When i study about effective potential in QFT. I found an equation for Fourier transform of $n$-vertex function, $\Gamma(x_1,x_1,....,x_n)$, given by: $$\Gamma(x_1,x_1,....,x_n)=\int d^4p_1d^4p_2...d^4p_n\Gamma(p_1,p_2,...,p_3)\delta(p_1+p_2+....+p_n)e^{-i p_1\cdot x_1-i p_2\cdot x_2-...-i p_n\cdot x_n}$$ I don't understand how the dirac delta appear in the integrand. When i look up from another source, it says that this is due to translational invariance. But, How do i prove this?
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1$\begingroup$ This has nothing to do with vertex functions; it is a simple property of Fourier transforms. Have you studied a simpler case first? Consider a function $f: \mathbb R \times \mathbb R\to \mathbb C$, s.t. $f(x_1,x_2)=f(x_1+a,x_2+a)$ for all $a\in \mathbb R$; make a suitable choice for $a$ and show that there exists a function $h:\mathbb R \to \mathbb C$ such that $f(x_1,x_2)=h(x_1-x_2)$. Now compute the (two-dimensional) Fourier transform of both sides and conclude. $\endgroup$– Tobias FünkeCommented Dec 8 at 8:50
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Let's look at the inverse Fourier transform. Define $$ {\hat \Gamma}(p_1,\cdots,p_n) = \int d^4 x_1 \cdots d^4 x_n e^{i p_1 \cdot x_1 + \cdots i p_n \cdot x_n } \Gamma (x_1 , \cdots , x_n ) . $$ For all $i \neq n$, we change the integration variable $x_i \to x_i + x_n$. Then, $$ {\hat \Gamma}(p_1,\cdots,p_n) = \int d^4 x_1 \cdots d^4 x_{n-1} d^4 x_n e^{i p_1 \cdot x_1 + \cdots i p_{n-1} \cdot x_{n-1} + i ( p_1 + \cdots + p_n ) \cdot x_n } \Gamma (x_1 + x_n , \cdots , x_{n-1} + x_n , x_n ) . $$ Now, translational invariance implies that $$ \Gamma (x_1 + x_n , \cdots , x_{n-1} + x_n , x_n ) = \Gamma(x_1,\cdots,x_{n-1},0). $$ We can now perform the integral over $x_n$ explicitly and find $$ {\hat \Gamma}(p_1,\cdots,p_n) = (2\pi)^4 \delta^4( p_1 + \cdots + p_n ) \int d^4 x_1 \cdots d^4 x_{n-1} e^{i p_1 \cdot x_1 + \cdots i p_{n-1} \cdot x_{n-1} } \Gamma (x_1 , \cdots , x_{n-1} , 0 ) . $$ We now define $$ \Gamma(p_1,\cdots,p_n) \equiv \int d^4 x_1 \cdots d^4 x_{n-1} e^{i p_1 \cdot x_1 + \cdots i p_{n-1} \cdot x_{n-1} } \Gamma (x_1 , \cdots , x_{n-1} , 0 ) $$ which implies $$ {\hat \Gamma}(p_1,\cdots,p_n) = (2\pi)^4 \delta^4( p_1 + \cdots + p_n ) \Gamma(p_1,\cdots,p_n) $$ Therefore $$ (2\pi)^4 \delta^4( p_1 + \cdots + p_n ) \Gamma(p_1,\cdots,p_n) = \int d^4 x_1 \cdots d^4 x_n e^{i p_1 \cdot x_1 + \cdots i p_n \cdot x_n } \Gamma (x_1 , \cdots , x_n ) . $$ We can now inverse this Fourier transform to obtain the equation in the post.
