Let $H$ be a positive definite Hermitian matrix. How to prove
$\int_{\mathbb R^n}\int_{\mathbb R^n}e^{-z^*Hz}dxdy=\frac{\pi^n}{\det H},$ where $z=x+iy$ and $dx$, $dy$ denote the integration over real $n$ dimensional region?
Let $H$ be a positive definite Hermitian matrix. How to prove
$\int_{\mathbb R^n}\int_{\mathbb R^n}e^{-z^*Hz}dxdy=\frac{\pi^n}{\det H},$ where $z=x+iy$ and $dx$, $dy$ denote the integration over real $n$ dimensional region?
Since $H$ is Hermitian, it admits Cholesky decomposition, i.e. $H = U^\dagger U$. Notice that $\det H = \vert \det U \vert^2$.
Then $-z^\ast H z = (U z)^\ast (U z)$. Introducing $w = U z$, and noticing that $\mathrm{d} x \mathrm{d} y = \frac{1}{\vert \det U \vert^2 } \mathrm{d}w_x \mathrm{d} w_y$ we have ( here $w_x = \mathop{Re}(U z)$ and $w_y = \mathop{Im}(U z)$):
$ \int_{\mathbb{R}^n \times \mathbb{R}^n} \mathrm{d} x \mathrm{d} y \,\, \exp(-z^\ast H z) = \int_{\mathbb{R}^n \times \mathbb{R}^n} \frac{\mathrm{d} w_x \mathrm{d} w_y}{\vert \det U \vert^2} \,\, \exp(-w^\ast w) = \frac{\pi^n}{\det H }. $