利用复本方法计算 Sherrington-Kirkpatrick（SK） 模型。该章节主要介绍 Parisi 解。
Reference: * Statistical Physics of Spin Glasses and Information Processing. Nishimori * Replica calculations for the SK model笔误有点多 * RS and RSB solutions for SK model with spin-S
Link: * Hopfield Model * Replica * Sherrington-Kirkpatrick Model
根据
$$\begin{align} -\beta[f]= & \lim _{n \rightarrow 0} \frac{\left[Z^n\right]-1}{n N}=\lim _{n \rightarrow 0}\left\{-\frac{\beta^2 J^2}{4 n} \sum_{\alpha \neq \beta} q_{\alpha \beta}^2-\frac{\beta J_0}{2 n} \sum_\alpha m_\alpha^2+\frac{1}{4} \beta^2 J^2+\frac{1}{n} \log \operatorname{Tr} \mathrm{e}^L\right\} \label{2.17} \end{align}$$
Multi-step replica symmetry breaking (RSB)
在复本对称假设$\eqref{RS}$中qαβ与mαβ并不依赖于αβ，为了打破对称需要考虑q矩阵的结构。现在考虑 n-RSB 的矩阵，首先对于 1-RSB有m1 < n，例如当n = 6, m1 = 3时候，q矩阵有（对角线为零的原因在于，当α = β时候没有定义）：
$$\begin{align} \left(\begin{array}{ccc|ccc} 0 & q_1 & q_1 & & & \\ q_1 & 0 & q_1 & & q_0 & \\ q_1 & q_1 & 0 & & & \\ \hline & & & 0 & q_1 & q_1 \\ & q_0 & & q_1 & 0 & q_1 \\ & & & q_1 & q_1 & 0 \end{array}\right)\label{1-RSB} \end{align}$$
可以执行相同的迭代操作，将粒度不断细化，例如再进行对称破缺$\eqref{1-RSB}$左上角的块变为： $$\begin{align} \begin{array}{ccc|ccc} 0 & q_2 & q_2 & & & \\ q_2 & 0 & q_2 & & q_1 & \\ q_2 & q_2 & 0 & & & \\ \hline & & & 0 & q_2 & q_2 \\ & q_1 & & q_2 & 0 & q_2 \\ & & & q_2 & q_2 & 0 \end{array} \label{2-RSB} \end{align}$$
同时应当满足条件：
$$\begin{align} n \geq m_1 \geq m_2 \geq \ldots \geq 1 \end{align}$$
定义函数： $$\begin{align} q(x)=q_i \quad\left(m_{i+1} \leq x \leq m_i\right) \end{align}$$
进行复本对称的极限变换n → 0，将不等关系进行任意的翻转（令人费解的反号）：
$$\begin{align} 0 \leq m_1 \leq m_2 . . \leq 1 \end{align}$$
First step RSB
再次回到计算[Zn]，结合$\eqref{19}$，在J0 = h = 0的假设下，将$\eqref{1-RSB}$融入进去有：
$$\begin{align} \sum_{\alpha<\beta} q_{\alpha \beta} S^\alpha S^\beta=\frac{1}{2}\left\{q_0\left(\sum_\alpha^n S^\alpha\right)^2+\left(q_1-q_0\right) \sum_{b=1}^{n / m_1}\left(\sum_{\alpha \in B_b}^{m_1} S^\alpha\right)^2-n q_1\right\} \label{67} \end{align}$$
其中Bb 表示 b-th block。第一项是没有进行复本破缺的，第二项是表示一阶复本破缺，第三项将对角线的贡献排除。同样的，对于$\eqref{28}$有：
$$\begin{align} \lim _{n \rightarrow 0} \frac{1}{n} \sum_{\alpha \neq \beta} q_{\alpha \beta}^2=\lim _{n \rightarrow 0} \frac{1}{n}\left\{n^2 q_0^2+\frac{n}{m_1} m_1^2\left(q_1^2-q_0^2\right)-n q_1^2\right\}=\left(m_1-1\right) q_1^2-m_1 q_0^2 \label{68} \end{align}$$
