물리:구면_p-스핀_유리_모형

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물리:구면_p-스핀_유리_모형 [2023/01/04 15:27] – [$n$개의 복제본에 대한 분배함수] jiwon물리:구면_p-스핀_유리_모형 [2023/09/05 15:46] (current) – external edit 127.0.0.1
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 \begin{align*} \begin{align*}
 \overline{Z^n}=&\int\prod_{\alpha<\beta}dq_{\alpha\beta}\int_{-i\infty}^{+i\infty}\prod_{\alpha<\beta}\frac N{2\pi i}d\lambda_{\alpha\beta}\int_{-i\infty}^{+i\infty}\prod_\alpha\frac{d\lambda_{\alpha\alpha}}{4\pi i}\int_{-\infty}^{+\infty}\prod_{i,\alpha}d\sigma_i^\alpha\\ \overline{Z^n}=&\int\prod_{\alpha<\beta}dq_{\alpha\beta}\int_{-i\infty}^{+i\infty}\prod_{\alpha<\beta}\frac N{2\pi i}d\lambda_{\alpha\beta}\int_{-i\infty}^{+i\infty}\prod_\alpha\frac{d\lambda_{\alpha\alpha}}{4\pi i}\int_{-\infty}^{+\infty}\prod_{i,\alpha}d\sigma_i^\alpha\\
-&\quad\times\exp\left[-\frac N2\sum_{\alpha\beta}\lambda_{\alpha\beta}q_{\alpha\beta}+\frac{(\beta J)}4N\sum_{\alpha\beta}q_{\alpha\beta}^p-\frac{(\beta J)^2}{8N}p(p-1)\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}\left(\sum_i(\sigma_i^\alpha\sigma_i^\beta)^2\right)+\frac12\sum_{\alpha\beta}\lambda_{\alpha\beta}\sum_i\sigma_i^\alpha\sigma_i^\beta+\beta h\sum_{i,\alpha}\sigma_i^\alpha\right]+&\quad\times\exp\left[-\frac N2\sum_{\alpha\beta}\lambda_{\alpha\beta}q_{\alpha\beta}+\frac{(\beta J)^2}4N\sum_{\alpha\beta}q_{\alpha\beta}^p-\frac{(\beta J)^2}{8N}p(p-1)\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}\left(\sum_i(\sigma_i^\alpha\sigma_i^\beta)^2\right)+\frac12\sum_{\alpha\beta}\lambda_{\alpha\beta}\sum_i\sigma_i^\alpha\sigma_i^\beta+\beta h\sum_{i,\alpha}\sigma_i^\alpha\right]
 \end{align*} \end{align*}
 와 같이 쓸 수 있다. 와 같이 쓸 수 있다.
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 \approx&\left[\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha \approx&\left[\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha
 e^{\beta H_{\text{eff}}}\left(1-\frac{(\beta J)^2}{8N}p(p-1)\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}(\sigma^\alpha\sigma^\beta)^2+\mathcal O(N^{-2})\right)\right]^N\\ e^{\beta H_{\text{eff}}}\left(1-\frac{(\beta J)^2}{8N}p(p-1)\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}(\sigma^\alpha\sigma^\beta)^2+\mathcal O(N^{-2})\right)\right]^N\\
-&\exp\left[N\log\left\{\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha+=&\exp\left[N\log\left\{\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha
 e^{\beta H_{\text{eff}}}-\frac{(\beta J)^2}{8N}p(p-1)\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha e^{\beta H_{\text{eff}}}-\frac{(\beta J)^2}{8N}p(p-1)\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha
