The Hamiltonian of the p-spherical spin glass model is 


$$H\_{N,p}(\sigma)=\frac{1}{N^{\frac{p-1}{2}}} \sum\_{i\_1,...,i\_p=1}^N X\_{i\_1,...,i\_p} \sigma\_{i\_1}\cdot...\cdot \sigma\_{i\_p}$$
where $\sigma \in \mathbb S^{N-1}(\sqrt{N}).$ The $X\_{i\_1,...,i\_p}$ are random i.i.d. standard centred Gaussians. 


In this paper [in eq. (2.10).](https://arxiv.org/pdf/1003.1129.pdf) it is claimed that this Hamiltonian is (if $p=2$!) equivalent to the quadratic form of a random GOE-matrix $M$, i.e.


$$\sigma \mapsto \langle M\sigma,\sigma \rangle.$$


However, I do not quite understand how the Hamiltonian can correspond to a symmetric matrix, because the $X\_{ij}$ are not symmetric, as they are uncorrelated. So how can I see this property?


My follow-up question is then: Is there a generalization using symmetric multilinear forms for arbitrary p?More precisely, can I have $H\_p = \sum\_{i\_1,...,i\_p=1}^N X\_{i\_1,...,i\_p} \sigma\_{i\_1} \cdot ...\cdot \sigma\_{i\_p}$ with $X\_{i\_1,...,i\_p}$ being invariant under cyclic permutation of indices.



Random matrix and spherical spin-glass
The spherical two-spin model has Hamiltonian
$$H=N^{-1/2}\sum\_{i\neq j}X\_{ij}\sigma\_i\sigma\_j$$
where the $\sigma\_i$, $i=1,2,\ldots N$, are continuous spin variables subject to the constraint $\sum\_i\sigma\_i^2=N$. The matrix $X$ is a symmetric $N\times N$ matrix of independently distributed Gaussian random variables of zero mean and unit variance.  

Any antisymmetric contribution to $X$ drops out of the sum, so we may assume $X$ is symmetric without loss of generality.


This ensemble of random symmetric matrices is called the Gaussian Orthogonal Ensemble (GOE). One may therefore immediately conclude that the statistics ground state of $H$ (the state with the lowest eigenvalue) is given by the distribution of the smallest eigenvalue of GOE, which is the [Tracy-Widom distribution](https://en.wikipedia.org/wiki/Tracy%E2%80%93Widom_distribution).


