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Unformatted text preview: Chemistry 120A, Spring 2009 Problem Set 9 Solutions Due May 1, 2009 Problems 1. Prove the ’Brillouin theorem’ which says that any single substitution, | Φ a i i , has a vanishing matrix element with the Hartree-Fock configuration, | Φ HF i . In other words: D Φ a i ˆ H Φ HF E = 0. You’ll do this in three easy pieces: (a) For the example of singlet H 2 (one electron in a bonding alpha orbital, one electron in the corresponding beta orbital), in a minimal basis, write out explicitly what the HF configuration is, and write down the expression for the single substitution which replaces the alpha spin bonding ( σ ) orbital by alpha spin antibonding ( σ * ) empty orbital. Φ HF = 1 √ 2 φ α σ φ β σ = 1 √ 2 φ α σ (1) φ β σ (2)- φ β σ (1) φ α σ (2) Φ σ * α σα = 1 √ 2 φ α σ * φ β σ = 1 √ 2 φ α σ * (1) φ β σ (2)- φ β σ (1) φ α σ * (2) (b) The Fock operator is the mean-field Hamiltonian from Hartree-Fock theory. Its matrix elements in the spin-orbital basis can be written as: D φ p ˆ f φ q E = D φ p ˆ h φ q E + occ X i h φ p φ i || φ q φ i i From the basic eigenvalue properties of the Fock operator, show that: D φ a ˆ f φ i E = 0 where level i is occupied and level a is empty. For the Fock operator, we have the eigenvalue equation: ˆ f | φ q i = q | φ q i Therefore, if we project with a different bra, h φ p | , we obtain D φ p ˆ f φ q E = h φ p | q | φ q i = q h φ p | φ q i = 0 1 because the eigenfunctions of the Fock operator are orthonormal. Therefore, we have 0 = D φ p ˆ f φ q E = D φ p ˆ h φ q E + occ X i h φ p φ i || φ q φ i i (c) Use the fundamental matrix element rules given in class to show that D Φ a i ˆ H Φ HF E = D φ a ˆ f φ i E and then, using your result from part (b), that this is zero, which establishes the theorem. We can expand the Hamiltonian into one and two electron components to obtain D Φ a i ˆ H Φ HF E = D Φ a i ˆ H 1 + ˆ H 2 Φ HF E Where we have the following relations from class: D Φ a i ˆ H 1 Φ HF E = D φ a ˆ...
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- Spring '09
- Atomic orbital, φA, ae Φe, ab ij, φa φb