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Evaluating the Coulomb Integral of Helium

  1. Jun 10, 2009 #1
    1. The problem statement, all variables, equations and given/known data
    ¨
    Hi guys. I am working on solving the stationary Schrôdinger equation of the Helium atom by the variational method using a Slater determinant constructed from Slater type 1s orbitals, and in that respect i need to solve the Coulomb integral:

    [tex] J = \int_ {R^3} \int_{R^3} \frac{e^{-\zeta r_1}e^{\zeta r_2}}{|\vec{r_1}-\vec{r_2}|}d^3r_1 d^3r_2[/tex]

    where [tex] \vec{r}_n [/tex] denotes the position of the n'th electron with respect to the nucleus and [tex] R^3 [/tex] denotes that the integral is over all three dimensional space wheras [tex]d^3 r[/tex] symbolises an infinitesemal volume element. [tex] r_n[/tex] is the distance between the n'th electron and the nucleus whereas [tex] \zeta [/tex] is the variational parameter ( a constant with respect to the integral).

    I know this might be a bit difficult to solve, but if any of you could give a hint or direct me to appropriate (preferably free) reading i would be very happy.

    Thanks - Mads

    Edit: I may have posted this in the wrong thread. Sorry if that is the case - just ignore.
     
    Last edited: Jun 10, 2009
  2. jcsd
  3. Jun 12, 2009 #2

    CompuChip

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    It looks like the integral diverges... hard.
    I don't know if this is anywhere near correct, but I would do the following.

    Note that if you combine the exponents, everything only depends on the distance [itex]r = r_1 - r_2[/itex], through
    [tex]J = \iint \frac{e^{-\zeta r}}{|r|} d^3r_1 d^3r_2[/tex]
    Now suppose for a moment that you knew that r > 0. Then you could switch to polar coordinates and get
    [tex]J = 4 \pi^2 V \int_0^\infty \frac{e^{-\zeta r}}{r} r^2 \, dr.[/tex]
    The integral converges if [itex]\operatorname{Re}(\zeta) > 0[/itex]. However, you have one integral left, which I have symbolically denoted by V: the volume over which you integrate. If that is R3, then
    [tex]V = \int_{\mathbb{R}^3} d\vec r = \infty.[/tex]
    Moreover, if we also allow r to be negative, then we have to do a similar integral:
    [tex]J = 4 \pi^2 V \int_0^\infty \frac{e^{\zeta r}}{r} r^2 \, dr[/tex]
    where the integral only converges if the real part of zeta is negative this time. So adding the two gives something infinite (V) times something which is only finite when zeta has no real part (and then goes like 1 / zeta^2).

    Haven't thought really deeply about it though, so maybe there is some little mistake somewhere which makes everything work out OK...
     
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