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Given the Euclidean action \begin{equation} S_E(\phi) = \int d^d x \frac{1}{2}\big(\nabla\phi\cdot\nabla\phi + m^2\phi^2\big)\end{equation} and the partition function \begin{equation}\mathcal{Z} = \int \mathcal{D}\phi(x)e^{-S_E(\phi) + \int d^d x J(x)\phi(x)} \end{equation} 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 \begin{equation} \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} \end{equation} 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?

Qmechanic
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Lies Van Dael
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  • You should include the calculations you've done (the most important steps). – Tobias Fünke Oct 06 '21 at 14:20
  • You answered the question yourself. The answer is Wick's theorem. –  Oct 06 '21 at 15:27
  • Wick's theorem would be a more straightforward calculation. If you are using Generating functional, you can refer to derivation for 4-point function in section 14.3.2 of Matthew D. Schwartz. Or you can also use the Schwinger-Dyson equations which will readily give you the desired expression for n-point function – KP99 Oct 06 '21 at 16:41
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    Schwinger-Dyson has been brought up a couple of times so you may want to check this PSE post, in particular equation 10. – AccidentalFourierTransform Oct 06 '21 at 20:20

1 Answers1

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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.

Qmechanic
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