Fourier transform of Coulomb potential

In summary, the Fourier transform of the Coulomb potential results in the screened Coulomb potential, which is solved by introducing a small finite photon mass and taking the limit as the mass goes to 0. This regularization is necessary due to the divergence of the integral at r=0. By using the "imaginary part" trick, the correct result can be obtained.
  • #1
IanBerkman
54
1
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.
 
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  • #2
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.
 
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  • #3
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.
 
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  • #4
vanhees71 said:
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.
 
Last edited:

1. What is the Fourier transform of the Coulomb potential?

The Fourier transform of the Coulomb potential is a mathematical operation that converts the Coulomb potential function, which describes the electric potential of a point charge, into a function of spatial frequency. It is commonly used in quantum mechanics to analyze the behavior of charged particles in an electromagnetic field.

2. How is the Fourier transform of the Coulomb potential calculated?

The Fourier transform of the Coulomb potential can be calculated using the integral formula: where q is the charge of the particle, r is the position vector, and k is the wave vector.

3. What is the physical significance of the Fourier transform of the Coulomb potential?

The Fourier transform of the Coulomb potential can help us understand the spatial distribution of the electric potential around a point charge. It reveals the different spatial frequencies present in the potential and how they contribute to the overall behavior of the charged particle.

4. How does the Fourier transform of the Coulomb potential relate to the Coulomb potential energy?

The Fourier transform of the Coulomb potential is related to the Coulomb potential energy by a mathematical inverse relationship. This means that the Fourier transform of the Coulomb potential can be used to calculate the Coulomb potential energy and vice versa.

5. What are some applications of the Fourier transform of the Coulomb potential?

The Fourier transform of the Coulomb potential has many applications in various fields such as quantum mechanics, electromagnetism, and signal processing. It is used to study the behavior of charged particles in an electromagnetic field, analyze the electric potential around a point charge, and understand the frequency components of a signal in signal processing.

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