The Hopfield Model
partition function:
$$ Z = \text{Tr}_{S}\exp{(-\beta H\{S\})} $$
energy function:
$$ H_{0} = -\frac{1}{2N}\sum_{\mu=1}^{p}\left( \sum_{i}S_{i}\xi_{i}^{\mu} \right)^{2} + \frac{p}{2} $$
Hopfield 网络的 energy function:
$$ H = -\frac{1}{2}\sum_{i,j}J_{ij}S_{i}S_{j} $$
Hebb’s rule:
$$ J_{ij} = \frac{1}{N}\sum_{\mu=1}^{p}\xi_{i}^{\mu}\xi_{j}^{\mu} $$
将 Hebb’s rule 代入 energy function:
$$ H = -\frac{1}{2}\sum_{i,j}\left(\frac{1}{N}\sum_{\mu=1}^{p}\xi_{i}^{\mu}\xi_{j}^{\mu}\right)S_{i}S_{j} = -\frac{1}{2N}\sum_{\mu=1}^{p}\left(\sum_{i}S_{i}\xi_{i}^{\mu}\right)^{2} $$
假设 $\xi_{i}^{\mu}$ 是独立同分布的随机变量, 取值为 $\pm 1$ 且概率相等. 那么定义随机变量
$$ X^{\mu} \equiv \sum_{i=1}^{N}S_{i}\xi_{i}^{\mu} $$
则
$$ (X^{\mu})^{2} = \sum_{ij}S_{i}S_{j}\xi_{i}^{\mu}\xi_{j}^{\mu} $$
期望
$$ \langle (X^{\mu})^{2}\rangle = \sum_{ij}S_{i}S_{j}\langle\xi_{i}^{\mu}\xi_{j}^{\mu}\rangle $$
- 当 $i\neq j$ 时, $\xi_{i}$ 和 $\xi_{j}$ 互相独立, 则 $\langle \xi_{i}^{\mu}\xi_{j}^{\mu}\rangle = \langle\xi_{i}^{\mu}\rangle\langle\xi_{j}^{\mu}\rangle = 0$;
- 当 $i = j$ 时, $\langle (\xi_{i}^{\mu})^{2}\rangle = 1$
因此合并得到 $\langle\xi_{i}^{\mu}\xi_{j}^{\mu}\rangle = \delta_{ij}$. 于是
$$ \langle (X^{\mu})^{2}\rangle = \sum_{ij}S_{i}S_{j}\delta_{ij} = N $$
所以
$$ \langle H\rangle = -\frac{1}{2N}\sum_{\mu=1}^{p}\langle (X^{\mu})^{2}\rangle = -\frac{1}{2N}\cdot p\cdot N = -\frac{p}{2} $$
在后面加上相抵的常数项 $\frac{p}{2}$ 使得 $\langle H_{0}\rangle = 0$.
定义 $\alpha = \frac{p}{N}$.
Mean Field Theory for $\alpha=0$
设外场 $h^{\mu}\xi_{i}^{\mu}$, 则
$$ H = H_{0} - \sum_{\mu}h^{\mu}\sum_{i}\xi_{i}^{\mu}S_{i} $$
partition function:
$$ Z = e^{-\beta p/2}\text{Tr}_{S}\exp{\left[ \frac{\beta}{2N}\sum_{\mu}{{\color{red}{\left(\sum_{i}S_{i}\xi_{i}^{\mu}\right)^{2}}}} + \beta\sum_{\mu}h^{\mu}\sum_{i}\xi_{i}^{\mu}S_{i} \right]} $$
由于存在二次项, 所以无法拆分为 $N$ 个 exp 乘积.
技巧:
$$ \int_{-\infty}^{\infty}e^{-ax^{2}\pm bx}\mathrm{d}x = \sqrt{\frac{\pi}{a}}e^{\frac{b^{2}}{4a}} $$
反向使用该公式, 将 $e^{x^{2}}$ 化为积分式:
$$ \frac{\beta}{2N}\left(\sum_{i}S_{i}\xi_{i}^{\mu}\right)^{2} = \frac{b^{2}}{4a} \Rightarrow \begin{cases} b &= \beta \sum_{i}S_{i}\xi_{i}^{\mu}\\ a &= \frac{\beta N}{2} \end{cases} $$
引入辅助积分变量 $m^{\mu}$, 即有
$$ \exp{\left[ \frac{\beta}{2N}\left(\sum_{i}S_{i}\xi_{i}^{\mu}\right)^{2} \right]} = \left(\frac{\beta N}{2\pi}\right)^{\frac{1}{2}}\int\mathrm{d}m^{\mu}\exp{\left[ -\frac{1}{2}\beta N(m^{\mu})^{2} + \beta m^{\mu}\sum_{i}\xi_{i}^{\mu}S_{i} \right]} $$
原指数中的一次项即可合并入各 $\mu$ 的指数积分中:
$$ Z = e^{-\beta p/2}\left(\frac{\beta N}{2\pi}\right)^{\frac{p}{2}} \times \text{Tr}_{S}\prod_{\mu}\int\mathrm{d}m^{\mu}\exp{\left[ -\frac{1}{2}\beta N(m^{\mu})^{2} + \beta (m^{\mu} + h^{\mu})\sum_{i}\xi_{i}^{\mu}S_{i} \right]} $$
将 $m^{\mu}, h^{\mu}, \xi_{i}^{\mu}$ 合并写作含 $p$ 个分量的矢量形式 $\vec{m},\vec{h},\vec{\xi}_{i}$.
注意: 在这里 $\xi_{i}^{\mu}$ 是指模式 $\mu$ 下, 在神经元/节点 $i$ 处的取值. 而 $\vec{x}_{i}$ 是指在神经元/节点 $i$ 处的所有模式取值构成的向量. 因为认为这些模式彼此独立, 因此在 $\mu$ 尺度上可以认为 $\xi_{i}^{\mu}$ 是 $\{\pm 1\}$ 随机变量, 即 $\vec{\xi}$ 一共有 $2^{p}$ 种可能取值.
