물리:구면_p-스핀_유리_모형

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물리:구면_p-스핀_유리_모형 [2023/01/04 17:07] – [$\lambda$ 적분] jiwon물리:구면_p-스핀_유리_모형 [2023/09/05 15:46] (current) – external edit 127.0.0.1
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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. 따라서 분배함수는 여기서 $(\tilde\Lambda)_{\alpha\beta}=\lambda_{\alpha\beta}$이다. 따라서 분배함수는
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 $$ $$
 가 된다. 여기서 가 된다. 여기서
-$$G[\mathbf q,\lambda] = -\frac 12\sum_{\alpha\beta}\lambda_{\alpha\beta}q_{\alpha\beta}+\frac{(\beta J)^2}4\sum_{\alpha\beta}q_{\alpha\beta}^p-n\log(2\pi)-\frac{(\beta J)^2}{8N}p(p-1)\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}\left\langle(\sigma^\alpha\sigma^\beta)^2\right\rangle+\log\det(-\tilde\Lambda)+\frac{(\beta J)^2}2\sum_{\alpha\beta}(\tilde\Lambda^{-1})_{\alpha\beta}$$+$$G[\mathbf q,\lambda] = \frac 12\sum_{\alpha\beta}\lambda_{\alpha\beta}q_{\alpha\beta}-\frac{(\beta J)^2}2\sum_{\alpha\beta}q_{\alpha\beta}^p-\frac n2\log(2\pi)+\frac{(\beta J)^2}{8N}p(p-1)\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}\left\langle(\sigma^\alpha\sigma^\beta)^2\right\rangle+\frac12\log\det(-\tilde\Lambda)+\frac{(\beta h)^2}2\sum_{\alpha\beta}(\tilde\Lambda^{-1})_{\alpha\beta}$$
 이다. 이다.
  
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 $$G_0[\mathbf q] = -\frac\mu{2p}\sum_{\alpha\beta}q_{\alpha\beta}^p-\frac{b^2}2\sum_{\alpha\beta}q_{\alpha\beta}-\frac12\log\det\mathbf q+\frac{b^4}4\left(\sum_{\alpha\beta}q_{\alpha\beta}\right)^2$$ $$G_0[\mathbf q] = -\frac\mu{2p}\sum_{\alpha\beta}q_{\alpha\beta}^p-\frac{b^2}2\sum_{\alpha\beta}q_{\alpha\beta}-\frac12\log\det\mathbf q+\frac{b^4}4\left(\sum_{\alpha\beta}q_{\alpha\beta}\right)^2$$
 $$G_1[\mathbf q] = \frac\mu4(p-1)\sum_{\alpha\beta}\left\langle(\sigma^\alpha\sigma^\beta)^2\right\rangle q_{\alpha\beta}^{p-2}+\log\det\mathbf q$$ $$G_1[\mathbf q] = \frac\mu4(p-1)\sum_{\alpha\beta}\left\langle(\sigma^\alpha\sigma^\beta)^2\right\rangle q_{\alpha\beta}^{p-2}+\log\det\mathbf q$$
-=====$q$ 적분=====+=====$q$ 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)