lecture27_umn

lecture27_umn - Electron-Electron Interactions and...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 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
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 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
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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) .
Background image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/14/2010 for the course CHEM 3502 taught by Professor Staff during the Fall '08 term at Minnesota.

Page1 / 30

lecture27_umn - Electron-Electron Interactions and...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online