Here is my reading note for the book ‘Statistical Physics of Spin Glasses and Information Processing: An Introduction’ by Hidetoshi Nishimori. I am still not quite clear about this field so please discuss your insights and understanding with me if you are also interested.
Prerequisites
Basic Notions in Statistical Physics
State distribution: \(\begin{align} \mathbb{P}\left( \text{state}_i \right)=\dfrac{e^{-\beta \mathcal{H}(\text{state}_i)}}{Z},\quad Z=\sum_{\text{all state}}e^{-\beta \mathcal{H}(\text{state}_i)} \end{align}\)
Free Energy: \(\begin{align} F(\beta )\equiv -\dfrac{1}{\beta }\log Z \end{align}\)
(Canonical) Entropy: \(\begin{align} S(\beta )=&-\sum_{i}\dfrac{e^{-\beta \mathcal{H}_i}}{Z}\log\dfrac{e^{-\beta \mathcal{H}_i}}{Z}\\ =&\sum_{i}\dfrac{e^{-\beta \mathcal{H}_i}}{Z}\left[\beta \mathcal{H}_i+\log Z \right]\\ =&-\sum_i \dfrac{\beta }{Z}\dfrac{\partial^{} e^{-\beta \mathcal{H}_i}}{\partial \beta ^{}}+\log Z\\ =&-\beta \dfrac{\partial^{} \log Z}{\partial \beta ^{}}+\log Z\\ =&-\beta ^2\dfrac{\partial^{} \dfrac{1}{\beta }\log Z}{\partial \beta ^{}}\\ =&\beta ^2\dfrac{\partial^{} F}{\partial \beta ^{}} \end{align}\)
Average energy (internal energy): \(\begin{align} U=\sum_{i}\mathcal{H}_i\dfrac{e^{-\beta \mathcal{H}_i}}{Z}=&-\dfrac{1}{Z}\sum_{i}\dfrac{\partial^{} e^{-\beta \mathcal{H}_i}}{\partial \beta ^{}}\\ =&-\dfrac{\partial^{} \log Z}{\partial \beta ^{}}\\ =&\dfrac{\partial^{} \beta F}{\partial \beta ^{}} \end{align}\)
Legendre transform bet. $U$ and $F$: \(\begin{align} U=&\dfrac{\partial^{} \beta F}{\partial \beta ^{}}\\ =&F+\beta \dfrac{\partial^{} F}{\partial \beta ^{}}\\ =&F+\dfrac{S}{\beta } \end{align}\)
Average free energy: \(\begin{align} f=\lim_{N\to \infty}\dfrac{1}{N}F(\beta ,N) \end{align}\)
Useful Lemmas
Lemma 1: Gaussian integral \(\begin{align} \int e^{-\alpha x^2+\beta x} \,\mathrm{d}x=&\sqrt{\dfrac{\pi}{\alpha }}e^{\frac{\beta ^2}{4\alpha }}\quad (\mathrm{Re}(\alpha )\geq 0)\\ e^{\frac{ax^2}{2}}=&\sqrt{\dfrac{a}{2\pi}}\int e^{-\frac{am^2}{2}+amx} \,\mathrm{d}m \end{align}\)
Lemma 2: delta function \(\begin{align} &\delta(x)=\begin{cases} 0,&x\neq 0\\ \infty,&x=0 \end{cases},\quad w.r.t. \int_\mathbb{R}\delta (x)\,\mathrm{d}x=1\\ &\int _\mathbb{R}f(x)\delta (x-a) \,\mathrm{d}x=f(a)\\ &\text{Fourier:}\begin{cases} 1=\int_{\mathbb{R}}e^{-ikx}\delta (x)\,\mathrm{d}x\\ \delta (x)=\dfrac{1}{2\pi}\int _\mathbb{R}e^{ikx} \,\mathrm{d}k \end{cases} \end{align}\)
Lemma 3: delta function + Gaussian integral = Fourier*2 \(\begin{align} f(a)=&\int _\mathbb{R}f(x)\delta (x-a) \,\mathrm{d}x\\ =&\int _\mathbb{R}\int_\mathbb{R}\dfrac{1}{2\pi}f(x)e^{ik(x-a)}\,\mathrm{d}k \,\mathrm{d}x \end{align}\)
Chapter 2 of the Book
Problem to study
Say we are solving some combinatorial problem $\mathop{\arg\min}\limits_{x} \mathcal{H}(x,\mathcal{C})$, where $\mathcal{C}$ denotes the configuration of the parameters of the problem.
