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3502_28_nove24

# 3502_28_nove24 - Chem 3502/5502 Physical Chemistry...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2010 Laura Gagliardi Lecture 28, November 24, 2010 (Some material in this lecture has been adapted from Cramer, C. J. Essentials of Computational Chemistry , Wiley, Chichester: 2002; pp. 118-119, 154-158.) Solved Homework Consider the diagonal and off-diagonal kinetic energy matrix elements T μμ = μ 1 2 2 μ T μ ν = μ 1 2 2 ν Let's think about what the Laplacian operator does to the function on which it is operating. It reports back the sum of second derivatives in all coordinate directions. That is, it is a measure of how fast the slope of the function is changing in various directions. If two functions μ and ν are far apart, then since good basis functions go to zero at least exponentially fast with distance, ν is likely to be very flat where μ is large. The second derivative of a flat function is zero. So, every point in the integration will be roughly the amplitude of μ times zero, and not much will accumulate. For the diagonal element, on the other hand, the interesting second derivatives will occur where the function has maximum amplitude (amongst other places) so the accumulation should be much larger. Thus, we expect diagonal elements in this case to be larger than off-diagonal elements. Analogous arguments can be made for the nuclear attraction integrals. V μμ = μ Z k r k k nuclei μ V μ ν = μ Z k r k k nuclei ν The 1/ r operator acting on ν will ensure that the largest contribution to the overall integral will come from the nucleus k on which basis function ν resides. Unless μ also has significant amplitude around that nucleus, it will multiply the result by roughly zero and the whole integral will be small. In this case, however, it is conceivable (i) that off- diagonal elements involving Z =100 might be larger than diagonal elements about Z =1 and (ii) that off-diagonal elements involving two basis functions on the same nuclear center might compete with diagonal elements involving a more diffuse basis function on that center (e.g., a 1s2s interaction on the same atom might be larger than a 5f interaction with itself, since the latter is so spread out that its amplitude isn't very large anywhere).

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28-2 The 2 J K terms are somewhat more subtle to evaluate, but it will still generally be the case that diagonal terms will generate larger magnitudes than off-diagonal terms. We will not explore the point closely. The Hartree-Fock Self-Consistent Field Procedure To find the HF MOs we need to solve the HF secular determinant F 11 ES 11 F 12 ES 12 F 1 N ES 1 N F 21 ES 21 F 22 ES 22 F 2 N ES 2 N F N 1 ES N 1 F N 2 ES N 2 F NN ES NN = 0 (28-1) and find its various roots. We know that we can compute overlap integrals and that Fock matrix elements are defined by F μ ν = μ 1 2 2 ν Z k k nuclei μ 1 r k ν + P λσ λσ μ ν λσ ( ) 1 2 μ λ νσ ( ) (28-2) So, now for the tricky part.
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3502_28_nove24 - Chem 3502/5502 Physical Chemistry...

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