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Integral calculation

  1. Apr 22, 2014 #1

    ChrisVer

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    1. The problem statement, all variables and given/known data
    I'm trying to calculate the integral:
    [itex]S= \int (d^{N}x) exp(x_{i} A_{ij} x_{j}) = (\frac{\pi^{N}}{detA})^{\frac{1}{2}}[/itex]
    where the integration is done over (-∞,+∞) , and [itex]A_{ij}=A_{ji}[/itex] (symmetric NxN matrix)


    2. Relevant equations



    3. 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??
     
  2. jcsd
  3. Apr 22, 2014 #2

    Ray Vickson

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    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.
     
  4. Apr 22, 2014 #3

    ChrisVer

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    In fact the question is introductory to get into calculating:
    [itex] \int d^{N}\theta d^{N}\bar{\theta} exp(-\bar{\theta}_{i}A_{ij} \theta_{j})[/itex]
    for [itex]\theta[/itex] being grassmann variables...
    My problem with that integral, is the case of finding the normal "gaussian" integral....for which I have:
    [itex]exp(-a \bar{\theta} \theta)= 1-a \bar{\theta} \theta[/itex]
    which gives as a result after integrating:
    [itex] \int d\theta d\bar{\theta} (1-a \bar{\theta} \theta))= -a [/itex]
    (or should I first anticommute the [itex]\theta[/itex]s?)
    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:
    [itex] \int d^{N}\theta d^{N}\bar{\theta} exp(\sum_{i}-\bar{\theta}_{i}A_{ii} \theta_{i})= ∏_{i} (-A_{ii}) ≠ detA[/itex]
    which I find everywhere as a result.... well it depends on the dimensions, because for N=even then indeed I get the [itex]detA[/itex] result...otherwise (if N is odd) I'm getting a minus overall sign...
     
    Last edited: Apr 22, 2014
  5. Apr 23, 2014 #4

    vanhees71

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    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!
     
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