In order to solve the Green’s function of the Helmholtz operator $$(\nabla^2+k^2)G(\vec r-\vec r’)=\delta^{(3)} (\vec r-\vec r’)$$ one can obtain four different Green’s functions corresponding to four different choices of avoiding the poles and choosing the contour. How to choose the correct Green’s function for a given problem?


Consider the case of a free scalar field, governed by the usual Lagrangian,

$$\mathcal{L} = \frac{1}{2}\partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2 \phi^2$$

The propagator, or equivalently Green's function for the theory is a function which can be though of as a response when we use a delta function as an input in the equations of motion, i.e.

$$\square \Delta_F(x-y) \sim \delta^{(4)}(x-y)$$

Explicitly, it is given by a Fourier integral over four-momentum,

$$\Delta_F (x-y)=\int \frac{d^4 p}{(2\pi)^4} \frac{i}{p^2-m^2} e^{-ip \cdot (x-y)}$$

We encounter a singularity at $p^0 = \pm \sqrt{p^2+m^2} = \pm E_{p}$. We can choose a contour which avoids these by dipping below the first, and then above the other. However, to apply the residue theorem, it must be closed. For $x^0 > y^0$, we close it in an anti-clockwise direction in the upper-half plane, and the opposite if $y^0 > x^0$. Alternatively, we can define the Feynman propagator,

$$\Delta_F = \int \frac{d^4 p}{(2\pi)^4} \, \frac{i}{p^2-m^2 + i\epsilon} e^{-ip\cdot (x-y)}$$

The $i\epsilon$ prescription due to Feynman shifts the poles by an infinitesimal amount away from the real axis; as a result the integral going straight through the real line is equivalent to the aforementioned contour.

Depending on your purpose, it may be useful to pick a particular contour, in which case we can define the retarded propagator $\Delta_R$ as the one which chooses to go over each pole on the real line, and the advanced contour going under. See the depiction below:

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To understand what they mean physically, consider in the context of response theory, the response function $\chi$ which determines how a system changes under the addition of a source $\phi$, i.e.

$$\delta \langle \mathcal{O}_i(t) \rangle = \int dt' \chi(t-t')\phi(t')$$

It's clear the aforementioned is in fact a convolution, $\chi \ast \phi$, and $\chi$ also has the interpretation of a Green's function. But we can't affect the past, so clearly,

$$\chi(t) = 0, \quad t < 0$$

For the Fourier transform, this is equivalent to stating,

$$\chi(\omega) \quad \mathrm{analytic}, \quad \mathrm{Im} \, \omega > 0$$

In other words, $\chi(\omega)$ has no poles in the upper-half plane. This object $\chi(t)$ is in fact our retarded Green's function, and is also called the causal Green's function, precisely because the above requirement is imposed.


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