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 L x , L y i = i hL z , h L x , L z i =- i hL y , h L y , L z i = i hL x . These three relations can be compactly summarized as h L i , L j i = i h ijk 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: L x =- i h y z- z y L y =- i h- x z + z x L z =- i h x y- y x Now, we just have to evaluate the commutators... h L x , L y i =- i h y z- z y ,- i h- x z + z x = ( i h ) 2 y z- z y- x z + z x - - x z + z x y z- z y = ( i h ) 2 y z- z y- x z + z x- - x z + z x y z- z y = ( i h ) 2- xy 2 z 2 + y x + z z x + xz y z- z 2 y x + xy 2 z 2- x y- z z y- yz x z + z 2 x y ! = ( i h ) 2- xy 2 z 2 + y x + yz z x ) + xz y z- z 2 y x + xy 2 z 2- x y- xz z y- yz x z + z 2 x y ! Now we can see that most of these terms will cancel, leaving us with: = ( i h ) 2 y x- x y 1 =- i h- i h y x- x y =- i h i h x y- y x = i h- i h x y- y x h L x , L y i = i hL z h L x , L y i = i hL z h L y , L z i = i h- x z + z x ,i h x y- y x = ( i h ) 2- x z + z x x y- y x - x y- y x- x z + z x = ( i h ) 2- x z + z x x y- y x- x y- y x- x z + z x = ( i h ) 2 (- x 2 z y + xy z x + z y + x x y- yz 2 x 2 + x 2 y x- xz y x- y z + x x z + yz 2 x 2 ) = ( i h ) 2 (- x 2 z y + xy z x + z y + xz x y- yz 2 x 2 + x 2 y x- xz y x- y z- xy x z + yz 2 x 2 ) again, canceling common terms leaves = ( i h ) 2 z y- y z =- i h- i h z y- y z =- i h i h y z- z y = i h- i h y z- z y h L y , L z i = i h L x h L...
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Problem Set 4 Solutions - Chemistry 120A, Spring 2009...

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