- Solve the problem for some specific/explicit $\mathcal{C}$
- If $\mathcal{C}$ has some distribution, we can solve this type of problem of $\mathcal{C}\sim f_\mathcal{C}$
Example:
- TSP (Travaling Salesman Problem), $\mathcal{C}$ for locations and path, its distribution represents `this kinds of map to travel’
- ML (Machine learning), $\mathcal{C}$ for training data, its distribution represents the ability to generalize the model
Self-averaging Property
Studying the ‘averaging property’ v.s. ‘specific problem’
- self averaging: when size of the system grows, the average of some observable $[O(x)]_\mathcal{C}$ ‘represents almost all typical cases’ of $\mathcal{C}$.
$f$ is a self-averaging quantity $\to$ evaluate $[f]$ instead of studying any explicit $f(x,\mathcal{C})$ \(\begin{align} [f]=&\lim_{N\to\infty}\dfrac{1}{N}[F]_\mathcal{C}\lim_{N\to\infty}-\dfrac{1}{N\beta }[\log Z]_\mathcal{C} \end{align}\)
Replica trick:
\(\begin{align} [\log Z]=[\dfrac{1}{n}\log Z^n]=\lim_{n\to 0}[\dfrac{Z^n-1}{n}]=\lim_{n\to 0}\dfrac{[Z^n]-1}{n}\tag{2.6} \end{align}\)
Idea: first evaluate $[Z^n]$ at $n\in\mathbb{N}^+$, then use some other trick to evaluate $n\to 0$ with ‘analytically continuation’
Comments on relica trick:
- Avoid averaging on $\log$, better calculation
- Validity of the continuation methods? commonly used method: Replica symmetry splution / Parisi equation
SK Model
Hamiltonian of Sherrington-Kirkpatrick Model : $i,j$ for sites \(\begin{align} \mathcal{H}=-\sum_{i<j}J_{ij}S_iS_j-h\sum_{i}S_i,\quad i,j=1,2,\ldots,N\tag{2.7} \end{align}\)
Distribution of bond $J_{ij}$, say normal distribution $J_{ij}\sim_{iid}\mathcal{N}(\dfrac{J_0}{N},\dfrac{J^2}{N})$ \(\begin{align} \mathbb{P}\left( J_{ij} \right) =\sqrt{\dfrac{N}{2\pi J^2}}\exp\left\{ -\dfrac{N(J_{ij}-J_0)^2}{2J^2} \right\}\tag{2.8} \end{align}\)
Comment: $\bar{J}_{ij}=J_0\big/N$ to ensures $\mathcal{H}$ grows linearly with $N$. ($\mathcal{H}$ should be extensive quantity) \(\begin{align} [F]=-\dfrac{1}{\beta }[\log Z]=-\dfrac{1}{\beta }\int \log Z \,\prod_{i<j}\mathbb{P}\left( J_{ij} \right)\,\mathrm{d}J_{ij}\tag{2.5} \end{align}\)
Replica trick: \(\begin{align} [Z^n]=\int \left(\sum_{\{S_{i=1}^N\}}e^{-\beta \mathcal{H}(S_{i=1}^N)}\right)^n \,\prod_{i<j}\mathbb{P}\left( J_{ij} \right)\,\mathrm{d}J_{ij}\tag{2.5} \end{align}\)
