## 3 Collective-Rotation Noise Analysis
### Multi-Photon Analysis
The probability of this event is  
\[\rho_{i}^{\text{even}}=C_{2k+2}^{i}(\varepsilon_{0})^{i}(\overline{\varepsilon _{0}})^{2k-i} \tag{12}\]  
The overall error rate \(\varepsilon^{\text{even}}\) is  
\[\varepsilon^{\text{even}}=\frac{1}{2}\rho_{k+1}+\sum_{i=k+2}^{i=2k+2}\rho_{i}^ {\text{even}}=\frac{1}{2}C_{2k+2}^{k+1}(\varepsilon_{0})^{k+1}(1-\varepsilon _{0})^{k+1}+\sum_{i=k+2}^{i=2k+2}C_{2k+2}^{i}(\varepsilon_{0})^{i}(1- \varepsilon_{0})^{2k+2-i} \tag{13}\]  
It can be re-written as  
\[\varepsilon^{\text{even}}= \frac{1}{2}C_{2k+2}^{k+1}(\frac{1}{4}\sin^{2}\theta+\frac{3}{4} \sin^{2}3\theta)^{k+1}(1-\frac{1}{4}\sin^{2}\theta-\frac{3}{4}\sin^{2}3\theta )^{k+1} \tag{14}\] \[+\sum_{i=k+2}^{i=2k+2}C_{2k+2}^{i}(\frac{1}{4}\sin^{2}\theta+ \frac{3}{4}\sin^{2}3\theta)^{i}(1-\frac{1}{4}\sin^{2}\theta-\frac{3}{4}\sin^{ 2}3\theta)^{2k+2-i}.\]Mathematically, Equation (10) is equivalent to Equation (13) when they have the same value of \(k\), which means that sending \(2k+1\) photons has the same mean error rate as sending \(2k+2\) photons. It is counter-intuitive. We briefly show the mathematical proof and explain the physical meaning. Suppose the mean error rate of sending an odd number of \(2k+1\) photons is \(\varepsilon_{2k+1}\). Suppose we add one more photon to make it even. We are interested in determining if having one more photon joining the transmission will affect the error rate. Denote the mean error rate of sending \(2k+2\) photons \(\varepsilon_{2k+2}\). To understand the effect of the added photon, we first exam the possible outcomes from the previous \(2k+1\) photons. Denote the number of photons reaching the correct photon detector \(N^{c}\), and denote the number of photons reaching the wrong photon detector \(N^{w}\). Based on our assumption, the information bit is correctly detected if \(N^{c}>N^{w}\); or if \(N^{c}=N^{w}\), there is 50% chance the information is correctly detected. We discuss the following possibilities. 1. More than \(k+1\) photons out of the \(2k+1\) photons hit the correct photon detector (\(N^{c}\geq k+2\)). In this case, it does not matter whether the added photon hits the correct or the wrong detector. In either cases, \(N^{w}\leq k\). Therefore, \(N^{c}>N^{w}\). The information bit is detected correctly. 2. More than \(k+1\) photons out of the \(2k+1\) photons hit the wrong photon detector (\(N^{w}\geq k+2\)). In this case, it again does not matter whether the added photon hits the correct or the wrong detector. In this case, the total number of photons reaching the correct detector \(N^{c}\leq k\). Therefore, \(N^{w}>N^{c}\). The information bit is detected incorrectly. 3. Out of \(2k+1\) photons, \(k+1\) photons reach the correct detector and \(k\) photons reach the wrong detector. In this case, there are two possibilities. If the added photon reaches the correct detector, this will make \(N^{c}=k+2\), and \(N^{w}=k\). Therefore, \(N^{c}>N^{w}\) and the information bit is detected correctly. On the other hand, if the added photon reaches the wrong detector, we have \(N^{c}=N^{w}=k+1\). Based on the assumption, the probability to detect a correct information bit is \(1/2\). 4. Out of \(2k+1\) photons, \(k\) photons reach the correct detector and \(k+1\) photons reach the wrong detector. In this case, there are also two possibilities. If the added photon reaches the wrong detector, this will make \(N^{w}=k+2\), and \(N^{c}=k\). Therefore, \(N^{w}>N^{c}\) and the information bit is detected incorrectly. If the added photon reaches the correct detector, this will make \(N^{c}=k+1\), and \(N^{w}=k+1\). Therefore, \(N^{c}=N^{w}\) and the probability of detecting a correct information bit is \(1/2\). The mean error rate \(\varepsilon_{2k+2}\) can be written as  
\[