Problem Set 8 Solutions

# Problem Set 8 Solutions - Chemistry 120A Spring 2009...

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Unformatted text preview: Chemistry 120A, Spring 2009 Problem Set 8 Solutions Due April 24, 2009 Problems 1. Let’s explore converting many-electron matrix elements into 1 and 2 electron matrix elements, using the H 2 molecule as an example. Let the wavefunction be a single determinant, Φ = 1 √ 2 | φ 1 φ 2 | (a) For a singlet H 2 (one electron in a bonding alpha orbital, and one electron in the corresponding beta orbital), intergrate the energy expression in terms of spin-orbitals over spin to obtain the energy in terms of spatial orbitals In this problem, the fact that we have a wavefunction comprised of a singe determinant means that we can use the Hartree-Fock For energy expression. E HF = occupied X i h φ i | ˆ h | φ i i + 1 2 occ X i occ X j h φ i φ j || φ i φ j i These equations are in terms of molecular orbitals which have two components; a spin portion and a spatial portion. | φ i i = | χ i i ⊗ | σ i ( α ) i ¯ φ i = | χ i i ⊗ | σ i ( β ) i Note that I will be using the notation that the bar over the molecular orbital signifies that the molecular orbital has beta spin character whereas the absence of a bar denotes alpha spin character. For the problem at hand, singlet H 2 , one of the molecular orbitals will have α spin character and the other will have β spin character. In order to have an overall anti-symmetric wavefunction the spins of the molecular orbitals will be anti-symmetric and the spatial portions will be symmetric. | φ 1 i = | χ 1 i| σ ( α ) i ¯ φ 2 = | χ 1 i| σ ( β ) i For part a, all we need to do is to plug in the orbitals in the E HF equation and split out the spin portion for each electron. Since every scalar product is simply shorthand for the integral over all space for both the spatial coordinates and the spin coordinates for each electron, then these two spaces can be separated because there are no terms that depend on both spin and spatial coordinates. E HF = h φ 1 | ˆ h | φ 1 i + ¯ φ 2 ˆ h ¯ φ 2 + 1 2 h φ 1 ¯ φ 2 || φ 1 ¯ φ 2 i + 1 2 h ¯ φ 2 φ 1 || ¯ φ 2 φ 1 i E HF = h χ 1 | ˆ h | χ 1 ih α | α i + h χ 1 | ˆ h | χ 1 ih β | β i + 1 2 h χ 1 χ 1 | χ 1 χ 1 ih α | α ih β | β i + 1 2 h χ 1 χ 1 | χ 1 χ 1 ih β | β ih α | α i- 1 2 h χ 1 χ 1 | χ 1 χ 1 ih α | β ih β | α i - 1 2 h χ 1 χ 1 | χ 1 χ 1 ih β | α ih α | β i 1 Since, h α | α i = h β | β i = 1 and h α | β i = h β | α i = 0, several terms can be eliminated. E HF = h χ 1 | ˆ h | χ 1 i + h χ 1 | ˆ h | χ 1 i + 1 2 h χ 1 χ 1 | χ 1 χ 1 i + 1 2 h χ 1 χ 1 | χ 1 χ 1 i = 2 h χ 1 | ˆ h | χ 1 i + h χ 1 χ 1 | χ 1 χ 1 i (b) For the case of two electrons in two different alpha orbitals (this is a triplet spin state), integrate over spin to obtain the energy in terms of spatial orbitals. Would you expect this state to be higher or lower in energy that the singlet, and why?...
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## This note was uploaded on 02/22/2010 for the course CHEM N/A taught by Professor Head-gordon during the Spring '09 term at Berkeley.

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Problem Set 8 Solutions - Chemistry 120A Spring 2009...

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