Key steps:
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replica: $(\sum_{}\,\cdot \,)^n\to (\sum_{\mathrm{replica} 1 })\times (\sum_{\mathrm{replica} 2 })\times \ldots\times (\sum_{\mathrm{replica} n }) $, use $S^\alpha $ to denote replica$\alpha $ \(\begin{align} [Z^n]=&\int \left(\sum_{\{S_{i=1}^N\}_{\alpha =1}^n}\prod_{\alpha =1}^ne^{-\beta \mathcal{H}(S_{i=1}^{\alpha N})}\right) \,\prod_{i<j}\mathbb{P}\left( J_{ij} \right)\,\mathrm{d}J_{ij}\tag{2.10}\\ =&\int \left(\sum_{\{S_{i=1}^N\}_{\alpha =1}^n}\exp\left\{ \beta \sum_{i<j}J_{ij}\sum_{\alpha =1}^nS_i^\alpha S_j^\alpha +\beta h\sum_{i}\sum_{\alpha =1}^nS_i^\alpha \right\}\right) \,\prod_{i<j}\mathbb{P}\left( J_{ij} \right)\,\mathrm{d}J_{ij}\\ \mathop{=}\limits_{Lemma1} &\sum_{\{S^\alpha |_{i=1}^N\}_{\alpha =1}^n}\exp\Bigg\{ \dfrac{1}{N}\sum_{i<j}\Bigg(\dfrac{1}{2}\beta ^2J^2\sum_{\alpha ,\beta }S_i^\alpha S_j^\alpha S_i^\beta S_j^\beta \\ &\qquad\qquad\qquad\quad+\beta J_0\sum_{\alpha }S_i^\alpha S_j^\alpha \Bigg)+\beta h\sum_{i}\sum_{\alpha }S_i^\alpha \Bigg\} \end{align}\)
- Expectation $\mathbb{E}$ using gaussian integral
\(\begin{align}
[Z^n]=\sum_{\{S^\alpha |_{i=1}^N\}_{\alpha =1}^n}\exp\Bigg\{ \dfrac{1}{N} &\sum_{i<j}\Bigg(\dfrac{1}{2}\beta ^2J^2\sum_{\alpha ,\beta }S_i^\alpha S_j^\alpha S_i^\beta S_j^\beta\\
& +\beta J_0\sum_{\alpha }S_i^\alpha S_j^\alpha \Bigg)+\beta h\sum_{i}\sum_{\alpha }S_i^\alpha \Bigg\} \tag{2.11}
\end{align}\)
Note:
- $S_i^\alpha \in{+1,-1}\Rightarrow S_i^\alpha S_i^\alpha =1$.
- $\sum_{\alpha ,\beta }S_i^\alpha S_j^\alpha S_i^\beta S_j^\beta=n+2\sum_{\alpha <\beta }S_i^\alpha S_j^\alpha S_i^\beta S_j^\beta$
- $\sum_{i<j}S_i^\alpha S_j^\alpha S_i^\beta S_j^\beta=-N+\dfrac{1}{2}\sum_{i,j}S_i^\alpha S_j^\alpha S_i^\beta S_j^\beta=-N+\dfrac{1}{2}\left(\sum_{i=1}^NS_i^\alpha S_i^\beta \right)^2$
- $\sum_{i<j}S_i^\alpha S_j^\alpha =-N+\dfrac{1}{2}\sum_{i,j}S_i^\alpha S_j^\alpha =-N+\dfrac{1}{2}\left(\sum_{i=1}^NS_i^\alpha \right)^2 $ \(\begin{align} [Z^n]=&\exp\left[ \dfrac{Nn\beta ^2J^2}{4} \right]\sum_{\{S^\alpha |_{i=1}^N\}_{\alpha =1}^n}\exp\Bigg\{ \dfrac{\beta ^2J^2}{2N}\sum_{\alpha<\beta}\left(\sum_iS^\alpha_iS^\beta _i \right)^2\\ &+\dfrac{\beta J_0}{2N}\sum_{\alpha }\left(\sum_i S_i^\alpha \right)^2+\beta h\sum_i\sum_\alpha S_i^\alpha \Bigg\} \end{align}\)
