I just wanted to check my result.

Let's define the Fourier transform as (the integral over whole real line):

$$g(k)=\frac{1}{\sqrt{2 \pi}} \int e^{-i k x} f(x) dx$$

We have the following function:


I want to Fourier it in all the Cartesian coordinates, which should be:

$$g(\mathbf{k}_1,\mathbf{k}_2)=\frac{1}{(2 \pi)^3} \int \int e^{-i \mathbf{k}_1 \mathbf{r}_1-i \mathbf{k}_2 \mathbf{r}_2} \frac{d\mathbf{r}_1 d\mathbf{r}_2}{|\mathbf{r}_1-\mathbf{r}_2|}=$$



$$=\frac{1}{(2 \pi)^3} \int \int e^{-i (\mathbf{k}_1+\mathbf{k}_2) \mathbf{r}_2-i \mathbf{k}_1 \mathbf{r}} \frac{d\mathbf{r}_2 d\mathbf{r}}{|\mathbf{r}|}=\delta(\mathbf{k}_1+\mathbf{k}_2) \int e^{-i \mathbf{k}_1 \mathbf{r}} \frac{d\mathbf{r}}{|\mathbf{r}|}=\frac{4 \pi \delta(\mathbf{k}_1+\mathbf{k}_2)}{\mathbf{k}_1^2}$$

The last integral is evaluated using spherical coordinates and the properties of Bessel functions.

What's confusing to me is that the transform should be symmetric in $1 \leftrightarrow 2$, meaning that if we exchange the indices, the result should be the same. But it doesn't seem to be symmetric. Where's my mistake?


$$\int e^{-i \mathbf{k} \mathbf{r}} \frac{d\mathbf{r}}{|\mathbf{r}|}=\int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{\infty} e^{-i \rho ( k_{x} \sin \theta \cos \phi + k_{y} \sin \theta \sin \phi+k_{z} \cos \theta)} \rho \sin \theta d \rho d \theta d \phi=$$

$$=2 \pi \int_{0}^{\pi} \int_{0}^{\infty} e^{-i \rho k_{z} \cos \theta} J_0 (\sqrt{k_x^2+k_y^2} \rho \sin \theta ) \rho \sin \theta d \rho d \theta=$$

$$=2 \pi \int_{0}^{\pi} \int_{0}^{\infty} e^{-i \rho k_{z} \cot \theta} J_0 (\sqrt{k_x^2+k_y^2} \rho ) \rho \csc \theta d \rho d \theta=$$

$$=4 \pi \int_{0}^{\infty} K_0 (k_z \rho) J_0 (\sqrt{k_x^2+k_y^2} \rho ) \rho d \rho=\frac{4 \pi}{k_x^2+k_y^2+k_z^2}$$


Since $\delta\ne0\implies\mathbf{k}_1=-\mathbf{k}_2$, an equivalent way to write the result would be the manifestly symmetric $\dfrac{4\pi\delta(\mathbf{k}_1+\mathbf{k}_2)}{|\mathbf{k}_1||\mathbf{k}_2|}$.

  • $\begingroup$ Could you please elaborate? How is this equivalent? Can we show it from the definition? $\endgroup$ – Yuriy S Apr 8 at 12:08
  • $\begingroup$ Oh wait, the delta-function factor helps here, am I right? Since we can (and should) have $\mathbf{k_1}=-\mathbf{k_2}$ $\endgroup$ – Yuriy S Apr 8 at 12:13
  • $\begingroup$ @YuriyS Exactly. This is a common thing that surprises people with Dirac deltas. $\endgroup$ – J.G. Apr 8 at 12:18

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