3
$\begingroup$

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?

$\endgroup$
1
5
$\begingroup$

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:

enter image description here


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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.