## 3 Secret Key Rate in Inter-Satellite Links
### Simulation and Discussion  
In this section, we will show the performance of THz-CVQSS. We assume that the transmission distance between the dealer and the farthest player is at least 1500 km. As shown in Figure 2, we plot the relationship between the SKR and the frequency (0.1 THz-50 THz) with \(n=2,5,8,10,12,15\), respectively in a 1500 km inter-satellite link. We observe that as the number of players increases, the required frequency increases. In the case of the same number of players, SKR increases with higher frequency. The required frequency for a \(10^{-3}\) bits/pulse SKR with \(n=2\) is about 18 THz and a \(10^{-4}\) bits/pulse SKR with \(n=8\) is about 22 THz. When \(n\geq 5\), the required frequency is at least 1 THz. When \(n=15\), the required frequency needs to reach 49 THz. Thus, it is necessary to extend traditional THz frequency (0.1 THz-10 THz) to the mid infrared (MIR) and the far infrared (FIR) bands to study the THz-CVQSS system with high SKR and multiple players. However, in a practical communication system, the higher the frequency, the smaller the beam radius. That is to say, high-frequency communication system requires a high-precision acquisition, pointing, and tracking (APT) subsystem to communicate with the secure receiver. Here, we consider extending the frequency to a relatively suitable range (up to 50 THz) to estimate the performance of THz-CVQSS in inter-satellite links. Figure 3 shows the SKR of inter-satellite THz-CVQSS versus transmission distance at different THz frequencies. We obtain that as the frequency increases, the performance of the THz-CVQSS system also improves. In Figure 3a, for \(n=8,f=1\) THz, the maximal transmission distance can only attain 70 km. This is far from enough for long-distance communication between satellites. In Figure 3b, when the frequency is increased to 10 THz, for \(n=8\), the maximal transmission distance can attain 715 km. This still cannot reach the assumed communication distance (1500 km). In Figure 3c, the maximal transmission distance can reach 1440 km for \(n=8,f=20\) THz. As the number of players increases to 10 (\(n=10\)), the maximal transmission distance decreases rapidly, which can reach 940 km. In Figure 3d,e, the transmission distance can reach 1980 km and 2640 km with a \(10^{-4}\) bits/pulse SKR for \(n=8\). The results indicate that a long-distance THz-CVQSS can be achieved in the inter-satellite channel. However, when the frequency is increased to 50 THz as shown in Figure 3f, the transmission distance for \(n=8,10,12,15\) can reach almost 3290 km, 2280 km, 1860 km and 1510 km, respectively with a \(10^{-4}\) bits/pulse SKR. In particular, for \(n=8\) and a \(10^{-3}\) bits/pulse SKR, the transmiss  
Figure 2: The relationship between the SKR and the frequency (0.1 THz–50 THz) with \(n=2,5,8,10,12,15\), respectively. The simulation parameters are \(\xi_{0}=0.001,\eta=0.6,\gamma_{el}=0.1\), \(\beta=0.98\), \(V_{M}=7\) and \(d=1500\) km. 1900 km. The results shown here again strongly support the feasibility of inter-satellite THz-CVQSS. Figure 3: The SKR of inter-satellite THz-CVQSS versus transmission distance (km) for (**a**) \(f=1\) THz, (**b**) \(f=10\) THz, (**c**) \(f=20\) THz, (**d**) \(f=30\) THz, (**e**) \(f=40\) THz, and (**f**) \(f=50\) THz. The simulation parameters are \(\xi_{0}=0.001,\eta=0.6,\upsilon_{el}=0.1,\beta=0.98,V_{M}=7\). According to the analysis results in Figure 3, we obtain that 50 THz is an optimal frequency for our system. In order to achieve the maximal value of SKR in different scenarios, we analyze the optimal domain and optimal value of the modulation variance \(V_{M}\) at 50 THz as illustrated in Figure 4. It can be clearly observed that the optimal domains of \(V_{M}\) are constricted with the increase of transmission distance. In Figure 4a, as the number of players increases, the optimal domains of \(V_{M}\) are also constricted. However, there is a common domain in these different optimal domains. Here, we can select the common symmetric point \(V_{M}=7\) as the common optimal value.