I'm new to information theory and i need to calculate $I(A,E)$. To calculate it I need conditional entropy $H(A|E)$.
I assume the BB84 protocol standard states $\{ |0\rangle,|1\rangle \},\{|+\rangle,|-\rangle\}$. I assume a standard intercept resend attack from Eve, that in a random fashion measures in one of the two basis and resends the measured state to Bob.
Unfortunately, when I calculate the conditional entropy, I obtain the value $\frac{1}{2}-\frac{3}{4}\log \frac{3}{4}$ and not $\frac{1}{2}$ like in Gisin original paper (https://arxiv.org/abs/quant-ph/0101098, section VI-D.
My first calculation begun with quantum states: i expressed conditional entropy like that \begin{equation} \begin{aligned} H(A \mid E = |0\rangle) &= - P(A = |0\rangle \mid E =|0\rangle) \log P((|0\rangle \mid E = |0\rangle)) \\ &- P(A = |1\rangle \mid E = |0\rangle) \log P((A = |1\rangle \mid E = |0\rangle)) \, .\\ \end{aligned} \end{equation} Calculating the probabilities with bayes theorem, i found this entropy to be equal to 0, and the others like that: \begin{equation} \begin{aligned} H(A \mid E = |1\rangle) &= 0 \, ,\\ H(A \mid E = |+\rangle) &= 1 \, ,\\ H(A \mid E = |-\rangle) &= 1 \, . \end{aligned} \end{equation} In this way the final entropy is numerically $$ H(A\mid E) =\frac{1}{2} \, . $$
But then i re-thought it: perhaps i have to consider the probability not of the states, but of the bits results, because the formula i'm using is classical! So i tried evaluating the following: \begin{equation} H(A \mid E) = P(E=0) H(A \mid E = 0) + P(E=1) H(A \mid E = 1) \, . \end{equation} but this leads me to the strange value i linked above, the same value i obtain for the conditional entropy of Alice and Bob, that is $H(A|B)$.
What am i not understanding correctly?
(If you can, give me references)
PS the full calculations for the first method are reported here: \begin{equation} \begin{alignedat}{2} P(E = |0\rangle \mid A = |0\rangle ) &= \frac{1}{2} \, , & \hspace{1in} P(E = |0\rangle \mid A = |1\rangle ) &= 0 \, ,\\ P(E = |1\rangle \mid A = |0\rangle ) &= 0 \, , & P(E = |1\rangle \mid A = |1\rangle ) &= \frac{1}{2} \, , \\ P(E = |+\rangle \mid A = |0\rangle ) &= \frac{1}{4} \, , & P(E = |+\rangle \mid A = |1\rangle ) &=\frac{1}{4} \, ,\\ P(E = |-\rangle \mid A = |0\rangle ) &= \frac{1}{4} \, , & P(E = |-\rangle \mid A = |1\rangle ) &= \frac{1}{4} \, , \\ P(A = |0\rangle) &= \frac{1}{2} \, , & P(A = |1\rangle) &= \frac{1}{2} \, . \\ \end{alignedat} \end{equation} and \begin{equation} \begin{aligned} P(E = |0\rangle) &= P(E = |0\rangle \mid A = |0\rangle ) P(A = |0\rangle) + P(E = |0\rangle \mid A = |1\rangle ) P(A = |1\rangle) \\ &= \frac{1}{2} \cdot \frac{1}{2} + 0 \cdot \frac{1}{2} = \frac{1}{4} \, , \\ P(E = |1\rangle) &= P(E = |1\rangle \mid A = |0\rangle ) P(A = |0\rangle) + P(E = |1\rangle \mid A = |1\rangle ) P(A = |1\rangle) \\ &= 0 \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \, , \\ P(E = |+\rangle) &= P(E = |+\rangle \mid A = |0\rangle ) P(A = |0\rangle) + P(E = |+\rangle \mid A = |1\rangle ) P(A = |1\rangle) \\ &= \frac{1}{4} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{4} \, , \\ P(E = |-\rangle) &= P(E = |-\rangle \mid A = |0\rangle ) P(A = |0\rangle) + P(E = |-\rangle \mid A = |1\rangle ) P(A = |1\rangle) \\ &= \frac{1}{4} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{4} \, . \\ \end{aligned} \end{equation} then with bayes \begin{equation} \begin{alignedat}{2} P(A= |0\rangle \mid E = |0\rangle ) &= 1 \, , & \hspace{1in} P(A= |1\rangle \mid E = |0\rangle ) &= 0 \, ,\\ P(A= |0\rangle \mid E = |1\rangle ) &= 0 \, , & \hspace{1in} P(A= |1\rangle \mid E = |1\rangle ) &= 1 \, ,\\ P(A= |0\rangle \mid E = |+\rangle ) &= \frac{1}{2} \, , & \hspace{1in} P(A= |1\rangle \mid E = |+\rangle ) &= \frac{1}{2}\, ,\\ P(A= |0\rangle \mid E = |-\rangle ) &= \frac{1}{2} \, , & \hspace{1in} P(A= |1\rangle \mid E = |-\rangle ) &= \frac{1}{2} \, . \end{alignedat} \end{equation} with all these numbers i have calculated the entropies reported above.
