[FeLiXe]

Hi, I have been trying to insert the atomic orbital functions into the Schrödinger equation but it never works. Actually I did it three times (with the help of Mathematica): with spherical coordinates, cartesian coordinates, and kind of spherical coordinates. It was the same result every time. But I did not get a constant electron energy. Could someone tell me how to do that?

The problem might be the potential energy, i.e. coulomb potential + centrifugal potential. Isn't the coulomb potential just as usual -e^2/(4 * pi * eps0 * r) and the centrifugal potential as it says here l*(l+1)*h^2/(8*pi^2*m*r^2) ? My thought is that the coulomb potential is different for some reason.

[Juan R.]

Hi, I have been trying to insert the atomic orbital functions into the Schrödinger equation but it never works. Actually I did it three times (with the help of Mathematica): with spherical coordinates, cartesian coordinates, and kind of spherical coordinates. It was the same result every time. But I did not get a constant electron energy. Could someone tell me how to do that?

The problem might be the potential energy, i.e. coulomb potential + centrifugal potential. Isn't the coulomb potential just as usual -e^2/(4 * pi * eps0 * r) and the centrifugal potential as it says here l*(l+1)*h^2/(8*pi^2*m*r^2) ? My thought is that the coulomb potential is different for some reason.

Well, i do not know exactly the details of your computation but

H Phy = E Phy

works perfectly for Hydrogen atom with a Coulomb interaction.

Since that we are considering nuclei at rest, one needs use reduced motion. The total Hamiltonian is

H = -[ [h(/2 pi)]^2 nabla / (2 mu) ] - Zee/(4 pi eps₀ r)

with Z = 1.

The solution is a direct product of RY whith R a radial function and Y a spherical armonic. The triple R, R', R'' satisfies a radial equation with centrifugal potential

l (l+1) [h(/2 pi)]^2 /(4 pi eps₀ mu r^2)

mu, reduced mass, is defined by

1/mu = (1/m_electron) + (1/m_nuclei)

[FeLiXe]

thanks for the help, but I still don't know what my mistake is though.

if you take time looking at my computation, there it is: http://stud4.tuwien.ac.at/~e0425252/Schroedinger/
It starts with defining constants then the actual computation starts at *** here it starts ***

In my physics script it says that you have to include the centrifugal potential into the Hamiltonian. Is that not true? Anyway it does not make a difference for the s-orbital anyway because it becomes 0.

[Juan R.]

i looked your link. It contains some errors.

In the initial Hamiltonian in cartesian coordinates there is not "centrifugal potential" (i dislike this name) . When you transform to spherical coordinates the Laplacian already contains the term "centrifugal potential". I dislike the name becauses the real potential is the Coulomb one, and the "centrifugal potential" follows from the Laplacian in spherical, therefore, is a kinetic term. Note that i already wrote above the centrifugal term. You do not need introduce it in the potential

Then one solve H phy = E phy. e.g.  H 1s = E 1s

Note that phy may be eigenfunction of L^2 operator and therefore phy = R · Y_l^m.

Note that reduced mass -no mass enter- in the centrifugal term (i already said that).

Note also that eigenfunction computed from Nabla in spherical is only valid for R, already also said. for obtaining the total wave function one may multiply after by the corresponding armonic.

your 1s is correct but you are using an incorrect Hamiltonian. Since you post "SetCoordinates[Spherical[r, th, ph]]" the Mathematica already add the centrifugal term from the Laplacian and you do not need introduce it again in the potential. Note also that it is the reduced mass that enter in the Hamiltonian.

[FeLiXe]

hey cool, it works now. Actually my mistake for the s-orbital was in the numerical evaluation. If I had just used constants it would have worked.

For the p-orbital of course I had to delete that "centrifugal potential". It's nice to know where that comes from. I was always wondering how they came up with the idea of just putting another term in there. Sounds nice that it is just a term in polar coordinate tensor calculus. Just math, no weird physical addition as I had thought.

Thanks a lot it feels good to finally be able to at least touch the Schrödinger equation.

Is the correct energy for the 1s orbital -13.5983 (the you get with corrected mass)? Because sometimes you also see -13,6057 that you have without corrected mass.

[Juan R.]

Congratulations!

The energy is with the reduced mass i.e. the -13.598...

But note that reduced mass mu = 0.995 electron mass. Then mu is aproximately the electron mass doing the approximation of infinitely weight nucleus. The error is of order of 1/2000 because the nucleus is very heavy.

The value -13.606... appears in some textbooks doing approximation of inifinitely heavy nucleus, but there is an experimental error with ionization energy of around 7/13000.