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복제 대칭성 깨짐 해의 경우, 어떤 복제본들을 선택하냐에 따라서 헤세 행렬의 성분이 달라질 수 있다. 따라서 각 성분에 복제본의 인덱스를 붙여주어야 한다. 가능한 경우의 수를 모두 고려해보면 각 성분들은 다음과 같이 나누어질 수 있다. $$A\equiv A_{\alpha\alpha}=1-\beta J_0(1-m^2)$$ $$B_{\alpha\beta} \rightarrow \begin{cases} B_1=-\beta J_0(q_1-m^2)&\text{if }[\alpha\beta]\\ B_0=-\beta J_0(q_0-m^2)&\text{if }[\alpha][\beta] \end{cases} $$ $$C_{\alpha(\alpha\beta)} \rightarrow \begin{cases} C_1=-\beta J\sqrt{\beta J_0}(mq_1-m)&\text{if }[\alpha\beta]\\ C_0=-\beta J\sqrt{\beta J_0}(mq_0-m)&\text{if }[\alpha][\beta] \end{cases}$$ $$D_{\gamma(\alpha\beta)}\rightarrow\begin{cases} D_{10} = \beta J\sqrt{\beta J_0}(mq_1-t_0)&\text{if }[\gamma\alpha\beta]\\ D_{11} = \beta J\sqrt{\beta J_0}(mq_1-t_1)&\text{if }[\gamma][\alpha\beta]\\ D_{01} = \beta J\sqrt{\beta J_0}(mq_0-t_1)&\text{if }[\gamma\alpha][\beta]\\ D_{02} = \beta J\sqrt{\beta J_0}(mq_0-t_2)&\text{if }[\gamma][\alpha][\beta] \end{cases}$$ $$P_{(\alpha\beta)(\alpha\beta)} \rightarrow \begin{cases} P_1=1-\beta^2J^2(1-q_1^2)&\quad\text{if }[\alpha\beta]\\ P_0=1-\beta^2J^2(1-q_0^2)&\quad\text{if }[\alpha][\beta] \end{cases}$$ $$Q_{(\alpha\beta)(\alpha\gamma)} \rightarrow \begin{cases} Q_{11}=-\beta^2J^2(q_1-q_1^2)&\quad\text{if }[\alpha\beta\gamma]\\ Q_{01}=-\beta^2J^2(q_0-q_1q_0)&\quad\text{if }[\alpha\beta][\gamma]\\ Q_{10}=-\beta^2J^2(q_1-q_0q_1)&\quad\text{if }[\alpha][\beta\gamma]\\ Q_{00}=-\beta^2J^2(q_0-q_0^2)&\quad\text{if }[\alpha][\beta][\gamma]\\ \end{cases}$$ $$R_{(\alpha\beta)(\alpha\gamma)} \rightarrow \begin{cases} R_4=-\beta^2J^2(r_4-q_1^2)&\quad\text{if }[\alpha\beta\gamma\delta]\\ R_{3:1}=-\beta^2J^2(r_{3:1}-q_1q_0)&\quad\text{if }[\alpha\beta\gamma][\delta]\\ R_{2:2,1}=-\beta^2J^2(r_{2:2}-q_1^2)&\quad\text{if }[\alpha\beta][\gamma\delta]\\ R_{2:2,0}=-\beta^2J^2(r_{2:2}-q_0^2)&\quad\text{if }[\alpha\gamma][\beta\delta]\\ R_{2:1:1,1}=-\beta^2J^2(r_{2:1:1}-q_1q_0)&\quad\text{if }[\alpha\beta][\gamma][\delta]\\ R_{2:1:1,0}=-\beta^2J^2(r_{2:1:1}-q_0^2)&\quad\text{if }[\alpha][\beta\gamma][\delta]\\ R_{1:1:1:1,0}=-\beta^2J^2(r_{1:1:1:1}-q_0^2)&\quad\text{if }[\alpha][\beta][\gamma][\delta]\\ \end{cases}$$ 여기서 $[\alpha\beta]$는 두 복제본 $\alpha$, $\beta$가 같은 블록에 속해있다는 뜻이고, $[\alpha][\beta]$는 서로 다른 블록에 속해있다는 의미이다. 그리고 각 $r$, $t$들의 의미는 다음과 같다. \begin{align*} t_0 &= \int Du\frac{\int Dv\cosh^{m_1}\Xi\tanh^3\Xi}{\int Dv\cosh^{m_1}}\\ t_1 &= \int Du\left(\frac{\int Dv\cosh^{m_1}\Xi\tanh^2\Xi}{\int Dv\cosh^{m_1}}\right)\left(\frac{\int Dv\cosh^{m_1}\Xi\tanh\Xi}{\int Dv\cosh^{m_1}}\right)\\ t_2 &= \int Du\left(\frac{\int Dv\cosh^{m_1}\Xi\tanh^3\Xi}{\int Dv\cosh^{m_1}}\right)^3 \end{align*} \begin{align*} r_4 &= \int Du\frac{\int