BB84 and B92 protocols: what is maximum and minimum bits that Alice and Bob might agree on?

Question

Say Alice wants to transmit a 10-bit key.

For the BB84 protocol, if we assume that Alice and Bob use the same basis to encode and decode 6 qubits (and different ones for the other 4), then what would be the maximum and minimum bits they might agree on?

Would the minimum number be 6 because they use the same basis 6 times and the maximum be 10? (in case Bob measures the same value even when using a different basis)

Also, since they disagree on the basis 4 times, are there $2^4$ possible runs of this protocol possible?

Extension: How would one go about doing the above analysis for B92?

Answer 1

I’m not exactly sure what is meant by “the bits they agree on”, but I would interpret it slightly differently to you. If Alice and Bob don’t agree on the same basis, they discard those bits, so they have a maximum of 6. If everything works perfectly, all 6 bits should be equal. Of course, in the real world, there are experimental imperfections and it could be that as few as 0 agree.

I’m also unsure why you’re counting the number of “possible runs” of the protocol. But I suppose I would count the number of possible basis choices. I forget if we’re talking about a 4-state or 6-state protocol here (2 bases or 3 bases) so let k be the number of bases. Each of the agreeing rounds has k different ways it could have occurred. The non-agreeing rounds have k(k-1) different ways the bases could have been chosen. This means that overall there are $$\binom{10}{6}k^6(k(k-1))^4$$ different ways that result in the described situation. That’s out of the $k^{20}$ different choices overall.