# Quantum Key Distribution in the Presence of the Intercept-Resend with Faked States Attack
## 4 Detecting the \(\mathit{IRFS}\) Attack
### Faking Double Detection Events  
Another scenario for the eavesdropper is to fake double detection events. After all, we might ask why [PERSON] cannot fake double detection events while she remains hidden in the channel. First of all, let us recall that [PERSON] knows which biqubits contain parallel or orthogonal states. Second, consider the cases depicted in Table 2. Suppose [PERSON] has sent the \(\left(\left|0_{Z}\right\rangle,\left|0_{Z}\right\rangle\right)\) biqubit to [PERSON]. The first pulse reaches [PERSON]'s station, who measures it with the \(X\) (or \(Z\)) basis, but the second pulse arrives as a vacuum state either by the effect of the quantum channel, the detection system or the photon source. As a result, [PERSON] gets a single detection event. Now, [PERSON] decides to fake the second state, but she realizes that there are six possibilities to fake the \(\left(\left|0_{Z}\right\rangle,\left|0_{Z}\right\rangle\right)\) biqubit; such cases are listed in Table 2. Additionally, one of those cases is erroneous because no orthogonal measurement can be derived from parallel states. In this example, \(\left(\left|1_{Z}\right\rangle,\left|0_{Z}\right\rangle\right)\) cannot be obtained from \(\left(\left|0_{Z}\right\rangle,\left|0_{Z}\right\rangle\right)\). Similarly, \(\left(\left|0_{Z}\right\rangle,\left|0_{Z}\right\rangle\right)\) cannot be derived from \(\left(\left|1_{Z}\right\rangle,\left|0_{Z}\right\rangle\right)\). As a consequence, if [PERSON] tries to fake a double detection event, she will produce a bit error of \(\frac{1}{6}\). In this case, a bit error is produced when [PERSON] announces a double matching event, but [PERSON] expects a double non-matching event or vice versa. According to [22], [PERSON]'s visibility of [PERSON]'s quantum state is computed as \(V_{AB}=\frac{P(signal)}{P(total)}\) where \(P(signal)=T_{AB}\times\eta\times V_{opt}\) and \(P(total)=T_{AB}\times\eta+(1-T_{AB}\times\eta)\times 2\times Y_{0}\). Here, \(V_{opt}\) is the optical visibility with a perfect source and detectors; \(\eta\) is the probability of detecting the photon when it arrives; \(T_{AB}\) is the transmittance between [PERSON] and [PERSON]; and \(Y_{0}\) is the background noise. For realistic experimental parameters: \(\alpha=0.25\) dB-km\({}^{-1}\), \(\eta=0.3\), \(Y_{0}=10^{-4}\) and \(V_{opt}=0.99\). Figure 4 shows the visibility as a function of the distance. On the other hand, the \(QBER\) in \(BB84\) can be computed as \(QBER=\frac{pe}{pe+pc}\), where \(pc\) (\(pe\)) is the probability to get, correctly or erroneously, the quantum bit sent by [PERSON], respectively. If we write such probabilities as a function of the optical visibility \(V\), we have \(pc=(1+V)/2\) and \(pe=(1-V)/2\). \begin{table}
\begin{tabular}{c c c c c c} \hline \hline
**[PERSON]’s Biqubit** & **[PERSON]’s Basis** & **[PERSON]’s Detection** & **Forwarded Dates** & **[PERSON]’s Result** \\ \hline \multirow{4}{*}{\((\ket{0_{Z}},\ket{0_{Z}})\)} & \multirow{2}{*}{\(Z\)} & \((-,\ket{0_{Z}})\) & \((\ket{0_{Z}},\ket{0_{Z}})\) & hidden \\  & & & \((-,\ket{0_{X}})\) & \((\ket{1_{X}},\ket{0_{X}})\) & hidden \\ \cline{3-6}  & & & \((-,\ket{1_{X}})\) & \((\ket{0_{X}},\ket{1_{X}})\) & hidden \\ \cline{3-6}  & & & \((-,\ket{1_{X}})\) & \((\ket{1_{X}},\ket{1_{X}})\) & hidden \\ \hline \multirow{4}{*}{\((\ket{1_{Z}},\ket{0_{Z}})\)} & \multirow{2}{*}{\(Z\)} & \((-,\ket{0_{Z}})\) & \((\ket{0_{Z}},\ket{0_{Z}})\) & detected \\  & & & \((1_{Z})\) & \((\ket{1_{Z}},\ket{0_{Z}})\) & hidden \\ \cline{3-6}  & & & \((-,\ket{0_{X}})\) & \((\ket{0_{X}},\ket{0_{X}})\) & hidden \\ \cline{3-6}  & & & \((-,\ket{1_{X}})\) & \((\ket{0_{X}},\ket{1_{X}})\) & hidden \\ \cline{3-6}  & & & \((-,\ket{1_{X}})\) & \((\ket{1_{X}},\ket{1_{X}})\) & hidden \\ \hline \hline \end{tabular}
\end{table} 
Table 2: After [PERSON] detects the first state of a biqubit, she tries to fake the second state. However, there exist six possible states, but one of them is erroneous, so she introduces an error probability of \(\frac{1}{6}\).