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Unformatted text preview: a b b a a b a i + =  + = ⇒ ⇒ = = ⇒ + = = ≠ The normalization will show you that 1 2 a = . Furthermore, from above relation we have / a b i or i = =(λ here is in fact the eigenvalue). You see that: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 1 ˆ  2  3  3  2  2  3 2 2 2 ˆ  2  3 1 1 ˆ  3  2  2  3  2  3 2 2 2 i O i i i O a b i i i O i  ++ + = = = + So you see that ( 29 1  2  3 2 i ± are the eigenvectors corresponding to eigenvalues – i and i . Thus we obtained the three eigenvectors and their corresponding eigenvalues. Note, however, that the operator O here is not Hermitian because its 3 eigenvalues are not all real numbers....
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 Spring '11
 CFLO
 Linear Algebra, mechanics, Hilbert space, basis vectors, independent basis vectors

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