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Unformatted text preview: 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 ( , ) as the eigenfunctions of L 2 and L z in the position (physical space) representation. Here we will further examine their physical significance. The Y lm ( , ) are probability amplitudes of finding a particle at a particular value of and . 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 q 1 4 , which implies that the probabilities of finding a particle at any point ( , ) are equal (spherically symmetric). For the states Y 1 , 1 , we had complex amplitudes: Y 1 , 1 = r 3 8 sin e i Y 1 , 1 = r 3 8 sin e i , for which the corresponding probabilities are P 1 , 1 ( , ) = 3 8 sin 2 . Here, the probability is 0 at = , (along the z axis) and increases until = 2 , i.e. in the xy plane. The probability for Y 1 , is P 1 , = 3 4 cos 2 , 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 Schrodinger 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 L 2 . (1) Chem 120A, Spring 2006, Lecture 23 1 Because h r , ~ L i = 0, we can solve the angular part of the Schrodinger equation immediately, irrespective of...
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