## 4 Security Analysis of the Intercept-and-Resend Attack
The matrix describing conditional probabilities to prepare particular states by [PERSON] and the measurements obtained by [PERSON] is given by  
\[P_{B|E}=\left( \begin{array}{cccc}\frac{1}{2}&0&\frac{1}{4}&\frac{1}{4}\\ 0&\frac{1}{2}&\frac{1}{4}&\frac{1}{4}\\ \frac{1}{4}&\frac{1}{2}&\frac{1}{2}&0\\ \frac{1}{4}&\frac{1}{4}&0&\frac{1}{2}\\ \end{array} \right). \tag{21}\]  
Suppose [PERSON] performs the attack presented in this strategy; this allows her to minimize her quantum \"footprint\" by replacing the collective-rotation noise channel with a lossless classical channel. However, this strategy fails to enhance her odds of going undetected. This is depicted in Figure 2. To find \(Q_{2}\), first, we evaluate the elements \(a-d\) in Table 2 by employing Equation (14). The matrices \(P_{E|A}\) and \(P_{B|E}\) come from Equations (7) and (21), respectively. As a result, the terms \(a-d\) are evaluated to be:  
\[a =\frac{1+2\cos^{2}\theta}{32}, \tag{22}\] \[b =\frac{1+2\sin^{2}\theta}{32},\] (23) \[c =\frac{2-\sin 2\theta}{32},\] (24) \[d =\frac{2+\sin 2\theta}{32}. \tag{25}\]  
We note that the terms \(c\) and \(d\) disappear during sifting. Thus, \(Q_{2}\) becomes:  
\[Q_{2}=\frac{1+2\sin^{2}\theta}{4}. \tag{26}\] \begin{table}
\begin{tabular}{c c c c c} \hline \hline  & \multicolumn{5}{c}{**[PERSON] Obtains**} \\ \cline{2-5}
**[PERSON] Sends** & \(|\mathbf{0}\rangle\) & \(|\mathbf{1}\rangle\) & \(|\mathbf{+}\rangle\) & \(|\mathbf{-}\rangle\) \\ \hline \(|0\rangle\) & \((\cos^{2}\phi)/2\) & \((\sin^{2}\phi)/2\) & \((1-\sin 2\phi)/4\) & \((1+\sin 2\phi)/2\) \\ \(|1\rangle\) & \((\sin^{2}\phi)/2\) & \((\cos^{2}\phi)/2\) & \((1+\sin 2\phi)/4\) & \((1-\sin 2\phi)/4\) \\ \(|+\rangle\) & \((1+\sin 2\phi)/4\) & \((1-\sin 2\phi)/4\) & \((\cos^{2}\phi)/2\) & \((\sin^{2}\phi)/2\) \\ \(|-\rangle\) & \((1-\sin 2\phi)/4\) & \((1+\sin 2\phi)/4\) & \((\sin^{2}\phi)/2\) & \((\cos^{2}\phi)/2\) \\ \hline \multicolumn{5}{c}{**[PERSON] Receives**} \\ \cline{2-5}
**[PERSON] Sends** & \(|\mathbf{0}\rangle\) & \(|\mathbf{1}\rangle\) & \(|\mathbf{+}\rangle\) & \(|\mathbf{-}\rangle\) \\ \hline \(|0\rangle\) & a & b & c & d \\ \(|1\rangle\) & b & a & d & c \\ \(|+\rangle\) & d & c & a & b \\ \(|-\rangle\) & c & d & b & a \\ \hline \hline \end{tabular}
\end{table} 
Table 2: A summary of joint probabilities for states prepared by [PERSON] and [PERSON]’s measurement outcomes corresponding to each of the four states when [PERSON]’s intercept-and-resend attack. Figure 2 illustrates the relationship between the QBER and rotational noise in the channel. It can be observed from Figure 2 that, when rotational angle \(\theta\leq\frac{\pi}{4}\), \(Q_{1}>Q_{2}\), [PERSON]'s eavesdropping results in a more significant bit error rate and she will be detected. We observe that, in a noise-free environment (i.e., \(\theta=0\)), [PERSON] will induce a 25% QBER. This corresponds to the well-known error bound for the BB84 protocol under the simple intercept-resend attack. The results also indicate that QBER \(Q_{1}\) reaches a maximum for every rotation angle \(\theta=n\pi+\frac{\pi}{4}\) (where \(n\) is an integer), and this corresponds to noise, which will flip the prepared quantum states into a state in the opposite basis (phase flip error). Furthermore, for curve \(Q_{1}\), we note that, for every \(\theta=n\frac{\pi}{2}\), the QBER is at its lowest value. Finally, we observe that when [PERSON] intercepts the qubits and prepares them at the site closer to [PERSON]'s measurement station, represented by the curve \(Q_{2}\), the noise in the channel does not heavily contribute to the QBER. Thus, according to the curve \(Q_{2}\), for every rotation angle \(\theta=n\pi+\frac{\pi}{2}\), the QBER is at its maximum and reaches its lowest at every \(\theta=n\pi\). Moreover, it is observed that for \(Q_{1}\), the maximum QBER is reached at \(\frac{\pi}{4}\), which occurs before \(Q_{2}\) reaches its maximum QBER at \(\theta=\frac{\pi}{2}\).