此时已经将自由能中含有qαβ的地方进行替换，接下来分析$\eqref{28}$。首先将$\eqref{68}$代入：
$$\begin{align} \beta\left[f_{1 R S B}\right]=\frac{\beta^2 J^2}{4}\left\{\left(m_1-1\right) q_1^2-m_1 q_0^2-1\right\}+\frac{\beta J_0}{2} m^2-\frac{1}{n} \log \sum_{\left\{S^\alpha, S^3, \ldots, S^n\right\}} \exp \left(L_{1 R S B}\right) \label{70} \end{align}$$
再将$\eqref{67}$代入$\eqref{19}$：
$$ \begin{gather} L\left(\left\{q_{\alpha \beta}, m_\alpha\right\}\right):=\beta^2 J^2 \sum_{\alpha<\beta} q_{\alpha \beta} S^\alpha S^\beta+\beta \sum_\alpha\left(J_0 m_\alpha+h\right) S^\alpha \label{71} \\ \Rightarrow L_{1 R S B}=\frac{\beta^2 J^2}{2}\left\{q_0\left(\sum_\alpha^n S^\alpha\right)^2+\left(q_1-q_0\right) \sum_{b=1}^{n / m_1}\left(\sum_{\alpha \in B_b}^{m_1} S^\alpha\right)^2-n q_1\right\}+\beta \sum_\alpha\left(J_0 m_\alpha+h\right) S^\alpha \label{72} \end{gather} $$
由于里面有$\left(\sum_\alpha^n S^\alpha\right)^2$和$\left(\sum_{\alpha \in B_b}^{m_1} S^\alpha\right)^2$，需要将平方项变为线性项，同样使用$\eqref{11}$的办法：
$$ \begin{align} & \frac{1}{n} \log \sum_{\left\{S^\alpha, S^\beta, \ldots, S^n\right\}} \exp \left(L_{1 R S B}\right) \\ & =\frac{1}{n} \log \sum_{\left\{S^\alpha, S^\beta, \ldots, S^n\right)} \int D u \exp \left(\beta J \sqrt{q_0} u \sum_\alpha S^\alpha-\frac{n}{2} \beta^2 J^2 q_1+\sum_\alpha S^\alpha \beta\left(J_0 m+h\right)+\left(q_1-q_0\right)\frac{\beta^2 J^2}{2} \sum_{b=1}^{n / m_1}\left(\sum_{\alpha \in B_b}^{m_1} S^\alpha\right)^2\right) \\ & =-\frac{1}{2} \beta^2 J^2 q_1+\underbrace{\frac{1}{n} \log \sum_{\left\{S^\alpha, S^\beta, \ldots, S^n\right)} \int D u \exp \left(\sum_\alpha S^\alpha\left(\beta J \sqrt{q_0} u+\beta\left(J_0 m+h\right)\right)+\left(q_1-q_0\right)\frac{\beta^2 J^2}{2} \sum_{b=1}^{n / m_1}\left(\sum_{\alpha \in B_b}^{m_1} S^\alpha\right)^2\right)}_{\Delta} \label{72_2} \end{align} $$
引入vb线性化$\left(\sum_{\alpha \in B_b}^{m_1} S^\alpha\right)^2$：
