4604_Solutions_Set6_fa07

# 4604_Solutions_Set6_fa07 - PHY4604 Fall 2007 Problem Set 6...

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PHY4604 Fall 2007 Problem Set 6 Solutions Department of Physics Page 1 of 9 PHY 4604 Problem Set #6 Solutions Problem 1 (20 points): The Pauli spin matrices are given by = 0 1 1 0 x σ = 0 0 i i y = 1 0 0 1 z (a) (10 points) Show that σ i = σ i , det( σ i ) = -1, Tr( σ i ) = 0, [ σ i , σ j ] = 2i ε ijk σ k , and { σ i , σ j } = 2 δ ij . Note that [A, B] = AB – BA and {A, B} = AB +BA. Solution: It is trivial to see that σ i = ( σ i ) T * = σ i . Also, 1 0 1 1 0 ) det( = = x 1 0 0 ) det( = = i i y 1 1 0 0 1 ) det( = = z And it is easy to see that Tr( σ i ) = 0. We see that z y x i i i i i i = = = = 1 0 0 1 0 0 0 0 0 1 1 0 z x y i i i i i i = = ⎛− = = 1 0 0 1 0 0 0 1 1 0 0 0 x z y i i i i i i = = = = 0 1 1 0 0 0 1 0 0 1 0 0 x y z i i i i i i = = = = 0 1 1 0 0 0 0 0 1 0 0 1 y x z i i i i = = = = 0 0 0 1 1 0 0 1 1 0 1 0 0 1 y z x i i i i = = = = 0 0 0 1 1 0 1 0 0 1 0 1 1 0 1 1 0 0 1 0 1 1 0 0 1 1 0 2 = = = = x x x 1 1 0 0 1 0 0 0 0 2 = = = = i i i i y y y 1 1 0 0 1 1 0 0 1 1 0 0 1 2 = = = = z z z Thus [ σ i , σ j ] = 2i ε ijk σ k , and { σ i , σ j } = 2 δ ij . (b) (10 points) Show that + = l l ijl ij j i i ε δ . Solution: From (a) we see that [ σ i , σ j ] = σ i σ j − σ j σ i = 2i ε ijk σ k and { σ i , σ j } = σ i σ j + σ j σ i = 2 δ ij and adding these two equations yields i σ j = 2 δ ij + 2i ε ijk σ k and dividing by two gives σ i σ j = δ ij + i ε ijk σ k .

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PHY4604 Fall 2007 Problem Set 6 Solutions Department of Physics Page 2 of 9 Problem 2 (20 points): An electron is in the spin state >= 4 3 | i A χ . (a) (4 points) Determine the normalization constant A. Answer: A = 1/5 Solution: We normalize | χ> as follows: () 2 2 2 25 ) 16 9 ( 4 3 4 3 | 1 A A i i A = + = >= =< and hence A = 1/5. (b) (8 points) Find the expectation value of S x , S y , and S z for the state | χ >, where σ r r h 2 = S .
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4604_Solutions_Set6_fa07 - PHY4604 Fall 2007 Problem Set 6...

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