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Lecture23Notes

# Lecture23Notes - Chem 120A Spring 2006 READING Angular...

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Chem 120A Angular Momentum, continued 03/15/06 Spring 2006 Lecture 23 READING: Engel (CS): Chapters 7.4-7.5 and 9; Appendices A.4.2 and A.6 Additional optional reading: McQuarrie and Simon, Physical Chemistry , pg. 191-206 Chapter 3 of Townsend is also on the course website. Notes regarding the spherical harmonics, Y lm In the last lecture, we introduced the spherical harmonic functions Y lm ( θ , φ ) as the eigenfunctions of ˆ L 2 and ˆ L z in the position (physical space) representation. Here we will further examine their physical significance. The Y lm ( θ , φ ) are probability amplitudes of finding a particle at a particular value of θ and φ . We mentioned in the last lecture that the state Y 00 is the spherical harmonic that corresponds to the s orbitals. Y 00 is equal to 1 4 π , which implies that the probabilities of finding a particle at any point ( θ , φ ) are equal (spherically symmetric). For the states Y 1 , ± 1 , we had complex amplitudes: Y 1 , 1 = 3 8 π sin θ e i φ Y 1 , 1 = 3 8 π sin θ e i φ , for which the corresponding probabilities are P 1 , ± 1 ( θ , φ ) = 3 8 π sin 2 θ . Here, the probability is 0 at θ = 0 , π (along the z axis) and increases until θ = π 2 , i.e. in the xy plane. The probability for Y 1 , 0 is P 1 , 0 = 3 4 π cos 2 θ , which is maximal along the z axis. When the functions Y 1 x , Y 1 y , and Y 1 z are each combined with the solution to the radial part of the Schr¨odinger equation for hydrogen, R n , 1 ( r ) , then these comprise the np orbitals of hydrogen. Back to the hydrogen atom For any single particle in 3D space, the kinetic energy depends on the Laplacian, 2 . In spherical polar coordinates, the Laplacian is 2 = 2 r 2 + 2 r r 1 r 2 ¯ h 2 ˆ L 2 . (1) Chem 120A, Spring 2006, Lecture 23 1

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Because r , L = 0, we can solve the angular part of the Schr¨odinger equation immediately, irrespective of
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