lecture27_umn

# lecture27_umn - Electron-Electron Interactions and...

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Review of spin and Slater determinants Electron-Electron interactions Coulomb integrals Exchange integrals Intro to Hartree-Fock Method 1 Electron-Electron Interactions and Hartree- Fock Method

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Recall how the self- consistent Feld (scf) method works guess wavefunction h i = - 1 2 2 i - M ± k =1 Z k r ik + V i { j } Give an approximate interaction potential ( ) felt by a 1s helium electron to use as a starting point in a Hartree calculation: V i solve h i new wavefunction converged? nope yep Done! ψ =? Solve ±aster! 2
Charge cloud is diffuse, integrate over 3D space E ( r 0 )= q ± 0 ± π 0 ± 2 π 0 1 | r - r 0 | | Ψ 100 ( r, θ , φ ) | 2 r 2 dr sin θ d θ d φ Wavefunction? Give an approximate interaction potential ( ) felt by one of the 1s 2 helium electrons to use as a starting point in a Hartree calculation: V i Ψ 100 ( θ , φ ± 8 π ² 1 / 2 e - 2 r Hydrogenic 1s orbital Remember Z=2 for He What does the total repulsion look like for a point charge? Probability density Charge 3 Distance

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Give an approximate interaction potential ( ) felt by one of the 1s 2 helium electrons to use as a starting point in a Hartree calculation: V i but. .. we aren’t dealing with a point charge, both electrons are delocalized Total repulsion will be a double integral: E = | Ψ 100 (1) | 2 1 r 12 | Ψ 100 (2) | 2 dr 1 dr 2 Two hydrogenic electrons 4
After the Frst step, the electrons will be pushed away from each other too far, so the second iteration will bring them closer together again. .. then push them away. .. then together. .. etc. .. If we were looking at one 2s electron interacting with a 1s electron the original repulsion would be much less (2s electrons are farther away from the nucleus) Give an approximate interaction potential ( ) felt by one of the 1s 2 helium electrons to use as a starting point in a Hartree calculation: V i 5

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Quick Spin Review Hartree wavefunctions are not rigorous for two reasons They’re not antisymmetric Ψ HP ( x 1 ,x 2 )= χ 1 ( x 1 ) χ 2 ( x 2 ) Ψ ( x 2 1 χ 1 ( x 2 ) χ 2 ( x 1 ) ± = - χ 1 ( x 1 ) χ 2 ( x 2 ) They don’t factor in spin 6
Need to somehow take relativistic effects into account, this is where spin comes in Spin Up Eigenvalue: Eigenfunction: + ± 2 Spin Down Eigenvalue: Eigenfunction: β - ± 2 Quick Spin Review 7

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Slater Determinants Can account for antisymmetry and spin by making a slater determinant wavefunction Ψ SD = 1 N ! ± ± ± ± ± ± ± ± χ 1 (1) χ 2 (1) · · · χ N (1) χ 1 (2) χ 2 (2) · · · χ N (2) .
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## This note was uploaded on 12/14/2010 for the course CHEM 3502 taught by Professor Staff during the Fall '08 term at Minnesota.

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lecture27_umn - Electron-Electron Interactions and...

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