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Functional Quantization of Scalar Fields

  1. Oct 16, 2009 #1
    Hi everyone,

    I'm reading section 9.2 of Peskin and Schroeder, and have trouble understanding the origin of a term in the transition from equation 9.26 to 9.27. Specifically, equation 9.26 is

    [tex]\frac{1}{V^2}\sum_{m,l}e^{-(k_m\cdot x_1 + k_l\cdot x_2)}\left(\prod_{k_{n}^{0}>0}\int d \Re \phi_{n}d \Im \phi_{n}\right)\times (\Re \phi_m + i\Im \phi_m)(\Re \phi_n + i\Im \phi_n)\times\exp{\left[-\frac{i}{V}\sum_{k_{n}^{0}>0}(m^2-k_{n}^2)\left[(\Re \phi_n)^2 + (\Im \phi_n)^2\right]\right]}[/tex]

    This splits into two cases:

    Case 1: [itex]k_l = k_m[/itex], when the integral is zero.
    Case 2: [itex]k_l = -k_m[/itex], when the integral is nonzero.

    So, evaluating the Gaussian integral one gets

    [tex]\mbox{Numerator} = \frac{1}{V^2}\sum_{m,l}e^{-(k_m\cdot x_1 + k_l\cdot x_2)}\left(\prod_{k_{n}^{0}>0}\frac{-i\pi V}{m^2 - k_{n}^2}\right)\frac{-iV}{m^2-k_{m}^2-i\epsilon}[/tex]

    Where does the factor

    [tex]\frac{-iV}{m^2-k_{m}^2-i\epsilon}[/tex]

    come from? I know this is like [itex]\int x^2 e^{-x^2}dx[/itex], but I can't seem to get this factor.)
     
  2. jcsd
  3. Oct 18, 2009 #2
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