 e^{\beta H_{\text{eff}}}(\sigma^\alpha\sigma^\beta)^2\right\}\right]\\ e^{\beta H_{\text{eff}}}(\sigma^\alpha\sigma^\beta)^2\right\}\right]\\
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 이고, 지수 위의 첫 번째 항은 가우스 적분 이고, 지수 위의 첫 번째 항은 가우스 적분
 $$\int\prod_\alpha d\sigma_\alpha\exp\left[-\frac12\vec\sigma\cdot\mathbf A\cdot\vec\sigma+\mathbf J\cdot\vec\sigma\right] = \sqrt{\frac{(2\pi)^n}{\det\Lambda}}\exp\left[-\frac12\mathbf J\cdot \mathbf A^{-1}\cdot\mathbf J\right]$$ $$\int\prod_\alpha d\sigma_\alpha\exp\left[-\frac12\vec\sigma\cdot\mathbf A\cdot\vec\sigma+\mathbf J\cdot\vec\sigma\right] = \sqrt{\frac{(2\pi)^n}{\det\Lambda}}\exp\left[-\frac12\mathbf J\cdot \mathbf A^{-1}\cdot\mathbf J\right]$$
-를 이용해  $J_i = \beta J$, $\mathbf A = -\tilde\Lambda$로 두고 다음과 같이 계산할 수 있다.+를 이용해  $J_i = \beta h$, $\mathbf A = -\tilde\Lambda$로 두고 다음과 같이 계산할 수 있다.
 \begin{align*} \begin{align*}
 &\log\left(\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha &\log\left(\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha
 e^{\beta H_{\text{eff}}}\right)=\log\left(\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha e^{\beta H_{\text{eff}}}\right)=\log\left(\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha
 \exp\left[\frac12\sum_{\alpha\beta}\lambda_{\alpha\beta}\sigma^\alpha\sigma^\beta+\beta h\sum_{\alpha}\sigma^\alpha\right]\right)\\ \exp\left[\frac12\sum_{\alpha\beta}\lambda_{\alpha\beta}\sigma^\alpha\sigma^\beta+\beta h\sum_{\alpha}\sigma^\alpha\right]\right)\\
-=&n\log(2\pi)-\log\det(-\tilde\Lambda)-\frac{(\beta J)^2}2\sum_{\alpha\beta}(\tilde\Lambda^{-1})_{\alpha\beta}+=&\frac n2\log(2\pi)-\frac12\log\det(-\tilde\Lambda)+\frac{(\beta h)^2}2\sum_{\alpha\beta}(\tilde\Lambda^{-1})_{\alpha\beta}
 \end{align*} \end{align*}
-여기서 $(\tilde\Lambda)_{\alpha\beta}=\lambda_{\alpha\beta}$이다.+여기서 $(\tilde\Lambda)_{\alpha\beta}=\lambda_{\alpha\beta}$이다. 따라서 분배함수는 
 +$$ 
 +\overline{Z^n}=\int\prod_{\alpha<\beta}dq_{\alpha\beta}\int_{-i\infty}^{+i\infty}\prod_{\alpha<\beta}\frac N{2\pi i}d\lambda_{\alpha\beta}\int_{-i\infty}^{+i\infty}\prod_\alpha\frac{d\lambda_{\alpha\alpha}}{4\pi i} 
 +e^{-NG[\mathbf q,\lambda]} 
 +$$ 
 +가 된다. 여기서 
 +$$G[\mathbf q,\lambda] = \frac 12\sum_{\alpha\beta}\lambda_{\alpha\beta}q_{\alpha\beta}-\frac{(\beta J)^2}2\sum_{\alpha\beta}q_{\alpha\beta}^p-\frac n2\log(2\pi)+\frac{(\beta J)^2}{8N}p(p-1)\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}\left\langle(\sigma^\alpha\sigma^\beta)^2\right\rangle+\frac12\log\det(-\tilde\Lambda)+\frac{(\beta h)^2}2\sum_{\alpha\beta}(\tilde\Lambda^{-1})_{\alpha\beta}$$ 
 +이다.
  