由于恒等式
$$ \begin{aligned} \prod_{\mu}\int\mathrm{d}m^{\mu} &\equiv \int\mathrm{d}m^{1}\cdots\mathrm{d}m^{p} \equiv \int\mathrm{d}\vec{m}\\ \sum_{\mu}(m^{\mu})^{2} &\equiv \vec{m}^{2}\\ \sum_{\mu}(m^{\mu} + h^{\mu})\xi_{i}^{\mu} &\equiv (\vec{m} + \vec{h})\cdot\vec{\xi}_{i}\\ \text{Tr}_{S}(*) &\equiv \sum_{S_{1}=\pm 1}\sum_{S_{2}=\pm 1}\cdots\sum_{S_{N}=\pm 1}(*) \end{aligned} $$
具体推导过程:
$$ \begin{aligned} Z &= e^{-\frac{\beta p}{2}} \left(\frac{\beta N}{2\pi}\right)^{\frac{p}{2}} \text{Tr}_{S}\prod_{\mu}\int\mathrm{d}m^{\mu}\exp{\left[ -\frac{1}{2}\beta N(m^{\mu})^{2} + \beta (m^{\mu} + h^{\mu})\sum_{i}\xi_{i}^{\mu}S_{i} \right]}\\ &= e^{-\frac{\beta p}{2}} \left(\frac{\beta N}{2\pi}\right)^{\frac{p}{2}} \text{Tr}_{S}\int\prod_{\mu}\mathrm{d}m^{\mu}\exp{\left\{\sum_{\mu}\left[ -\frac{1}{2}\beta N(m^{\mu})^{2} + \beta (m^{\mu} + h^{\mu})\sum_{i}\xi_{i}^{\mu}S_{i} \right]\right\}}\\ &= e^{-\frac{\beta p}{2}} \left(\frac{\beta N}{2\pi}\right)^{\frac{p}{2}} \text{Tr}_{S}\int\mathrm{d}\vec{m}\exp{\left[ -\frac{1}{2}\beta N \sum_{\mu}(m^{\mu})^{2} + \beta \sum_{i\mu}(m^{\mu}+h^{\mu})\xi_{i}^{\mu}S_{i} \right]}\\ &= e^{-\frac{\beta p}{2}} \left(\frac{\beta N}{2\pi}\right)^{\frac{p}{2}} \text{Tr}_{S}\int\mathrm{d}\vec{m}\exp{\left[ -\frac{1}{2}\beta N \vec{m}^{2} + \beta \sum_{i}(\vec{m} + \vec{h})\cdot\vec{\xi}_{i}S_{i} \right]}\\ &= e^{-\frac{\beta p}{2}} \left(\frac{\beta N}{2\pi}\right)^{\frac{p}{2}} \int\mathrm{d}\vec{m}e^{-\frac{1}{2}\beta N \vec{m}^{2}}\text{Tr}_{S}e^{\beta \sum_{i}(\vec{m} + \vec{h})\cdot\vec{\xi}_{i}S_{i}}\\ &= e^{-\frac{\beta p}{2}} \left(\frac{\beta N}{2\pi}\right)^{\frac{p}{2}} \int\mathrm{d}\vec{m}e^{-\frac{1}{2}\beta N \vec{m}^{2}}\prod_{i}\text{Tr}_{S_{i}}e^{\beta (\vec{m} + \vec{h})\cdot\vec{\xi}_{i}S_{i}}\\ &= e^{-\frac{\beta p}{2}} \left(\frac{\beta N}{2\pi}\right)^{\frac{p}{2}} \int\mathrm{d}\vec{m}e^{-\frac{1}{2}\beta N \vec{m}^{2}}\prod_{i}\{2\cosh{[\beta(\vec{m} + \vec{h})\cdot\vec{\xi}_{i}]}\}\\ &= e^{-\frac{\beta p}{2}} \left(\frac{\beta N}{2\pi}\right)^{\frac{p}{2}} \int\mathrm{d}\vec{m}e^{-\frac{1}{2}\beta N \vec{m}^{2}} \prod_{i}e^{\ln\{2\cosh{[\beta(\vec{m} + \vec{h})\cdot\vec{\xi}_{i}]}\}}\\ &= e^{-\frac{\beta p}{2}} \left(\frac{\beta N}{2\pi}\right)^{\frac{p}{2}} \int\mathrm{d}\vec{m}e^{-\frac{1}{2}\beta N \vec{m}^{2}} e^{\sum_{i}\ln\{2\cosh{[\beta(\vec{m} + \vec{h})\cdot\vec{\xi}_{i}]}\}}\\ &= e^{-\beta N {\color{red}{\frac{p}{2N}}}} \left(\frac{\beta N}{2\pi}\right)^{\frac{p}{2}} \int\mathrm{d}\vec{m} e^{-\beta N {\color{red}{\frac{\vec{m}^{2}}{2}}}} e^{-\beta N {\color{red}{\left(-\frac{1}{\beta N}\sum_{i}\ln{(2\cosh{[\beta(\vec{m} + \vec{h})\cdot\vec{\xi}_{i}]})}\right)}}}\\ &= \left(\frac{\beta N}{2\pi}\right)^{p/2}\int\mathrm{d}\vec{m}e^{-\beta N{\color{red}{f(\beta,\vec{m})}}} \end{aligned} $$
其中 $$ f(\beta,\vec{m}) = \frac{1}{2}\alpha + \frac{1}{2}\vec{m}^{2} - \frac{1}{\beta N}\sum_{i}\ln{(2\cosh{[\beta(\vec{m} + \vec{h})\cdot\vec{\xi}_{i}]})} $$
$$ Z = \left(\frac{\beta N}{2\pi}\right)^{p/2}\int\mathrm{d}\vec{m}e^{-\beta Nf(\beta,\vec{m})} $$
在热力学极限($N\to\infty$)下可对该积分进行估算. 提取其一维形式积分
$$ \begin{aligned} I &= \sqrt{N}\int\mathrm{d}x e^{-Ng(x)}\\ &= \sqrt{N}\int\mathrm{d}x \exp\left\{-N\left[ g(x_{0}) + \frac{1}{2}g^{\prime\prime}(x_{0})(x - x_{0})^{2} + \cdots \right]\right\}, \quad \begin{cases}g^{\prime}(x_{0}) &= 0\\ g^{\prime\prime}(x_{0}) &> 0\end{cases}\\ &\approx \sqrt{N}e^{-Ng(x_{0})}\int\mathrm{d}x \exp\left\{-\frac{Ng^{\prime\prime}(x_{0})}{2}(x - x_{0})^{2}\right\}\\ &= \sqrt{N}e^{-Ng(x_{0})}\sqrt{\frac{2\pi}{Ng^{\prime\prime}(x_{0})}}\\ &= e^{-Ng(x_{0})}\sqrt{\frac{2\pi}{g^{\prime\prime}(x_{0})}} \end{aligned} $$
可发现
$$ \begin{aligned} -\frac{1}{N}\ln{I} &= g(x_{0}) + \frac{1}{2N}\left[\ln{g^{\prime\prime}(x_{0})}-\ln{2\pi}\right]\\ &\overset{N\to\infty}{\longrightarrow} g(x_{0}) \end{aligned} $$
即知道了 $g(x)$ 最小值, 即可估算积分值 $I$. 回到原 $p$ 维积分, 同样的处理技巧, 即 $g(\cdot) = \beta f(\beta, \vec{m})$:
$$ -\frac{1}{N}\ln{Z} = \beta\underset{\vec{m}}{\min}f(\beta,\vec{m}) $$
而自由能
$$ F = -\frac{1}{\beta}\ln{Z} = N\underset{\vec{m}}{\min}f(\beta,\vec{m})\Rightarrow \frac{F}{N} = \underset{\vec{m}}{\min}f(\beta,\vec{m}) $$
接下来寻找 $f(\beta,\vec{m})$ 的最小值.