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Change summation indices \(\begin{align} [Z^n]=&\exp\left[ \dfrac{Nn\beta ^2J^2}{4} \right]\cdot \sum_{\{S^\alpha |_{i=1}^N\}_{\alpha =1}^n}\exp\Bigg\{ \dfrac{\beta ^2J^2}{2N}\sum_{\alpha<\beta}\left(\sum_iS^\alpha_iS^\beta _i \right)^2\\ &+\dfrac{\beta J_0}{2N}\sum_{\alpha }\left(\sum_i S_i^\alpha \right)^2+\beta h\sum_i\sum_\alpha S_i^\alpha \Bigg\}\\ \tag{2.12} \end{align}\)
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Use lemma 1 to linearize $\left( \sum_{i } \right)^2$ term, integral dummy variable taken as $q_{\alpha \beta }$ and $m_\alpha $ \(\begin{align} [Z^n]=&\exp\left\{ \dfrac{Nn\beta ^2J^2}{4} \right\}\int \prod_{\alpha <\beta }\,\mathrm{d}q_{\alpha \beta }\prod_{\alpha } \,\mathrm{d}m_\alpha \\ \cdot&\exp\left[ -\dfrac{N\beta ^2J^2}{2}\sum_{\alpha <\beta }q_{\alpha \beta }^2-\dfrac{N\beta J_0}{2}\sum_\alpha m_\alpha ^2 \right]\\ \cdot&\sum_{\{S^\alpha |_{i=1}^N\}_{\alpha =1}^n}\exp\left[ \beta ^2J^2\sum_{\alpha <\beta }\sum_iS_i^\alpha S_i^\beta+\beta \sum_\alpha (J_0m_\alpha +h)\sum_iS_i^\alpha \right]\tag{2.13} \end{align}\)
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Reduction of \(\sum_{\{S^\alpha |_{i=1}^N\}_{\alpha =1}^n}\): equivalent for all \(i\), \(\sum_{\{S^\alpha |_{i=1}^N\}_{\alpha =1}^n}\to (\sum_{\alpha })^N\) \(\begin{align} [Z^n]=&\exp\left\{ \dfrac{Nn\beta ^2J^2}{4} \right\}\int \prod_{\alpha <\beta }\,\mathrm{d}q_{\alpha \beta }\prod_{\alpha } \,\mathrm{d}m_\alpha \\ \cdot&\exp\left[ -\dfrac{N\beta ^2J^2}{2}\sum_{\alpha <\beta }q_{\alpha \beta }^2-\dfrac{N\beta J_0}{2}\sum_\alpha m_\alpha ^2 \right]\\ \cdot&\exp\left[ N\log \sum_{\{S^\alpha \}_{\alpha =1}^n}\exp\left[ \beta ^2J^2\sum_{\alpha <\beta }q_{\alpha \beta }S^\alpha S^\beta +\beta \sum_\alpha (J_0m_\alpha +h)S^\alpha \right] \right]\tag{2.15}\\ =&\exp\left\{ \dfrac{Nn\beta ^2J^2}{4} \right\}\int \prod_{\alpha <\beta }\,\mathrm{d}q_{\alpha \beta }\prod_{\alpha } \,\mathrm{d}m_\alpha\\ \cdot&\exp N\Big\{ -\dfrac{\beta ^2J^2}{2}\sum_{\alpha <\beta }q^2_{\alpha \beta }-\dfrac{\beta J_0}{2}\sum_\alpha m_\alpha ^2 \\ &+\log \sum_{\{S^\alpha \}_{\alpha =1}^n}\exp\left[ \beta ^2J^2\sum_{\alpha <\beta }q_{\alpha \beta }S^\alpha S^\beta +\beta \sum_\alpha (J_0m_\alpha +h)S^\alpha \right]\Big\}\\ :=&\int \prod_{\alpha <\beta }\,\mathrm{d}q_{\alpha \beta }\prod_{\alpha } \,\mathrm{d}m_\alpha \exp\left\{ N\Xi(q_{\alpha \beta },m_\alpha ) \right\}\\ \Xi\equiv&\dfrac{n\beta ^2J^2}{4} -\dfrac{\beta ^2J^2}{2}\sum_{\alpha <\beta }q^2_{\alpha \beta }-\dfrac{\beta J_0}{2}\sum_\alpha m_\alpha ^2 \\ &+\log \sum_{\{S^\alpha \}_{\alpha =1}^n}\exp\left[ \beta ^2J^2\sum_{\alpha <\beta }q_{\alpha \beta }S^\alpha S^\beta +\beta \sum_\alpha (J_0m_\alpha +h)S^\alpha \right] \end{align}\)
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Thermal limit $N\to\infty$, the integral is dominated by $\max_{q_{\alpha \beta },m_\alpha }\Xi$ term, i.e. \(\begin{align} [Z^n]\propto \exp\left\{ N\Xi \right\},\quad w.r.t. q_{\alpha \beta },m_\alpha =\mathop{\arg\max}\limits_{q_{\alpha \beta },m_\alpha }\Xi(q_{\alpha \beta },m_\alpha ) \end{align}\)
- Average free energy should be finite: $[f]=-\dfrac{1}{\beta }\lim_{N\to\infty}\lim_{n\to 0}\dfrac{[Z^n]-1}{nN}<\infty$ \(\begin{align} [f]=&-\dfrac{1}{\beta }\left\{ \lim_{n\to 0}\lim_{N\to\infty}\dfrac{[Z^n]-1}{nN} \right\}\\ =&-\dfrac{1}{\beta }\left\{ \lim_{n\to 0}\lim_{N\to\infty}\dfrac{\exp\left\{ N\Xi \right\}-1}{nN} \right\}\\ \mathop{=}\limits_{N\Xi\to 0}^{} &-\dfrac{1}{\beta }\lim_{n\to 0}\dfrac{\Xi}{n}\\ =&\dfrac{1}{\beta }\left\{ \dfrac{\beta ^2J^2}{4n}\sum _{\alpha \neq \beta }q^2_{\alpha \beta }+\dfrac{\beta J_0}{2n}\sum_\alpha m_\alpha ^2-\dfrac{1}{4}\beta ^2J^2-\dfrac{1}{n}\log\sum_{\{S^\alpha \}_{\alpha =1}^n} e^L \right\}\\ L=& \beta ^2J^2\sum_{\alpha <\beta }q_{\alpha \beta }S^\alpha S^\beta +\beta \sum_\alpha (J_0m_\alpha +h)S^\alpha \tag{2.15-17}\\ w.r.t.&\,q_{\alpha \beta },m_\alpha =\mathop{\arg\max}\limits_{q_{\alpha \beta },m_\alpha }\Xi(q_{\alpha \beta },m_\alpha ) /n \end{align}\)
Comment:
- Now free energy depends on order parameters $q_{\alpha \beta },m_\alpha $, instead of all ${S_{i=1}^N}_{\alpha =1}^n$
- Exchangablity of $\lim_{n}$ and $\lim_N$? I believe that’s what results in replica breaking.
