3502_29_nove20_LG - Chem 3502/5502 Physical Chemistry II...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 29, November 20, 2009 Solved Homework Water, H 2 O, involves 2 hydrogen atoms and an oxygen atom. To minimally represent hydrogen is simple—one needs only a 1s function. As there are two H atoms, we will need two such functions, one centered on each atom. Oxygen has electrons in the second principal quantum level, so we will need one 1s, one 2s, and three 2p functions (one each of p x , p y , and p z ). So, that's five functions total on O, plus one each on the H atoms which is then 7 total. The total number of one-electron integrals we need to solve is determined by considering that we need kinetic-energy integrals T μ ! " 1 2 # 2 ! and potential energy integrals V μ " = μ # Z k r k k nuclei $ " Since both μ and ν can be any one of the 7 basis functions, there are 7 x 7 possible kinetic energy integrals (49) and the same number of potential energy integrals for each nucleus . As there are 3 nuclei, that is 7 x 7 x 3 = 147 integrals. The grand total of one- electron integrals is thus 196. The two-electron integrals are μ ! "# ( ) = $ μ 1 ( ) $ ! 1 ( ) 1 r 12 $ " 2 ( ) $ # 2 ( ) dr 1 dr 2 % where now μ , ν , λ , and σ can be any one of the seven basis functions (note that letting them be any function covers both all Coulomb and all exchange integrals). There are thus 7 x 7 x 7 x 7 = 2401 two-electron integrals. The numbers computed above involve the contracted basis functions, each of which, since the basis is STO-3G, is composed of 3 primitive functions. Thus, for any individual one-electron integral, there will be 3 x 3 = 9 separate integrals involving the primitives. There are thus 9 x 196 = 1764 individual primitive one-electron integrals.
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29-2 As for the two-electron integrals, again, every individual integral will require considering every possible combination of constituent primitives which is 3 x 3 x 3 x 3 = 81. Thus, the total number of primitive two-electron integrals is 81 x 2401 = 194,481 (gulp!) Notice that even for this small molecule the number of two-electron integrals totally dominates the number of one-electron integrals. The disparity only increases with molecular size. All this to find 5 occupied molecular orbitals from which to form a final Slater determinant (10 electrons, two to an orbital, so 5 orbitals). The situation sounds horrible, but it should be recognized that by using gaussians, the solutions to all of the integrals are known to be analytic formulae involving only interatomic distances, cartesian exponents, and α values in the gaussians. So, the total number of floating-point operations to solve the almost 200,000 grand-total integrals may be about 1,000,000. In computer speak that's one megaflop (megaflop = million FLoating-point OPerations). A modern digital computer processor can achieve giga flop per second performance, so the computer can accomplish all these calculations in under one second. In fact, modern computers are so fast at calculations that it can be faster to recompute the integral values than to take the time to write them onto a storage device to retrieve the number later!
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This note was uploaded on 12/14/2010 for the course CHEM 3502 taught by Professor Staff during the Spring '08 term at Minnesota.

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3502_29_nove20_LG - Chem 3502/5502 Physical Chemistry II...

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