$$ \begin{align} & \frac{1}{n} \log \sum_{\left\{S^\alpha, S^\beta, \ldots,\right\}} \int D u \prod_{b=1}^{n / m_1}\left(\int D v_b\right) \exp \left(\sum_\alpha S^\alpha \beta J \sqrt{q_0} u+\sum_\alpha S^\alpha \beta\left(J_0 m+h\right)+\beta J \sqrt{q_1-q_0} \sum_{b=1}^{n / m_1} v_b \sum_{\alpha \in B_b}^{m_1} S^\alpha\right) \\ = & \frac{1}{n} \log \int D u \sum_{\left\{S^\alpha, S^\beta, \ldots, S^n\right\}} \prod_{b=1}^{n / m_1}\left\{\int v e \exp \left(\sum_\alpha S^\alpha \beta\left(J_0 m+h+\beta J \sqrt{q_0} u\right)\right) \exp \left(\beta J \sqrt{q_1-q_0} v \sum_{\alpha \in B_b}^{m_1} S^\alpha\right)\right\} \\ = & \frac{1}{n} \log \int D u \prod_{b=1}^{n / m_1}\left\{\int D v_b \sum_{\left\{S^\alpha\right\} \in B_b} \exp \left(\sum_{\left\{S^\alpha\right\} \in B_b} S^\alpha\beta\left(J_0 m+h+\beta J \sqrt{q_0} u+J \sqrt{q_1-q_0} v_b\right)\right)\right\} \\ = & \frac{1}{n} \log \int D u \prod_{b=1}^{n / m_1}\left\{\int D v_b\left\{\sum_{S= \pm 1} \exp \left(S \beta\left(J_0 m+h+\beta J \sqrt{q_0} u+J \sqrt{q_1-q_0} v_b\right)\right)\right\}^{m_1}\right\} \\ = & \frac{1}{n} \log \int D u\left\{\int v\left\{\sum_{S= \pm 1} \exp \left(S \beta\left(J_0 m+h+\beta J \sqrt{q_0} u+J \sqrt{q_1-q_0} v\right)\right)\right\}^{m_1}\right\}^{n / m_1} \\ = & \frac{1}{n} \log \int D u\left\{\int D v\left\{2 \cosh \left(\beta\left(J_0 m+h+\beta J \sqrt{q_0} u+J \sqrt{q_1-q_0} v\right)\right)\right\}^{m_1}\right\}^{n / m_1} \\ = & \frac{1}{n} \log \int D u \exp \left\{\frac{n}{m_1} \log \left\{\int D v\left\{2 \cosh \left(\beta\left(J_0 m+h+\beta J \sqrt{q_0} u+J \sqrt{q_1-q_0} v\right)\right)\right\}^{m_1}\right\}\right\} \\ \approx & \frac{1}{n} \log \left\{1+\frac{n}{m_1} \int D u \log \left\{\int D v\left\{2 \cosh \left(\beta\left(J_0 m+h+\beta J \sqrt{q_0} u+J \sqrt{q_1-q_0} v\right)\right)\right\}^{m_1}\right\}\right\} \text { expand exponent around } 0 \\ \approx & \frac{1}{n} \frac{n}{m_1} \int D u \log \left\{\int D v\left\{2 \cosh \left(\beta\left(J_0 m+h+\beta J \sqrt{q_0} u+J \sqrt{q_1-q_0} v\right)\right)\right\}^{m_1}\right\} \text { expand exponent around } 1 \\ = & \log 2+\frac{1}{m_1} \int D u \log \left\{\int D v\left\{\cosh \left(\beta\left(J_0 m+h+\beta J \sqrt{q_0} u+J \sqrt{q_1-q_0} v\right)\right)\right\}^{m_1}\right\} \text { pull the factor of } 2 \text { out } \label{82} \end{align} $$
令： $$\begin{align} \Xi=\beta\left(J_0 m+h+\beta J \sqrt{q_0} u+J \sqrt{q_1-q_0} v\right) \label{86} \end{align}$$
有： $$\begin{align} \Delta=\log 2+\frac{1}{m_1} \int D u \log \left\{\int D v\{\cosh \Xi\}^{m_1}\right\} \label{87} \end{align}$$
