Problem Set 4 Solutions

Problem Set 4 Solutions - Chemistry 120A Spring 2009...

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Unformatted text preview: Chemistry 120A, Spring 2009 Problem Set 4 Solutions Due March 4, 2009 Problems 1. Working with angular momentum operators (a) Given the definition of the angular momentum operators in Cartesian coordinates, explicitly show that h ˆ L x , ˆ L y i = i ¯ hL z , h ˆ L x , ˆ L z i =- i ¯ hL y , h ˆ L y , ˆ L z i = i ¯ hL x . These three relations can be compactly summarized as h ˆ L i , ˆ L j i = i ¯ h ijk ˆ L k , where the permutation tensor ijk is 1 for a cyclic permutation (xyz, yzx, zxy), and -1 for a permutation that needs one swap of indicies to be cyclic (i.e. xzy), and 0 otherwise. For this problem, we start with: ˆ L x =- i ¯ h y ∂ ∂z- z ∂ ∂y ˆ L y =- i ¯ h- x ∂ ∂z + z ∂ ∂x ˆ L z =- i ¯ h x ∂ ∂y- y ∂ ∂x Now, we just have to evaluate the commutators... h ˆ L x , ˆ L y i ψ =- i ¯ h y ∂ ∂z- z ∂ ∂y ,- i ¯ h- x ∂ ∂z + z ∂ ∂x ψ = ( i ¯ h ) 2 y ∂ ∂z- z ∂ ∂y- x ∂ ∂z + z ∂ ∂x ψ- - x ∂ ∂z + z ∂ ∂x y ∂ ∂z- z ∂ ∂y ψ = ( i ¯ h ) 2 y ∂ ∂z- z ∂ ∂y- x ∂ψ ∂z + z ∂ψ ∂x- - x ∂ ∂z + z ∂ ∂x y ∂ψ ∂z- z ∂ψ ∂y = ( i ¯ h ) 2- xy ∂ 2 ψ ∂z 2 + y ∂ψ ∂x + z ∂ ∂z ∂ψ ∂x + xz ∂ ∂y ∂ψ ∂z- z 2 ∂ ∂y ∂ψ ∂x + xy ∂ 2 ψ ∂z 2- x ∂ψ ∂y- z ∂ ∂z ∂ψ ∂y- yz ∂ ∂x ∂ψ ∂z + z 2 ∂ ∂x ∂ψ ∂y ! = ( i ¯ h ) 2- xy ∂ 2 ψ ∂z 2 + y ∂ψ ∂x + yz ∂ ∂z ∂ψ ∂x ) + xz ∂ ∂y ∂ψ ∂z- z 2 ∂ ∂y ∂ψ ∂x + xy ∂ 2 ψ ∂z 2- x ∂ψ ∂y- xz ∂ ∂z ∂ψ ∂y- yz ∂ ∂x ∂ψ ∂z + z 2 ∂ ∂x ∂ψ ∂y ! Now we can see that most of these terms will cancel, leaving us with: = ( i ¯ h ) 2 y ∂ψ ∂x- x ∂ψ ∂y 1 =- i ¯ h- i ¯ h y ∂ψ ∂x- x ∂ψ ∂y =- i ¯ h i ¯ h x ∂ψ ∂y- y ∂ψ ∂x = i ¯ h- i ¯ h x ∂ψ ∂y- y ∂ψ ∂x h ˆ L x , ˆ L y i ψ = i ¯ hL z ψ ⇒ h ˆ L x , ˆ L y i = i ¯ hL z h ˆ L y , ˆ L z i ψ = i ¯ h- x ∂ ∂z + z ∂ ∂x ,i ¯ h x ∂ ∂y- y ∂ ∂x = ( i ¯ h ) 2- x ∂ ∂z + z ∂ ∂x x ∂ ∂y- y ∂ ∂x ψ- x ∂ ∂y- y ∂ ∂x- x ∂ ∂z + z ∂ ∂x ψ = ( i ¯ h ) 2- x ∂ ∂z + z ∂ ∂x x ∂ψ ∂y- y ∂ψ ∂x- x ∂ ∂y- y ∂ ∂x- x ∂ψ ∂z + z ∂ψ ∂x = ( i ¯ h ) 2 (- x 2 ∂ ∂z ∂ψ ∂y + xy ∂ ∂z ∂ψ ∂x + z ∂ψ ∂y + x ∂ ∂x ∂ψ ∂y- yz ∂ 2 ψ ∂x 2 + x 2 ∂ ∂y ∂ψ ∂x- xz ∂ ∂y ∂ψ ∂x- y ∂ψ ∂z + x ∂ ∂x ∂ψ ∂z + yz ∂ 2 ψ ∂x 2 ) = ( i ¯ h ) 2 (- x 2 ∂ ∂z ∂ψ ∂y + xy ∂ ∂z ∂ψ ∂x + z ∂ψ ∂y + xz ∂ ∂x ∂ψ ∂y- yz ∂ 2 ψ ∂x 2 + x 2 ∂ ∂y ∂ψ ∂x- xz ∂ ∂y ∂ψ ∂x- y ∂ψ ∂z- xy ∂ ∂x ∂ψ ∂z + yz ∂ 2 ψ ∂x 2 ) again, canceling common terms leaves = ( i ¯ h ) 2 z ∂ψ ∂y- y ∂ψ ∂z =- i ¯ h- i ¯ h z ∂ψ ∂y- y ∂ψ ∂z =- i ¯ h i ¯ h y ∂ψ ∂z- z ∂ψ ∂y = i ¯ h- i ¯ h y ∂ψ ∂z- z ∂ψ ∂y h ˆ L y , ˆ L z i ψ = i ¯ h ˆ L x ψ ⇒ h ˆ L...
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Problem Set 4 Solutions - Chemistry 120A Spring 2009...

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