# Mutual information between Alice and Eve in a BB84 intercept resend attack

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{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} Calculating the probabilities with bayes theorem, i found this entropy to be equal to 0, and the others like that: \begin{aligned} H(A \mid E = |1\rangle) &= 0 \, ,\\ H(A \mid E = |+\rangle) &= 1 \, ,\\ H(A \mid E = |-\rangle) &= 1 \, . \end{aligned} 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: $$$$H(A \mid E) = P(E=0) H(A \mid E = 0) + P(E=1) H(A \mid E = 1) \, .$$$$ but this leads me to the strange value i linked above, the same value i obtain for the conditional entropy of Alice and Bob, that is $$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{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} and \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} then with bayes \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} with all these numbers i have calculated the entropies reported above.

• You need to add more details for this to make sense. What are the states sent, what are the measurements of Eve. It is a function of the probability distribution, you can tell us how you are calculating it. Mar 6 at 18:28
• Thanks, i tried to explain a bit more Mar 7 at 1:24