结合$\eqref{72_2}\eqref{87}$将$\eqref{70}$写为： $$\begin{align} \beta\left[f_{1 R S B}\right]=\frac{\beta^2 J^2}{4}\left\{\left(m_1-1\right) q_1^2-m_1 q_0^2+2 q_1-1\right\}+\frac{\beta J_0}{2} m^2-\log 2-\frac{1}{m_1} \int D u \log \left\{\int D v\{\cosh \Xi\}^{m_1}\right\} . \end{align}$$
得到极值点为： $$\begin{gather} m^*=\int D u \frac{\int D v\{\cosh \Xi\}^{m_1} \tanh \Xi}{\int D v\{\cosh \Xi\}^{m_1}} \\ q_0^* =\int D u\left(\frac{\int D v\{\cosh \Xi\}^{m_1} \tanh \Xi}{\int D v\{\cosh \Xi\}^{m_1}}\right)^2 \\ q_1^* =\int D u \frac{\int D v\{\cosh \Xi\}^{m_1}\{\tanh \Xi\}^2}{\int D v\{\cosh \Xi\}^{m_1}} \end{gather}$$
Full RSB Solution
由于熵还是负数，接下来进行更高阶的复本对称破缺（k-RSB）计算$\eqref{28}$。与一阶复本对称$\eqref{67}$类似:
$$\begin{align} \sum_{\alpha \neq \beta} q_{\alpha \beta}^l & =q_0^l n^2+\left(q_1^l-q_0^l\right) m_1^2 \cdot \frac{n}{m_1}+\left(q_2^l-q_1^l\right) m_2^2 \cdot \frac{m_1}{m_2} \cdot \frac{n}{m_1}+\cdots-q_K^l \cdot n \\ & =n \sum_{j=0}^K\left(m_j-m_{j+1}\right) q_j^l \label{3.37} \end{align}$$
其中l是任意整数，m0 = n, mK + 1 = 1，在极限情况n → 0下，使用mj − mj + 1 → −dx，可以得到：
$$\begin{align} \frac{1}{n} \sum_{\alpha \neq \beta} q_{\alpha \beta}^l \rightarrow-\int_0^1 q^l(x) \mathrm{d} x \label{3.38} \end{align}$$
在J0 = 0, h = 0的条件下，结合$E=-\frac{\partial \log Z}{\partial \beta}$、$\chi=-\frac{\partial^2 F}{\partial h^2}$与$\eqref{28}$，可得：
没有证明
$$\begin{gather} E=-\frac{\beta J^2}{2}\left(1+\frac{2}{n} \sum_{\alpha<\beta} q_{\alpha \beta}^2\right) \rightarrow-\frac{\beta J^2}{2}\left(1-\int_0^1 q^2(x) \mathrm{d} x\right) \\ \chi=\beta\left(1+\frac{1}{n} \sum_{\alpha \neq \beta} q_{\alpha \beta}\right) \rightarrow \beta\left(1-\int_0^1 q(x) \mathrm{d} x\right) \end{gather}$$
parisi equation
接下来需要将高阶复本破缺技巧应用在自由能$\eqref{2.17}$，和计算一阶复本对称的操作一样，首先计算q0然后计算第一次分块q1的贡献并且排除q0的影响，以此类推，在主对角线上是qK。首先计算如下项：
$$ \begin{align} G & =\operatorname{Tr} \exp \left(\frac{1}{2} \sum_{\alpha, \beta=1}^n q_{\alpha \beta} S^\alpha S^\beta+h \sum_\alpha^n S^\alpha\right) \\ & =\left.\exp \left(\frac{1}{2} \sum_{\alpha, \beta} q_{\alpha \beta} \frac{\partial^2}{\partial h_\alpha \partial h_\beta}\right) \prod_\alpha 2 \cosh h_\alpha\right|_{h_\alpha=h} \label{B.1} \end{align} $$
