I am reading the one lecture note [Dynamics for Spherical Models of Spin-Glass and Aging](https://doi.org/10.1007/978-3-540-40908-3_5).


On page 126. In the Sherrington-Kirkpatrick (SK) model, we suppose that there are $N$ people labeled as $[N]:=\{1,2,\ldots,N\}$, $\sigma \in \{+1,-1\}^N$. Let $\mathbf{J}=(J\_{ij})\_{1\le i, j\le N}$ be symmetric and let $J\_{ij}$ be a centered independent random variable such that
$$
E[J\_{ij}^2]=1/n, \, E[J\_{ii}^2]=2/n,
$$
often assumed to be Gaussian for simplicity. Assume that $x\_0$ is independent of $\mathbf{J}$.


If $\{\lambda\_i\}\_{1\le i\le N}$ denotes the eigenvalue of $\mathbf{J}=\{J\_{ij}\}$, then for a vector $\mathbf{x}\_0$,
$$
\frac{1}{N}\langle \mathbf{x}\_0, e^{\mathbf{J}}\mathbf{x}\_0\rangle=^d\frac{1}{N}\langle \mathbf{x}\_0, \mathbf{x}\_0\rangle\sum\_{i=1}^N e^{\lambda\_i}u\_i^2
$$
where $u$ is independent of $\lambda\_i$ and $\mathbf{x}\_0$, and follows the law of the sphere $S\_{\sqrt{N}}^{N-1}$ with radius $\sqrt{N}$.


***(I think $u$ follows the law of the unit sphere?)***




---


From the Weak law of large numbers, the first term
$$
\frac1N\langle x\_0, x\_0\rangle\to \int x^2d\mu(x)
$$
as $n\to \infty$.


The main question is as follows.



> 
> Question: Why do we have
> $$
> \lim\_{N\to \infty} \sum\_{i=1}^N e^{\lambda\_i}u\_i^2=\int e^{\lambda}d\sigma(\lambda)
> $$
> 
> 
> 


From semi-circle law, we have the empirical measure $\frac{1}{N}\sum \delta\_{\lambda\_i}\to \sigma(dx)=C\sqrt{4-x^2}dx$. But how to get the above limit? I think there is missing the scaling $1/N$ on the left-hand side.




---


I try to show that
$$
law(\frac{1}{N}\langle \mathbf{x}\_0, e^{\mathbf{J}}\mathbf{x}\_0\rangle)=law(\frac{1}{N}\langle \mathbf{x}\_0, \mathbf{x}\_0\rangle\sum\_{i=1}^N e^{\lambda\_i}u\_i^2)
$$


Let $\mathbf{J}=G^\*DG$ where $D$ is the diagonal matrix with eigenvalues $\lambda\_i$. Then $e^\mathbf{J}$ can be expressed via spectral decomposition as
$$
e^\mathbf{J}=G^Te^{D}G=\sum e^{\lambda\_i} s\_is\_i^T
$$
(the second one is wrong?)
where $s\_i\in R^N$ are eigenvectors of $\mathbf{J}$ and $G=[s\_1,\dots, s\_N]$.


I am confused about where $u\_i^2$ comes from? If we plug in the spectral decomposition, it becomes that
$$
\frac{1}{N}\langle \mathbf{x}\_0, e^{\mathbf{J}}\mathbf{x}\_0\rangle=\frac{1}{N}\langle \mathbf{x}\_0, \sum e^{\lambda\_i} s\_is\_i^T\mathbf{x}\_0\rangle
$$


Here
$$
E[\sum e^{\lambda\_i} s\_is\_i^T]=E[\frac{1}{N}I]
$$



How to get $\lim_{N\to \infty} \sum_{i=1}^N e^{\lambda_i}u_i^2=\int e^{\lambda}d\sigma(\lambda)$?
The eigenvector components $u\_i$ have zero mean and variance $1/N$ (since $\sum\_i u\_i^2=1$); they are independent of the eigenvalues $\lambda\_i$. We therefore have the expectation value
$$\lim\_{N\to \infty}\mathbb{E}\left[ \sum\_{i=1}^N e^{\lambda\_i}u\_i^2\right]=\lim\_{N\to \infty}\mathbb{E}\left[\frac{1}{N} \sum\_{i=1}^N e^{\lambda\_i}\right]=\int e^{\lambda}d\sigma(\lambda)=I\_2(2),$$
with $p(\lambda)=d\sigma(\lambda)/d\lambda=(2\pi)^{-1}\sqrt{4-\lambda^2}\,\mathbb{1}\_{[-2,2]}(\lambda)$ the semi-circular probability distribution of the eigenvalues (normalized to unity).


Fluctuations arond the expectation value are smaller by a factor $1/\sqrt N$ and may be neglected in the large-$N$ limit.