$$ f(\beta,\vec{m}) = \frac{1}{2}\alpha + \frac{1}{2}\vec{m}^{2} - \frac{1}{\beta}{\color{red}{\frac{1}{N}\sum_{i}}}\ln{(2\cosh{[\beta(\vec{m} + \vec{h})\cdot\vec{\xi}_{i}]})} $$
$$ \begin{aligned} \frac{\partial f}{\partial m^{\mu}} &= m^{\mu} - \frac{1}{N}\sum_{i}\xi_{i}^{\mu}\tanh{[\beta(\vec{m} + \vec{h})\cdot\vec{\xi}_{i}]} = 0\\ \Rightarrow m^{\mu} &= {\color{red}{\frac{1}{N}\sum_{i}}}\xi_{i}^{\mu}\tanh{[\beta(\vec{m} + \vec{h})\cdot\vec{\xi}_{i}]} = {\color{red}{\langle}} \xi^{\mu}\tanh{[\beta(\vec{m} + \vec{h})\cdot\vec{\xi}]}{\color{red}{\rangle}} \end{aligned} $$
提示:
$$ (\vec{m} + \vec{h})\cdot\vec{\xi} = \sum_{\nu}(m^{\nu} + h^{\nu})\xi^{\nu} $$
所以在取平均时, 求和符合内隐含固定某个 $\mu$ 的假设, 但是 $\tanh$ 函数里计算时的的确确是对所有 $\mu$ 都进行了计算.
也许更精准的写法是
$$ m^{\mu} = E_{\{\xi^{\nu}\}_{\nu=1}^{p}}\left\{\xi^{\mu}\tanh{\left[\beta\sum_{\nu}(m^{\nu}+h^{\nu})\xi^{\nu}\right]}\right\} $$
同理, 求和写作期望形式:
$$ f \overset{\alpha\rightarrow 0}{=} \frac{1}{2}\vec{m}^{2} -\frac{1}{\beta}\bigg\langle \ln\{2\cosh{[\beta(\vec{m} + \vec{h})\cdot\vec{\xi}]}\}\bigg\rangle $$
注意 $m^{\mu}$ 是作为辅助变量引入的, 它实际上可具有实际的物理含义.
$$ \begin{aligned} \frac{\partial F}{\partial h^{\mu}} &= \frac{\partial }{\partial h^{\mu}}\left[-\frac{1}{\beta}\ln{Z}\right] = -\frac{1}{\beta}\frac{1}{Z}\frac{\partial Z}{\partial h^{\mu}} = -\frac{1}{\beta Z}\frac{\partial}{\partial h^{\mu}} [\text{Tr}_{S}e^{-\beta H}]\\ &= -\frac{1}{\beta Z} \text{Tr}_{S}\left(-\beta \frac{\partial H}{\partial h^{\mu}}\right) e^{-\beta H}\\ &= \frac{e^{-\beta H}}{Z} \left(-\sum_{i}\xi_{i}^{\mu}S_{i}\right)\\ &= -\sum_{i} \xi_{i}^{\mu}\langle S_{i}\rangle \end{aligned} $$
而另一方面,
$$ f \overset{\alpha\rightarrow 0}{=} \frac{1}{2}\vec{m}^{2} -\frac{1}{\beta}\bigg\langle \ln\{2\cosh{[\beta(\vec{m} + \vec{h})\cdot\vec{\xi}]}\}\bigg\rangle $$
$$ \begin{aligned} \frac{\partial f}{\partial h^{\mu}} = -\frac{1}{\beta} \bigg\langle \beta \xi^{\mu}\tanh{[\beta(\vec{m} + \vec{h})\cdot\vec{\xi}]} \bigg\rangle = -\bigg\langle \xi^{\mu}\tanh{[\beta(\vec{m} + \vec{h})\cdot\vec{\xi}]} \bigg\rangle = -m^{\mu} \end{aligned} $$
而我们又有 $F/N = \underset{\vec{m}}{\min}f(\beta,\vec{m})$, 因此
$$ \begin{aligned} \frac{\partial F}{\partial h^{\mu}} &= -\sum_{i}\xi_{i}^{\mu}\langle S_{i}\rangle\\ =N\frac{\partial f}{\partial h^{\mu}} &= -Nm^{\mu} \end{aligned} $$
即
$$ m^{\mu} = \frac{1}{N}\sum_{i}\xi_{i}^{\mu}\langle S_{i}\rangle $$
所以使得自由能最小的 $m^{\mu}$ 就是网络对于模式 $\mu$ 的平均 overlap.
在 $h^{\mu} = 0$ 时, 自洽方程(self-consistency equation) 形式为
$$ m^{\mu} = \bigg\langle \xi^{\mu}\tanh{\left(\beta \vec{m} \cdot \vec{\xi}\right)} \bigg\rangle $$
- memory state: 只和某个特定的 $\mu$ 有 correlation 的 $\vec{m}$ 解. 如 $\vec{m} = (m,0,0,\cdots)$, 则自洽方程进一步化为
$$ m^{\mu} = \bigg\langle \xi^{\mu}\tanh{(\beta m \xi^{1})}\bigg\rangle = \langle \xi^{\mu}\xi^{1}\rangle \tanh{(\beta m)} = \delta_{\mu 1}\tanh{(\beta m)} $$
当 $\mu = 1$ 时, 有
$$ m = \tanh{(\beta m)} $$
-
spurious state.