这里将qαβ贡献的项目视为h项的偏移，因此使用$\exp(\frac{\partial^2}{\partial h_\alpha \partial h_\beta})$的形式表示这个偏移量。如果使用复本对称的假设，则：
$$\begin{align} G=\exp \left(\frac{q}{2} \frac{\partial^2}{\partial h^2}\right)(2 \cosh h)^n \label{B.2} \end{align}$$
其中使用到：
$$ \left.\sum_\alpha \frac{\partial f\left(h_1, \ldots, h_n\right)}{\partial h_\alpha}\right|_{h_\alpha=h}=\frac{\partial f(h, \ldots, h)}{\partial h} $$
从$\eqref{67}$中可以看出，当时的思想是从复本对称假设出发，然后将整个矩阵分为几部分（取决于$\frac{n}{m}$大小），在主对角线的块上为一阶复本对称q1；那么同样的二阶复本对称可以将每一个一阶复本对称的块分为几部分，然后将主对角线的块作为二阶复本q2；这个流程以此类推。现在可以反过来，从K阶复本逐渐变为0阶级复本：
- 对于K − 1阶复本有(qK − qK − 1)I(mK)，表示(mK × mK)大小。
- 生长出(K − 2)复本，将K阶复本通过DiagK − 1扩张 (qK − qK − 1)DiagK − 1[I(mK)]，然后K − 2阶写为(qK − 1 − qK − 2)I(mK − 1)矩阵的大小为(mK − 1 × mK − 1)。整体为: (qK − qK − 1)DiagK − 1[I(mK)] + (qK − 1 − qK − 2)I(mK − 1)
- 重复步骤2，得到K − 3复本。矩阵的大小为(mK − 2 × mK − 2) (qK − qK − 1)DiagK − 2[I(mK)] + (qK − 1 − qK − 2)DiagK − 2[I(mK − 1)] + (qK − 2 − qK − 3)I(mK − 2)
在清楚以上步骤之后，接下来首先处理(mK × mK)大小矩阵Tr ，并将结果标注为g(mK, h)，由于这是复本破缺的最后一次，因此这个矩阵元是复本对称的，由$\eqref{B.2}$得到：
$$\begin{align} g\left(m_K, h\right)=\exp \left\{\frac{1}{2}\left(q_K-q_{K-1}\right) \frac{\partial^2}{\partial h^2}\right\}(2 \cosh h)^{m_K} \label{B.4} \end{align}$$
对于K − 1阶，是考虑将g(mK, h)复制到对角部分，然后每一个元素相加qK − 1 − qK − 2，从$\eqref{B.2}$可以得到：
$$\begin{align} g\left(m_{K-1}, h\right)=\exp \left\{\frac{1}{2}\left(q_{K-1}-q_{K-2}\right) \frac{\partial^2}{\partial h^2}\right\}\left[g\left(m_K, h\right)\right]^{m_{K-1} / m_K} \label{B.5} \end{align}$$
一直重复这个过程，直到复本对称项： $$\begin{align} G=g(n, h)=\exp \left\{\frac{1}{2} q(0) \frac{\partial^2}{\partial h^2}\right\}\left[g\left(m_1, h\right)\right]^{n / m_1} \label{B.6} \end{align}$$
考虑到极限n → 0下有mj − mj − 1 = −dx，将$\eqref{B.5}$写为：
$$\begin{align} g(x+\mathrm{d} x, h)=\exp \left\{-\frac{1}{2} \mathrm{~d} q(x) \frac{\partial^2}{\partial h^2}\right\} g(x, h)^{1+\mathrm{d} \log x} \label{B.7} \end{align}$$
对于$\eqref{B.4}$，当K → ∞有mk → 1，因为存在g(1, h) = 2cosh h。将$\eqref{B.7}$转化为微分形式（对dq与dx进行展开）：
$$ \frac{\partial g}{\partial x}=-\frac{1}{2} \frac{\mathrm{d} q}{\mathrm{~d} x} \frac{\partial^2 g}{\partial h^2}+\frac{1}{x} g \log g $$
利用注记f0(x, h) = (1/x)log g(x, h)改写为：
$$\begin{align} \frac{\partial f_0}{\partial x}=-\frac{1}{2} \frac{\mathrm{d} q}{\mathrm{~d} x}\left\{\frac{\partial^2 f_0}{\partial h^2}+x\left(\frac{\partial f_0}{\partial h}\right)^2\right\} \label{B.8} \end{align}$$
