hws05-1

# hws05-1 - PHYS 115B Homework 5 Solutions Spring 2008...

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PHYS 115B Homework 5 Solutions Spring 2008 Problem 1 (a) Recall the one dimensional harmonic oscillator ψ n ( x ) = A n H n ( β x ) e - β 2 x 2 / 2 , where β = μ ω 0 / . In RECTANGULAR coordinates, separation of variables gives: ψ = ψ n x ( x ) ψ n y ( y ) ψ n z ( z ) = A n x H n x ( β x ) e - β 2 x 2 / 2 A n y H n y ( β y ) e - β 2 y 2 / 2 A n z H n z ( β z ) e - β 2 z 2 / 2 = A n x A n y A n z H n x ( β x ) H n y ( β y ) H n z ( β z ) e - β 2 r 2 / 2 A = A n x A n y A n z if you are using Equation (2.35) of revised notes 2 Note that ψ is normalized because ψ n x ( x ) , ψ n y ( y ) and ψ n z ( z ) are normalized. We have n x = 1 and n y = n z = 0: A = A 1 A 0 A 0 = 1 2 μ ω 0 π 3 / 4 = 1 2 β 2 π 3 / 4 H 0 ( β y ) = H 0 ( β z ) = 1 H 1 ( β x ) = 2 β x x = 2 π 3 r [ Y 1 , - 1 ( θ , φ ) - Y 1 , +1 ( θ , φ )] as we showed in Homework 2 Therefore we have: ψ = A H 1 ( β x ) H 0 ( β y ) H 0 ( β z ) e - β 2 r 2 / 2 = 1 2 β 2 π 3 / 4 e - β 2 r 2 / 2 2 β x = 1 2 β 5 / 2 π 3 / 4 e - β 2 r 2 / 2 2 x = 1 2 β 5 / 2 π 3 / 4 e - β 2 r 2 / 2 2 2 π 3 r [ Y 1 , - 1 - Y 1 , +1 ] = 1 2 β 5 / 2 π 1 / 4 e - β 2 r 2 / 2 8 3 r [ Y 1 , - 1 - Y 1 , +1 ] = 1 2 ( ψ 1 , 1 , - 1 - ψ 1 , 1 , +1 ) using Equation 2.30 in revised notes 2 1

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ψ = 1 2 ( ψ 1 , 1 , - 1 - ψ 1 , 1 , +1 ) (b) Recall: ψ n, ,m is an eigenstate of ˆ L 2 with eigenvalue ( + 1) 2 .
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