3502_20_octo28_LG

3502_20_octo28_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 20, October 28, 2009 Solved Homework We determined that the two coefficients in our two-gaussian wave function were c 1 = 0.3221 and c 2 = 0.7621. We also determined that " 1 s r , # , $ ; c 1 , % 1 , c 2 , % 2 ( ) 1 2 2 " 1 s r , # , $ ; c 1 , % 1 , c 2 , % 2 ( ) = 1.5 c 1 2 + 0.6435 c 1 c 2 + 0.3 c 2 2 Using the normalized coefficient values, we have < T > = 0.4873. Since < H > was 0.4819, < V > must be 0.9692. We could also simply plug the coefficients into " 1 s r , # , $ ; c 1 , % 1 , c 2 , % 2 ( ) 1 r " 1 s r , # , $ ; c 1 , % 1 , c 2 , % 2 ( ) = 1.5958 c 1 2 1.5908 c 1 c 2 0.7136 c 2 2 which is also (as it must be) 0.9692. If we take 2< T > we have –0.9746. So, although the virial theorem is almost satisfied, it is not quite. The error is less than 1%. Exact wave functions must satisfy the virial theorem, but that is not true of approximate wave functions. As such, one may examine the virial ratio to assess the quality of an approximate wave function. As one gets closer to the exact wave function, one should get closer to a ratio of –2. Antisymmetry Consider a quantum mechanical system consisting of two indistinguishable particles, e.g., two electrons. If we wanted to compute the probability that we would find electron 1 in some volume of space characterized by r a r r b , θ a θ θ b , and φ a φ φ b , and at the same time find electron 2 in some volume of space characterized by r c r r d , θ c θ θ d , and φ c φ φ d , we would compute this probability as P V 1 1 ( ) , V 2 2 ( ) [ ] = ! " 1 1 ( ) , " 2 2 ( ) [ ] # c # d $ % c % d $ r c r d $ # a # b $ % a % b $ r a r b $ 2 r 1 2 dr 1 sin % 1 d % 1 d # 1 r 2 2 dr 2 sin % 2 d % 2 d # 2 (20-1) where the cumbersome notation is meant to emphasize that the probability has to do with electron 1 being in volume 1 and electron 2 being in volume 2, that Ψ depends (in an unspecified way) on two individual electron wave functions ψ , the first of which is
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20-2 occupied by electron 1 and the second by electron 2, and that the first set of integration coordinates corresponds to electron 1 and the second set to electron 2. Now, since the electrons are indistinguishable, we are not permitted to really label them. Put differently, the probability of finding electron 2 in volume 1 and electron 1 in volume 2 must be the same as the probability we’ve already discussed up to this point (that is, there is a single probability of finding “an” electron in volume one and “another” electron in volume 2). We would write the second probability just mentioned as P V 1 2 ( ) , V 2 1 ( ) [ ] = ! " 2 1 ( ) , " 1 2 ( ) [ ] # c # d $ % c % d $ r c r d $ # a # b $ % a % b $ r a r b $ 2 r 1 2 dr 1 sin % 1 d % 1 d # 1 r 2 2 dr 2 sin % 2 d % 2 d # 2 (20-2) where the change is simply that we have swapped the coordinates that we used to use for
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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_20_octo28_LG - Chem 3502/5502 Physical Chemistry II...

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