分析$\eqref{B.6}$的极限情况，根据之前的讨论知道，当n → 0的时候，m1 → 0：
$$\begin{align} \frac{1}{n} \log \operatorname{Tr} \mathrm{e}^L &= \left. \exp \left(\frac{1}{2} q(0) \frac{\partial^2}{\partial h^2}\right) \frac{1}{x} \log g(x, h)\right|_{x, h \rightarrow 0} \\ & =\left.\exp \left(\frac{1}{2} q(0) \frac{\partial^2}{\partial h^2}\right) f_0(0, h)\right|_{h \rightarrow 0} \\ & =\int \mathrm{D} u f_0(0, \sqrt{q(0)} u) \label{B.10} \end{align}$$
结合$\eqref{3.38},\eqref{B.10}$将$\eqref{2.17}$写为q(1)这一项怎么来的？：
$$\begin{align} \beta f=-\frac{\beta^2 J^2}{4}\left\{1+\int_0^1 q(x)^2 \mathrm{~d} x-2 q(1)\right\}-\int \mathrm{D} u f_0(0, \sqrt{q(0)} u) \label{3.41} \end{align}$$
其中f0应当满足 Parisi equation，即$\eqref{B.9}$:
$$\begin{align} \frac{\partial f_0(x, h)}{\partial x}=-\frac{J^2}{2} \frac{\mathrm{d} q}{\mathrm{~d} x}\left\{\frac{\partial^2 f_0}{\partial h^2}+x\left(\frac{\partial f_0}{\partial h}\right)^2\right\} \label{3.42} \end{align}$$
有 f0(1, h) = log 2cosh βh。
Order parameter near the critical point
虽然求解$\eqref{3.41}$十分困难，但是在临界点附近的性质是可以研究的。
在J0 = h = 0的条件下，$\eqref{2.17}$展为4阶可以是：
未经计算，直接抄的，并且十分简略
$$ \begin{align} \beta f= & \lim _{n \rightarrow 0} \frac{1}{n}\left\{\frac{1}{4}\left(\frac{T^2}{T_{\mathrm{f}}^2}-1\right) \operatorname{Tr} Q^2-\frac{1}{6} \operatorname{Tr} Q^3 - \frac{1}{8} \operatorname{Tr} Q^4+\frac{1}{4} \sum_{\alpha \neq \beta \neq \gamma} Q_{\alpha \beta}^2 Q_{\alpha \gamma}^2-\frac{1}{12} \sum_{\alpha \neq \beta} Q_{\alpha \beta}^4\right\}, \end{align} $$
其中Qαβ = (βJ)2qαβ，令$\theta=\frac{T_f-T}{T_f}$并且考虑复本极限n → 0有：
$$ \beta f=\frac{1}{2} \int_0^1 \mathrm{~d} x\left\{|\theta| q^2(x)-\frac{1}{3} x q^3(x)-q(x) \int_0^x q^2(y) \mathrm{d} y+\frac{1}{6} q^4(x)\right\} $$
对q(x)求导：
$$ 2|\theta| q(x)-x q^2(x)-\int_0^x q^2(y) \mathrm{d} y-2 q(x) \int_x^1 q(y) \mathrm{d} y+\frac{2}{3} q^3(x)=0 $$
持续求微分得到结果： |θ|−xq(x) − ∫x1q(y)dy + q2(x) = 0 or q′(x) = 0
$$ q(x)=\frac{x}{2} \quad \text { or } \quad q^{\prime}(x)=0 $$
最后得到的解与|θ|有关： $$ \begin{align} & q(x)=\frac{x}{2} \quad\left(0 \leq x \leq x_1=2 q(1)\right) \\ & q(x)=q(1) \quad\left(x_1 \leq x \leq 1\right) \\ & q(1)=|\theta|+\mathcal{O}\left(\theta^2\right) \end{align} $$
当接近相变点时，q(x = 0)。