-
- 最简单的是 mixture state, 即
$$ \vec{m} = (\underbrace{m,m,\cdots,m}_{n},\underbrace{0,0,\cdots,0}_{p-n}) $$
一共有 $C_{p}^{n}$ 种方式. 那么自洽方程化为
$$ \begin{aligned} m^{\mu} &= \bigg\langle \xi^{\mu}\tanh{\left(\beta m\sum_{\nu=1}^{n}\xi^{\nu}\right)}\bigg\rangle\\ m &= \frac{1}{n}\bigg\langle z\tanh{(\beta mz)}\bigg\rangle, \quad z = \sum_{\mu=1}^{n}\xi^{\mu} \end{aligned} $$
除了要求 self-consistency, 还需要该解使得自由能 $F$ 或者 $f$ 最小. 这要求
$$ A_{\mu\nu} = \frac{\partial^{2}f}{\partial m^{\mu}\partial m^{\nu}} $$
的特征值均为正.
在驻点附近,
$$ f(\vec{m}^{*}+\delta\vec{m}) = f(\vec{m}^{*}) + \frac{1}{2}\sum_{\mu\nu}\delta m^{\mu} A_{\mu\nu}\delta m^{\nu} + \mathcal{O}(\delta m^{3}),\quad A_{\mu\nu} = \frac{\partial^{2}f}{\partial m^{\mu}\partial m^{\nu}}\bigg|_{\vec{m}=\vec{m}^{*}} $$
- 若 $A_{\mu\nu}$ 特征值均为正, 即为二次型正定, 驻点 $\vec{m}^{*}$ 为极小值点;
- 若 $A_{\mu\nu}$ 存在负特征值, 则存在方向使得 $f\downarrow$, 驻点 $\vec{m}^{*}$ 不稳定;
结论: 只有 $n$ 为奇数时才有可能正定, 且温度 $T = \frac{1}{\beta}$ 需要低于临界温度(critical temperature) $T_{c}$.
-
- asymmetric mixture state:
$$ m = \left(\frac{1}{2},\frac{1}{2},\frac{1}{4},\frac{1}{4},\frac{1}{4},0,0,0\cdots\right) $$
Mean Field Theory for $\alpha\neq 0$
计算 $\langle\log{Z}\rangle$. replica method:
$$ \ln{Z} = \lim_{n\to 0}\frac{Z^{n}-1}{n} $$
这样就可以通过计算 $\langle Z^{n}\rangle$ 来得到 $\langle\ln{Z}\rangle$. 先取 $n$ 为整数, 最后再取 $n\to 0$.
这样 $n$ 个 partition function 复制品即被称为 replicas, 通过 reolica index (下公式)
$$ Z^{n} = \prod_{\rho=1}^{n}Z^{(\rho)} = \text{Tr}_{S^{1}}\text{Tr}_{S^{2}}\cdots\text{Tr}_{S^{n}}\exp{\left[ -\beta (E\{S_{i}^{1}\}+\cdots+E\{S_{i}^{n}\}) \right]} $$
和前面 Gaussian integral trick 类似, 只不过多了 $n$ 次(取 $h^{\mu}=0$.):
$$ \begin{aligned} \langle Z^{n}\rangle &= e^{-\beta pn/2}\left\langle \text{Tr}_{S}\prod_{\mu=1}^{p}\prod_{\rho=1}^{n}\left\{\int\mathrm{d}m_{\rho}^{\mu}\left(\frac{\beta N}{2\pi}\right)^{\frac{1}{2}}\exp{\left[ {\color{green}{-\frac{1}{2}\beta N (m_{\rho}^{\mu})^{2}}} + \beta m_{\rho}^{\mu}\sum_{i}\xi_{i}^{\mu}S_{i}^{\rho} \right]} \right\}\right\rangle\\ \overset{\prod (ab) = \prod a\cdot \prod b}{=}& e^{-\beta pn/2} \text{Tr}_{S}\prod_{\mu=1}^{p}\int\left\{\left[\prod_{\rho=1}^{n}\mathrm{d}m_{\rho}^{\mu}\left(\frac{\beta N}{2\pi}\right)^{\frac{1}{2}}\right]{\color{green}{\exp{\left[ -\frac{\beta N}{2}\sum_{\rho=1}^{n}(m_{\rho}^{\mu})^{2} \right]}}}{\color{red}{\left\langle \prod_{\rho=1}^{n}\exp{\left( \beta m_{\rho}^{\mu}\sum_{i=1}^{N}\xi_{i}^{\mu}S_{i}^{\rho} \right)} \right\rangle}}\right\}\\ \overset{\mu>s}{\Rightarrow}& e^{-\beta pn/2}\text{Tr}_{S}\int\left[\prod_{\rho=1}^{n}\mathrm{d}m_{\rho}^{\mu}\left(\frac{\beta N}{2\pi}\right)^{\frac{1}{2}}\right]E \end{aligned} $$
每个 replica 有自己的自旋 $S_{i}^{\rho}$, 但是所有 replica 使用的是同一组 patterns $\xi_{i}^{\mu}$.
condensed patterns: 在 $p$ 个存储的模式中, 只有 $s$ 个(且 $s$ 与 $N$ 无关) 会与网络有较大的 overlap.
recall:
$$ m^{\mu} = \frac{1}{N}\sum_{i}\xi_{i}^{\mu}\langle S_{i}\rangle $$
对这些模式根据 overlap 大小排序, 也可得出上述定义的等效表述:
- 对 $\mu\leq s$(少数), $m^{\mu} = \mathcal{O}(1)$;
- 对 $\mu > s$(绝大多数), $m^{\mu}\sim\mathcal{O}(N^{-1/2})\ll 1$.
