## 3 Collective-Rotation Noise Analysis
### A Single-Photon Analysis  
In the three-stage protocol, a single photon is transmitted through the quantum channel three times between [PERSON] and [PERSON], each of which is subject to collective-rotation noise. In this section, we analyze the state of the photon in each of the stages, and derive the error rate due to collective-rotation noise.  
Based on the protocol described in Section 2, a rotation operator is applied to the qubit in each round to map the qubit to a non-orthogonal state. Therefore, the collective-rotation noise changes the non-orthogonal quantum state by either \(\theta\) or \(-\theta\), respectively. Since [PERSON] and [PERSON] will reverse their rotation operations eventually, the actual value of the rotation operator does not affect the results of the analysis. In order to make the derivation concise, the rotation operators are not shown in the derivation.  
After the first stage, the deflection angle can be either \(\theta\) or \(-\theta\) with probability 1/2, as shown in Table 1.  
The states of the qubit can be written as  
\begin{table}
\begin{tabular}{c c c} \hline
**Deflection Angle** & \(\theta\) & \(-\theta\) \\ \hline Probability & 1/2 & 1/2 \\ \hline \end{tabular}
\end{table}
Table 1: Probabilities of each angle after first round of transmission.  
\[\left|\,0\right\rangle\rightarrow\begin{cases}\left|\,0+\theta\right\rangle=\cos \theta\left|\,0\right\rangle+\sin\theta\left|\,1\right\rangle\\ \left|\,0-\theta\right\rangle=\cos\theta\left|\,0\right\rangle-\sin\theta\left| \,1\right\rangle\end{cases},\,\,\left|\,1\right\rangle\rightarrow\begin{cases} \left|\,1+\theta\right\rangle=-\sin\theta\left|\,0\right\rangle+\cos\theta \left|\,1\right\rangle\\ \left|\,1-\theta\right\rangle=\sin\theta\left|\,0\right\rangle+\cos\theta\left| \,1\right\rangle\end{cases}. \tag{4}\]  
Because there are two possible deflection angles at the end of the first stage of transmission, there are three possible angles after the second stage, namely, \(2\theta\), \(0\) and \(-2\theta\). The probabilities of the angles are shown in Table 2.  
The possible qubit states can be written as  
\[\left|\,0\right\rangle\rightarrow\begin{cases}\left|\,0+2\theta\right\rangle= \cos 2\theta\left|\,0\right\rangle+\sin 2\theta\left|\,1\right\rangle\\ \left|\,0+0\right\rangle=\left|\,0\right\rangle\end{cases},\,\,\left|\,1 \right\rangle\rightarrow\begin{cases}\left|\,1+2\theta\right\rangle=-\sin 2 \theta\left|\,0\right\rangle+\cos 2\theta\left|\,1\right\rangle\\ \left|\,1+0\right\rangle=\left|\,1\right\rangle\\ \left|\,1-2\theta\right\rangle=\sin 2\theta\left|\,0\right\rangle+\cos 2 \theta\left|\,1\right\rangle\end{cases}. \tag{5}\]  
After the third stage, there are four possible angles: \(3\theta\), \(\theta\), \(-\theta\) and \(-3\theta\). The probabilities of the angles are show in Table 3.  
The possible qubit states can be written as  
\[\left|\,0\right\rangle\rightarrow\begin{cases}\left|\,0+3\theta\right\rangle= \cos 3\theta\left|\,0\right\rangle+\sin 3\theta\left|\,1\right\rangle\\ \left|\,0+\theta\right\rangle=\cos\theta\left|\,0\right\rangle+\sin\theta\left| \,1\right\rangle\\ \left|\,0-\theta\right\rangle=\cos\theta\left|\,0\right\rangle-\sin\theta\left| \,1\right\rangle\end{cases},\,\,\,\left|\,1\right\rangle\rightarrow\begin{cases} \left|\,1+3\theta\right\rangle=-\sin 3\theta\left|\,0\right\rangle+\cos 3\theta \left|\,1\right\rangle\\ \left|\,1+\theta\right\rangle=-\sin\theta\left|\,0\right\rangle+\cos\theta \left|\,1\right\rangle\\ \left|\,1-\theta\right\rangle=\sin\theta\left|\,0\right\rangle+\cos\theta\left| \,1\right\rangle\\ \left|\,1-\theta\right\rangle=\sin\theta\left|\,0\right\rangle+\cos\theta\left| \,1\right\rangle\end{cases}. \tag{6}\]  
Because each case in (6) occurs with probability \(1/8\), we can derive the mean error probability as  
\[\varepsilon_{0}=\frac{1}{8}\sin^{2}3\theta+\frac{3}{8}\sin^{2}\theta+\frac{3 }{8}\sin^{2}(-\theta)+\frac{1}{8}\sin^{2}(-3\theta)=\frac{1}{4}\sin^{2}\theta +\frac{3}{4}\sin^{2}3\theta \tag{7}\]