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Gaussian Path Integral

  1. Sep 5, 2007 #1
    Einstein summation convention employed throughout

    We want to calculate

    [tex]\hbar \ln \int D x_i \exp[\frac{1}{32 \pi^3} \int ds \int d^3 r x_i(-is,r) M_{ij}(s,r) x_j(is,r)][/tex]

    The answer is

    [tex]\hbar \int \frac{ds}{2\pi} \ln \det[M_{ij}\delta^3(r-r')][/tex]

    I know that

    [tex]\int d^3 x_i e^{\frac{1}{2}x_i B_{ij} x_j} = \sqrt{\frac{(2\pi)^n}{\det B_{ij}}}[/tex]

    and that standard logarithmic properties will be used. Also the [tex]\delta^3(r-r')[/tex] means that a Fourier transform involving that delta function will be employed at some point.

    Beyond that I'm at a complete loss as to how to continue. One question is why we don't need to employ a Fourier transform involving a [tex]\delta(s-s')][/tex]. Any help would be much appreciated.
     
    Last edited: Sep 5, 2007
  2. jcsd
  3. Sep 6, 2007 #2

    Avodyne

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    I was hoping someone else would tackle this one, but here goes ...

    The answer doesn't make sense to me. I don't see why s should be fundamentally different than r. I think you should have a [tex]\delta(s+s')[/tex], with + rather than - because the arguments have opposite sign, and that there should be no integral over s.

    But perhaps there is something about the definition of x(is,r) that you haven't told us that would change this ...
     
  4. Sep 7, 2007 #3
    Thanks.

    What kind of things might make s different to r in that way?

    Perhaps these? "is" is wick rotated frequency; and started out as the fourier transform of time. r is real space. r is a vector and s is a scalar.
     
    Last edited: Sep 7, 2007
  5. Sep 7, 2007 #4

    Avodyne

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    No, none of that should matter.

    Is this from a book? If so, which one?
     
  6. Sep 7, 2007 #5
  7. Sep 7, 2007 #6

    Avodyne

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    He's just being sloppy. His eq.(1) is wrong, and should be what I said. Then he trades log det for Tr log, and the integral over zeta in eq.(3) is part of the trace, just like the integrals over r.
     
  8. Sep 7, 2007 #7
    Ahh brilliant. Thanks again. :)
     
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