## 3 Collective-Rotation Noise Analysis  
### Collective-Rotation Noise Model  
In this section, we analyze the effect of collective-rotation noise [14] on the three-stage protocol. In a real situation, the noise will fluctuate with time and space. In order to analyze conveniently, the environment noise is set as a constant using the maximum value of the noise as a upper bound to analyze the performance of the protocol under the collective-rotation noise model.  
Based on the collective-rotation noise model, each photon is deflected to either counterclockwise or clockwise by an angle of \(\theta\), with a probability of 1/2 respectively. For a random qubit state \(\big{|}x\big{\rangle}\), clockwise deflection is denoted as \(\big{|}x-\theta\big{\rangle}\) and counterclockwise deflection is denoted as \(\big{|}x+\theta\big{\rangle}\). The state \(\big{|}1\big{\rangle}\), \(\big{|}x\big{\rangle}\), \(\big{|}0\big{\rangle}\) and their corresponding deflection states are shown in Figure 2.  
Parameter \(\theta\) denotes the noise in the collective-rotation channel. In the following analysis, \(\theta\) is considered as a constant for all three stages. However, the impact of different values of \(\theta\) is analyzed.  
Based on the value of information bit \(X\), the initial qubit Alice prepares can be in one of the two states: \(\left|0\right\rangle\) and \(\left|1\right\rangle\). Without considering the rotation operator, let us first consider the case where \(\left|0\right\rangle\) is sent. Because of the collective-rotation channel noise, the qubit \(\left|0\right\rangle\) has a probability of 1/2 to become \(\left|0+\theta\right\rangle\) and a probability of 1/2 to become \(\left|0-\theta\right\rangle\), which can be written as  
\[\left|0\right\rangle\rightarrow\begin{cases}\left|0+\theta\right\rangle= \cos\theta\left|0\right\rangle+\sin\theta\left|1\right\rangle\\ \left|0-\theta\right\rangle=\cos\theta\left|0\right\rangle-\sin\theta\left|1 \right\rangle\end{cases} \tag{1}\]  
The probability that qubit \(\left|0\right\rangle\) is recognized as 0 is \(1/\,2\cos^{2}\theta+1/\,2\cos^{2}\theta=\cos^{2}\theta\) and the probability that qubit \(\left|0\right\rangle\) is recognized as 1 is \(1/\,2\sin^{2}\theta+1/\,2(-\sin)^{2}\theta=\sin^{2}\theta\). The error rate is given by \(\sin^{2}\theta\).  
Let us consider the case where \(\left|1\right\rangle\) is sent. Similarly, qubit \(\left|1\right\rangle\) has a probability of 1/2 to become \(\left|1+\theta\right\rangle\) and a probability of 1/2 to become \(\left|1-\theta\right\rangle\), which can be written as  
\[\left|1\right\rangle\rightarrow\begin{cases}\left|1+\theta\right\rangle=- \sin\theta\left|0\right\rangle+\cos\theta\left|1\right\rangle\\ \left|1-\theta\right\rangle=\sin\theta\left|0\right\rangle+\cos\theta\left|1 \right\rangle\end{cases} \tag{2}\]  
So the probability that qubit \(\left|1\right\rangle\) is recognized as 1 is \(1/\,2\cos^{2}\theta+1/\,2\cos^{2}\theta=\cos^{2}\theta\) and the probability that qubit \(\left|1\right\rangle\) is recognized as 0 is \(1/\,2(-\sin)^{2}\theta+1/\,2\sin^{2}\theta=\sin^{2}\theta\). The error rate is therefore given by \(\sin^{2}\theta\).  
For each qubit sent in a noisy quantum channel, the qubit error rate \(\sigma_{0}\) is  
\[\sigma_{0}=\frac{1}{2}\sin^{2}\theta+\frac{1}{2}\sin^{2}\theta=\sin^{2}\theta \tag{3}\]