# Delta function from poles of Green's function

In quantum mechanical scattering theory, we often use Green's functions which contain poles. For example, in Schroedinger quantum mechanics the free Green's function is given by

$$G_0(\vec{p}) = \frac{1}{E-\frac{p^2}{2m}+i\epsilon}$$

in momentum space where the infinitesimal constant $\epsilon >0$ has been introduced to take care of the pole at $E = \frac{p^2}{2m}$. The imaginary part of $G_0$ is then

$$\text{Im}G_0 = -\frac{\epsilon}{(E-\frac{p^2}{2m})^2+\epsilon^2}$$

and letting $\epsilon$ go to zero, we get (using the Sokhotski–Plemelj formula)

$$\lim_{\epsilon\to0}\text{Im}\,G_0=-\pi\delta\left(E-\frac{p^2}{2m}\right).$$

The full Green's function is then given by the Dyson equation

$$G = G_0 + G_0 VG_0 + G_0VG_0VG_0+\ldots$$

with the scattering potential $V$. Looking at the second term in the Dyson equation, we see that the free Green's function appears twice giving rise to an expression proportional to $\frac{1}{(E-p^2/(2m)+i\epsilon)^2}$. I wonder what the imaginary part of this expression is. Phyiscally, there should still be some delta function because the physical particle fulfills the relation $E=\frac{p^2}{2m}$ even after elastic scattering but I don't see how the delta enters the game. So, how can I get a delta function from the fraction

$$\frac{1}{(E-\frac{p^2}{2m}+i\epsilon)^2}?$$

• The assumption that the momentum does not get changed by the interaction is unjustified. – Emilio Pisanty Aug 14 '18 at 9:38

The Green functions $G$ and $G_0$, as well as the scattering potential $V$ are operators. Therefore, if we choose to work in the momentum space, the string of operators, e.g. $G_0 V G_0$, have to be written as a convolution and all the dummy variables have to be integrated out. Doing this, you will never obtain a square of the free Green function as you claim. You shouldn't interpret Dyson equation a mere multiplication of functions.

• In position space $G_0VG_0$ corresponds to convolution; in momentum space, to multiplication. This is a basic property of the Fourier transform: it maps convolutions to point-wise products and vice-versa. – AccidentalFourierTransform Aug 14 '18 at 14:19
• It depends on the operator. Potentials are usually multiplicative in position space and convolutive in the momentum space. – Fizikus Aug 14 '18 at 17:23