Rényi information that Eve derives about the sift events for which Alice and Bob do not suffer errors. Let $B = \{0,1\}$ and $E = \{0,1\}$ denote the ensembles of possible bit values that Bob and Eve receive on a sift event in which Bob’s bit value agrees with Alice’s. The Rényi information (in bits) that Eve learns about each Alice/Bob error-free sift event is $$\begin{aligned}
I_R &\equiv&
-\log_2\!\left(\sum_{b= 0}^1P^2(b)\right) \nonumber \\
&+&\sum_{e = 0}^1P(e)\log_2\!\left(\sum_{b = 0}^1
P^2(b\mid e)\right),\end{aligned}$$ where $\{P(b), P(e)\}$ are the prior probabilities for Bob’s and Eve’s bit values, and $P(b\mid e)$ is the conditional probability for Bob’s bit value to be $b$ given that Eve’s is $e$. Alice’s bits are equally likely to be 0 or 1, and Eve’s conditional error probabilities satisfy [@Helstrom] $$\begin{aligned}
\lefteqn{P(e = 1\mid b = 0) = P(e = 0\mid b = 1)} \\ 
&=& \frac{1}{2}\!\left(1 - \sqrt{1 - \frac{|\langle T_+|T_-\rangle|^2}{\langle T_+|T_+\rangle \langle T_-|T_-\rangle}}\right) \\
&=& \frac{1}{2}\!\left(1- \frac{\sqrt{4P_E(1-2P_E)}}{1-P_E}\right).\end{aligned}$$ These results imply that $b$ is also equally likely to be 0 or 1, and that $P(b\mid e) = P(e\mid b)$, whence $$I_R = \log_2\!\left(1 + \frac{4P_E(1-2P_E)}{(1-P_E)^2}\right),$$ which we have plotted in Fig. 4.

Figure 4 reveals several noteworthy performance points for the 