4-point correlation in phi 4 to first order

[jfy4]

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.

 

[vanhees71]

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.

 

[jfy4]

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?

 

[arojo]

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.