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Unformatted text preview: Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 37, December 14, 2009 (Some material in this lecture has been adapted from Cramer, C. J. Essentials of Computational Chemistry , Wiley, Chichester: 2002; pp. 201203; 207210; 216; 220 222.) Solved Homework With only two CSFs to consider, we determine the energies from H 11 " E H 12 H 21 H 22 " E = The twobytwo determinant is easy to expand: it is E 2 " H 11 + H 22 ( ) E + H 11 H 22 " H 12 2 ( ) = where weve replaced H 21 with H 12 (they are equal since H is hermitian). This is a quadratic equation in E , so it has roots determined from the quadratic formula, namely E = H 11 + H 22 ( ) H 11 + H 22 ( ) 2 " 4 H 11 H 22 " H 12 2 ( ) 2 = H 11 + H 22 ( ) H 11 2 " 2 H 11 H 22 + H 22 2 + 4 H 12 2 2 = H 11 + H 22 ( ) H 11 " H 22 ( ) 2 + 4 H 12 2 2 Now consider what this equation says. The two energies are the average of the individual singledeterminantal state energies H 11 and H 22 (the part before the ) plus or minus onehalf the square root of the square of the difference between the two state energies plus something that must be nonnegative. Note that if H 12 is zero, then the two roots are just H 11 and H 22 , i.e., if there is no offdiagonal coupling between the states, there is no advantage to doing a CI with the two states. But, when it is positive, it will cause the root derived from subtracting the square root term from the average state energy to fall below H 11 . 372 So, what is H 12 ? Its an easy matter to determine given the simplicity of the ground and excitedstate determinants. We evaluate it as H 12 = " 1 ( ) " 2 ( ) H " * 1 ( ) " * 2 ( ) = " 1 ( ) " 2 ( ) 1 r 12 " * 1 ( ) " * 2 ( ) = K "" * All of the oneelectron integrals in the expectation value of the Hamiltonian must be zero since if it is a oneelectron term for electron 1, the integral over electron 2 is the overlap integral of and orthogonal *, and if it is a oneelectron term for electron 2, the integral over electron 1 must be zero for the same reason. The only term that survives is the electronelectron repulsion integral, which in this case corresponds to the exchange integral between the two orbitals. Limited Configuration Interaction We left off last time considering the structure of the CI matrix H which must be diagonalized to find the improved groundstate energy compared to the HF energy. What if we only keep single excitations? In that case, we see from the last figure in Lecture 37 that the CI matrix will be socalled block diagonal. One block will be the HF energy, H 11 , and the other will be the singles/singles region. Since a block diagonal matrix can be fully diagonalized block by block, and since the HF result is already a block by itself, it is apparent that the lowest energy root, i.e., the groundstate HF root, is unaffected by inclusion of single excitations. Indeed, one way to think about the HF process is that it is an optimization of orbitals subject to the constraint that single...
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 Spring '08
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