# One-Photon-Interference Quantum Secure Direct Communication
## Appendix B Details of Security Analysis
\begin{split}|\varphi_{1}\rangle\equiv&|-_{A}\rangle( \sqrt{\lambda_{1}}|E_{1}\rangle+\sqrt{\lambda_{3}}|E_{3}\rangle)+|+_{A} \rangle(\sqrt{\lambda_{2}}|E_{2}\rangle+\sqrt{\lambda_{4}}|E_{4}\rangle),\\ |\varphi_{2}\rangle\equiv&|-_{A}\rangle(-\sqrt{ \lambda_{2}}|E_{2}\rangle+\sqrt{\lambda_{4}}|E_{4}\rangle)+|+_{A}\rangle(- \sqrt{\lambda_{1}}|E_{1}\rangle+\sqrt{\lambda_{3}}|E_{3}\rangle).\end{split}  \tag{15}\]  
Then, [PERSON] measures her pulse in X basis, and obtains the bit value \(V_{A}\). The states of [PERSON] become  
\[ \begin{split}\rho^{V_{A}=+}_{E}&=\mathit{Tr}_{A}(| +_{A}\rangle\langle+_{A}|\rho_{AE}|+_{A}\rangle\langle+_{A}|)\\ &=|\psi_{1}\rangle\langle\psi_{1}|+|\psi_{2}\rangle\langle\psi_{ 2}|,\\ \rho^{V_{A}=-}_{E}&=\mathit{Tr}_{A}(|-_{A}\rangle \langle-_{A}|\rho_{AE}|-_{A}\rangle\langle-_{A}|)\\ &=|\psi_{3}\rangle\langle\psi_{3}|+|\psi_{4}\rangle\langle\psi_{ 4}|,\end{split}  \tag{16}\]  
where  
\[ \begin{split}|\psi_{1}\rangle\equiv&\sqrt{\lambda_{2}}| E_{2}\rangle+\sqrt{\lambda_{4}}|E_{4}\rangle,\\ |\psi_{2}\rangle\equiv&-\sqrt{\lambda_{1}}|E_{1} \rangle+\sqrt{\lambda_{3}}|E_{3}\rangle,\\ |\psi_{3}\rangle\equiv&\sqrt{\lambda_{1}}|E_{1} \rangle+\sqrt{\lambda_{3}}|E_{3}\rangle,\\ |\psi_{4}\rangle\equiv&-\sqrt{\lambda_{2}}|E_{2} \rangle+\sqrt{\lambda_{4}}|E_{4}\rangle.\end{split}  \tag{17}\]We obtain the mutual information \(I(A:E)\) of the joint system \(AE\)  
\[I(A:E)=\ S\!\left(\sum_{k}p_{k}\rho_{E}^{V_{A}=k}\right)-\sum_{k}p_{k}S\!\left( \rho_{E}^{V_{A}=k}\right), \tag{100}\]  
where \(S(\cdot)\) is the von Neumann entropy, and \(p_{k}\) is the distribution of \(\rho_{E}^{V_{A}=k}\). Supporting that [PERSON] sends the same number of bits 0 and 1, so \(p_{k}=\frac{1}{2}\). The upper bound on \(I(A:E)\) is  
\[ \begin{split}& I(A:E)\\ =& H(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4})-h( \lambda_{1}+\lambda_{3})\\ =&(\lambda_{1}+\lambda_{3})h\!\left(\frac{\lambda_{1 }}{\lambda_{1}+\lambda_{3}}\right)+(\lambda_{2}+\lambda_{4})h\!\left(\frac{ \lambda_{2}}{\lambda_{2}+\lambda_{4}}\right)\\ \leq&\ h(\lambda_{3}+\lambda_{4})\\ =&\ h(E_{u}^{Z}),\end{split}  \tag{101}\]  
where \(H(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4})\equiv-\lambda_{1}\log_{2} \lambda_{1}-\lambda_{2}\log_{2}\lambda_{2}-\lambda_{3}\log_{2}\lambda_{3}- \lambda_{4}\log_{2}\lambda_{4}\). Therefore, if we estimate the phase-error rate \(E_{u}^{Z}\), we can calculate the amount of information leakage. Next, we analyze the process of our protocol to estimate the phase-error rate \(E_{u}^{Z}\). In the **Step 2**, the beam splitter transforms the joint state \(|\phi_{AaBb}\rangle=|\Phi_{Aa}^{+}(\alpha)\rangle\otimes|\Phi_{Bb}^{+}(\alpha)\rangle\) into the following state  
\[ \begin{split}|\phi_{ABCD}\rangle=&\ \frac{1}{2}(|++ \rangle_{AB}|\sqrt{2}\alpha,0\rangle_{CD}+|+-\rangle_{AB}|0,\sqrt{2}\alpha \rangle_{CD}\\ &+|-+\rangle_{AB}|0,-\sqrt{2}\alpha\rangle_{CD}+|--\rangle_{AB} |-\sqrt{2}\alpha,0\rangle_{CD}).\end{split}  \tag{102}\]  
Here, we assume \(|\alpha\rangle_{a}=|\alpha\rangle_{b}=|\alpha\rangle\). And we have  
\[ \begin{split}|\sqrt{2}\alpha\rangle=&\ e^{-|\alpha| ^{2}}\sum_{n=0}^{\infty}\frac{(\sqrt{2}\alpha)^{n}}{\sqrt{n!}}|n\rangle\\ =&\ \sum_{n=0}^{\infty}C_{2n}|2n\rangle+\sum_{n=0}^{ \infty}C_{2n+1}|2n+1\rangle,\\ |-\sqrt{2}\alpha\rangle=&\ e^{-|\alpha|^{2}}\sum_{n=0 }^{\infty}\frac{(-\sqrt{2}\alpha)^{n}}{\sqrt{n!}}|n\rangle\\ =&\ \sum_{n=0}^{\infty}C_{2n}|2n\rangle-\sum_{n=0}^{ \infty}C_{2n+1}|2n+1\rangle,\end{split}  \tag{103}\]  
where  
\[ \begin{split} C_{2n}\equiv&\ e^{-|\alpha|^{2}}\frac{( \sqrt{2}\alpha)^{2n}}{\sqrt{(2n)!}},\\ C_{2n+1}\equiv&\ e^{-|\alpha|^{2}}\frac{(\sqrt{2} \alpha)^{2n+1}}{\sqrt{(2n+1)!}}.\end{split}  \tag{104}\]  
After arranging the Equation (102), we obtain  
\[