Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Fourier transform of Coulomb potential

  1. Aug 2, 2016 #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,
    Ian


    EDIT: It is already solved, ##r## and ##q## need to be taken as vectors. This thread can be deleted.
     
  2. jcsd
  3. Aug 2, 2016 #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.
     
  4. Aug 3, 2016 #3

    vanhees71

    User Avatar
    Science Advisor
    2016 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.
     
  5. Aug 3, 2016 #4
    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: Aug 3, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Fourier transform of Coulomb potential
  1. Fourier Transform (Replies: 8)

Loading...