I Fourier transform of Coulomb potential

[IanBerkman]

Dear all,

In my quantum mechanics book it is stated that the Fourier transform of the Coulomb potential
$$\frac{e^2}{4\pi\epsilon_0 r}$$
results in
$$\frac{e^2}{\epsilon_0 q^2}$$

Where $r$ is the distance between the electrons and $q$ is the difference in wave vectors.

What confuses me, is how the Fourier transform of the first term is taken since the integral diverges at r = 0.
I hope anyone can clear this up for me.

Thanks,
IanEDIT: It is already solved, $r$ and $q$ need to be taken as vectors. This thread can be deleted.

 

[absalonsen]

In condensed matter applications, the divergence problem is solved by introducing screened Coulomb potential (known as Yukawa Potential):

\begin{equation}
V(r) = \frac{e^2}{4\pi\epsilon_{0}} \frac{e^{-mr}}{r}
\end{equation}

One can get the usual (long-range) Coulomb potential back if one takes the limit where $m \rightarrow 0$.

Thus, one takes the Fourier transform of the screened Coulomb potential and takes the limit $m \rightarrow 0$ to obtain the correct result.

 

[vanhees71]

The Fourier transform doesn't diverge at $r=0$ in 3D. It's however UV divergent. The latter is solved by regularizing the integral with a small finite photon mass as mentioned in #2 and then make $m \rightarrow 0$. Here I'll introduce the regularization a bit later in the calculation, leading to the same result.

So let's do it. I just transform $1/r$ to save typing the constants (the $\epsilon_0=1$ in physical units anyway). So let's evaluate
$$\tilde{V}(\vec{p})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{1}{|\vec{x}|} \exp(-\mathrm{i} \vec{x} \cdot \vec{p}).$$
The first step is to introduce spherical coordinates with $\vec{p} \neq 0$ in the polar direction. Then the integral over $\varphi \in [0,2 \pi]$ is trivial, and for the integral over $\vartheta$ we introduce $u=\cos \vartheta$ and use $\mathrm{d} \vartheta \sin \vartheta=\mathrm{d} u$, which leads to
$$\tilde{V}(\vec{p})=2 \pi \int_0^{\infty} \mathrm{d} r \int_{-1}^{1} \mathrm{d} u r \exp(-\mathrm{i} u r p).$$
The $u$ integral is easy:
$$\tilde{V}(\vec{p})=2 \pi \mathrm{i} \int_0^{\infty} \mathrm{d} r \frac{1}{p} [\exp(-\mathrm{i} r p)-\exp(\mathrm{i} r p)].$$
Now to make sense of this integral we have to regularize it by introducing a small negative (positive) imaginary part to $p$ in the left (right). Then both integrals give the same result, and together you have (taking the imaginary parts in $p$ to $0$ again)
$$\tilde{V}(\vec{p})=\frac{4 \pi}{p^2}.$$
QED.

 

[IanBerkman]

The $u$ integral is easy:
$$\tilde{V}(\vec{p})=2 \pi \mathrm{i} \int_0^{\infty} \mathrm{d} r \frac{1}{p} [\exp(-\mathrm{i} r p)-\exp(\mathrm{i} r p)].$$
Now to make sense of this integral we have to regularize it by introducing a small negative (positive) imaginary part to $p$ in the left (right). Then both integrals give the same result, and together you have (taking the imaginary parts in $p$ to $0$ again)
$$\tilde{V}(\vec{p})=\frac{4 \pi}{p^2}.$$
QED.

I came to this part and found the solution to this integral somewhere, not knowing I had to use the "imaginary part" trick. I tried it on my own with this trick and got to the same conclusion.

Thanks.