물리:구면_p-스핀_유리_모형

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물리:구면_p-스핀_유리_모형 [2023/06/30 16:30] – [스핀에 대한 대각합] 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)\\
-=&\frac n2\log(2\pi)-\frac12\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
-$$G[\mathbf q,\lambda] = \frac 12\sum_{\alpha\beta}\lambda_{\alpha\beta}q_{\alpha\beta}-\frac{(\beta J)^2}2\sum_{\alpha\beta}q_{\alpha\beta}^p-\frac n2\log(2\pi)+\frac{(\beta J)^2}{8N}p(p-1)\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}\left\langle(\sigma^\alpha\sigma^\beta)^2\right\rangle+\frac12\log\det(-\tilde\Lambda)+\frac{(\beta J)^2}2\sum_{\alpha\beta}(\tilde\Lambda^{-1})_{\alpha\beta}$$+$$G[\mathbf q,\lambda] = \frac 12\sum_{\alpha\beta}\lambda_{\alpha\beta}q_{\alpha\beta}-\frac{(\beta J)^2}2\sum_{\alpha\beta}q_{\alpha\beta}^p-\frac n2\log(2\pi)+\frac{(\beta J)^2}{8N}p(p-1)\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}\left\langle(\sigma^\alpha\sigma^\beta)^2\right\rangle+\frac12\log\det(-\tilde\Lambda)+\frac{(\beta h)^2}2\sum_{\alpha\beta}(\tilde\Lambda^{-1})_{\alpha\beta}$$
 이다. 이다.
  
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