# 4-point correlation in phi 4 to first order

1. Aug 1, 2012

### jfy4

Hi,

Look at
\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}
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
\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}
now if I directly make the substitution $d^4 x_i \rightarrow d^4 y_i$ and integrate over $y_i$ I get
$$(-i\lambda) D(p_1)D(p_2)D(p_3)D(p_4) \int d^4 x \exp[i(p_1 + p_2 - k_1 -k_2)x]$$
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.

2. Aug 2, 2012

### 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.

3. Aug 2, 2012

### jfy4

Thanks, I guess then my question is why is it okay for us to hold $x$ in $y_i = x_i +x$ fixed?

4. Aug 2, 2012

### 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.