先考虑 $\mu > s$ 部分, 对于 ${\color{red}{\langle\cdots\rangle}}$ 部分:
我们可以引入矩阵 $\widetilde{\Lambda}_{\rho\sigma}$, 使表达式简化:
$$ E = \exp{\left( -\frac{\beta N}{2}\sum_{\rho\sigma}\widetilde{\Lambda}_{\rho\sigma}m_{\rho}^{\mu}m_{\sigma}^{\mu} \right)} $$
其中
$$ \widetilde{\Lambda}_{\rho\sigma} \equiv \delta_{\rho\sigma} - \frac{\beta}{N}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma} $$
注意
$$ \begin{aligned} \frac{\beta^{2}}{2}\sum_{\rho\sigma}m_{\rho}^{\mu}m_{\sigma}^{\mu}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma} &= -\frac{\beta N}{2}\sum_{\rho\sigma}\left(-\frac{\beta}{N}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma}\right)m_{\rho}^{\mu}m_{\sigma}^{\mu} \end{aligned} $$
在总表达式中还有 $\begin{aligned}{\color{green}{-\frac{\beta N}{2}(m_{\rho}^{\mu})^{2}}}\end{aligned}$, 将其改写为
$$ -\frac{\beta N}{2}\sum_{\rho}(m_{\rho}^{\mu})^{2} = -\frac{\beta N}{2}\sum_{\rho\sigma}\delta_{\rho\sigma}m_{\rho}^{\mu}m_{\sigma}^{\mu} $$
于是合并得到
$$ -\frac{\beta N}{2}\sum_{\rho\sigma}\left[ \delta_{\rho\sigma} - \frac{\beta}{N}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma} \right]m_{\rho}^{\mu}m_{\sigma}^{\mu} $$
那么
$$ \begin{aligned} \int\left[ \prod_{\rho=1}^{n}\mathrm{d}m_{\rho}^{\mu}\left( \frac{\beta N}{2\pi} \right)^{\frac{1}{2}} \right]\exp{\left( -\frac{\beta N}{2}\sum_{\rho\sigma}\widetilde{\Lambda}_{\rho\sigma}m_{\rho}^{\mu}m_{\sigma}^{\mu} \right)} = \left(\frac{\beta N}{2\pi}\right)^{\frac{n}{2}}\sqrt{\frac{\pi^{n}}{\det{\left(\frac{1}{2}\beta N\widetilde{\Lambda}\right)}}} = \left(\det{\widetilde{\Lambda}}\right)^{-\frac{1}{2}} \end{aligned} $$
因为 condensed patterns 数量少, 不妨假定 $s\ll p$, 那么 $\mu>s$ 的个数为 $p-s\sim p$, 即以上积分结果可近似为相乘 $p$ 次, 即
$$ \begin{aligned} \left(\det{\widetilde{\Lambda}}\right)^{-\frac{p}{2}} &= \exp{\left[-\frac{p}{2}\ln{(\det{\widetilde{\Lambda}})}\right]} = \exp{\left[ -\frac{p}{2}\ln{\left(\prod_{\rho=1}^{n}\widetilde{\lambda}_{\rho}\right)} \right]} = \exp{\left[ -\frac{\alpha N}{2}\sum_{\rho=1}^{n}\ln{\widetilde{\lambda}_{\rho}} \right]} \end{aligned} $$
接下来需要进行 $\text{Tr}_{S}$ 的计算, 而 $S_{i}^{\rho}$ 相关的信息被合并入 $\widetilde{\Lambda}$ 的特征值 $\widetilde{\lambda}_{\rho}$ 中, 因此需要另寻方法.
定义广义 $\widetilde{\Lambda}_{\rho\sigma}$:
$$ \Lambda_{\rho\sigma} \equiv (1-\beta)\delta_{\rho\sigma} - \beta q_{\rho\sigma} $$
若 $\begin{aligned}q_{\rho\sigma} = \frac{1-\delta_{\rho\sigma}}{N}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma}\end{aligned}$, 则 $\Lambda_{\rho\sigma} = \widetilde{\Lambda}_{\rho\sigma}$.
那么任意对 $\widetilde{\lambda}_{\rho}$ 的函数 $G\{\widetilde{\lambda}_{\rho}\}$ 都可转写为形式
$$ G\{\widetilde{\lambda}_{\rho}\} = \int \left[ \prod_{(\rho\sigma)}\mathrm{d}q_{\rho\sigma}{\color{green}{\delta\left( q_{\rho\sigma} - \frac{1}{N}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma} \right)}} \right]G\{\lambda_{\rho}\} $$
$(\rho\sigma)$ 代表对所有的 $\rho-\sigma$ 对, 总共有 $n(n-1)/2$ 个(因为 $q_{\rho\sigma}=q_{\sigma\rho}$ 且 $q_{\rho\rho} = 0$).
使用该公式, 即可将原依赖于 $\widetilde{\lambda}_{\rho}$ 的表达式 $\begin{aligned}\left(\det{\widetilde{\Lambda}}\right)^{-\frac{p}{2}}\end{aligned}$ 改写为依赖于 $\lambda_{\rho}$ 的形式.
