The quantum key establishment protocol

Question

Consider the quantum key establishment protocol. Assume that bit values, emitter basis, and receiver lter orientation are selected independently and uniformly at random (i.e., 50% chance for each of the bit values/ bases/orientations).

1)What is the joint entropy of the emitter bit value and receiver bit value?

My assumption is: H(X=0,Y=0) = 3/4 H(X=0,Y=1) = 1/4 H(X=1,Y=0) = 1/4 H(X=1,Y=1) = 3/4

H(Y=0,X=0) = 3/4 H(Y=0,X=0) = 1/4 H(Y=0,X=0) = 1/4 H(Y=0,X=0) = 3/4

I could not go further here.

2)What is the mutual information between the emitter bit value and the receiver bit value?

In my opinion, the filters can measure the polarasation of a photon if the filter and the photon do not agree the outcome is uniformly random. So, with respect to receiver bit value and emitter bit value we can find the sifted key.

Note:It is not a homework question.

Answer 1

The protocol is formally known as BB84 protocol.

Your "assumption" are conditional probabilities, and they should be calculated, not "assumed". For example, let Alice sends bit $0$; what is the probability that Bob receives bit $0$? Bob has 2 equally likely choices: either his basis coincides with the Alice's basis or not; in the 1st case Bob receives $0$ bit with probability $1$, in the 2nd case Bob receives $0$ bit with probability $1/2$. So the conditional probability that Bob receives $0$ bit given Alice sent $0$ bit is $$ P_{X,Y}(0,0|X=0) = 1/2\cdot 1 + 1/2\cdot 1/2 = 3/4$$

Since Alice sends $0$ and $1$ bits equally likely, $$ P_{X,Y}(0,0) = 1/2\cdot 3/4 = 3/8$$ and this I believe is one of the 4 probabilities that should be inserted into the Joint Entropy Formula