- Condition to $\arg\max \Xi/n$
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$\dfrac{\partial^{}\Xi/n }{\partial (q_{\alpha \beta },m_\alpha )^{}}=0$: \(\begin{align} \begin{cases} q_{\alpha \beta }=\langle S^\alpha S^\beta \rangle_L&=[\langle S_i^\alpha S_i^\beta \rangle]=[\langle \dfrac{1}{N}\sum_{i=1}^N S_i^\alpha S_i^\beta \rangle _{\mathcal{H}_\mathrm{replica}}]\\ m_\alpha =\langle S^\alpha \rangle_L&=[\langle S_i^\alpha \rangle]=[\langle \dfrac{1}{N}\sum_{i=1}^NS_i^\alpha \rangle _{\mathcal{H}_\mathrm{replica}}] \end{cases} \end{align}\) Intuition: $q_{\alpha \beta }$ is the overlap between repica $\alpha $ and $\beta $. $\langle\, \cdot \, \rangle_{\mathcal{H}_\mathrm{replica}}$ is the distribution of the whole replicas system, while $S^\alpha_i S^\beta _i$ focuses on part of the whole replica system
Thermal equilibrium: $\mathrm{replica1}\leftrightarrow\mathrm{replica2}\leftrightarrow\cdots\leftrightarrow\mathrm{replica}\alpha $ at $t\to \infty, \beta \to \infty$. Spin glass equilibrium: $t$ is large but $t<\infty$, which means that each replica is trapped in its valley $\to$ replica symmetry breaking
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Semi-positive definite of Hessian $\dfrac{\partial^{2} \Xi/n}{\partial \partial ^T (q_{\alpha \beta },m_\alpha )}$
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Another intuition from ensemble theory:
To study how states changes when reaching equilibrium at $t\to \infty$, you can either:
- obsereve one system till $t\to \infty$
- or observe $n\to \infty$ systems simultanously
Perhaps replica would be similar to ensemble? then $n\to 0$ would be similar to $t<\infty$, which is the spin glass state time scale. But I haven’t seen any work talking about such analogy.
Appendix B – Parisi Solution
(B.1)
Lemma (1): \(\begin{align} \exp\left[ \dfrac{\partial^{} }{\partial h_\beta } \right]\exp\left[ \sum_{\alpha }h_\alpha S^\alpha \right]=&\sum_{i=0}^\infty \dfrac{1}{i!}\dfrac{\partial^{i} }{\partial h_\beta ^i}\exp\left[ \sum_\alpha h_\alpha S^\alpha \right]\\ =&\sum_{i=0}^\infty\dfrac{1}{i!}(S^\beta )^i\exp\left[ \sum_{\alpha }h_\alpha S^\alpha \right]\\ =&\exp\left[ S^\beta \right]\exp\left[ \sum_{\alpha }h_\alpha S^\alpha \right]\\ \Rightarrow \exp\left[ q_{\alpha \beta }\dfrac{\partial^{2} }{\partial h_\alpha h_\beta } \right]\exp\left[ \sum_{\alpha }S^\alpha \right]=&\exp\left[ q_{\alpha \beta }S^\alpha S^\beta+\sum_{\alpha }h_\alpha S^\alpha \right] \end{align}\) \(\begin{align} G=&\mathrm{Tr}\exp\left[ \dfrac{1}{2}\sum_{\alpha ,\beta }^nq_{\alpha \beta }S^\alpha S^\beta +h\sum_{\alpha }^nS^\alpha \right]\\ =&\exp\left[ \dfrac{1}{2}\sum_{\alpha ,\beta }q_{\alpha \beta}\dfrac{\partial^{2} }{\partial h_\alpha h_\beta } \right]\mathrm{Tr}\exp\left[ \sum_{\alpha }h_\alpha S^\alpha \right]\Bigg|_{h_\alpha =h}\\ =&\exp\left[ \dfrac{1}{2}\sum_{\alpha ,\beta }q_{\alpha \beta}\dfrac{\partial^{2} }{\partial h_\alpha h_\beta } \right]\prod_\alpha \cosh h_\alpha S^\alpha \Bigg|_{h_\alpha =h}\\ \end{align}\)
