# Fourier transform of Coulomb potential

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## 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.

## Answers and Replies

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):

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

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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