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3890ps7_09

# 3890ps7_09 - Physical Chemistry I Chem 3890 Fall 2009...

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Chem 3890 Physical Chemistry I Fall 2009 Chemistry 3890 Problem Set 7 Fall 2009 Due: in class, Friday, November 6 Last revised: November 1, 2009 Additions/corrections: Problem 1 Let ˆ A and ˆ B be two commuting Hermitian operators, [ ˆ A, ˆ B ] = 0. Show that the product operator ˆ A ˆ B is Hermitian; that is, verify that integraldisplay d * ( ˆ A ˆ ) = bracketleftbiggintegraldisplay d * ( ˆ A ˆ ) bracketrightbigg * (1) for all φ and ψ . 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 (2a) ˆ L z ψ λm = m planckover2pi1 ψ λm . (2b) 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 (3a) ˆ L + ˆ L - = ˆ L 2 - ˆ L 2 z + planckover2pi1 ˆ L z (3b) ˆ L 2 - ˆ L 2 z = 1 2 parenleftBig ˆ L - ˆ L + + ˆ L + ˆ L - parenrightBig . (3c) 2. Verify the commutation relations [ ˆ L z , ˆ L ± ] = ± planckover2pi1 ˆ L ± (4a) [ ˆ L 2 , ˆ L ± ] = 0 (4b) Hence show that ˆ L 2 ( ˆ L ± ψ λm ) = λ planckover2pi1 2 ( ˆ L ± ψ

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3890ps7_09 - Physical Chemistry I Chem 3890 Fall 2009...

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