  
 +=====$\lambda$ 적분(작성중)=====
 +$b = \beta h$, $\mu=b^2p/2$.
  
-=====$\lambda$ 적분=====+$(\tilde b^2)_{\alpha\beta} = b^2$: $n\times n$ 행렬 
 +\begin{align*} 
 +\log\det(-\tilde\Lambda-\tilde b^2)=&\log\det\left[-\tilde\Lambda(I+\tilde b^2\cdot\tilde \Lambda^{-1})\right]\\ 
 +=&\log\det(-\tilde \Lambda)+\log\det(I+\tilde b^2\cdot\tilde\Lambda^{-1})\\ 
 +=&\log\det(-\tilde \Lambda)+\text{Tr}\log(I+\tilde b^2\cdot\tilde\Lambda^{-1})\\ 
 +\approx&\log\det(-\tilde\Lambda)+\text{Tr }\left(\tilde b^2\cdot\tilde\Lambda^{-1}-\frac{(\tilde b^2\cdot\tilde\Lambda^{-1})^2}2+\mathcal O(n^3)\right) 
 +\end{align*}
  
-=====$q$ 적분=====+극값 조건 
 +$$\langle\sigma^\alpha\sigma^\beta\rangle = q_{\alpha\beta}$$ 
 +$$\lambda_{\alpha\beta}+b^2+(\mathbf q^{-1})_{\alpha\beta} = \mathcal O(n)$$ 
 + 
 +분배함수 
 +$$\overline{Z^n} = e^{nS(\infty)}\int\prod_{\alpha<\beta}\sqrt{\frac N{2\pi}}dq_{\alpha\beta}\exp\left[-NG_0[\mathbf q]-G_1[\mathbf q]+\mathcal O(N^{-1})\right]$$ 
 + 
 +$$G_0[\mathbf q] = -\frac\mu{2p}\sum_{\alpha\beta}q_{\alpha\beta}^p-\frac{b^2}2\sum_{\alpha\beta}q_{\alpha\beta}-\frac12\log\det\mathbf q+\frac{b^4}4\left(\sum_{\alpha\beta}q_{\alpha\beta}\right)^2$$ 
 +$$G_1[\mathbf q] = \frac\mu4(p-1)\sum_{\alpha\beta}\left\langle(\sigma^\alpha\sigma^\beta)^2\right\rangle q_{\alpha\beta}^{p-2}+\log\det\mathbf q$$ 
 +=====$q$ 적분(작성중)===== 
 + 
 +극값 조건 
 +$$\mu q_{\alpha\beta}^{p-1}+b^2+(\mathbf q^{-1})_{\alpha\beta} = 0\qquad\alpha\neq\beta$$ 
 + 
 +2차항까지 전개 
 +$$\delta^2G_0 = -\frac{\mu(p-1)}2\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}(\delta q_{\alpha\beta})^2+\text{Tr}(\mathbf q^{-1}\delta\mathbf q)^2+b^4\left(\sum_{\alpha\beta}q_{\alpha\beta}\right)^2$$ 
 + 
 +\begin{align*} 
 +\text{Tr}(\mathbf q^{-1}\delta\mathbf q)^2=&\text{Tr}(\mathbf q^{-1}\delta\mathbf q\mathbf q^{-1}\delta\mathbf q)\\ 
 +=&\sum_\alpha(\mathbf q^{-1}\delta\mathbf q\mathbf q^{-1}\delta\mathbf q)_{\alpha\alpha}\\ 
 +=&\sum_{\alpha\beta\gamma\epsilon}(A\delta_{\alpha\beta}+B)\delta q_{\beta\gamma}(A\delta_{\gamma\epsilon}+B)\delta q_{\epsilon\alpha}\\ 
 +=&\sum_{\alpha\beta\gamma\epsilon}\left[A^2\delta_{\alpha\beta}\delta_{\gamma\epsilon}+AB(\delta_{\alpha\beta}+\delta_{\gamma\epsilon})+B^2\right]\delta q_{\beta\gamma}\delta q_{\epsilon\alpha}\\ 
 +=&A^2\sum_{\alpha\beta}\delta q_{\alpha\beta}^2 +2AB\sum_{\alpha\beta\gamma}\delta_{\alpha\gamma}\delta_{\gamma\beta}+B^2\left(\sum_{\alpha\beta}\delta q_{\alpha\beta}\right)^2 
 +\end{align*} 
 +$$\delta^2G_0 = -\frac{\mu(p-1)}2\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}(\delta q_{\alpha\beta})^2+A^2\sum_{\alpha\beta}\delta q_{\alpha\beta}^2 +2AB\sum_{\alpha\beta\gamma}\delta_{\alpha\gamma}\delta_{\gamma\beta}+\left(B^2+b^4\right)\left(\sum_{\alpha\beta}q_{\alpha\beta}\right)^2$$
  
 +=====복제 대칭 해=====
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  • Last modified: 2023/09/05 15:46
  • (external edit)