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

Question

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 $n>2$?

Comments

We are not answering homework questions. Once you know how to show this for $n=2$ you can immediately generalize the argument to $n>2$.

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It's not a homework question.

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@ArnoldNeumaier: I think we should apply the term homework only in its proper narrow meaning to actual assignments ...

So this seems to be an ordinary (but maybe a bit too trivial?) technical question. 

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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.) 

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I am not good at expressing myself in English. I derived it for $n=2$, and I cannot get a delta function prefactor for $n=3$. Maybe I am too stupid, but it is not trivial to me. That's why I ask this question here. Saying that it is a homework question is totally clueless. If you think that my questions are inappropriate here, I would stop asking more questions.  

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The least you need to do is to add the derivation for $n=2$, as requested. Then one can give hints on how to modify your derivation to work in general.

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

$\tilde{G}(k_{1},k_{2})=\int dx \int dy e^{ik_{1}x+ik_{2}y}G(x,y)$

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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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$G(x,y)=G(x-y,0)=D(x-y)$.

$\tilde{G}(k_{1},k_{2})=\int dx \int dy e^{ik_{1}x+ik_{2}y}G(x,y)=$

$=\int dx \int dy e^{ik_{1}x+ik_{2}y}D(x-y)=\int dz \int dy e^{ik_{1}z} e^{i(k_{1}+k_{2})y}D(z)$. $U(y)=\exp(i\hat{P}y)$

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

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

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