Gaussian integration of $Z(h) = \int e^{x.\Gamma x+hx} dx$

Question

I am working on the function defined as:

$$Z(h) = \int e^{-\frac{1}{2}x\Gamma x+hx}, dx$$

where $\Gamma$ is a $n\times n$ symmetric matrix, $h$ and $x$ are $n$-component vectors : $x =( x_1, x_2,\dots, x_n)$.

We define the scalar product $$\langle g(x)\rangle =\int g(x) P(x)dx$$ where

$$P(x)=\frac{1}{Z(0)}e^{-\frac{1}{2}x\Gamma x}.$$

Any ideas on how to show that $$\langle x_i,x_j \rangle = \left. \frac{\partial^2 \ln(Z(h))}{\partial h_i \partial h_j} \right|_{h=0} ?$$

Answer 1

You mean $\Gamma$ is positive definite, then $\Gamma = M^\top M$ so that $$Z(h) = \int_{\Bbb{R}^n} e^{-\frac12 \| M x-M^{-\top} h\|^2+\frac12\|M^{-\top}h\|^2}d^n x= \int_{\Bbb{R}^n} e^{- \frac12\| M M^{-1} y-M^{-\top} h\|+\frac12\|M^{-\top}h\|^2}d^n (M^{-1} y) \\ = \int_{\Bbb{R}^n} e^{-\frac12 \| y-M^{-\top} h\|^2+\frac12\|M^{-\top}h\|^2}|\det(M)|^{-1} d^n y= \int_{\Bbb{R}^n} e^{- \frac12\| y\|^2+\frac12\|M^{-\top}h\|^2}|\det(M)|^{-1} d^n y\\ = |\det(M)|^{-1} \sqrt{\pi/2}^ne^{\frac12\|M^{-\top}h\|^2}= |\det(\Gamma)|^{-1/2} \sqrt{\pi/2}^ne^{\frac12 h^\top \Gamma^{-1} h}$$