Dv\cosh^{m_1}\Xi\tanh^4\Xi}{\int Dv\cosh^{m_1}}\\ r_{3:1} &= \int Du\left(\frac{\int Dv\cosh^{m_1}\Xi\tanh^3\Xi}{\int Dv\cosh^{m_1}}\right)\left(\frac{\int Dv\cosh^{m_1}\Xi\tanh\Xi}{\int Dv\cosh^{m_1}}\right)\\ r_{2:2} &= \int Du\left(\frac{\int Dv\cosh^{m_1}\Xi\tanh^2\Xi}{\int Dv\cosh^{m_1}}\right)^2\\ r_{2:1:1} &= \int Du\left(\frac{\int Dv\cosh^{m_1}\Xi\tanh^2\Xi}{\int Dv\cosh^{m_1}}\right)\left(\frac{\int Dv\cosh^{m_1}\Xi\tanh\Xi}{\int Dv\cosh^{m_1}}\right)^2\\ r_{1:1:1:1} &= \int Du\left(\frac{\int Dv\cosh^{m_1}\Xi\tanh\Xi}{\int Dv\cosh^{m_1}}\right)^4 \end{align*} 예를 들어 $n=4$, $m_1=4$인 경우 헤세 행렬은 $$\mathbf G= \left( \begin{array}{cccccccccc} A & B_1 & B_1 & B_1 & C_1 & C_1 & C_1 & D_{10} & D_{10} & D_{10} \\ B_1 & A & B_1 & B_1 & C_1 & D_{10} & D_{10} & C_1 & C_1 & D_{10} \\ B_1 & B_1 & A & B_1 & D_{10} & C_1 & D_{10} & C_1 & D_{10} & C_1 \\ B_1 & B_1 & B_1 & A & D_{10} & D_{10} & C_1 & D_{10} & C_1 & C_1 \\ C_1 & C_1 & D_{10} & D_{10} & P_1 & Q_{11} & Q_{11} & Q_{11} & \ Q_{11} & R_4 \\ C_1 & D_{10} & C_1 & D_{10} & Q_{11} & P_1 & Q_{11} & Q_{11} & R_4 & \ Q_{11} \\ C_1 & D_{10} & D_{10} & C_1 & Q_{11} & Q_{11} & P_1 & R_4 & Q_{11} & \ Q_{11} \\ D_{10} & C_1 & C_1 & D_{10} & Q_{11} & Q_{11} & R_4 & P_1 & Q_{11} & \ Q_{11} \\ D_{10} & C_1 & D_{10} & C_1 & Q_{11} & R_4 & Q_{11} & Q_{11} & P_1 & \ Q_{11} \\ D_{10} & D_{10} & C_1 & C_1 & R_4 & Q_{11} & Q_{11} & Q_{11} & \ Q_{11} & P_1 \\ \end{array} \right) $$ 로 주어지고, 고윳값들은 다음과 같다. $$ \left( \begin{array}{c} P_1-2 Q_{11}+R_4 \\ P_1-2 Q_{11}+R_4 \\ \frac{1}{2} \left(-\sqrt{(A+3 B_1+P_1+4 Q_{11}+R_4)^2-4 (A+3 B_1) \ (P_1+4 Q_{11}+R_4)+24 C_1^2+48 C_1 D_{10}+24 D_{10}^2}+A+3 B_1+P_1+4 \ Q_{11}+R_4\right) \\ \frac{1}{2} \left(\sqrt{(A+3 B_1+P_1+4 Q_{11}+R_4)^2-4 (A+3 B_1) \ (P_1+4 Q_{11}+R_4)+24 C_1^2+48 C_1 D_{10}+24 D_{10}^2}+A+3 B_1+P_1+4 \ Q_{11}+R_4\right) \\ \frac{1}{2} \left(-\sqrt{(A-B_1+P_1-R_4)^2-4 (A-B_1) (P_1-R_4)+8 \ C_1^2-16 C_1 D_{10}+8 D_{10}^2}+A-B_1+P_1-R_4\right) \\ \frac{1}{2} \left(-\sqrt{(A-B_1+P_1-R_4)^2-4 (A-B_1) (P_1-R_4)+8 \ C_1^2-16 C_1 D_{10}+8 D_{10}^2}+A-B_1+P_1-R_4\right) \\ \frac{1}{2} \left(-\sqrt{(A-B_1+P_1-R_4)^2-4 (A-B_1) (P_1-R_4)+8 \ C_1^2-16 C_1 D_{10}+8 D_{10}^2}+A-B_1+P_1-R_4\right) \\ \frac{1}{2} \left(\sqrt{(A-B_1+P_1-R_4)^2-4 (A-B_1) (P_1-R_4)+8 \ C_1^2-16 C_1 D_{10}+8 D_{10}^2}+A-B_1+P_1-R_4\right) \\ \frac{1}{2} \left(\sqrt{(A-B_1+P_1-R_4)^2-4 (A-B_1) (P_1-R_4)+8 \ C_1^2-16 C_1 D_{10}+8 D_{10}^2}+A-B_1+P_1-R_4\right) \\ \frac{1}{2} \left(\sqrt{(A-B_1+P_1-R_4)^2-4 (A-B_1) (P_1-R_4)+8 \ C_1^2-16 C_1 D_{10}+8 D_{10}^2}+A-B_1+P_1-R_4\right) \\ \end{array} \right) $$ 복제 대칭해의 경우에 얻었던 $n(n-1)2$의 중복도를 가지는 replicon mode를 똑같이 얻어내었다.