이 모형의 해밀토니안은 무작위 에너지 모형의 해밀토니안과 같다. 단 한가지 차이는 이 모형에서는 스핀 변수가 $-\infty$부터 $\infty$까지 실수의 값을 가질 수 있다는 것이다. 여기서도 복제 방법을 사용해서 모형을 분석해보자.

무작위 에너지 모형을 참고하면 $n$개의 복제본에 대한 분배함수는 다음과 같이 쓸 수 있다. $$\overline{Z^n}=\text{Tr}_{\sigma}\exp\left[\frac{(\beta J)^2}{4}N\sum_{\alpha\beta}\frac{p!}{N^p}\sum_{i_1<\cdots<i_p}\sigma_{i_1}^\alpha\sigma_{i_1}^\beta\cdots\sigma_{i_p}^\alpha\sigma_{i_p}^\beta+\beta h\sum_{i,a}\sigma_i^a\right]$$ $i_1,\cdots,i_p$에 대한 합을 다음과 같이 나누어 쓰자. $$p!\sum_{i_1<\cdots<i_p} = \sum_{i_1,\cdots,i_p}-\frac{p(p-1)}2\sum_{i_1,i_1\neq i_3,\cdots}$$ 이렇게 두고 분배함수를 다시 쓰면 $$\overline{Z^n}=\text{Tr}_{\sigma}\exp\left[\frac{(\beta J)^2}{4}N\sum_{\alpha\beta}\left\{\frac1{N^p}\left(\sum_i\sigma_i^\alpha\sigma_i^\beta\right)^p-\frac{p(p-1)}2\frac1{N^{p-2}}\left(\sum_i\sigma_i^\alpha\sigma_i^\beta\right)^{p-2}\left(\sum_i(\sigma_i^\alpha\sigma_i^\beta)^2\right)\right\}+\beta h\sum_{i,a}\sigma_i^a\right]$$ 가 된다. $q_{\alpha\beta}=N^{-1}\sum_i\sigma_i^\alpha\sigma_i^\beta$로 정의하고, 이를 만족하기 위한 구속조건 $$\begin{align*} 1=&\int\prod_{\alpha<\beta}dq_{\alpha\beta}\delta\left(Nq_{\alpha\beta}-\sum_{i=1}^N\sigma_i^\alpha\sigma_i^\beta\right)\\ =&\int\prod_{\alpha<\beta}dq_{\alpha\beta}\int_{-i\infty}^{+i\infty}\prod_{\alpha<\beta}\frac N{2\pi i}d\lambda_{\alpha\beta}\exp\left[-\frac12\sum_{\alpha\neq\beta}\lambda_{\alpha\beta}\left(Nq_{\alpha\beta}-\sum_{i=1}^N\sigma_i^\alpha\sigma_i^\beta\right)\right] \end{align*}$$ 과 스핀에 대한 대각합 $$\begin{align*} \text{Tr}_\sigma =& \int_{-\infty}^{+\infty}\prod_{i,\alpha}d\sigma_i^\alpha\prod_\alpha\delta\left(N-\sum_{i=1}^N(\sigma_i^\alpha)^2\right)\\ =&\int_{-\infty}^{+\infty}\prod_{i,\alpha}d\sigma_i^\alpha\int_{-i\infty}^{+i\infty}\prod_\alpha\frac{d\lambda_{\alpha\alpha}}{4\pi i}\exp\left[-\frac12\sum_\alpha\lambda_{\alpha\alpha}\left(Nq_{\alpha\alpha}-\sum_{i=1}^N(\sigma_i^\alpha)^2\right)\right] \end{align*}$$ 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.

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.

$b = \beta h$, $\mu=b^2p/2$.

$(\tilde b^2)_{\alpha\beta} = b^2$: $n\times n$ 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 $$\overline{Z^n} = e^{nS(\infty)}\int\prod_{\alpha<\beta}\sqrt{\frac N{2\pi}}dq_{\alpha\beta}\exp\left[-NG_0[\mathbf q]-G_1[\mathbf q]+\mathcal O(N^{-1})\right]$$

$$G_0[\mathbf q] = -\frac\mu{2p}\sum_{\alpha\beta}q_{\alpha\beta}^p-\frac{b^2}2\sum_{\alpha\beta}q_{\alpha\beta}-\frac12\log\det\mathbf q+\frac{b^4}4\left(\sum_{\alpha\beta}q_{\alpha\beta}\right)^2$$ $$G_1[\mathbf q] = \frac\mu4(p-1)\sum_{\alpha\beta}\left\langle(\sigma^\alpha\sigma^\beta)^2\right\rangle q_{\alpha\beta}^{p-2}+\log\det\mathbf q$$

극값 조건 $$\mu q_{\alpha\beta}^{p-1}+b^2+(\mathbf q^{-1})_{\alpha\beta} = 0\qquad\alpha\neq\beta$$

2차항까지 전개 $$\delta^2G_0 = -\frac{\mu(p-1)}2\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}(\delta q_{\alpha\beta})^2+\text{Tr}(\mathbf q^{-1}\delta\mathbf q)^2+b^4\left(\sum_{\alpha\beta}q_{\alpha\beta}\right)^2$$

$$\begin{align*} \text{Tr}(\mathbf q^{-1}\delta\mathbf q)^2=&\text{Tr}(\mathbf q^{-1}\delta\mathbf q\mathbf q^{-1}\delta\mathbf q)\\ =&\sum_\alpha(\mathbf q^{-1}\delta\mathbf q\mathbf q^{-1}\delta\mathbf q)_{\alpha\alpha}\\ =&\sum_{\alpha\beta\gamma\epsilon}(A\delta_{\alpha\beta}+B)\delta q_{\beta\gamma}(A\delta_{\gamma\epsilon}+B)\delta q_{\epsilon\alpha}\\ =&\sum_{\alpha\beta\gamma\epsilon}\left[A^2\delta_{\alpha\beta}\delta_{\gamma\epsilon}+AB(\delta_{\alpha\beta}+\delta_{\gamma\epsilon})+B^2\right]\delta q_{\beta\gamma}\delta q_{\epsilon\alpha}\\ =&A^2\sum_{\alpha\beta}\delta q_{\alpha\beta}^2 +2AB\sum_{\alpha\beta\gamma}\delta_{\alpha\gamma}\delta_{\gamma\beta}+B^2\left(\sum_{\alpha\beta}\delta q_{\alpha\beta}\right)^2 \end{align*}$$ $$\delta^2G_0 = -\frac{\mu(p-1)}2\sum_{\alpha\beta}q_{\alpha\beta}^{p-2}(\delta q_{\alpha\beta})^2+A^2\sum_{\alpha\beta}\delta q_{\alpha\beta}^2 +2AB\sum_{\alpha\beta\gamma}\delta_{\alpha\gamma}\delta_{\gamma\beta}+\left(B^2+b^4\right)\left(\sum_{\alpha\beta}q_{\alpha\beta}\right)^2$$