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

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

  • $\begingroup$ The assumption that the momentum does not get changed by the interaction is unjustified. $\endgroup$ Commented Aug 14, 2018 at 9:38

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Commented Aug 14, 2018 at 14:19
  • $\begingroup$ It depends on the operator. Potentials are usually multiplicative in position space and convolutive in the momentum space. $\endgroup$
    – Fizikus
    Commented Aug 14, 2018 at 17:23

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