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I need to carry out an average over a random matrix $J$ where its entries are gaussian random variables that are correlated by $\tau$. Typically the average is computed by evaluating the integral:

$$\langle\; f(J)\;\rangle\propto \int \left(\prod_{ij}\mathrm{d}J_{ij}\right)f(J)\exp\left \{ -\frac{N}{2(1-\tau^2)}\sum_{ij}J_{ij}^2+\frac{N}{2(1-\tau^2)}\sum_{ij}J_{ij}J_{ji}\right\} $$

My problem is how to decouple the $J_{ij}J_{ji}$ in these kind of integrals.

Usually I can express it as a square, call it $b^2$ and then can decouple it by adding a Gaussian integral: $$\int \frac{\mathrm{d} x}{\sqrt{2 \pi / a}} \exp \left\{-\frac{a}{2} x^{2} \pm b x\right\}=\exp \left\{\frac{b^{2}}{2 a}\right\}$$ By adding the integral, the quadratic terms indeed temporarily vanish.

However in this case I do not know how to represent $\sum_{ij}J_{ij}J_{ji}$ as a square of sums.

Any ideas? thanks !

Edit: To give some context I am computing the following integral:

\begin{equation} =\int_\mathbf{J}\exp{ \left\{-\epsilon \sum_{i}\left|z_{i}\right|^{2}-|\omega|^2\sum_i|z_i|^2-\sum_{ij}z_i^*z_j\omega^*J_{ij}-\sum_{ij}z_i^*z_j\omega J_{ji}+\sum_{ijk}J_{ki}J_{kj}z_i^*z_j-\frac{N}{2(1-\tau)}\sum_{ij}J_{ij}^2+\frac{N}{2(1-\tau)}\sum_{ij}J_{ij}J_{ji}\right\}} \end{equation}

where $z_i$ and $\omega$ are complex numbers and $\omega^*$ represents the complex conjugate. As you can see, I also have trouble with decoupling $\sum_{ijk}J_{ki}J_{kj}z_i^*z_j$ which would be written as: $$ \sum_k\left|\sum_{i}J_{ki}z_i\right|^2$$ But that would be the result of a complex gaussian integral, and $1)$ I don't know how to do that, $2)$ that would probably deserve a separate question. My current concern is how to uncouple $J_{ij}J_{ji}$ which is something that would be useful to know in the future. Thanks!

(To give all the context, I am trying to reproduce the results of the following paper... Sommers, H. J., Crisanti, A., Sompolinsky, H., & Stein, Y. (1988). Spectrum of large random asymmetric matrices. Physical review letters, 60(19), 1895. )

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  • $\begingroup$ Is the first term supposed to be $\left(\prod_{ij} dJ_{ij}\right)$? $\endgroup$
    – Kajelad
    Jul 29 '19 at 17:52
  • $\begingroup$ Yes indeed! Thank you for noticing it. I corrected it in an edit. $\endgroup$
    – Matt
    Jul 29 '19 at 17:55
  • $\begingroup$ A few other points of confusion: The exponential term does not seem to be a probability distribution, for instance it vanishes when $f(J)=1$ in the $1\times 1$ case. Are you looking for a change of variables which transforms the integral into an expectation for an uncorrelated Gaussian matrix, or something else? What, if anything, do you know about the function $f$? $\endgroup$
    – Kajelad
    Jul 29 '19 at 18:09
  • $\begingroup$ Sorry, I realise it is the case, I know my function $f$ but I discarded it because I thought it would bring more confusion. I have added it now, for completeness. $\endgroup$
    – Matt
    Jul 29 '19 at 18:27
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It may be helpful to write out the integral in matrix notation. I'll omit the prefactor $\frac{N}{2(1-\tau)}$. $$ \int_J\exp\left\{\text{Tr}(J^2-J^T J)\right\} $$ Here, one useful trick is to change variables so that the symmetric and antisymmetric parts of $J$ (which I'll call $A$ and $B$ respectively) are separated. Or, to be more precise: $$ A_{ij}=\frac{1}{2}(J_{ij}+J_{ji}),\ \ \ i\le j $$ $$ B_{ij}=\frac{1}{2}(J_{ij}-J_{ji}),\ \ \ i<j $$ $$ \prod_{ij}dJ_{ij}=2^{\frac{-N(N-1)}{4}}\left(\prod_{i\le j}dA_{ij}\right)\left(\prod_{i<j}dB_{ij}\right) $$ $$ J_{ij}=\begin{cases} A_{ij}+B_{ij} & i<j \\ A_{ij} & i=j \\ A_{ji}-B_{ji} & i>j \end{cases} $$ Wewriting the integral, $$ \int_J\exp\left\{\text{Tr}(2B^2+2BA)\right\} $$ Using the cyclic property of the trace, as well as the symmetry/antisymmetry of $A$/$B$, the second term vanishes. $$ \int_J\exp\left\{\text{Tr}(-2B^TB)\right\} $$ Or, in terms of elements, $$ 2^{\frac{-N(N-1)}{4}}\int\left(\prod_{i\le j}dA_{ij}\right)\left(\prod_{i<j}dB_{ij}\right)\exp\left(-4\sum_{i<j}B_{ij}^2\right) $$ Fortunately, the integral is completely separated; unfortunately, it diverges, since there is nothing constraining the symmetric part of $J$.

For your full integral, this approach wouldn't mess anything else up, but it wouldn't take care of the other quadratic term $z^\dagger J^T Jz$. Other tricks are needed for that one; perhaps a unitary change of variables to rotate $z$ onto one of the coordinate axes would do the trick.

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  • $\begingroup$ This is a great trick and very clearly explained, thank you. $\endgroup$
    – Matt
    Jul 30 '19 at 8:45

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