137A_HW06_solutions.pdf - Physics 137A Homework 6 Solutions...

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Physics 137A: Homework 6 Solutions March 14, 2017 Problem 1 (a) Note the matrix is real, so Hermitian is the same as symmetric here. At any rate, R ( θ ) = cos θ - sin θ sin θ cos θ 6 = R ( θ ) . So R ( θ ) is not Hermitian . (b) R ( θ ) R ( θ ) = cos θ - sin θ sin θ cos θ cos θ sin θ - sin θ cos θ = 1 0 0 1 , where I used that cos 2 θ + sin 2 θ = 1. So R ( θ ) is unitary . (c) We solve the characteristic polynomial equation: 0 = cos θ - λ sin θ - sin θ cos θ - λ = (cos θ - λ ) 2 + sin 2 θ = ( λ - (cos θ + i sin θ ))( λ - (cos θ - i sin θ )) = ( λ - e )( λ - e - ) I used the fact that e = cos θ + i sin θ . So the eigenvalues of R ( θ ) are complex, and equal to e ± . (d) Let | v 1 i = x y 1
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be the eigenvector for the eigenvalue e . The eigenvector equation R ( θ ) | v 1 i = e | v 1 i can also be written as ( R ( θ ) - e I ) | v 1 i = 0 (where I is the 2 x 2 Identity matrix), so we can write: cos θ - e sin θ - sin θ cos θ - e x y = 0 Let us use the fact once again that e = cos θ + i sin θ . The above eigenvector equation then becomes: - i sin θ sin θ - sin θ - i sin θ x y = 0 This system gives only one equation in the two variables x and y . The first equation, say, gives: - i x sin θ + y sin θ = 0 So the equation is satisfied for instance for x = 1 and y = i . So | v 1 i = 1 i . This is not a unit vector, so we need to normalize it: h v 1 | v 1 i = ( 1 - i ) 1 i = 2 . So we have the normalized eigenvector for the eigenvalue e : | v 1 i = 1 2 1 i . Now we repeat the exercise for the eigenvalue e - . Let | v 2 i = x y be the corresponding eigenvector. Then: cos θ - e - sin θ - sin θ cos θ - e - x y = 0 Just as before, we can rewrite the above as: i sin θ sin θ - sin θ i sin θ x y = 0 The first equation gives: i x sin θ + y sin θ = 0 So the equation is satisfied for instance for x = 1 and y = - i . So | v 2 i = 1 - i . This once again is not a unit vector, so we need to normalize it: h v 2 | v 2 i = ( 1 i ) 1 - i = 2 . So we have the normalized eigenvector for the eigenvalue e - : | v 1 i = 1 2 1 - i . 2
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(e) We have checked already that | v 1 i and | v 2 i have unique length, and furthermore they span the space C 2 , but we need to check that they are orthogonal: h v 1 | v 2 i = 1 2 ( 1 - i ) 1 - i = 1 - 1 = 0 .
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