22 - Chem 3502/5502 Physical Chemistry II(Quantum Mechanics...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2010 Laura Gagliardi Lecture 22, November 05, 2010 (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 α 1 ( ) α 2 ( ) S 2 α 1 ( ) α 2 ( ) = α 1 ( ) α 2 ( ) S 2 1 ( ) α 1 ( ) α 2 ( ) d ω 1 ( ) d ω 2 ( ) + α 1 ( ) α 2 ( ) α 1 ( ) S 2 2 ( ) α 2 ( ) d ω 1 ( ) d ω 2 ( ) + 2 α 1 ( ) α 2 ( ) S x 1 ( ) α 1 ( ) S x 2 ( ) α 2 ( ) d ω 1 ( ) d ω 2 ( ) + 2 α 1 ( ) α 2 ( ) S y 1 ( ) α 1 ( ) S y 2 ( ) α 2 ( ) d ω 1 ( ) d ω 2 ( ) + 2 α 1 ( ) α 2 ( ) S z 1 ( ) α 1 ( ) S z 2 ( ) α 2 ( ) d ω 1 ( ) d ω 2 ( ) = α 1 ( ) α 2 ( ) 1 2 1 2 + 1 α 1 ( ) α 2 ( ) d ω 1 ( ) d ω 2 ( ) + α 1 ( ) α 2 ( ) α 1 ( ) 1 2 1 2 + 1 α 2 ( ) d ω 1 ( ) d ω 2 ( ) + 2 α 1 ( ) α 2 ( ) 1 2 β 1 ( ) 1 2 β 2 ( ) d ω 1 ( ) d ω 2 ( ) + 2 α 1 ( ) α 2 ( ) i 2 β 1 ( ) i 2 β 2 ( ) d ω 1 ( ) d ω 2 ( ) + 2 α 1 ( ) α 2 ( ) 1 2 α 1 ( ) 1 2 α 2 ( ) d ω 1 ( ) d ω 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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 β 1 ( ) β 2 ( ) S 2 β 1 ( ) β 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 Ψ V S 2 Ψ V = 1 2 a 1 ( ) α 1 ( ) b 2 ( ) β 2 ( ) S 2 a 1 ( ) α 1 ( ) b 2 ( ) β 2 ( ) a 1 ( ) α 1 ( ) b 2 ( ) β 2 ( ) S 2 a 2 ( ) α 2 ( ) b 1 ( ) β 1 ( ) a 2 ( ) α 2 ( ) b 1 ( ) β 1 ( ) S 2 a 1 ( ) α 1 ( ) b 2 ( ) β 2 ( ) + a 2 ( ) α 2 ( ) b 1 ( ) β 1 ( ) S 2 a 2 ( ) α 2 ( ) b 1 ( ) β 1 ( ) (22-2) Note that the second and third
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/18/2010 for the course CHEM 3502 taught by Professor Staff during the Fall '08 term at Minnesota.

Page1 / 9

22 - Chem 3502/5502 Physical Chemistry II(Quantum Mechanics...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online