PHY356_PS1_solutions_2019.pdf - PHY356 Problem Set#1 Solutions 1 Exponential of 2x2 matrices I will prove the relation in part(c then the results for(a

# PHY356_PS1_solutions_2019.pdf - PHY356 Problem Set#1...

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PHY356 Problem Set #1 Solutions September 28, 2019 1 Exponential of 2x2 matrices I will prove the relation in part (c), then the results for (a) and (b) can be found by choosing λ = 1 , μ = 0 and μ = 1 , λ = 0 , respectively. (c) We want to determine an equivalent expression for exp [ ( λσ x + μσ y )] in terms of sines and cosines, with λ 2 + μ 2 = 1 . We start by expanding the exponential as a series, as was hinted: exp [ ( λσ x + μσ y )] = 1 + ( λσ x + μσ y ) - α 2 2! ( λσ x + μσ y ) 2 - 3 3! ( λσ x + μσ y ) 3 + ... Consider: ( λσ x + μσ y ) 2 = λ 2 σ 2 x + λμ ( σ x σ y + σ y σ x ) + μ 2 σ 2 y You can check the matrix multiplication to see: σ 2 x = σ 2 y = I = 1 0 0 1 ! σ x σ y + σ y σ x = 0 Thus: ( λσ x + μσ y ) 2 = ( λ 2 + μ 2 ) I = I = ( λσ x + μσ y ) n = I n even ( λσ x + μσ y ) n odd Therefore, our series expansion can be reduced to the even terms series and the odd terms series: exp [ ( λσ x + μσ y )] = X n =0 i n α n n ! ( λσ x + μσ y ) n = X n even i n α n n ! ( λσ x + μσ y ) n + X n odd i n α n n ! ( λσ x + μσ y ) n = I X n even i n α n n ! + ( λσ x + μσ y ) X n odd i n α n n ! We now recognize that the even terms form the Taylor series for cos( α ) while the odd terms form the Taylor series for i sin( α ) . Thus: exp [ ( λσ x + μσ y )] = I cos α + i ( λσ x + μσ y ) sin α 1
(d) Using the results from (a), (b) and (c): e 2 x = I cos 2 + x sin 2 = cos 2 i sin 2 i sin 2 cos 2 ! ( e x ) 2 = cos 1 i sin 1 i sin 1 cos 1 ! cos 1 i sin 1 i sin 1 cos 1 ! = cos 2 1 - sin 2 1 2 i cos 1 sin 1 2 i cos 1 sin 1 cos 2 1 - sin 2 1 ! = cos 2 i sin 2 i sin 2 cos 2 ! e i ( σ x + σ y ) = e i 2( σ x / 2+ σ y / 2) = I cos 2 + i σ x + σ y 2 sin 2 = cos 2 i +1 2 sin 2 i - 1 2 sin 2 cos 2 ! e x e y = cos 1 i sin 1 i sin 1 cos 1 ! cos 1 sin 1 - sin 1 cos 1 ! = cos 2 1 - i sin 2 1 cos 1 sin 1 + i cos 1 sin 1 i cos 1 sin 1 - cos 1 sin 1 cos 2 1 + i sin 2 1 ! 6 = e i ( σ x + σ y ) 2 Matrix algebra a) Find the eigenvalues first: M 1 = 1 2 - 2 0 2 0 0 3 0 M 1 | m i = m | m i = ( M 1 - m I ) | m i = 0 = det ( M 1 - m I ) = 0 = (1 - m )[(2 - m )( - m )] = 0 = m = 1 , 2 , 0 Now the eigenvectors. m = 1 : M 1 | 1 i = 1 | 1 i a + 2 b - 2 c 2 b 3 b = a b c = b = c = 0 = ⇒ | 1 i = 1 0 0 2
m = 2 : M 1 | 2 i = 2 | 2 i a + 2 b - 2 c 2 b 3 b = 2 a b c = b = 2 c/ 3 a = - 2 c/ 3 = ⇒ | 2 i = q 1 17 2 - 2 3 m = 0 : M 1 | 0 i = 0 | 0 i a + 2 b - 2 c 2 b 3 b = 0 a b c = b = 0 , c = a/ 2 = ⇒ | 0 i = 1 5 2 0 1 The matrix is clearly not Hermitian: M 1 = 1 2 - 1 0 2 0 0 3 0 6 = 1 0 0 2 2 3 - 1 0 0 = M 1 The eigenvectors are not othorgonal either h 1 | 2 i = 2 / 17 6 = 0 h 1 | 0 i = 2 / 5 6 = 0 h 2 | 0 i = 7 / 85 6 = 0 b) Find the eigenvalues first: M 2 = 0 0 1 + 3 i 0 0 0 1 - 3 i 0 0 M 2 | m i = m | m i = ( M 2 - m I ) | m i = 0 = det ( M 2 - m I ) = 0 = m 3 - 10 m = 0 = m = 0 , ± 10 3
Now the eigenvectors. m = 0 : M 2 | 0 i = 0 | 0 i (1 + 3 i ) c 0 (1 - 3 i ) a = 0 a b c = a = c = 0 , b unconstrained = ⇒ | 0 i = 0 1 0 m = 10 : M 2 | 10 i = 10 | 10 i (1 + 3 i ) c 0 (1 - 3 i ) a = 10 a b c

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