## 5 Improved Setup with a Fully Active Base Selection
### Attack Analysis  
From the discussion above, it would naively seem that [PERSON] can successfully eavesdrop on the receiver in Figure 4 using blinding techniques. However, this particular setup enables yet another option available to [PERSON] and [PERSON] for defeating [PERSON] which relies on monitoring in-gate and out-of-gate coincidences. Let us first discuss a regular quantum communication using single qubits in which [PERSON] knows the arrival time of [PERSON]'s qubit and opens a short detection time window \(t_{G}\), called the \"gate\", around it to detect the qubit. In order for the BB84 protocol to have a non-zero secret key rate, the bit error rate of the first phase must be \(<\)11.5% [29]. The dark count occurrence probability during the gate for a single detector is  
\[p_{DCR}=(1-e^{-t_{G}f_{DCR}})\cong t_{G}f_{DCR} \tag{2}\]  
where \(f_{DCR}\) is the dark count rate (DCR) of the detector and \(t_{G}\) is the duration of the gate. The right-hand approximation is valid for \(t_{G}f_{DCR}\ll 1\), which is often the case. [PERSON] can confidently determine the dark count rates for all four detectors by disconnecting the receiver from the quantum channel and performing measurements in the dark. To simplify the calculation, I will henceforth assume equal DCR values for all detectors. A DCR-caused \"accidental\" coincident detection between a pair of detectors belonging to the same base is a Poissonian random event. Its probability during the gate-open period \(t_{G}\), or its \"in gate\" probability, is given by  
\[p_{AIn}=p_{DCR}^{2}\cong t_{G}^{2}f_{DCR}^{2} \tag{3}\]  
When receiving qubits, the coincidence rate between two detectors forming a base is described by the Hong-Ou-Mandel (HOM) effect, which characterizes the probability of simultaneous photon detection at the two outputs of a beam splitter [30]. By considering a coherent input state qubit emitted from Alice with an average photon number \(\langle n\rangle\) within a gate-open period \(t_{G}\), the per-gate coincidence probability measured by [PERSON] in any of its two bases is given by [31]:  
\[p_{QIn}=\frac{1}{2}\bigg{(}1-e^{-\frac{\langle n\rangle e_{B}e_{QCh}}{2}} \bigg{)}^{2}\cong\frac{\langle n\rangle^{2}e_{B}^{2}e_{QCh}^{2}}{8} \tag{4}\]  
where \(e_{B}\) is Bob's system quantum efficiency, which includes the effects of the finite detector's quantum efficiency, optics imperfections, and other losses in Bob, while the quantum channel efficiency \(e_{QCh}\) is a fraction of the qubits that survive through the quantum channel connecting Alice and Bob. The right-hand approximation is valid for \(\big{(}\langle n\rangle e_{B}e_{QCh}\big{)}/2\ll 1\)which holds true in a practical QKD in which typically \(\langle n\rangle\sim 0.1\) and \(\epsilon_{S}\sim 0.1-0.5\), while losses can be anywhere between a few dB and several tens of dB. The overall factor \(1/2\) equals the probability that the qubit is in a \"wrong\" base. In the case in which the base is \"right\" for the given qubit, there will be no coincidence in principle because the photon will end up in only one detector. Here, I neglect a possible imperfection in Bob's polarizers which in may allow for a small probability of coincidence (typically less than 0.01). The coincidence probability is composed of both the accidental and qubit-related coincidences. As both are Poisson random processes, the total in-gate coincidence probability \(p_{Cin}\) for the regular BB84 protocol in setup in Figure 4 is given by  
\[p_{CIn}=p_{AIn}+p_{QIn}-p_{AIn}p_{QIn}\cong t_{G}^{2}f_{DCR}^{2}+\frac{ \langle n\rangle^{2}\epsilon_{B}^{2}\epsilon_{QCH}^{2}}{8} \tag{5}\]  
As an example, assuming the following realistic values of a gate time \(t_{G}=5\) ns, a dark count rate \(f_{DCR}=1000\) cps, an average photon number in a qubit \(\langle n\rangle=0.1\), [PERSON]'s system quantum efficiency \(\epsilon_{B}=0.25\), and a quantum channel transfer efficiency \(e_{QCH}=0.16\) (\(-8\) dB), one arrives at \(p_{CIn}=2\times 10^{-6}\).