Why are 2-point functions Green's functions?

Question

I asked a question about this earlier but I think it was unfocused so I have rephrased it and asked it again.

The propagator/two-point function $\langle \phi(x_1)\phi(x_2)\rangle$ for any theory can be found by finding the Green's function $G$ to the free theory, see e.g. this Phys.SE post. That is, if $L$ denotes the differential operator of the free theory then we would like to find $G$ such that $LG = \delta$.

I would like to know why the above is true. To be clear, I understand that one can directly compute the 2-point function and substitute the answer in the equations of motion to verify that it is indeed a Green's function. However, why do we expect this to be true? In other words, is there any reason why the 2-point function should be a Green's function to the free theory? Or is this just a nice coincidence that turns out to be true?

Answer 1

It follows from the Schwinger-Dyson equations. Let's prove it for the case you're interested in (though the equations are more general). I will do it for a scalar field and let you work out the generalisations to Dirac spinors and gauge fields.

Let $S[\phi]$ be the action for some theory and consider a one-point function. By definition $$ \langle \phi(x) \rangle = \int [d\phi] e^{i S[\phi]} \phi(x) = \int [d\phi] e^{i S[\phi+ \epsilon]} ( \phi(x) + \epsilon(x)) $$ In the second equality, we have shifted the integration variable $\phi(x) \to \phi(x) + \epsilon(x)$. Expanding RHS to first order in $\epsilon$, we find \begin{align} \langle \phi(x) \rangle &= \int [d\phi] e^{i S[\phi]} \left( 1 + i \int d^d y E(\phi(y)) \epsilon(y) \right) ( \phi(x) + \epsilon(x) ) \\ &= \int [d\phi] e^{i S[\phi]} \left( \phi(x) + \epsilon(x) + i \int d^d y \phi(x) E(\phi(y)) \epsilon(y) \right) \end{align} where $E(\phi(x))$ is the Euler-Lagrange equation for the action $S[\phi]$. The $O(1)$ term cancels on both sides. The $O(\epsilon)$ gives the equation $$ \int d^d z \langle E(\phi(z)) \phi(y) \rangle \epsilon(z) = i \epsilon(y) $$ This is true for any $\epsilon$, so we can set $\epsilon(z) = \delta^d(z-x)$ and we find $$ \langle E(\phi(x)) \phi(y) \rangle = i \delta^d(x-y). $$ This is true in ALL theories for any action. Now consider a Gaussian theory with action $S[\phi] = - \frac{1}{2} \int d^d x [ ( \partial \phi )^2 + m^2 \phi^2 ] $ for which $E(\phi(x)) = ( \Box - m^2 ) \phi(x)$. In this case, $$ ( \Box - m^2 ) \langle \phi(x) \phi(y) \rangle = i \delta^d(x-y). $$ But this is precisely what defines a Green's function (up to a factor of $i$. Therefore, $$ \langle \phi(x) \phi(y) \rangle = i G(x-y) $$