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Question on integration, in the Hartree-Fock theory for free electron gas.

  1. Jan 13, 2010 #1
    Hi all,

    I am trying to learn Hartree-Fock theory on free electron gas. But I am stumbled on one integration that I cannot seem to figure out. Here is the integral:

    [tex]\int_{k'<k_{F}}\frac{d\textbf{k}'}{(2\pi)^3}\frac{4\pi e^2}{\left|\textbf{k}-\textbf{k}'\right|^2}[/tex]

    I cannot figure out on how the solution becomes like this:

    [tex]\frac{2 e^2}{\pi} k_{F} F\left(\frac{k}{k_{F}}\right)[/tex]



    Well, up to this point, I know that I can change the coordinate system to spherical. So I can change



    [tex]\int^{k_{F}}_{0}}4\pi k'^2 dk'[/tex]

    but I am hopeless with this component.


    I learn that I can change the latter into

    [tex]\frac{1}{\left|\textbf{k}-\textbf{k}'\right|^2}=\frac{1}{k \sqrt{1+k'^2/k^2-2\textbf{k}.\textbf{k}'/k^2}}[/tex]

    But what should I do with the [tex]\textbf{k}.\textbf{k}'[/tex] part? Because I want to turn all these vectors into scalars, so I can integrate them. Or is there other way to change [tex]\frac{1}{\left|\textbf{k}-\textbf{k}'\right|^2}[/tex] to spherical cordinate system that make it to scalar?

    Any help, advice, or suggestion will be very much appreciated. Thanks in advance.
  2. jcsd
  3. Jan 14, 2010 #2


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    Homework Helper

    Going one step back, you can change the integral to
    \int_0^{2\pi} d\phi \int_{0}^\pi \sin\theta d\theta \int^{k_{F}}_{0}} (k')^2 dk'
    Now let's choose the orientation of the axes such that theta denotes the angle between k and k'.
    Carrying out the phi integral gives a factor of 2pi, and
    k . k' = k k' cos(theta).

    \frac{1}{\left|\textbf{k}-\textbf{k}'\right|^2}=\frac{1}{k \sqrt{1+k'^2/k^2-2\textbf{k}.\textbf{k}'/k^2}} = \frac{1}{\sqrt{k^2 + k'^2 + k k' \cos\theta}}

    Now let x = cos(theta) and try to carry out the integral :)
  4. Jan 20, 2010 #3
    Hi thanks a lot. I've done the integral, and I also found a mistake that I made. The procedure you told me is right, but my first inception of the definition is wrong, it should be like this:

    \frac{1}{\left|\textbf{k}-\textbf{k}'\right|}=\frac{1}{k \sqrt{1+k'^2/k^2-2\textbf{k}.\textbf{k}'/k^2}} = \frac{1}{\sqrt{k^2 + k'^2 + k k' \cos\theta}}

    so without the square on the [tex]\frac{1}{\left|\textbf{k}-\textbf{k}'\right|}[/tex] denominator.

    Thanks again.
  5. Jan 20, 2010 #4


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    Science Advisor
    Homework Helper

    Whoops, sorry, I also overlooked that.
    Of course |v|2 = v . v and the norm |v| of v is the square root of that.
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