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4-point correlation in phi 4 to first order

  1. Aug 1, 2012 #1
    Hi,

    Look at
    [tex]
    \begin{align}
    & \int d^4 x d^4 x_1 d^4 x_2 d^4 x_3 d^4 x_4 \exp [i(p_1 x_1 + p_2 x_2 -k_1 x_3 -k_2 x_4)] \\
    & \times (-i\lambda)D(x_1-x)D(x_2-x)D(x_3-x)D(x_4-x)
    \end{align}
    [/tex]
    for first order in lambda for 2-2 scattering. In Maggiore I am told to substitute [itex]y_i=x_i-x[/itex] as a variable substitution and then carry out the [itex]y_i[/itex] integrals. Fair enough, but what about the measures? [itex]y_i=x_i -x \rightarrow dy_i = dx_i -dx [/itex]. Surely [itex]x[/itex] isn't a constant, since the coupling point could be anywhere, then I should have a whole separate integral with [itex]-d^4 x[/itex], right? If I make the substitution while ignoring the measure for emphesis I get
    [tex]
    \begin{align}
    & \int d^4 x d^4 x_1 d^4 x_2 d^4 x_3 d^4 x_4 \exp [i(p_1 (y_1+x) + p_2 (y_2+x) -k_1 (y_3+x) -k_2 (y_4+x))] \\
    & \times (-i\lambda)D(y_1)D(y_2)D(y_3)D(y_4)
    \end{align}
    [/tex]
    now if I directly make the substitution [itex]d^4 x_i \rightarrow d^4 y_i [/itex] and integrate over [itex]y_i[/itex] I get
    [tex]
    (-i\lambda) D(p_1)D(p_2)D(p_3)D(p_4) \int d^4 x \exp[i(p_1 + p_2 - k_1 -k_2)x]
    [/tex]
    which is precisely the next line of Maggiore, except I cheated, didn't I...? (Here [itex]D(p_i)[/itex] are the momentum space representation of the Feynman propagator.) What happened to my [itex]-d^4 x[/itex] when I make the variable substitution above? A point in the right direction would be lovely, thanks.
     
  2. jcsd
  3. Aug 2, 2012 #2

    vanhees71

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    2016 Award

    You have not cheated. Why do you come to this idea? You first integrate over the [itex]x_j[/itex] with [itex]j \in \{1,2,3,4\}[/itex] at fixed [itex]x[/itex]. Then indeed [itex]\mathrm{d}^4 x_j=\mathrm{d}^4 y_j[/itex]. After that the [itex]x[/itex] integral gives you the energy-momentum conserving Dirac [itex]\delta[/itex], i.e., [itex](2 \pi)^4 \delta^{(4)}(p_1+p_2-k_1-k_2)[/itex] as it should be. In this way you can derive the Feynman-diagram rules, which are directly in four-momentum space and much more convenient than always doing these Fourier integrals.:smile:
     
  4. Aug 2, 2012 #3
    Thanks, I guess then my question is why is it okay for us to hold [itex]x[/itex] in [itex]y_i = x_i +x[/itex] fixed?
     
  5. Aug 2, 2012 #4
    Hi,

    You do not hold x fixed, it is just a change of variables, if you calculate explicitly the Hessian you will see.

    Actually you must do this using the 5 variables, x, y1, y2, y3 and y4.
     
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