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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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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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