Evaluating the Coulomb Integral of Helium

In summary, the conversation discusses solving the Coulomb integral using the variational method for the Helium atom. The integral involves calculating the distance between the electrons and the nucleus using a variational parameter. The integral appears to diverge and the speaker is looking for guidance or resources to solve it. The conversation also includes a potential solution using polar coordinates, but there may be some errors in the calculations.
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1. The problem statement, all variables, equations and given/known data
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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.
 
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  • #2
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...
 

1. What is the Coulomb integral of Helium?

The Coulomb integral of Helium is a mathematical term used to describe the potential energy between two electrons in the Helium atom. It is a measure of the Coulombic attraction between the two negatively charged electrons and is an important factor in understanding the electronic structure of Helium.

2. Why is evaluating the Coulomb integral of Helium important?

Evaluating the Coulomb integral of Helium is important because it allows us to understand the electronic structure and properties of the Helium atom. It is also a key factor in predicting the behavior of Helium in chemical reactions and in various physical processes.

3. How is the Coulomb integral of Helium calculated?

The Coulomb integral of Helium is typically calculated using complex mathematical equations and advanced computational methods. It involves solving the Schrödinger equation, which describes the behavior of quantum particles, for the two electrons in the Helium atom.

4. What factors affect the value of the Coulomb integral of Helium?

The value of the Coulomb integral of Helium is affected by various factors such as the distance between the two electrons, the charge of the nucleus, and the distribution of electron density in the atom. It can also be influenced by external factors, such as the presence of other atoms or molecules nearby.

5. What is the significance of the Coulomb integral in studying Helium?

The Coulomb integral is significant in studying Helium because it helps us understand the stability and reactivity of Helium in different chemical and physical environments. It also plays a crucial role in determining the energy levels and behavior of electrons in the Helium atom, which is essential for understanding its properties and applications in various fields.

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