Download Pauli Exclusion Principle - Advanced Quantum Chemistry and Spectroscopy - Lecture Slides and more Slides Chemistry in PDF only on Docsity! Next consider Li. Assume because Li has 3 electrons that its ground state electronic configuration is given by 1s3. Slater Determinant becomes: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )331331331 221221221 111111111 !3 13,2,1 αβα αβα αβα ψ sss sss sss = where the 3rd e - has either α (or β) spin However, the 1st and 3rd columns in this determinant are identical, so Ψ(1,2,3) =0. Thus, the 3rd e- must go into the next higher energy orbital with n = 2. ⇔ Each orbital have a maximum occupancy of two electrons. = Pauli Exclusion Principle. docsity.com Use of Slater Determinants is a course in itself. Will only consider first excited state of He (N = 2) to introduce the concepts of exchange and Coulomb energies. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]122122112 1 2221 1211 !2 1 0 αααααα αα ssss ss ss −==Φ This is one of 3 degenerate possibilities for the excited 3S triplet state (both spins up). Calculate <Φ0|V|Φ0> ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) )4(2211||1221 2 1 )3(1221||2211 2 1 )2(1221||1221 2 1 )1(2211||2211 2 1 12212211||12212211 2 1 |||ˆ| 12 2 12 2 12 2 12 2 12 2 0 12 2 000 ><− ><− ><+ ><= ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ >−−<= >ΦΦ>=<ΦΦ< αααα αααα αααα αααα αααααααα ss r ess ss r ess ss r ess ss r ess ssss r essss r eV docsity.com The correct form of the excited singlet state of He can be written in Slater Determinant form as: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − 2)2(22)2(1 1)1(21)1(1 2 1 2)2(22)2(1 1)1(21)1(1 2 1 2 1 αβ αβ βα βα ss ss ss ss [ ][ ]( ) before as(1)(2)-(2)(1))1(2)2(1)2(2)1(1 βαβαssss += Can then show that E = E0+J+K What are the signs of J and K? The integrand of J is positive everywhere, and so J is positive. Makes sense since the electrostatic interaction between two electron cloud distributions is repulsive. The integrand of K has both positive and negative contributions. For K to be negative, the electron radius r1 or r2 must be inside the 2s orbital nodal surface while the other is outside this distance. docsity.com e- (2) or (1) here e- (1) or (2) here Then integrand of K is negative docsity.com However, the 1/r12 term weights those regions with smaller r12. Therefore K overall is a positive quantity. In addition to always being positive the exchange integral, K, always enters with a negative sign. Therefore, it always acts to diminish the electron-electron repulsion energy. The exchange energy is always much smaller than the associated Coulomb repulsion energy. In He exchange interaction splits the degeneracy of the singlet and triplet states by an amount = 2K Note: Hund’s Rules OK here. The triplet lies lower than the singlet. no electron interaction exchange interactionCoulomb interaction 3E = E0+J-K 1E = E0+J+K 2K E0+J J E0 docsity.com
