3890ps8_10 - Chem 3890 Physical Chemistry I Fall 2010...

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Unformatted text preview: Chem 3890 Physical Chemistry I Fall 2010 Chemistry 3890 Problem Set 8 Fall 2010 Due: in class, Friday, November 5 Last revised: October 29, 2010 Problem 1 Recall from PS 5, Problem 6 that, if A and B are two Hermitian operators, then their product A B is Hermitian provided that they commute, [ A, B ] = 0. Hence show that L x , L y , L z and L 2 are Hermitian. Problem 2 In this problem we apply an algebraic approach to determine the spectrum of eigenvalues for the commuting operators L 2 and L z . Define operators L L x i L y . Let m be common eigenstates of L 2 and L z : L 2 m = planckover2pi1 2 m (1a) L z m = m planckover2pi1 m . (1b) At this point, do not assume any knowledge of the allowed values for or m , except that they are real (both L 2 and L z are Hermitian), and that 0 ( L 2 is a sum of squares of Hermitian operators [see below]). 1. Show that L- L + = L 2- L 2 z- planckover2pi1 L z (2a) L + L- = L 2- L 2 z + planckover2pi1 L z (2b) L 2- L 2 z = 1 2 parenleftBig L- L + + L + L- parenrightBig . (2c) 2. Verify the commutation relations [ L z , L ] = planckover2pi1 L (3a) [ L 2 , L ] = 0 (3b) Hence show that L 2 ( L m ) = planckover2pi1 2 ( L m ) (4a) L z ( L m ) = ( m 1) planckover2pi1 ( L m ) (4b) That is, the operators L step up/down the m quantum number ladder, leaving the eigenvalue unchanged, i.e., L m m...
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This note was uploaded on 08/25/2011 for the course CHEM 3890 at Cornell University (Engineering School).

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3890ps8_10 - Chem 3890 Physical Chemistry I Fall 2010...

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