## 3 Collective-Rotation Noise Analysis
### Multi-Photon Analysis  
Since the three-stage protocol is multi-photon tolerant [11], multiple photons can be transmitted simultaneously to indicate one bit of information to improve the success rate of the transmission. We assume each photon is independent from each other and all photons transmitted at the same time are affected by the same collective-rotation noise. As discussed in Section 3.2, the probability that a single photon is recognized as incorrect information is \(\varepsilon_{0}\), which is the mean error rate for a single photon transmitted in the noisy channel, as shown in Equation (7). The probability that a single photon is recognized correctly is \begin{table}
\begin{tabular}{c c c c} \hline
**Deflection Angle** & \(2\theta\) & \(0\) & \(-2\theta\) \\ \hline Probability & \(1/4\) & \(1/2\) & \(1/4\) \\ \hline \end{tabular}
\end{table} 
Table 2: Probabilities of each angle after second round of transmission. \begin{table}
\begin{tabular}{c c c c} \hline
**Deflection Angle** & \(3\theta\) & \(\theta\) & \(-\theta\) & \(-3\theta\) \\ \hline Probability & \(1/8\) & \(3/8\) & \(3/8\) & \(1/8\) \\ \hline \end{tabular}
\end{table} 
Table 3: Probabilities of each angle after third round of transmission. \[\overline{\varepsilon_{0}}=1-\varepsilon_{0} \tag{8}\]  
We also assume that the channel is lossless. Therefore, all photons sent by [PERSON] reach the photon detector on [PERSON]'s side. [PERSON] will register a correct bit if a majority number of photons reach the correct photon detector. In the case of a tie, we assume that there is 50% of chance to register a correct bit. Let the total number of photons used in transmitting one bit be \(N\), where \(N=2k+1\),\(k\geq 0\) or \(N=2k+2,k\geq 0\). Note that we purposely use the notation \(2k+2\) to indicate that an even number of photons are used in the communication, instead of the standard notation of \(2k\). This allows us to establish a relationship between the even and odd numbers of photons under the same value of \(k\). First, we derive the error rate under the odd number and the even number of photons, respectively. _Case 1_: \(N=2k+1,k\geq 0\). In this situation, odd number photons are sent simultaneously. If more than \(k\) photons reach the correct photon detector, [PERSON] will receive the correct bit (either 0 or 1) sent by [PERSON]. Let \(i\) be the number of photons that reach the correct photon detector. The probability of this event is  
\[\rho_{i}^{\text{odd}}=C_{2k+1}^{i}(\varepsilon_{0})^{i}(\overline{\varepsilon _{0}})^{2k+1-i} \tag{9}\]  
So the overall error rate \(\varepsilon^{\text{odd}}\) is  
\[\varepsilon^{\text{odd}}=\sum_{i=k+1}^{i=2k+1}\rho_{i}^{\text{odd}}=\sum_{i=k+ 1}^{i=2k+1}C_{2k+1}^{i}(\varepsilon_{0})^{i}(1-\varepsilon_{0})^{2k+1-i} \tag{10}\]  
Using the results from Equation (7), the overall error rate \(\varepsilon_{\text{odd}}\) can be re-written as  
\[\varepsilon^{\text{odd}}=\sum_{i=k+1}^{i=2k+1}C_{2k+1}^{i}(\frac{1}{4}\sin^{2} \theta+\frac{3}{4}\sin^{2}3\theta)^{i}(1-\frac{1}{4}\sin^{2}\theta-\frac{3}{4 }\sin^{2}3\theta)^{2k+1-i} \tag{11}\]  
_Case 2_: \(N=2k+2,k\geq 0\). In this situation, even number photons are sent simultaneously. If more than \(k+1\) photons reach the correct photon detector, the receiver will receive the correct bit. In the case of a tie, where \(k+1\) photons reach the correct detector and \(k+1\) photons reach the wrong detector, we assume that the system will randomly choose between 1 or 0 with probability 1/2. Suppose the number of photons that reach the correct photon detector is \(i\).