# How do I show that the $n$-point correlator $\left\langle\phi(x_1)\phi(x_2)...\phi(x_n)\right\rangle$ is equal to this expression?

Given the Euclidean action $$$$S_E(\phi) = \int d^d x \frac{1}{2}\big(\nabla\phi\cdot\nabla\phi + m^2\phi^2\big)$$$$ and the partition function $$$$\mathcal{Z} = \int \mathcal{D}\phi(x)e^{-S_E(\phi) + \int d^d x J(x)\phi(x)}$$$$ I need to show that the $$n$$-point correlator $$\left\langle\phi(x_1)\phi(x_2)...\phi(x_n)\right\rangle$$ satisfies $$\left\langle\phi(x_1)\phi(x_2)...\phi(x_n)\right\rangle = \sum_{i = 2}^nG(x_1 - x_i)\left\langle\phi(x_2)...\phi(x_{i-1})\phi(x_{i+1})...\phi(x_n)\right\rangle$$ where $$G(x_1 - x_i) = \left\langle\phi(x_1)\phi(x_i)\right\rangle$$.

I tried getting there by using that $$$$\left\langle\phi(x_1)\phi(x_2)...\phi(x_n)\right\rangle = {\frac{1}{\mathcal{Z}}\frac{\delta}{\delta J(x_1)}...\frac{\delta}{\delta J(x_n)}\mathcal{Z}}_{J = 0}$$$$ by calculating the derivatives directly and I also tried doing it by induction, but both if these methods did not get me anywhere.

I also thought about proving that this statement is equivalent to Wick's theorem, but I'm not sure why.

Could someone help me out with this please?

• You should include the calculations you've done (the most important steps). Oct 6, 2021 at 14:20
Hint: Instead of Wick's theorem for free fields, one can alternatively use the Schwinger-Dyson equation on a $$(n\!-\!1)$$-point function $$\left\langle\phi(x_2)\ldots\phi(x_n)\right\rangle$$. Next apply $$(-\nabla_1^2+m^2)^{-1}$$ on both sides.