# Gaussian Integral over matrix elements with correlation

$$\mathbf{J}$$ is a random matrix where $$J_{ij}$$ follows a Gaussian distribution.

Consider the following integral:

$$I=\int\left(\prod_{ij}\mathrm{d}J_{ij}\right) \exp\left\{-\frac{N}{2} \sum_{i, j, k} J_{k i} A_{i j} J_{k j}+N\sum_{k, j} B_{k j} J_{k j}\right\}$$

Where $$\mathbf{A}$$ and $$\mathbf{B}$$ are Hermitian. This is a regular Gaussian integral and by completing the square I can obtain (if not mistaken?):

$$I=(2 \pi)^{\frac{N^2}{2}}(\operatorname{det} \mathbf{A})^{-N / 2} \exp \left\{\sum_{i,j,k}^{n} \frac{1}{2} B_{ki}\left( A^{-1}\right)_{i j} B_{jk}\right\}$$

However if the elements $$J_{ij}$$ are correlated and my integral $$I$$ now becomes:

$$I=\int\left(\prod_{ij}\mathrm{d}J_{ij}\right) \exp\left\{-\frac{N}{2} \sum_{i, j, k} J_{k i} A_{i j} J_{k j}+N\sum_{k, j} B_{k j} J_{k j} +\tau N\sum_{ij}J_{ij}J_{ji}\right\}$$

with $$-1<\tau<1$$.

How can I deal with the $$\sum_{ij}J_{ij}J_{ji}$$ terms?

Any remark or advice is always appreciated. Thanks.

• Replacing $A_{ij}$ with $A_{ij}+\tau N\delta_{ij}$ should do the trick. Feb 16 '20 at 6:09
• It actually does not: with this substitution we obtain $\tau N \sum_{ij}J_{ij}^2$. I aim for $\tau N \sum_{ij}J_{ij}J_{ji}$
– Matt
Feb 25 '20 at 5:17

We assume that we are integrating over Hermitian matrices. Completing the square gives \begin{aligned} &I=(2\pi)^{\frac{N^2}2}(\det A)^{-\frac N2}\int\prod_{i,j}dJ_{ij}\exp\Bigg(-\frac N2\text{tr}\bigg[((A-2\tau1_N)J-B)^\dagger(A-2\tau1_N)^{-1}((A-2\tau1_N)J-B) - \frac N2B(A-2\tau1_N)^{-1}B\bigg]\Bigg) \end{aligned} Doing a linear shift in $$J$$ by $$(A-2\tau 1_N)^{-1}B$$ gives us \begin{aligned} &=(2\pi)^{\frac{N^2}2}(\det A)^{-\frac N2}\int\prod_{i,j}dJ_{ij}\exp\Bigg(-\frac N2\text{tr}\bigg[J^\dagger(A-2\tau1_N)J - \frac N2B(A-2\tau1_N)^{-1}B\bigg]\Bigg). \end{aligned}

So we need now to evaluate \begin{aligned} Z=\int\prod_{i,j}dJ_{ij}\exp\Bigg(-\frac N2\text{tr}\bigg[J^\dagger AJ\bigg]\Bigg). \end{aligned} Since $$A$$ is Hermitian, there exists a unitary $$U$$ such that $$A=UDU^\dagger$$, for $$D=\text{diag}(\lambda_1,\dots,\lambda_N)$$ and we assume $$\lambda_i\in\mathbb R_{>0}$$. We do the change of variables $$U^\dagger MU= J$$. With this change of variables \begin{aligned} \text{tr}(JAJ)&=\sum_i\lambda_iM_{ii}^2+\sum_{i\neq j}(\lambda_i+\lambda_j)\left((M_{ij}^{(r)})^2+(M_{ij}^{(im)})^2\right)\\ &=\sum_i\lambda_iM_{ii}^2+2\sum_{i where $$M_{ij}^{(r)}$$ is the real part of $$M_{ij}$$ and $$M_{ij}^{(im)}$$ is the imaginary part.

Since $$\det U$$ has determinant 1, this means we can write \begin{aligned} Z&=\int\prod_{i,j}dM_{ij}\exp\Bigg(-\frac N2\bigg[\sum_i\lambda_iM_{ii}^2+2\sum_{i

Going back to $$I$$, if we assume that the eigenvalues of $$A$$ are all greater than $$2\tau$$, then the integral is convergent and we obtain $$$$I=(2\pi/\sqrt{N})^{N^2}\exp\left(\frac N2\text{tr}\left[B(A-2\tau1_N)^{-1}B\right]\right)\frac{1}{\sqrt{\det(A-2\tau 1_N)}(\det A)^{N/2}\prod_{i

I'm going to guess that the normalisation constant for the integral is incorrect. If the normalisation constant was $$C=\left(\frac{2\pi}{N^2/2}\right)^{-N^2}\sqrt{\det A}\prod_{i then the integral would be 1 at $$\tau=0$$.

As you pointed out, this is not the problem you had. You had $$J_{ij}$$ where $$J$$ is real valued, and the term you added was Tr$$(J^2)$$. Now we can decompose $$J=J^{(s)} + J^{(a)}$$ where $$J^{(s)}$$ is symmetric and $$J^{(a)}$$ is antisymmetric. Then \begin{align} \text{Tr}(J^2)&=\text{Tr}((J^{(s)})^2 + (J^{(a)})^2+2J^{(s)}J^{(a)})\\ &=\text{Tr}((J^{(s)})^2 + (J^{(a)})^2)\\ &=\text{Tr}((J^{(s)})^TJ^{(s)} - (J^{(a)})^TJ^{(a)})\\ \text{Tr}(JAJ^T)&=\text{Tr}(J^{(s)}AJ^{(s)} - J^{(a)}AJ^{(a)})\\ &=\text{Tr}(J^{(s)}AJ^{(s)} + (J^{(a)})^TAJ^{(a)}). \end{align} This uses the fact that the trace of an antihermitian matrix is zero.

Now change variables to $$K$$ where $$K^{(s)}=J^{(s)}$$ and $$K^{(a)}=iJ^{(a)}$$. Then $$K$$ is Hermitian and this should reduce to the problem at the top of my answer, except for the issue now that the pure imaginary part multiply $$A$$ has a negative sign. This means that this integral will actually diverge. This can be saved if $$\tau$$ is bigger than the absolute value of the eigenvalues of $$A$$.

• Pardon me if I am wrong, but I believe you represented $\tau N \sum_{i j} J_{i j} J_{j i}$ as $N\tau \text{Tr}\left[J^TJ\right]$, which is incorrect, when you complete the square.
– Matt
Feb 25 '20 at 7:41
• Also, J is not Hermitian.
– Matt
Feb 25 '20 at 7:53
• Your derivation was helpful, it taught me a new method so thank you. However, unless I misunderstood, my initial question has not been answered in the end since your derivation misrepresented the $\tau \sum_{i j} J_{i j} J_{j i}$terms...
– Matt
Feb 25 '20 at 7:55
• In what context is this problem appearing? Because, I'm not sure you can do the integral if $J$ is not hermitian. It won't converge. Feb 25 '20 at 19:43
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