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It is known that the Coulomb Potential in 2D is $V(\mathbf{x})=-\frac{e^2}{\epsilon_0}\log |\mathbf{x}|$. It is claimed that the Fourier Transform of this potential has the form: $$ V(\mathbf{k}) = -\frac{2\pi e^2}{\epsilon_0}\frac{1}{\mathbf{k}^2} $$

I can't find a source for this. How does one take the 2D Fourier Transform of $\log |\mathbf{x}|$, i.e.

$$\int_{\mathbb R^2}d^2\mathbf{x} \log{|\mathbf{x}|} e^{i\mathbf{k}\cdot \mathbf{x}}$$

This is not to be confused with the 2D Coulomb Potential in 3D, which has $V(\mathbf{x})=-\frac{e^2}{\epsilon_0}\frac{1}{|\mathbf{x}|}$ and $V(\mathbf{k}) = -\frac{2\pi e^2}{\epsilon_0}\frac{1}{|\mathbf{k}|}$.

Daphne
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  • Related: https://math.stackexchange.com/questions/847706 – Nihar Karve Nov 25 '20 at 15:29
  • duplicate of https://physics.stackexchange.com/questions/28565/coulomb-potential-in-2d?rq=1 ? – fgoudra Nov 25 '20 at 15:29
  • @fgoudra I don't see an answer to my question in that thread. – Daphne Nov 25 '20 at 15:32
  • @NiharKarve Thank you for linking this thread. I have seen it but I'm not entirely convinced by the explanation given in the answer. – Daphne Nov 25 '20 at 15:34
  • @Daphne if you look into the links, I think it is stated that this logarithmic form comes from a renormalization technique that is usual in QFT. You cannot Fourier transform the log function since it does not converge as is! You need renormalization techniques to do it (like for the Yukawa potential) – fgoudra Nov 25 '20 at 15:40

1 Answers1

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The Fourier transform of the Coulomb potential created by a charge +e at the origin is easily obtained from the Poisson equation $$\Delta V(\vec r)=-{e\over\varepsilon_0}\delta(\vec r)$$ Using the integral representation of the Dirac distribution $$\delta(\vec r)={1\over (2\pi)^2}\int e^{i\vec k.\vec r}d^2\vec k$$ the Fourier transform $$V(\vec r)={1\over 2\pi}\int V(\vec k)e^{i\vec k.\vec r}d^2\vec k$$ of the Poisson equation is $$-k^2V(\vec k)=-{e\over 2\pi\varepsilon_0}$$ whose solution is $$V(\vec k)={e\over 2\pi\varepsilon_0k^2}$$ The presence of the factor $2\pi$ is due to my choice of the constant prefactor of the Fourier transform.

In three dimensions, the same calculation leads to $$V(\vec k)={e\over (2\pi)^{3/2}\varepsilon_0k^2}$$

Christophe
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  • I'm not sure the Fourier Transform of the Poisson Equation is evidently $-k^2V(\vec{k})$. When integrating by parts we get a term $\left.[e^{ikx}V(x)]\right|_{x=-\infty}^{x=\infty} $ which I'm not convinced is $0$, given the $\log |\vec{x}|$ scaling of $V(\vec{x})$. – Daphne Nov 25 '20 at 16:50
  • Since $$\Delta V(\vec r)={1\over 2\pi}\int V(\vec k)\Delta\big(e^{i\vec k.\vec r}\big)d^3\vec k={1\over 2\pi}\int V(\vec k)\times -k^2e^{i\vec k.\vec r}d^3\vec k$$ the Fourier transform of $\Delta V(\vec r)$ is $ -k^2V(\vec k)$. – Christophe Nov 25 '20 at 20:36
  • Thank you, I understand now. I was using the FT rather than the Inverse FT, but this approach is better. – Daphne Nov 25 '20 at 23:09