This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHYS 455 Answers to Homework I 1. The eigenvalue equation. Consider the following 3 Ã— 3 matrix. A = ï£® ï£° 0 0 1 0 1 0 1 0 0 ï£¹ ï£» (a) Characteristic polynomial p ( Î» ) = det( Î»I A ) = ( Î» 1) 2 ( Î» + 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 = ï£® ï£° 1 ï£¹ ï£» , u 2 = 1 âˆš 2 ï£® ï£° 1 1 ï£¹ ï£» , u 3 = 1 âˆš 2 ï£® ï£° 1 1 ï£¹ ï£» . 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 Ã— 2 matrix U = â€¢ a b c d â€š is a unitary matrix. There are many nontrivial 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 iÎ± vu 3 for some Î± . ( Another way to see this: All 1 Ã— 1 unitary matrices are of the form U = [ e iÎ± ] for some real Î± . ) (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 = ï£® ï£° 1 ï£¹ ï£» , u 2 = 1 âˆš 3 ï£® ï£° 1 1 1 ï£¹ ï£» , 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...
View
Full Document
 Spring '11
 starg
 Linear Algebra, Hilbert space, Orthonormal basis

Click to edit the document details