# Mehta integral and orthogonality

The Mehta integral is the following expression:

$$\frac{1}{(2\pi)^{n/2}}\int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \prod_{i=1}^n e^{-t_i^2/2} \prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} dt_1 \cdots dt_n =\prod_{j=1}^n\frac{\Gamma(1+j\gamma)}{\Gamma(1+\gamma)}.$$

To simplify the notation, we introduce a measure $$d\mu_{n,\gamma}(t):=\prod_{i=1}^n e^{-t_i^2/2} \prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} \ dt.$$ It is easy to see that $$1$$ is orthogonal to $$(t^2-1)$$ when weighted with a Gaussian measure $$$$\label{eq:ortho} \frac{1}{(2\pi)^{1/2}}\int_{-\infty}^{\infty} e^{-t^2/2} (t^2-1)dt =0.$$$$

Now, it seems that again for different scalings of $$\gamma$$ we find interesting phenomena for the value of

$$\nu:=\lim_{n \rightarrow \infty}\frac{\int (t_1^2-1) \ d\mu_{n,\gamma}(t)}{\sqrt{\int (t_1^2-1)^2 \ d\mu_{n,\gamma}(t)}\sqrt{\int 1 \ d\mu_{n,\gamma}(t)}}$$

where I admit that I use the limit without actually knowing whether it exists.

Case 1:

As one can guess from this thread, if we choose $$\gamma=1/n^2$$ it seems that the product $$F(t)$$ does not contribute to the value of the above limit and we have $$\nu=0.$$

Case 2:

If we choose $$\gamma=1/n$$ then Carlo Beenakker's answer that treats the case 3 rigorously suggests that we find $$\nu=0$$ in this case, too.

Case 3:

If we choose $$\gamma=1$$ then it seems like we get that $$\nu$$ is of order one, which is confirmed by Carlo Beenakker's answer.

My question is: Consider Case 2 with scaling $$\gamma=1/n$$, then find the value of $$\nu$$ for large $$n$$?

• @CarloBeenakker you are right, simplifying the question I accidentally deleted it. Feb 15, 2020 at 22:44

Case 3:

Let me define $$t=x\sqrt{2\gamma}$$, then it is known from random-matrix theory (see, for example, Forrester's book) that for a fixed $$\gamma$$ the probability distribution $$P(x_1)$$ of a single eigenvalue $$x_1$$ tends in the limit $$n\rightarrow\infty$$ to the $$\gamma$$-independent semicircle $$P(x)=\frac{1}{\pi n}\sqrt{2n-x^2},\;\;|x|\leq\sqrt{2n}.$$ The desired ratio $$\nu$$ then evaluates to $$\nu=\frac{\int (2\gamma x^2-1)P(x)\,dx}{\left[\int (2\gamma x^2-1)^2P(x)\,dx\right]^{1/2}}=\frac{\gamma n-1}{\sqrt{2 \gamma n (\gamma n-1)+1}}\rightarrow \frac{1}{\sqrt 2}\;\;\text{for}\;\;n\rightarrow\infty.$$

Case 2:

The case that $$n\rightarrow\infty$$, $$\gamma\rightarrow 0$$ at fixed $$\gamma n=\alpha>0$$ has been studied in The mean spectral measures of random Jacobi matrices related to Gaussian beta ensembles (2014), see also arXiv:1611.09476. The probability distribution $$P_\alpha(t)$$ is given in this limit by $$P_\alpha(t)=\frac{e^{-t^2/2}}{\alpha\sqrt{2\pi}}\frac{\Gamma(\alpha)} {|f(t)|^2},\;\;f(t)=\int_0^\infty x^{\alpha-1}e^{ix t-x^2/2}\,dx.$$ From this the desired $$\nu$$ can be readily computed, $$\nu_\alpha=\frac{\int (t^2-1)P_\alpha(t)\,dt}{\left[\int (t^2-1)^2P_\alpha(t)\,dt\right]^{1/2}}=\frac{\alpha}{\sqrt{\alpha (2 \alpha+3)+2}},$$ so for $$\alpha=1$$ I find $$\nu_1=1/\sqrt 7$$. The value $$\nu=1/\sqrt 2$$ of case 3 is reached for $$\alpha\gg 1$$.

• thank you, that's interesting. Do you have any conjectures about case 2 whether the result will be of "order 1" or "tend to zero"? Feb 15, 2020 at 15:13
• my "conjecture" would be to substitute $\gamma n=\text{constant}$ in the formula for $\nu$, which would give $\nu=0$ in case 2; incidentally, the plots you show do not take into account the repulsion between the $x_i$'s with $i$ unequal to 1, so I don't think one can draw any conclusion from those. Feb 15, 2020 at 15:35
• thank you for your prompt response. This sounds very plausible. I will update the question accordingly. Feb 15, 2020 at 15:48
• very interesting, so in Case 3: we do not see $\gamma$ at all in the limit and in this regime $2$, although the value is of order $1$, it vanishes as $\alpha$ tends to zero. Feb 16, 2020 at 15:16
• yes, this somehow makes sense if you think of case 2 approaching case 3 when $\alpha\rightarrow\infty$ (hence $\nu\rightarrow 1/\sqrt 2$), and approaching case 1 if $\alpha\rightarrow 0$ (hence $\nu\rightarrow 0$). Feb 16, 2020 at 15:30