물리:짚고_넘어가야_할_점

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물리:짚고_넘어가야_할_점 [2022/12/02 14:18] – created jiwon물리:짚고_넘어가야_할_점 [2023/09/05 15:46] (current) – external edit 127.0.0.1
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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.+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.
 $$A\equiv A_{\alpha\alpha}=1-\beta J_0(1-m^2)$$ $$A\equiv A_{\alpha\alpha}=1-\beta J_0(1-m^2)$$
 $$B_{\alpha\beta} \rightarrow \begin{cases} $$B_{\alpha\beta} \rightarrow \begin{cases}
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  \right)  \right)
 $$ $$
-복제 대칭해의 경우에 얻었던 $n(n-1)2$의 중복도를 가지는 replicon mode를 똑같이 얻어내었다. +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.
  
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  • Last modified: 2023/09/05 15:46
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