(B.8)
\(\begin{align} g(x+\mathrm{d}x,h )&=\exp\left[ -\dfrac{1}{2}\,\mathrm{d}q(x)\dfrac{\partial^{2} }{\partial h^{2}} \right]g(x,h)^{1+\mathrm{d}\log x }\\ g(x+\mathrm{d}x,h )-g(x,h ) &=\left\{ 1-\dfrac{1}{2}\mathrm{d}q(x)\dfrac{\partial^{2} }{\partial h^{2}} \right\}g(x,h)^{1+\mathrm{d}\log x }-g(x,h )\\ \dfrac{\mathrm{d}^{} g(x,h)}{\mathrm{d}x^{}}=&-\dfrac{1}{2}\dfrac{\mathrm{d}^{} q(x)}{\mathrm{d}x^{}}\dfrac{\partial^{2} }{\partial h ^{2}}g(x,h)+\left( g(x,h)^{1+\mathrm{d}\log x }-g(x,h) \right)\Big/ \mathrm{d}x\\ \dfrac{\mathrm{d}^{} g(x,h)}{\mathrm{d}x^{}}=&-\dfrac{1}{2}\dfrac{\mathrm{d}^{} q(x)}{\mathrm{d}x^{}}\dfrac{\partial^{2} }{\partial h ^{2}}g(x,h)+g(x,h)\dfrac{g^{\,\mathrm{d}\log x}-g^0}{x\,\mathrm{d}\log x}\\ \dfrac{\mathrm{d}^{} g(x,h)}{\mathrm{d}x^{}}=&-\dfrac{1}{2}\dfrac{\mathrm{d}^{} q(x)}{\mathrm{d}x^{}}\dfrac{\partial^{2}g(x,h) }{\partial h ^{2}}+\dfrac{g(x,h)}{x}\log g(x,h)\\ \end{align}\)
(B.10)
\(\begin{align} \dfrac{1}{n}\log\mathrm{Tr}e^L=&\exp\left[ \dfrac{1}{2}q(0)\dfrac{\partial^{2} }{\partial h^{2}} \right] \dfrac{1}{n}\log [g(m_1,h)]^{n/m_1}\Big|_{h\to 0,m_1\to 0,m_1-0=\mathrm{d}x=x\to 0 }\\ =&\exp\left[ \dfrac{1}{2}q(0)\dfrac{\partial^{2} }{\partial h^{2}} \right]\dfrac{1}{x}\log g(x,h)\Big|_{x,h\to 0}\\ =&\exp\left[ \dfrac{1}{2}q(0)\dfrac{\partial^{2} }{\partial h^{2}} \right]f_0(0,h)\Big|_{h\to 0}\\ =&\exp\left[ \dfrac{1}{2}q(0)\dfrac{\partial^{2} }{\partial h^{2}} \right]\int_{w\in\mathbb{R}}f_0(0,w)\delta(h-w)\,\mathrm{d}w\Bigg|_{h\to 0}\\ =&\exp\left[ \dfrac{1}{2}q(0)\dfrac{\partial^{2} }{\partial h^{2}} \right]\int_{w\in\mathbb{R}}f_0(0,w)\int_{v\in\mathbb{R}}\dfrac{1}{2\pi}\exp\left[ iv(h-w) \right]\,\mathrm{d}v\,\mathrm{d}w\Bigg|_{h\to 0}\\ =&\int_{w\in\mathbb{R}}\int_{v\in\mathbb{R}}f_0(0,w)\dfrac{1}{2\pi}\exp\left[ \dfrac{1}{2}q(0)\dfrac{\partial^{2} }{\partial h^{2}} \right]\exp\left[ iv(h-w) \right]\,\mathrm{d}v \,\mathrm{d}w\Bigg|_{h\to 0}\\ (\text{Lemma 1})=&\int_{w\in\mathbb{R}}\int_{v\in\mathbb{R}}f_0(0,w)\dfrac{1}{2\pi}\exp\left[ -\dfrac{1}{2}q(0)v^2+iv(h-w) \right]\,\mathrm{d}v \,\mathrm{d}w\Bigg|_{h\to 0}\\ =&\int_{w\in\mathbb{R}}f_0(0,w)\dfrac{1}{2\pi}\sqrt{\dfrac{\pi}{q(0)/2}}\exp\left[ \dfrac{-(w-h)^2}{2q(0)} \right]\,\mathrm{d}w\Bigg|_{h\to 0}\\ =&\int_{w\in\mathbb{R}}f_0(0,w)\dfrac{1}{2\pi}\sqrt{\dfrac{\pi}{q(0)/2}}\exp\left[ \dfrac{-w^2}{2q(0)} \right]\,\mathrm{d}w\\ (w=\sqrt{q_0}u)=&\int _uf(0,u) \,\mathrm{D}u \end{align}\)