3502_22_nove02_LG

3502_22_nove02_LG - Chem 3502/5502 Physical Chemistry...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 22, November 02, 2009 (Some material in this lecture has been adapted from Cramer, C. J. Essentials of Computational Chemistry , Wiley, Chichester: 2002; pp. 514-518.) Solved Homework When we expand the two-electron S 2 operator into its individual components, we have 1 ( ) ! 2 ( ) S 2 ! 1 ( ) ! 2 ( ) = ! 1 ( ) ! 2 ( ) " S 2 1 ( ) ! 1 ( ) ! 2 ( ) d # 1 ( ) d # 2 ( ) + ! 1 ( ) ! 2 ( ) " ! 1 ( ) S 2 2 ( ) ! 2 ( ) d # 1 ( ) d # 2 ( ) + 2 ! 1 ( ) ! 2 ( ) " S x 1 ( ) ! 1 ( ) S x 2 ( ) ! 2 ( ) d # 1 ( ) d # 2 ( ) + 2 ! 1 ( ) ! 2 ( ) " S y 1 ( ) ! 1 ( ) S y 2 ( ) ! 2 ( ) d # 1 ( ) d # 2 ( ) + 2 ! 1 ( ) ! 2 ( ) " S z 1 ( ) ! 1 ( ) S z 2 ( ) ! 2 ( ) d # 1 ( ) d # 2 ( ) = ! 1 ( ) ! 2 ( ) " 1 2 1 2 + 1 $ % & ( ) ! 1 ( ) ! 2 ( ) d # 1 ( ) d # 2 ( ) + ! 1 ( ) ! 2 ( ) " ! 1 ( ) 1 2 1 2 + 1 $ % & ( ) ! 2 ( ) d # 1 ( ) d # 2 ( ) + 2 ! 1 ( ) ! 2 ( ) " 1 2 * 1 ( ) 1 2 * 2 ( ) d # 1 ( ) d # 2 ( ) + 2 ! 1 ( ) ! 2 ( ) " i 2 * 1 ( ) i 2 * 2 ( ) d # 1 ( ) d # 2 ( ) + 2 ! 1 ( ) ! 2 ( ) " 1 2 ! 1 ( ) 1 2 ! 2 ( ) d # 1 ( ) d # 2 ( ) = 1 2 1 2 + 1 $ % & ( ) + 1 2 1 2 + 1 $ % & ( ) + 0 + 0 + 1 2 = 2 Note how the terms deriving from S x and S y become zero since these operators transform the α spin function to the β spin function, and an integration over α ( i ) β ( i ) d ω ( i ) gives zero owing to the orthogonality of the spin functions.
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22-2 More Spin Algebra The final result derived in the homework above, 2, is a proper eigenvalue of S 2 corresponding to S = 1. That means that α (1) α (2) is an eigenfunction of S 2 . Since S = 1, it must be the case that M S can equal 1, 0, or 1. Recall that evaluation of M S is straightforward. One simply adds one-half for each α electron and subtracts one-half for each β electron. So, for the case of α (1) α (2), we have M S = 1. The other two cases, 1 and 0, will be degenerate in energy in the absence of a magnetic field, so this state is triply degenerate. Such states are called “triplet” states, and indicated by a superscript 3 to the left of the wave function. Technically, we did not prove that the expectation value of S 2 over the full wave function Ψ III is 2, we only showed it for the spin part of the wave function. It is a trivial matter, however, to show that the spatial portion of the wave function is normalized (it’s obviously a 2 x 2 determinant of orthonormal spatial functions preceded by the square root of 2!, so that’s that). So, we should indicate the triplet character of Ψ III by writing 3 Ψ III . Some quick consideration of the symmetric nature of the β (1) β (2) case should convince one that 1 ( ) ! 2 ( ) S 2 ! 1 ( ) ! 2 ( ) = 2 (22-1) and that this is the M S = 1 component of the triplet, 3 Ψ IV (the homework to prove this will also, no doubt, be convincing…) We now turn to Ψ V . Evaluation of S 2 proceeds in the fashion to which we should, by now, be accustomed ! V S 2 ! V = 1 2 1 ( ) " 1 ( ) b 2 ( ) # 2 ( ) S 2 a 1 ( ) " 1 ( ) b 2 ( ) # 2 ( ) $ a 1 ( ) " 1 ( ) b 2 ( ) # 2 ( ) S 2 a 2 ( ) " 2 ( ) b 1 ( ) # 1 ( ) $ a 2 ( ) " 2 ( ) b 1 ( ) # 1 ( ) S 2 a 1 ( ) " 1 ( ) b 2 ( ) # 2 ( ) + a 2 ( ) " 2 ( ) b 1 ( ) # 1 ( ) S 2 a 2 ( ) " 2 ( ) b 1 ( ) # 1 ( ) % & ( ) * * * * * * * * * * (22-2) Note that the second and third integrals on the r.h.s. are zero because of the orthonormality of the spatial orbitals a and b , whose products appear over the same electronic coordinate in those integrals. The spatial functions integrate to one in the first and fourth integrals, and the remaining spin expectation values are just those of eqs. 21- 14 and 21-15. Thus, the expectation value of eq. 22-2 is 1/2(1 0 0 + 1) = 1. There is no integer value of S for which S ( S + 1) = 1, so evidently Ψ V is
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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_22_nove02_LG - Chem 3502/5502 Physical Chemistry...

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