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--- 물리:구면_p-스핀_유리_모형 [2023/01/04 14:11] – created jiwon
+++ 물리:구면_p-스핀_유리_모형 [2023/09/05 15:46] (current) – external edit 127.0.0.1
@@ Line 20: / Line 20: @@
 \end{align*}
 을 넣어서 쓰면 분배함수를
+\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\\
+&\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*}
+와 같이 쓸 수 있다.
+=====스핀에 대한 대각합=====
+분배함수 중 스핀 변수와 관련된 부분을 모으면
+\begin{align*}
+&\int_{-\infty}^{+\infty}\prod_{i,\alpha}d\sigma_i^\alpha
+\exp\left[-\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]\\
+=&\left[\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha
+\exp\left[-\frac{(\beta J)^2}{8N}p(p-1)\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}(\sigma^\alpha\sigma^\beta)^2+\frac12\sum_{\alpha\beta}\lambda_{\alpha\beta}\sigma^\alpha\sigma^\beta+\beta h\sum_{\alpha}\sigma^\alpha\right]\right]^N
+\end{align*}
+로 쓸 수 있고, $\beta H_{\text{eff}} = \frac12\sum_{\alpha\beta}\lambda_{\alpha\beta}\sigma^\alpha\sigma^\beta+\beta h\sum_{\alpha}\sigma^\alpha$로 두고 위 식을 전개하면
+\begin{align*}
+&\left[\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha
+e^{\beta H_{\text{eff}}}\exp\left(-\frac{(\beta J)^2}{8N}p(p-1)\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}(\sigma^\alpha\sigma^\beta)^2\right)\right]^N\\
+\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\\
+=&\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}}}(\sigma^\alpha\sigma^\beta)^2\right\}\right]\\
+=&\exp\left[N\log\left(\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha
+e^{\beta H_{\text{eff}}}\right)+N\log\left\{1-\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\right\}\right]\\
+\approx&\exp\left[N\log\left(\int_{-\infty}^{+\infty}\prod_{\alpha}d\sigma^\alpha
+e^{\beta H_{\text{eff}}}\right)-N\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\right]
+\end{align*}
+이고, 지수 위의 첫 번째 항은 가우스 적분
+$$\int\prod_\alpha d\sigma_\alpha\exp\left[-\frac12\vec\sigma\cdot\mathbf A\cdot\vec\sigma+\mathbf J\cdot\vec\sigma\right] = \sqrt{\frac{(2\pi)^n}{\det\Lambda}}\exp\left[-\frac12\mathbf J\cdot \mathbf A^{-1}\cdot\mathbf J\right]$$
+를 이용해  $J_i = \beta h$, $\mathbf A = -\tilde\Lambda$로 두고 다음과 같이 계산할 수 있다.
+\begin{align*}
+&\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)\\
+=&\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*}
+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)=====
+$b = \beta h$, $\mu=b^2p/2$.
+$(\tilde b^2)_{\alpha\beta} = b^2$: $n\times n$ 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*}
+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
+$$\mu q_{\alpha\beta}^{p-1}+b^2+(\mathbf q^{-1})_{\alpha\beta} = 0\qquad\alpha\neq\beta$$
+차항까지 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=====