hw1a1 - PHYS 455 Answers to Homework I 1. The eigenvalue...

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Unformatted text preview: PHYS 455 Answers to Homework I 1. The eigenvalue equation. Consider the following 3 × 3 matrix. A = 0 0 1 0 1 0 1 0 0 (a) Characteristic polynomial p ( λ ) = det( λI- A ) = ( λ- 1) 2 ( λ + 1). The eigenvalues are +1 (degeneracy 2) and- 1 (nondegenerate). ( Note that the degeneracy and algebraic multiplicity are equal when the matrix is diagonalizable. ) (b) Associated linear equations have to be solved for finding the eigenvectors. The fol- lowing is a possible set of eigenvectors which are chosen to form an orthonormal basis. u 1 = 1 , u 2 = 1 √ 2 1 1 , u 3 = 1 √ 2 1- 1 . Here u 1 and u 2 have eigenvalues +1 and therefore these two vectors span the 2D eigensubspace for that eigenvalue. Finally, u 3 has eigenvalue- 1. (i.) { u 1 ,u 2 } is an orthonormal basis for the eigensubspace corresponding to eigenvalue +1. (ii.) We can check that h u 1 | u 3 i = h u 2 | u 3 i = 0. (iii.) { u 1 ,u 2 ,u 3 } is an orthonormal basis for the whole Hilbert space C 3 . (iv.) Let { v 1 ,v 2 ,v 3 } be another orthonormal basis for the whole Hilbert space C 3 such that each v k is an eigenvector of A . Let v 1 and v 2 be the vectors for eigenvalue +1. In that case, we have v 1 = au 1 + bu 2 , v 2 = cu 1 + du 2 , where a , b , c and d are four complex numbers satisfying | a | 2 + | b | 2 = | c | 2 + | d | 2 = 1 , c * a + d * b = 0 . The conditions above is the same thing as saying that the 2 × 2 matrix U = • a b c d ‚ is a unitary matrix. There are many non-trivial possibilities for U . For the degenerate eigenvector v 3 things are simple as this must be parallel to u 3 because the degeneracy is 1. Due to normalization requirement, v 3 differ by a phase factor: v 3 = e iα vu 3 for some α . ( Another way to see this: All 1 × 1 unitary matrices are of the form U = [ e iα ] for some real α . ) (v.) We have explicitly chosen u 1 and u 2 to be orthogonal. We could have chosen to be nonorthogonal if we wanted to. For example u 1 = 1 , u 2 = 1 √ 3 1 1 1 , u 3 = u 3 . 1 is also a basis of the Hilbert space such that each vector is a normalized eigen- vector of A . But they are not mutually orthogonal. ( Note that u 3 is always orthogonal to the u 1 and u 2 because A is hermitian. ) (c) Projection operators for the two eigensubspaces are P + = u 1...
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hw1a1 - PHYS 455 Answers to Homework I 1. The eigenvalue...

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