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

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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 ˆ A and ˆ B are two Hermitian operators, then their product ˆ A ˆ B is Hermitian provided that they commute, [ ˆ A, ˆ B ] = 0. Hence show that ˆ L x , ˆ L y , ˆ L z and ˆ L 2 are Hermitian. Problem 2 In this problem we apply an algebraic approach to determine the spectrum of eigenvalues for the commuting operators ˆ L 2 and ˆ L z . Define operators ˆ L ± ˆ L x ± i ˆ L y . Let ψ λm be common eigenstates of ˆ L 2 and ˆ L z : ˆ L 2 ψ λm = λ planckover2pi1 2 ψ λm (1a) ˆ L z ψ λm = m planckover2pi1 ψ λm . (1b) At this point, do not assume any knowledge of the allowed values for λ or m , except that they are real (both ˆ L 2 and ˆ L z are Hermitian), and that λ 0 ( ˆ L 2 is a sum of squares of Hermitian operators [see below]). 1. Show that ˆ L - ˆ L + = ˆ L 2 - ˆ L 2 z - planckover2pi1 ˆ L z (2a) ˆ L + ˆ L - = ˆ L 2 - ˆ L 2 z + planckover2pi1 ˆ L z (2b) ˆ L 2 - ˆ L 2 z = 1 2 parenleftBig ˆ L - ˆ L + + ˆ L + ˆ L - parenrightBig . (2c) 2. Verify the commutation relations [ ˆ L z , ˆ L ± ] = ± planckover2pi1 ˆ L ± (3a) [ ˆ L 2 , ˆ L ± ] = 0 (3b) Hence show that ˆ L 2 ( ˆ L ± ψ λm ) = λ planckover2pi1 2 ( ˆ L ± ψ λm ) (4a) ˆ L z ( ˆ L ± ψ λm ) = ( m ± 1) planckover2pi1 ( ˆ L ± ψ λm ) (4b) That is, the operators ˆ L ± step up/down the m quantum number ladder, leaving the eigenvalue λ unchanged, i.e., ˆ L ± ψ λm ψ λm ± 1 .
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