Show pageOld revisionsBacklinksFold/unfold allBack to top This page is read only. You can view the source, but not change it. Ask your administrator if you think this is wrong. 복제 대칭성 깨짐 해의 경우, 어떤 복제본들을 선택하냐에 따라서 헤세 행렬의 성분이 달라질 수 있다. 따라서 각 성분에 복제본의 첨자를 같이 표기해주어야 하는데, 각 복제본들끼리 짝지어지는 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. 만약 $n=4$, $m_1=1$이라면 $$ \mathbf G= \left( \begin{array}{cccccccccc} A & B_0 & B_0 & B_0 & C_0 & C_0 & C_0 & D_{02} & D_{02} & D_{02} \\ B_0 & A & B_0 & B_0 & C_0 & D_{02} & D_{02} & C_0 & C_0 & D_{02} \\ B_0 & B_0 & A & B_0 & D_{02} & C_0 & D_{02} & C_0 & D_{02} & C_0 \\ B_0 & B_0 & B_0 & A & D_{02} & D_{02} & C_0 & D_{02} & C_0 & C_0 \\ C_0 & C_0 & D_{02} & D_{02} & P_0 & Q_{00} & Q_{00} & Q_{00} & \ Q_{00} & R_{1:1:1:1} \\ C_0 & D_{02} & C_0 & D_{02} & Q_{00} & P_0 & Q_{00} & Q_{00} & \ R_{1:1:1:1} & Q_{00} \\ C_0 & D_{02} & D_{02} & C_0 & Q_{00} & Q_{00} & P_0 & R_{1:1:1:1} & \ Q_{00} & Q_{00} \\ D_{02} & C_0 & C_0 & D_{02} & Q_{00} & Q_{00} & R_{1:1:1:1} & P_0 & \ Q_{00} & Q_{00} \\ D_{02} & C_0 & D_{02} & C_0 & Q_{00} & R_{1:1:1:1} & Q_{00} & Q_{00} \ & P_0 & Q_{00} \\ D_{02} & D_{02} & C_0 & C_0 & R_{1:1:1:1} & Q_{00} & Q_{00} & Q_{00} \ & Q_{00} & P_0 \\ \end{array} \right) $$ 고윳값들은 $$ \left( \begin{array}{c} P_0-2 Q_{00}+R_{1:1:1:1} \\ P_0-2 Q_{00}+R_{1:1:1:1} \\ \frac{1}{2} \left(-\sqrt{(A+3 B_0+P_0+4 Q_{00}+R_{1:1:1:1})^2-4 (A+3 \ B_0) (P_0+4 Q_{00}+R_{1:1:1:1})+24 C_0^2+48 C_0 D_{02}+24 \ D_{02}^2}+A+3 B_0+P_0+4 Q_{00}+R_{1:1:1:1}\right) \\ \frac{1}{2} \left(\sqrt{(A+3 B_0+P_0+4 Q_{00}+R_{1:1:1:1})^2-4 (A+3 \ B_0) (P_0+4 Q_{00}+R_{1:1:1:1})+24 C_0^2+48 C_0 D_{02}+24 \ D_{02}^2}+A+3 B_0+P_0+4 Q_{00}+R_{1:1:1:1}\right) \\ \frac{1}{2} \left(-\sqrt{(A-B_0+P_0-R_{1:1:1:1})^2-4 (A-B_0) \ (P_0-R_{1:1:1:1})+8 C_0^2-16 C_0 D_{02}+8 \ D_{02}^2}+A-B_0+P_0-R_{1:1:1:1}\right) \\ \frac{1}{2} \left(-\sqrt{(A-B_0+P_0-R_{1:1:1:1})^2-4 (A-B_0) \ (P_0-R_{1:1:1:1})+8 C_0^2-16 C_0 D_{02}+8 \ D_{02}^2}+A-B_0+P_0-R_{1:1:1:1}\right) \\ \frac{1}{2} \left(-\sqrt{(A-B_0+P_0-R_{1:1:1:1})^2-4 (A-B_0) \ (P_0-R_{1:1:1:1})+8 C_0^2-16 C_0 D_{02}+8 \ D_{02}^2}+A-B_0+P_0-R_{1:1:1:1}\right) \\ \frac{1}{2} \left(\sqrt{(A-B_0+P_0-R_{1:1:1:1})^2-4 (A-B_0) \ (P_0-R_{1:1:1:1})+8 C_0^2-16 C_0 D_{02}+8 \ D_{02}^2}+A-B_0+P_0-R_{1:1:1:1}\right) \\ \frac{1}{2} \left(\sqrt{(A-B_0+P_0-R_{1:1:1:1})^2-4 (A-B_0) \ (P_0-R_{1:1:1:1})+8 C_0^2-16 C_0 D_{02}+8 \ D_{02}^2}+A-B_0+P_0-R_{1:1:1:1}\right) \\ \frac{1}{2} \left(\sqrt{(A-B_0+P_0-R_{1:1:1:1})^2-4 (A-B_0) \ (P_0-R_{1:1:1:1})+8 C_0^2-16 C_0 D_{02}+8 \ D_{02}^2}+A-B_0+P_0-R_{1:1:1:1}\right) \\ \end{array} \right) $$ 가 되며, replicon mode의 중복도는 2가 된다. $n=4$, $m_1=2$라면 $$\mathbf G= \left( \begin{array}{cccccccccc} A & B_1 & B_0 & B_0 & C_1 & C_0 & C_0 & D_{01} & D_{01} & D_{11} \\ B_1 & A & B_0 & B_0 & C_1 & D_{01} & D_{01} & C_0 & C_0 & D_{11} \\ B_0 & B_0 & A & B_1 & D_{11} & C_0 & D_{01} & C_0 & D_{01} & C_1 \\ B_0 & B_0 & B_1 & A & D_{11} & D_{01} & C_0 & D_{01} & C_0 & C_1 \\ C_1 & C_1 & D_{11} & D_{11} & P_1 & Q_{01} & Q_{01} & Q_{01} & \ Q_{01} & R_{2:2,1} \\ C_0 & D_{01} & C_0 & D_{01} & Q_{01} & P_0 & Q_{10} & Q_{10} & \ R_{2:2,0} & Q_{01} \\ C_0 & D_{01} & D_{01} & C_0 & Q_{01} & Q_{10} & P_0 & R_{2:2,0} & \ Q_{10} & Q_{01} \\ D_{01} & C_0 & C_0 & D_{01} & Q_{01} & Q_{10} & R_{2:2,0} & P_0 & \ Q_{10} & Q_{01} \\ D_{01} & C_0 & D_{01} & C_0 & Q_{01} & R_{2:2,0} & Q_{10} & Q_{10} & \ P_0 & Q_{01} \\ D_{11} & D_{11} & C_1 & C_1 & R_{2:2,1} & Q_{01} & Q_{01} & Q_{01} & \ Q_{01} & P_1 \\ \end{array} \right) $$ 고윳값들은 $$ \left( \begin{array}{c} P_0-2 Q_{10}+R_{2:2,0} \\ \frac{1}{2} \left(-\sqrt{(A-B_1+P_0-R_{2:2,0})^2-4 (A-B_1) \ (P_0-R_{2:2,0})+8 C_0^2-16 C_0 D_{01}+8 D_{01}^2}+A-B_1+P_0-R_{2:2,0}\ \right) \\ \frac{1}{2} \left(-\sqrt{(A-B_1+P_0-R_{2:2,0})^2-4 (A-B_1) \ (P_0-R_{2:2,0})+8 C_0^2-16 C_0 D_{01}+8 D_{01}^2}+A-B_1+P_0-R_{2:2,0}\ \right) \\ \frac{1}{2} \left(\sqrt{(A-B_1+P_0-R_{2:2,0})^2-4 (A-B_1) \ (P_0-R_{2:2,0})+8 C_0^2-16 C_0 D_{01}+8 D_{01}^2}+A-B_1+P_0-R_{2:2,0}\ \right) \\ \frac{1}{2} \left(\sqrt{(A-B_1+P_0-R_{2:2,0})^2-4 (A-B_1) \ (P_0-R_{2:2,0})+8 C_0^2-16 C_0 D_{01}+8 D_{01}^2}+A-B_1+P_0-R_{2:2,0}\ \right) \\ \frac{1}{2} \left(-\sqrt{(A-2 B_0+B_1+P_1-R_{2:2,1})^2-4 \ (P_1-R_{2:2,1}) (A-2 B_0+B_1)+8 C_1^2-16 C_1 D_{11}+8 D_{11}^2}+A-2 \ B_0+B_1+P_1-R_{2:2,1}\right) \\ \frac{1}{2} \left(\sqrt{(A-2 B_0+B_1+P_1-R_{2:2,1})^2-4 \ (P_1-R_{2:2,1}) (A-2 B_0+B_1)+8 C_1^2-16 C_1 D_{11}+8 D_{11}^2}+A-2 \ B_0+B_1+P_1-R_{2:2,1}\right) \\ \cdots \end{array} \right) $$ 로 중복도가 2가 아니라 1이 된다. 물리/짚고_넘어가야_할_점.txt Last modified: 2026/02/26 16:23by admin