$\delta(x)$ 函数拥有性质
$$ \int \mathrm{d}q \delta[q - g(S)]F(q) = F[g(S)] $$
特别地, 定义 $\begin{aligned}g_{\rho\sigma}(S) = \frac{1}{N}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma}\end{aligned}$, 则上式改写为多维形式
$$ \int \prod_{(\rho\sigma)}\mathrm{d}q_{\rho\sigma}\delta[q_{\rho\sigma}-g_{\rho\sigma}(S)]F(\{q_{\rho\sigma}\}) = F(\{g_{\rho\sigma}(S)\}) $$
$\Lambda_{\rho\sigma} \equiv (1-\beta)\delta_{\rho\sigma} - \beta q_{\rho\sigma}$
即存在函数关系 $\widetilde{\Lambda}_{\rho\sigma} = \Lambda_{\rho\sigma}(q_{\rho\sigma})$. 那么
$$ G\{\widetilde{\lambda}_{\rho}\} = G\{\lambda_{\rho}(q)\} $$
代回即有
$$ G\{\widetilde{\lambda}_{\rho}\} = G\{\lambda_{\rho}(\frac{1}{N}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma})\} = \int\prod_{(\rho\sigma)}\mathrm{d}q_{\rho\sigma}\delta\left[ q_{\rho\sigma} - \frac{1}{N}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma} \right]G\{\lambda_{\rho}\} $$
对于 $\delta(x)$, 存在其傅里叶变换
$$ \delta(x) = \int_{-i\infty}^{+i\infty}\frac{\mathrm{d}\widetilde{r}}{2\pi i}e^{-\widetilde{r}x} $$
$$ {\color{green}{\delta\left( q_{\rho\sigma} - \frac{1}{N}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma} \right)}} = \int \frac{\mathrm{d}\widetilde{r}_{\rho\sigma}}{2\pi i}\exp{\left[ -\widetilde{r}_{\rho\sigma}\left( q_{\rho\sigma} - \frac{1}{N}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma} \right) \right]} $$
代回原 $G$ 表达式:
$$ \begin{aligned} G\{\widetilde{\lambda}_{\rho}\} =& \int \left[ \prod_{(\rho\sigma)}\mathrm{d}q_{\rho\sigma}{\color{green}{\delta\left( q_{\rho\sigma} - \frac{1}{N}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma} \right)}} \right]G\{\lambda_{\rho}\}\\ =& \int\left\{ \prod_{(\rho\sigma)}\mathrm{d}q_{\rho\sigma}\int\frac{\mathrm{d}\widetilde{r}_{\rho\sigma}}{2\pi i} \exp{\left[ -\widetilde{r}_{\rho\sigma}\left( q_{\rho\sigma} - \frac{1}{N}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma} \right) \right]} \right\}G\{\lambda_{\rho}\}\\ \propto& \int\left\{ \prod_{\rho\sigma}\mathrm{d}q_{\rho\sigma}\mathrm{d}\widetilde{r}_{\rho\sigma}\exp{\left[ -\widetilde{r}_{\rho\sigma}q_{\rho\sigma} + \frac{\widetilde{r}_{\rho\sigma}}{N}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma} \right]} \right\}G\{\lambda_{\rho}\}\\ \overset{\widetilde{r}_{\rho\sigma}\equiv N\alpha\beta^{2}r_{\rho\sigma}}{\propto}& \int\left[ \prod_{(\rho\sigma)}\mathrm{d}q_{\rho\sigma}\mathrm{d}r_{\rho\sigma}\exp{\left( -N\alpha\beta^{2}r_{\rho\sigma}q_{\rho\sigma} + \alpha\beta^{2}r_{\rho\sigma}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma} \right)} \right]G\{\lambda_{\rho}\} \end{aligned} $$
$$ \begin{aligned} G\{\widetilde{\lambda}_{\rho}\} \propto& \int\left[ \prod_{(\rho\sigma)}\mathrm{d}q_{\rho\sigma}\mathrm{d}r_{\rho\sigma}\exp{\left( -N\alpha\beta^{2}r_{\rho\sigma}q_{\rho\sigma} + \alpha\beta^{2}r_{\rho\sigma}\sum_{i}S_{i}^{\rho}S_{i}^{\sigma} \right)} \right]G\{\lambda_{\rho}\}\\ =& \int\left[ \prod_{(\rho\sigma)}\mathrm{d}q_{\rho\sigma}\mathrm{d}r_{\rho\sigma} \exp{\left( -N\alpha\beta^{2}r_{\rho\sigma}q_{\rho\sigma} \right)} \exp{\left( {\color{orange}{\alpha\beta^{2}\sum_{i}r_{\rho\sigma}S_{i}^{\rho}S_{i}^{\sigma}}} \right)} \right]G\{\lambda_{\rho}\}\\ =& \int\left\{ \left[\prod_{(\rho\sigma)}\mathrm{d}q_{\rho\sigma}\mathrm{d}r_{\rho\sigma}\right] \exp{\left( {\color{blue}{-{\boxed{\frac{1}{2}}}N\alpha\beta^{2}\sum_{\rho\sigma}r_{\rho\sigma}q_{\rho\sigma}}} \right)} \exp{\left( {\color{orange}{{\boxed{\frac{1}{2}}}\alpha\beta^{2}\sum_{i\rho\sigma}r_{\rho\sigma}S_{i}^{\rho}S_{i}^{\sigma}}} \right)} \right\}{\color{purple}{G\{\lambda_{\rho}\}}} \end{aligned} $$
${\color{orange}{\frac{1}{2}\cdots}}$ 中的 $\frac{1}{2}$ 是因为 $\rho\sigma$ 对应矩阵元素重复计算了一次. 那么 $\sum_{\rho<\sigma} = \frac{1}{2}\sum_{\rho\neq\sigma}$.
令 $\begin{aligned}G\{\widetilde{\lambda}_{\rho}\} = \exp{\left[-\frac{1}{2}\alpha N\sum_{\rho=1}^{n}\ln{\widetilde{\lambda}_{\rho}}\right]}\end{aligned}$, 则 $\langle Z^{n}\rangle$ 可写作
$$ \begin{aligned} \langle Z^{n}\rangle &= e^{-\beta pn/2}\left\langle \text{Tr}_{S}\prod_{\mu=1}^{p}\prod_{\rho=1}^{n}\left\{\int\mathrm{d}m_{\rho}^{\mu}\left(\frac{\beta N}{2\pi}\right)^{\frac{1}{2}}\exp{\left[ {\color{green}{-\frac{1}{2}\beta N (m_{\rho}^{\mu})^{2}}} + \beta m_{\rho}^{\mu}\sum_{i}\xi_{i}^{\mu}S_{i}^{\rho} \right]} \right\}\right\rangle\\ \overset{\prod (ab) = \prod a\cdot \prod b}{=}& e^{-\beta pn/2} \text{Tr}_{S}\prod_{\mu=1}^{p}\int\left\{\left[\prod_{\rho=1}^{n}\mathrm{d}m_{\rho}^{\mu}\left(\frac{\beta N}{2\pi}\right)^{\frac{1}{2}}\right]{\color{green}{\exp{\left[ -\frac{\beta N}{2}\sum_{\rho=1}^{n}(m_{\rho}^{\mu})^{2} \right]}}}{\color{red}{\left\langle \prod_{\rho=1}^{n}\exp{\left( \beta m_{\rho}^{\mu}\sum_{i=1}^{N}\xi_{i}^{\mu}S_{i}^{\rho} \right)} \right\rangle}}\right\}\\ \propto& e^{-\beta pn/2}\int \left[\prod_{\mu\rho}\mathrm{d}m_{\rho}^{\mu}\left(\frac{\beta N}{2\pi}\right)^{\frac{1}{2}}\right]\\ \times& \left[ \prod_{(\rho\sigma)}\mathrm{d}q_{\rho\sigma}\mathrm{d}r_{\rho\sigma} \right] \exp{\left[ {\color{green}{-\frac{\beta N}{2}\sum_{\mu\rho}(m_{\rho}^{\mu})^{2}}} {\color{purple}{- \frac{\alpha N}{2}\sum_{\rho}\ln{\lambda_{\rho}}}} {\color{blue}{- \frac{1}{2}N\alpha\beta^{2}\sum_{\rho\sigma}r_{\rho\sigma}q_{\rho\sigma}}} \right]}\\ \times& \left\langle \text{Tr}_{S}\exp{\left( \beta\sum_{\mu\rho}m_{\rho}^{\mu}\sum_{i}\xi_{i}^{\mu}S_{i}^{\rho} + {\color{orange}{\frac{1}{2}}\alpha\beta^{2}\sum_{i\rho\sigma}r_{\rho\sigma}S_{i}^{\rho}S_{i}^{\sigma}} \right)} \right\rangle \end{aligned} $$
其中 $\mu = \{1,2,\cdots,s\}$. 这也可以解释为什么某些标记颜色的部分在经过处理之后仍然会保留下来: 因为与 condensed patterns 相关.
