dec 14 '20 at 8:54
• @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?
– Rasha rashed
Dec 15 '20 at 5:20
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: $$QBER = \frac{5}{60} + \frac{7}{60}.$$