How to get $\lim_{N\to \infty} \sum_{i=1}^N e^{\lambda_i}u_i^2=\int e^{\lambda}d\sigma(\lambda)$?

Question

I am reading the one lecture note Dynamics for Spherical Models of Spin-Glass and Aging.

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?)

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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.

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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 $$ E[\sum e^{\lambda_i} s_is_i^T]=E[\frac{1}{N}I] $$

Answer 1

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.