# Fourier transform of Coulomb potential

• I

## Main Question or Discussion Point

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,
Ian

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

Related Quantum Physics News on Phys.org
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.

• IanBerkman
vanhees71
Gold Member
2019 Award
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.

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