We construct an $N\times N$ matrix $J$ whose elements are drawn from Gaussian distribution with zero mean and variance $\frac{1}{N}$. Since we want to have different variances for different columns, so that we express the elements as $J_{ij}\sigma_j$ such that the matrix satisfies $\mathbb{Var}[J_{ij}\sigma_j] = \frac{\sigma_j^2}{N}$.

Given a Green function (analogous electrostatic potential) $$\Phi(\omega) = -\frac{1}{N}\mathbb{E}_J\big[ \ln\det((I\omega^* - J^T)(I\omega - J)) \big]$$, where $\omega\in \mathbb{C}$ and $J$ is an $N\times N$ random matrix. Since $J$ is positive semi-definite, by adding diagonal matrix $\epsilon\delta_{ij}$, determinant can be represented by a Gaussian integral over complex variables, we have $$ \Phi(\omega) = \frac{1}{N}\ln\mathbb{E}_J\big[ \int\big( \prod_i\frac{d^2 z_i}{\pi}\big)\exp\big\{ -\epsilon\sum_i|z_i|^2 - \sum_{i, j, k} z_i^*(\omega^*\delta_{ik}-J_{ik}^T)(\omega\delta_{kj}-J_{kj})z_j \big\} \big] $$

Then the potential $\phi$ is determined by

$$ \exp(-N\Phi) = \int\prod_{i = 1}^N \frac{dz_i^* dz_i}{\pi}\exp(-N\hat{Q}) $$ with the average over the distribution of $J$ values given by $$ \exp(-N\hat{Q}) = \int\prod_{i,j = 1}^N \sqrt{\frac{N}{2\pi}}dJ_{ij}\exp(-NQ)$$ where $$ Q = \sum_i\big( \frac{|\omega|^2}{\sigma_i^2} + \epsilon_i \big)\frac{z_i^*z_i}{N} + \frac{1}{2}\sum_{i, j, k} J_{ki}A_{ij}J_{kj} - \sum_{k, j}B_{kj}J_{kj} $$ and $$ A_{ij} = \frac{z_i^*z_j}{N} + \frac{z_j^*z_i}{N} + \delta_{ij}\text{ and } B_{kj} = \frac{\omega^*z_k^*z_j}{\sigma_kN}+ \frac{\omega z_j^*z_k}{\sigma_kN} $$


It confuses me how this later expression expends the previous formula for $\Phi(\omega)$ ? Could someone help me to show some detailed calculations to explain ?

How the formula later with $Q, \hat{Q}, A, B$ etc. to be derived from previous formula of $\Phi(\omega)$ ? They are supposed to be equivalent.

Remark: The question is extracted from two papers for understanding of gap calculation.

Sommers, H. J., et al. "Spectrum of large random asymmetric matrices." Physical Review Letters 60.19 (1988): 1895-1898.

Rajan, Kanaka, and L. F. Abbott. "Eigenvalue spectra of random matrices for neural networks." Physical review letters 97.18 (2006): 188104.


1 Answer 1


Your first expression for the potential $$ \Phi(\omega)=\frac{1}{N}\log E_J \int d^2z...$$ is equivalent to $$ e^{N\Phi(\omega)}=E_J \int d^2z...$$ Ok? Writing the expectation in $J$ according to its definition gives $$ e^{N\Phi(\omega)}=\int dJ e^{-N\sum_{ij}J_{ij}^2} \int \prod_i\frac{d^2 z_i}{\pi}e^{-\epsilon z^\dagger z}e^{-z^\dagger(\omega^*-(J\sigma)^T)(\omega-J\sigma)z}$$ Notice that the matrix they are interested in is $J\sigma$ (in the second paper you mentioned).

Now redefine the complex integration variables as $z_i\to z_i/\sigma_i$. This will lead to your final expressions.


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