# Prefactor $\delta(\Sigma_{i}^{n}k_{i})$ of the $n$-point correlation function

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It can be shown that the $2$-point function $\tilde{G}(k_{1},k_{2})$ of a Poincare-invariant QFT has a prefactor $\delta(k_{1}+k_{2})$ for translational invariance. How to show this for an $n$-point function, where $n>2$?

It's not a homework question.

My comment more or less contained already the answer. Expand your question by showing how you do it for $n=2$ instead of just saying ''it can be shown''. (The general case can also be shown, which tells you that ''It can be shown'' is an empty phrase unless you can show it.)

Why is $G(x,y)=G(x-y,0)$? You need this argument to be able to get the answer to your question.

$<\Omega|\phi(x)\phi(y)|\Omega>=<\Omega|U^{\dagger}(y)U(y)\phi(x)U^{\dagger}(y)U(y)\phi(y)U^{\dagger}(y)U(y)|\Omega>=$
$=<\Omega|\phi(x-y)\phi(0)|\Omega>$

Now do the same for the $n$-point function!

$G(x,y)=<\Omega|\phi(x)\phi(y)|\Omega>=<\Omega|U^{\dagger}U\phi(x)U^{\dagger}U\phi(y)U^{\dagger}U|\Omega>$,
$\tilde{G}(k_{1},k_{2})=\int dx \int dy e^{ik_{1}x+ik_{2}y}G(x,y)$
I edited the useful part of your derivation. You didn't specify what $U$ is and you didn't give the details that showed how you can see that $G(x,y)$ depends only on the difference. Thus you didn't actually show so far more than what I kept in your comment.
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