Greens functions
# Mathematical definition
Definition: Green’s function
The Green’s function $G(x,s)$ of a linear differential operator $L=L(x)$ is any solution of $$L G(x,s) = \delta(s-x)$$
If a Green’s function is found it can be used to find the solution to the inhomogenous differential equation $$L u(x) = f(x),$$ as convolution $$u(x) = \int G(x,s)f(s)ds,$$ since $$L \int G(x,s)f(s)ds = \int L G(x,s)f(s)ds = \int \delta(x-s)f(s)ds = f(x).$$
# Physical meaning
In QFT the Green’s function is used to define the Propagator which describes how a particle propagates from $x$ to $y$ as given by the equation of motion.
Example: Klein-Gordon equation
For the Klein-Gordon equation we e.g. have $$(\square + m^2)G(x,y) = - \delta(x-y).$$ Using the Fourier transformation we find the solution $$G(x,y) = \frac{1}{(2\pi)^4} \int d^4p \frac{e^{-ip(x-y)}}{p^2-m^2\pm i\epsilon}.$$ The $\pm i\epsilon$ arises from the correct solving for $G(p)$ in the sense that it is a distribution (?).
In general given some lagrangian $\mathcal{L}$ we can derive the equations of motion using the euler lagrange equations. The Green’s function then corresponds to the solution of the free kinetic part, and is also called 2-point Green’s function. Solving for a field in an interacting theory involves pertubation theory, which leads to higher order terms involving multiple propagators. This translates to the picture of feynman graphs.
Note: Sometimes people write the Green’s function for a free massless scalar field with $\square h$ just as $\Pi = -\frac{1}{\square}$, which is the inverse of the wave operator. This is of course equivalent since the inverse of the operator is just defined as the solution to $$\square \Pi(x,y) = - \delta^4(x-y).$$
Example: Graviton
Heuristic lagrangian for the graviton (assuming it was a scalar field $h$) $$\mathcal{L}=-\frac{1}{2}h\Box h + \frac{1}{3}\lambda h^3 + Jh.$$ Here we have a self-interaction $h^3$ and $J$ is again some source (image a mass sitting around somewhere). The equation of motion gives $$\Box h -\lambda h^2 - J =0$$ which can be solved perturbatively for small $\lambda$. At leading order we have the normal equation like before $$h_0 = \frac{1}{\Box}J$$. At next to leading order in $\lambda$ ($\mathcal{O}(\lambda)$), we have $$h = h_0 + h_1,$$ with $h_0$ given above and $h_1\in\mathcal{O}(\lambda)$. We can plug this into the equation of motion and solve for $h_1$: $$\Rightarrow \Box h_1 = \lambda h^2_0 + \mathcal{O}(\lambda^2),$$ where we left out terms of order $\lambda^2$ like e.g. $\lambda h_1$. Solving this again and pluggin gin the solution for $h_0$ gives $$h = \frac{1}{\Box} + \lambda \frac{1}{\Box}\left(\frac{1}{\Box}J\right)\left(\frac{1}{\Box}J\right)+\mathcal{O}(\lambda^2).$$
This corresponds to the mathematical picture of the Greensfunction given above. $-\frac{1}{\Box}$ is given as the solution to $$\Box_x \Pi(x,y)=-\delta^4(x-y),$$ Corresponding the the Propagator between $x$ and $y$. Then the solution to $$\Box_x h_0(x) = J(x)$$ can be written in terms of an integral over the propagator as $$h_0(x)=-\int d^4y \Pi(x,y)J(y),$$ like before where now our linear operator $L=\Box$. The same can be done for the order $\lambda$ term. To solve $\Box_x h_1 = \lambda h^2_0(x)$ we would insert the integral above to get $$h_1 = -\lambda\int d^4w \int d⁴y\int d^4z \Pi(x,w)\Pi(w,y)\Pi(w,z)J(y)J(z).$$ Here we have three propagators, corresponding to the three $\frac{1}{\Box}$ we had in the perturbative solution. They can be thought of three lines in a Feynman diagram with a single 3 line vertex (this is the order $\lambda$ contribution) and two sources $J(z)$ and $J(y)$.
Our reasoning here could be extendend to calculate higher order terms in $\lambda$, which naturally leads to Position space Feynman rules.