I need help finding the Fourier transform of the function

$$ \rho(\vec{r}) = \alpha \delta_{\vec{r},0} \left(\lambda\lambda' J_1 (\beta |\vec{r}|)Y_1(\beta |\vec{r}|) - \pi^2 J_0 (\beta |\vec{r}|)Y_0(\beta |\vec{r}|) \right), $$

where $\alpha,\beta \in \mathcal{R}$ and $\lambda,\lambda'=\pm1$. The $J_i(x)$ are the $i$-th Bessel functions of the first, the $Y_i(x)$ are the $i$-th Bessel functions of the second kind. $\delta_{\vec{r},0}$ denotes the Kronecker delta. This function represents the density of states after a scattering event off a delta potential on a discrete lattice in position space.

I would like to obtain the density of states in momentum space now and tried something like

$$ \rho(\vec{k}) = \sum_\vec{r} e^{-i\vec{k}\cdot\vec{r}}\rho(\vec{r}), $$

which should in fact not be too difficult because the only vector that contributes to the sum is $\vec{r} = \vec{0}$ due to the Kronecker delta in $\rho(\vec{r})$. The problem is that $\rho(\vec{r})$ diverges at this point. Is there any way to compute the Fourier transform of $\rho(\vec{r})$?

EDIT: In order to make clear where the Kronecker delta comes from, I'll explain briefly where it enters the scene.

I have the Green's function $G_0(\vec{r})$ needed for the description of scattering processes. This Green's function is a matrix and so is the scattering potential which is of the form

$$ V(\vec{r}) = \begin{pmatrix} V_0 \delta_{\vec{r},0} & 0 \\ 0 & 0 \end{pmatrix} $$

with $V_0$ constant. Since I'm working on a discrete lattice, the vector $\vec{r}$ is discrete, hence the Kronecker delta instead of a delta distribution. The density of states after one scattering event is then given by

$$ \rho(\vec{r}) = -\frac{1}{\pi} \text{Im Tr}\left[ G_0(\vec{r})V(\vec{r})G_0(\vec{r}) \right]. $$

So, the Kronecker delta comes from the potential.

  • $\begingroup$ The Kronecker delta makes extremely little sense with a continuous input $-$ and, basically, its integral is exactly zero, as you can modify it point-wise to just the zero function without changing the value of the integral. Your problems are upstream of the $\rho(\vec r)$ that you've posted, at the very least going back to the first point where that non-kosher $\delta_{\vec r,0}$ was introduced. $\endgroup$ – Emilio Pisanty Aug 10 '18 at 11:51
  • $\begingroup$ @EmilioPisanty: I've edited my post and explained where the Kronecker delta comes from. Could you tell me what should be done differently? $\endgroup$ – MeMeansMe Aug 10 '18 at 12:08
  • $\begingroup$ That still doesn't explain where the Kronecker delta comes from. The Dirac delta is useful because it is a distribution with a nontrivial effect, but when acting on the continuum the Kronecker delta is the distribution that takes any given function $f$ to zero (i.e. it is trivial). You need to do the distribution theory correctly upstream of that $V(\vec r)$. $\endgroup$ – Emilio Pisanty Aug 10 '18 at 12:15
  • $\begingroup$ To be honest, the only reason for the Kronecker delta is that I want to have some potential on the lattice point sitting in the origin. All other lattice points are unaffected by the potential. So, it's not clear to me what I'm supposed to do "before" $V(\vec{r})$ because I introduce the Kronecker delta by hand. Maybe I'm completely missing the point, though... $\endgroup$ – MeMeansMe Aug 10 '18 at 12:24
  • $\begingroup$ Introducing a Kronecker delta by hand is the same as introducing $V(\vec r) \equiv 0$ by hand, and you can just solve your problem right then and there. We cannot tell you what you're "supposed" to do because we do not have enough context of the analytical environment you're working with or what you want to describe with it. It sounds like you want to describe a contact interaction in some two-component spinor, which are normally given by Dirac deltas (and which you can read up on in any cold-atoms textbook), but again, we're not here to tell you what to do. $\endgroup$ – Emilio Pisanty Aug 10 '18 at 12:36

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