Atomic physics Part PHYS261

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Hartree-Fock:
The total energy and sum of orbital energies
Configuration Mixing
Spectroscopic Terms

For the PROJECT WORK - LINKS: SCF-Herman-Skillman-Exercise.html
The zip-file of the code and basic instructions are at this address:
../2010.10.14/ ( Mac OSX - 2009+, Windows, Linux )
1997 Project: folk.uib.no/AMOS/Hartree/lindex.html
folk.uib.no/AMOS/Hartree/

First we re-visited the way we have described the non-local interaction from the exchange-term of interaction.
For this we need the identity - sum over all "projectors", each | k > < k | is a projector on | k > - the state k,
where "k" can be an eigenstate of any operator ( we needed the eigenstate of position operator x with a hat)
(using hats to denote operators was usual in the earlier stages of QM and textbooks; to distinguish operators
from numbers)

1a_unity_op_and_Orbital_energies_vs_total_energy.png

1a_unity_op_and_Orbital_energies_vs_total_energy.png

The main point is summarized above: if we sum all the filled orbitals' energies, we do not get the total energy
of the system. This is shown in the slides below.

It can be seen that this is true already for a product function (without Slater determinant)

This result is shown and formulated in the following two frames

1b_Orbital_energies_vs_total_energy.png

1b_Orbital_energies_vs_total_energy.png
Sum over all Sum over all is twice Sum over pairs is highlighted above

Sum over all Sum over all is twice Sum over pairs is also repeated below
1c_Sum_Orbital_energies_EQU_total_energy_-_Pair_interaction.png

1c_Sum_Orbital_energies_EQU_total_energy_-_Pair_interaction.png

So when the orbital energies are summed, one has counted the PAIR INTERACTION TWICE.
Thus the whole PAIR INTERACTION must be taken away - subtracted as formulated above

More about atomic systems - it does not stop by independent electrons

2_L-S-coupling_etc.png

2_L-S-coupling_etc.png

CONFIGURATION MIXING

- "derived" from general expansion theorem - for two electrons. One can see that this can be generalized to
any number of electrons.

3_Configuration_Mixing_from stepwise-basis-expansion.png

3_Configuration_Mixing_from stepwise-basis-expansion.png

In the above drawing we represent the terms - and indicate the expansion coefficients.
The two electrons are thus in various configurations.
But how to find the expansion coefficients?

How to find the expansion coefficients in configuration mixing?
They must be found as eigenstates (eigenvectors) of the total hamiltonian (total energy)

So we must take the various configurations as a basis - construct a matrix and diagonalize it
Diagonalization - and finding the eigenvectors - are the same procedures (see e.g. MATLAB
or any similar methods of linear algebra)

General construction of the matrix is shown below.

4_Configuration_Mixing--diagonalize-eigenvalues.png

4_Configuration_Mixing--diagonalize-eigenvalues.png

Spectroscopic terms as a result of configurations (and the mixing of them)

5_Configuratins_TERMS_spectroscopy_notatio.png

5_Configuratins_TERMS_spectroscopy_notatio.png

Numerical work with the SELF-CONSISTENT FIELDS introduced
The Herman.Skillman historical code and links below.

6_Hartree-method_applicatin_fortran_1962.png

6_Hartree-method_applicatin_fortran_1962.png

   PROJECT WORK LINKS:    SCF-Herman-Skillman-Exercise.html
   The zip-file of the code and basic instructions are at this address:
    ../2010.10.14/   ( Mac OSX - 2009+, Windows, Linux )
    1997 Project: https://folk.uib.no/AMOS/Hartree/lindex.html     https://folk.uib.no/AMOS/Hartree/