4-point correlation in phi 4 to first order

[jfy4]

Hi,

Look at
<br /> \begin{align}<br /> &amp; \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)] \\<br /> &amp; \times (-i\lambda)D(x_1-x)D(x_2-x)D(x_3-x)D(x_4-x)<br /> \end{align}<br />
for first order in lambda for 2-2 scattering. In Maggiore I am told to substitute y_i=x_i-x as a variable substitution and then carry out the y_i integrals. Fair enough, but what about the measures? y_i=x_i -x \rightarrow dy_i = dx_i -dx. Surely x isn't a constant, since the coupling point could be anywhere, then I should have a whole separate integral with -d^4 x, right? If I make the substitution while ignoring the measure for emphesis I get
<br /> \begin{align}<br /> &amp; \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))] \\<br /> &amp; \times (-i\lambda)D(y_1)D(y_2)D(y_3)D(y_4)<br /> \end{align}<br />
now if I directly make the substitution d^4 x_i \rightarrow d^4 y_i and integrate over y_i I get
<br /> (-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]<br />
which is precisely the next line of Maggiore, except I cheated, didn't I...? (Here D(p_i) are the momentum space representation of the Feynman propagator.) What happened to my -d^4 x 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 x_j with j \in \{1,2,3,4\} at fixed x. Then indeed \mathrm{d}^4 x_j=\mathrm{d}^4 y_j. After that the x integral gives you the energy-momentum conserving Dirac \delta, i.e., (2 \pi)^4 \delta^{(4)}(p_1+p_2-k_1-k_2) 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 x in y_i = x_i +x 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.