Show That an Intercept and Resend Attack on all qubits reduces mutual information

Question

Exercise 5.3.3 With no eavesdropping the mutual information $H(A:B)$ between the substring of $\mathcal R$ held by Alice and $\mathcal R'$ held by Bob is $1$ bit. Show that if Eve is performing an intercept and resend attack on all qubits that she reduces this mutual information to $H(A:B)=0.189$ bits.

This is based on the BB84 protocol, where we know that the "1 bit" mutual information comes from the fact that Bob will measure the correct basis and therefore bit 50% of the time. And with the 50% incorrect basis will get the "correct" bit 25% of the time. After relaying his string back to Alice, they will confirm what errors there are and discard the wrong basis bits, even if they are the "correct" bit value. Thus there will be a discrepancy of 25% on average in their results. To detect Eve’s presence Alice and Bob choose some portion n of their shared bit. These revealed bits are useful for finding out whether extra errors are occurring in the protocol

From the equation: $$P_d = 1 -(\frac34)^n $$

If after sacrificing a few hundred of their bits Alice and Bob find no unexplained errors then they can be confident that Eve is not intercepting and resending the qubits. The remainder of the key is therefore secure.

Now, I know the equation for Mutual Information is given as: $$H(X:Y)=H(X) +H(Y) -H(X,Y)$$ I'm generally just unsure how to impliment this. The exercise details that the attack is performed on ALL qubits, I assume this is a clue?

Answer 1

You've started the analysis along the right lines. In the case of no eavesdropper, Alice and Bob get perfectly correlated outcomes (note that we're post-selecting on the cases where Alice and Bob choose the same measurement bases). You could think of this as a probability table $$ \begin{array}{ccc} \text{Alice bit} & \text{Bob bit} & \text{probability}\\ 0 & 0 & 1/2 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1/2 \end{array} $$ from which you can calculate $H(X)$, $H(Y)$ and $H(X,Y)$.

So, the trick is to do something similar for Eve's attack. Again, we post-select on Alice and Bob choosing the same measurement bases. Half the time, Eve chooses the same basis, and the probability table is the same as above. The other half, Eve chooses the other basis. Bob's outcomes are completely uncorrelated, i.e. all 4 probabilities are $1/4$. Combining these two cases, the overall probability table is $$ \begin{array}{ccc} \text{Alice bit} & \text{Bob bit} & \text{probability}\\ 0 & 0 & 3/8 \\ 0 & 1 & 1/8 \\ 1 & 0 & 1/8 \\ 1 & 1 & 3/8 \end{array} $$ So, again, calculate the entropies. You'll find $$ H(X:Y)=\frac34\log_23-1. $$