Integral Calculation: Solving for S with A as Symmetric Matrix

[ChrisVer]

Homework Statement

I'm trying to calculate the integral:
S= \int (d^{N}x) exp(x_{i} A_{ij} x_{j}) = (\frac{\pi^{N}}{detA})^{\frac{1}{2}}
where the integration is done over (-∞,+∞) , and A_{ij}=A_{ji} (symmetric NxN matrix)

Homework Equations

The Attempt at a Solution

I am not sure how am I supposed to start calculating... Please don't give explicit answer, just a starting hint??

 

[Ray Vickson]

Homework Statement

I'm trying to calculate the integral:
S= \int (d^{N}x) exp(x_{i} A_{ij} x_{j}) = (\frac{\pi^{N}}{detA})^{\frac{1}{2}}
where the integration is done over (-∞,+∞) , and A_{ij}=A_{ji} (symmetric NxN matrix)

Homework Equations

The Attempt at a Solution

I am not sure how am I supposed to start calculating... Please don't give explicit answer, just a starting hint??

I assume you meant to write $\exp(-\sum_i \sum_j A_{ij} x_i x_j)$ instead of $\exp(x_i A_{ij} x_j)$; note the sign difference, among other things. If so, look at 'Cholesky Decomposition'; see, eg., http://en.wikipedia.org/wiki/Cholesky_decomposition. That reduces the quadratic form to a sum of squares and thus reduces your integral to a sequence of standard Gaussians. Also: there are symmetric matrices A that make your so-called result false, so you had better find out what the true question really is.

 

[ChrisVer]

In fact the question is introductory to get into calculating:
\int d^{N}\theta d^{N}\bar{\theta} exp(-\bar{\theta}_{i}A_{ij} \theta_{j})
for \theta being grassmann variables...
My problem with that integral, is the case of finding the normal "gaussian" integral...for which I have:
exp(-a \bar{\theta} \theta)= 1-a \bar{\theta} \theta
which gives as a result after integrating:
\int d\theta d\bar{\theta} (1-a \bar{\theta} \theta))= -a
(or should I first anticommute the \thetas?)
If I use the same procedure as for the normal multidimensional gaussian integral (I'm asking about) -after diagonalizing the A etc- I will get:
\int d^{N}\theta d^{N}\bar{\theta} exp(\sum_{i}-\bar{\theta}_{i}A_{ii} \theta_{i})= ∏_{i} (-A_{ii}) ≠ detA
which I find everywhere as a result... well it depends on the dimensions, because for N=even then indeed I get the detA result...otherwise (if N is odd) I'm getting a minus overall sign...

 

[vanhees71]

I guess that's not a Grassmann integral but a usual real integral. The trick is to realize that you can diagonalize the matrix with an SO(N) transformation. Then everything splits in a product of single Gaussians, and this product can be written in terms of the determinant. Note that there should be the sign change as indicated in posting #2. The sum symbols are not necessary, if the Einstein summation convention is used.

Of course, you should also check for which matrices the integral exists at all!