# Estimating Interception Density in the BB84 Protocol: A Study with a Noisy Quantum Simulator
## 5 Intrusion Detection Method
### Probability of Detecting [PERSON]'s Presence  
Let us consider the situation in which it is known that [PERSON] eavesdrops all qubits. When assuming the ideal scenario of a lack of system noise, the probability of each event can be easily determined, as follows.  
* Case 1: [PERSON] selects the same basis as [PERSON]. The probability of this event is 0.5. In this case, the data are successfully exchanged between [PERSON] and [PERSON], and [PERSON] intercepts the bit without introducing any error.
* Case 2: [PERSON] selects a basis that is different from [PERSON]'s. The probability of this event is 0.5. However, two different subcases can be defined, depending on whether [PERSON] correctly receives the qubit transmitted by [PERSON] or not. In the first subcase, [PERSON]'s interception is not detected, and [PERSON] will successfully receive the bit sent by [PERSON]. The second subcase considers the possibility that [PERSON] fails to measure the qubit transmitted by [PERSON]. This leads to an error in the qubit reception and, consequently, to the decoded bit. Both subcases have the same conditional probability, i.e., 0.5.  
Considering all cases, the probability that [PERSON] correctly receives the bit transmitted by [PERSON] conditioned to the event that [PERSON] intercepts all qubits is 0.75. This value is obtained by summing the probability of Case 1 (0.5) and the probability of Case 2 (0.5), multiplied by the corresponding conditional probability of the subcase 1 (0.5). In such an event, [PERSON] and [PERSON] are not aware of [PERSON]'s interception because [PERSON] correctly decodes the qubit regenerated by [PERSON]. However, [PERSON] is present. This probability is, therefore, linked to the scenario where [PERSON] remains undetected because [PERSON] and [PERSON] have no actionable information to detect [PERSON]'s presence.  
This reasoning can be extended to determine the probability of [PERSON]'s non-detection when [PERSON] and [PERSON] use \(m\) bits for comparison. When assuming the statistical independence of [PERSON]'s interception process, the probability of the event \(EU\) ([PERSON]'s interception is undetected) is as follows:  
\begin{table}
\begin{tabular}{l l} \hline \hline
**Symbol** & **Meaning** \\ \hline \(EU\) & Event: [PERSON] is undetected. \\ \hline \(CB\) & Event: [PERSON] selects the correct basis. \\ \hline \(Sb\) & Event: [PERSON] decodes the same bit transmitted by [PERSON]. \\ \hline \(p\) & Interception density. \\ \hline \(n\) & Number of transmitted bits. \\ \hline \(k\) & Number of key bits measured on the same basis between [PERSON] and [PERSON]. \\ \hline \(m\) & Number of shared key bits publicly compared between [PERSON] and [PERSON]. \\ \hline \(d\) & Number of erroneous bits detected at [PERSON]’s side in the shared key bits. \\ \hline \(Q\dot{B}ER\) & Estimated QBER at [PERSON]’s side. \\ \hline \(QBER_{SN}\) & QBER due to system noise. \\ \hline \(QBER_{SN}\) & Estimated QBER due to system noise. \\ \hline \(\dot{p}\) & Estimated \(p\). \\ \hline \(MD\) & Event: missed [PERSON]’s detection. \\ \hline \hline \end{tabular}
\end{table}
Table 2: Summary of the meanings of the symbols used.  
\[\mathbb{P}\{EU\}=0.75^{m}.\]  
This equation demonstrates that the probability of detecting [PERSON]'s interception, equal to \(1-\mathbb{P}\{EU\}\), increases rapidly with the number of compared bits in the case where [PERSON] intercepts all qubits. For instance, in the case of 15 bits, \(\mathbb{P}\{EU\}=0.013\)[36]. For longer keys, comparing half of the bits becomes unnecessary and merely wastes useful bits that could be retained for the final key. For this reason, there is a clear need for the development of a more effective approach for practical applications.