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Unformatted text preview: CHAPTER 10 1. We need to solve h 2 i i u v = ± h 2 u v For the + eigenvalue we have u =  iv , s that the normalized eigenstate is χ + = 1 2 1 i The – eigenstate can be obtained by noting that it must be orthogonal to the + state, and this leads to χ = 1 2 1 i . 2, We note that the matrix has the form σ z cos α + σ x sin α cos β + σ y sin α sin β ≡ σ • n n = (sin α cos β ,sin α sin β ,cos α ) This implies that the eigenvalues must be ± 1. We can now solve cos α sin α e i β sin α e i β cos α u v = ± u v For the + eigenvalue we have u cos α + v sin α eI β = u . We may rewrite this in the form 2 v sin α 2 cos α 2 e i β = 2 u sin 2 α 2 From this we get χ + = cos α 2 e i β sin α 2 The – eigenstate can be obtained in a similar way, or we may use the requirement of orthogonality, which directly leads to χ = e i β sin α 2 cos α 2 c The matrix U = cos α 2 e i β sin α 2 e i β sin α 2 cos α 2 has the property that U + cos α sin α e i β sin α e i β cos α U = 1 0 1 as is easily checked. d The construction is quite simple. S z = h 3 / 2 0 0 0 0 1/ 2 0 0 0 0 1/ 2 0 0 0 0 3 / 2 To construct S + we use ( S + ) mn = h δ m , n + 1 ( l m + 1)( l + m ) and get S + = h 0 3 0 0 0 0 2 0 0 0 0 3 0 0 0 0 We can easily construct S = ( S + ) + . We can use these to construct S x = 1 2 ( S + + S ) = h 2 0 3 0 0 3 0 2 0 0 2 0 3 0 0 3 and S y = i 2 ( S S + ) = h 2 i 3 0 0 i 3 2 i 0 2 i i 3 0 0 i 3 The eigenstates in the above representation are very simple: χ 3/ 2 = 1 ; χ 1/ 2 = 1...
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This note was uploaded on 03/09/2008 for the course PHYS 401 taught by Professor Mokhtari during the Spring '05 term at UCLA.
 Spring '05
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