再次提示 $\prod(ab) = \prod a\cdot \prod b$, 这一点针对 $\mathrm{d}x$ 和 $\exp$ 都是成立的. 所以最外层的 $\prod$ 向内移动时需要将其分配至各乘子.
接下来需要去除 $i$ 指标.
$$ \begin{aligned} X \equiv& \text{Tr}_{S}\exp{\left( \sum_{i}F\{\xi_{i},S_{i}\} \right)} \overset{\prod e^{A}=e^{\sum A}}{=} \prod_{i}\text{Tr}_{S_{i}}\exp{F\{\xi_{i},S_{i}\}}\\ =& \exp{\left\{ \sum_{i}\ln{\left[ \text{Tr}_{S_{i}}e^{F\{\xi_{i},S_{i}\}} \right]} \right\}},\quad \begin{cases}\mu &= 1,2,\cdots,s; \\ \rho &= 1,2,\cdots,n\end{cases}\\ \overset{N\gg 2^{s}}{=}& \exp{\left\{ N\left\langle \ln{\left[ \text{Tr}_{S}e^{F\{\xi_{i},S_{i}\}} \right]}\right\rangle\right\}} \end{aligned} $$
那么要做的相当于在单个节点 $i$ 上进行 $\rho$ 和 $\mu$ 的平均. 那么 $\langle\cdots\rangle$ 求期望操作已经在内部通过 $\sum_{i}$ 自动完成了. 于是进一步化简:
$$ \begin{aligned} \langle Z^{n}\rangle &\propto e^{-\beta pn/2}\int\left[ \prod_{\mu\rho}\mathrm{d}m_{\rho}^{\mu}\left(\frac{\beta N}{2\pi}\right)^{\frac{1}{2}} \right]\left[ \prod_{(\rho\sigma)}\mathrm{d}q_{\rho\sigma}\mathrm{d}r_{\rho\sigma} \right]e^{-N\beta f\{m,q,r\}}\\ f\{m,q,r\} &= {\color{green}{\frac{1}{2}\sum_{\mu\rho}(m_{\rho}^{\mu})^{2}}} {\color{purple}{+ \frac{\alpha}{2\beta}\sum_{\rho}\ln{\lambda_{\rho}}}} {\color{blue}{+ \frac{1}{2}\alpha\beta\sum_{\rho\sigma}r_{\rho\sigma}q_{\rho\sigma}}}\\ &- \frac{1}{\beta}\left\langle \ln{\text{Tr}_{S}\exp{\left( \beta\sum_{\mu\rho}m_{\rho}^{\mu}\xi^{\mu}S^{\rho} {\color{orange}{+ \frac{1}{2}\alpha\beta^{2}\sum_{\rho\sigma}r_{\rho\sigma}S^{\rho}S^{\sigma}}} \right)}} \right\rangle \end{aligned} $$
那么每个节点的自由能
$$ \begin{aligned} \frac{F}{N} &= \frac{1}{N}\left(-\frac{1}{\beta}\langle \ln{Z}\rangle\right) = -\frac{1}{\beta N}\lim_{n\to 0}\frac{\langle Z^{n}\rangle - 1}{n} \\ \overset{\lim_{x\to 1}x-1\approx \ln{x}}{=}& -\frac{1}{\beta N}\lim_{n\to 0}\frac{\log{\langle Z^{n}\rangle}}{n} = \frac{\alpha}{2} + \lim_{n\to 0}\frac{1}{n}\min{f\{m,q,r\}} \end{aligned} $$
寻找 $f$ 的最小值, 即各变量的一阶导数为 $0$:
$$ \begin{aligned} \frac{\partial f}{\partial m_{\rho}^{\mu}} = \frac{\partial f}{\partial q_{\rho\sigma}} = \frac{\partial f}{\partial r_{\rho\sigma}} = 0 \end{aligned} $$
解得
$$ \begin{aligned} m_{\rho}^{\mu} = \frac{1}{N}\sum_{i}\xi_{i}^{\mu}\langle S_{i}^{\rho}\rangle,\quad q_{\rho\sigma} = \left\langle\frac{1}{N}\sum_{i}\langle S_{i}^{\rho}\rangle\langle S_{i}^{\sigma}\rangle\right\rangle,\quad r_{\rho\sigma} = \frac{1}{\alpha}\sum_{\mu>s}\langle m_{\rho}^{\mu}m_{\sigma}^{\mu}\rangle \end{aligned} $$
replica symmetry ansatz(对称假设): 鞍点处的序参量和 replica index 无关.
$$ \begin{aligned} m_{\rho}^{\mu} = m^{\mu},\quad q_{\rho\sigma} = q,\quad r_{\rho\sigma} = r \end{aligned} $$
它们也具有相应的物理意义: $m^{\mu}$ 是网络对 pattern $\mu$ 的 overlap(重合度); $q$ 是 mean squared magnetization(磁化强度的均方); $\alpha r$ 是 uncondensed patterns 的 overlap 均方.
