Problem Set 9 Solutions - Chemistry 120A, Spring 2009...

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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, | a i i , has a vanishing matrix element with the Hartree-Fock configuration, | HF i . In other words: D a i H HF E = 0. Youll 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 ( ) orbital by alpha spin antibonding ( * ) empty orbital. HF = 1 2 = 1 2 (1) (2)- (1) (2) * = 1 2 * = 1 2 * (1) (2)- (1) * (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 p f q E = D p h q E + occ X i h p i || q i i From the basic eigenvalue properties of the Fock operator, show that: D a f i E = 0 where level i is occupied and level a is empty. For the Fock operator, we have the eigenvalue equation: f | q i = q | q i Therefore, if we project with a different bra, h p | , we obtain D p f q E = h p | q | q i = q h p | q i = 0 1 because the eigenfunctions of the Fock operator are orthonormal. Therefore, we have 0 = D p f q E = D p h q E + occ X i h p i || q i i (c) Use the fundamental matrix element rules given in class to show that D a i H HF E = D a f 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 a i H HF E = D a i H 1 + H 2 HF E Where we have the following relations from class: D a i H 1 HF E = D a...
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Problem Set 9 Solutions - Chemistry 120A, Spring 2009...

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