利用该假设, 将序参量写作矢量形式
$$ \begin{aligned} f(\vec{m},q,r) =& {\color{green}{\frac{1}{2}n\vec{m}^{2}}} + {\color{purple}{\frac{\alpha}{2\beta}\sum_{\rho}\ln{\lambda_{\rho}}}} + {\color{blue}{\frac{1}{2}\alpha\beta n(n-1)rq}} + {\color{red}{\frac{1}{2}n\alpha\beta r}}\\ &- \frac{1}{\beta}\left\langle \ln{\text{Tr}_{S}\exp{\left[ \beta\vec{m}\cdot\vec{\xi}\sum_{\rho}S^{\rho} {\color{orange}{+ \frac{1}{2}\alpha\beta^{2}r\left(\sum_{\rho}S^{\rho}\right)^{2}}} \right]}} \right\rangle \end{aligned} $$
其中 ${\color{red}{\frac{1}{2}n\alpha\beta r}}$ 是为了消除 $\langle {\color{orange}{\cdots}} \rangle$ 中的 $\rho=\sigma$ 项.
接下来需要计算本征值 $\lambda_{\rho}$.
$$ \Lambda_{\rho\sigma} = \begin{cases} 1-\beta, & \rho = \sigma\\ -\beta q, & \rho \neq \sigma \end{cases} $$
这个矩阵的本征值为
$$ \begin{aligned} \lambda_{1} &= 1 - \beta - (n-1)\beta q\\ \lambda_{2} &= 1 - \beta(1 - q) \end{aligned} $$
其中 $\lambda_{1}$ 的简并度为 $1$, $\lambda_{2}$ 的简并度为 $n-1$. 那么
$$ \begin{aligned} \frac{1}{n}\sum_{\rho}\ln{\lambda_{\rho}} =& \frac{1}{n}\left\{1\cdot \ln{[1-\beta - (n-1)\beta q]} + (n-1)\cdot\ln{[1-\beta(1-q)]}\right\}\\ \overset{n\to 0}{=} & \ln{[1-\beta(1-q)]} - \frac{\beta q}{1-\beta(1-q)} \end{aligned} $$
对于 ${\color{orange}{\frac{1}{2}\alpha\beta^{2}r\cdots}}$ 部分, 再次使用 Gaussian integral trick:
$$ \begin{aligned} \exp{\left[\frac{1}{2}\left(\sqrt{\alpha r}\beta \sum_{\rho}S^{\rho}\right)^{2}\right]} = \int\frac{\mathrm{d}z}{\sqrt{2\pi}}\exp{\left( -\frac{1}{2}z^{2} + \sqrt{\alpha r}\beta z\sum_{\rho}S^{\rho} \right)} \end{aligned} $$
将其与 $\beta\vec{m}\cdot\vec{\xi}\sum_{\rho}S^{\rho}$ 合并, 则
$$ \begin{aligned} X =& \text{Tr}_{S}\int\frac{\mathrm{d}z}{\sqrt{2\pi}}\exp{\left[-\frac{1}{2}z^{2} + \beta\left(\sqrt{\alpha r} z + \vec{m}\cdot\vec{\xi}\right)\sum_{\rho}S^{\rho}\right]}\\ \overset{e^{a}+e^{-a}=2\cosh{a}}{=}& \int\frac{\mathrm{d}z}{\sqrt{2\pi}}e^{-\frac{z^{2}}{2}}\left[2\cosh{\beta\left(\sqrt{\alpha r} z + \vec{m}\cdot\vec{\xi}\right)}\right]^{n}\\ \overset{a^{n}=e^{n\ln{a}}}{=}& \int\frac{\mathrm{d}z}{\sqrt{2\pi}}e^{-\frac{z^{2}}{2}}\exp{\left\{ n\ln{\left[ 2\cosh{\beta\left(\sqrt{\alpha r} z + \vec{m}\cdot\vec{\xi}\right)} \right]} \right\}} \end{aligned} $$
最后取 $n\to 0$ 时会有 $1/n$ 因子:
$$ \begin{aligned} \frac{\langle\ln{X}\rangle}{n} \overset{e^{\epsilon}\sim 1+\epsilon}{=} & \frac{1}{n}\left\langle \ln{\int\frac{\mathrm{d}z}{\sqrt{2\pi}}e^{-\frac{z^{2}}{2}}\left[ 1 + n\ln{\left[ 2\cosh{\beta\left(\sqrt{\alpha r} z + \vec{m}\cdot\vec{\xi}\right)} \right]} + \mathcal{O}(n^{2}) \right]} \right\rangle\\ =& \frac{1}{n} \left\langle n\int\frac{\mathrm{d}z}{\sqrt{2\pi}}e^{-\frac{z^{2}}{2}}\ln{\left[ 2\cosh{\beta\left(\sqrt{\alpha r} z + \vec{m}\cdot\vec{\xi}\right)} \right]} \right\rangle\\ =& \left\langle\left\langle \ln{\left[ 2\cosh{\beta\left(\sqrt{\alpha r} z + \vec{m}\cdot\vec{\xi}\right)} \right]} \right\rangle_{\mu<s}\right\rangle_{z} \end{aligned} $$
综上所述, 得到自由能密度
$$ \begin{aligned} \frac{F}{N} =& \frac{1}{2}\alpha + \frac{1}{2}\vec{m}^{2} + \frac{\alpha}{2\beta}\left\{ \ln{[1-\beta(1-q)]} - \frac{\beta q}{1-\beta(1-q)} \right\}\\ &+ \frac{1}{2}\alpha\beta r(1-q) - \frac{1}{\beta} \left\langle\left\langle \ln{\left[ 2\cosh{\beta\left(\sqrt{\alpha r} z + \vec{m}\cdot\vec{\xi}\right)} \right]} \right\rangle_{\mu<s}\right\rangle_{z} \end{aligned} $$
对其求导得到鞍点条件方程, 得到
$$ \begin{aligned} m^{\mu} &= \left\langle \xi^{\mu}\tanh{\beta(\sqrt{\alpha r}z + \vec{m}\cdot\vec{\xi})} \right\rangle\\ q &= \left\langle \tanh^{2}{\beta(\sqrt{\alpha r}z + \vec{m}\cdot\vec{\xi})} \right\rangle\\ r &= \frac{q}{[1-\beta(1-q)]^{2}} \end{aligned} $$