# ch10 - CHAPTER 10 1 We need to solve 0 2 i i u 0 v u 2 v 1...

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CHAPTER 10 1. We need to solve h 2 0 - i i 0 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 co s α + σ x s in α co s β + σ y s in α s in β σ n n = (s in α co s β ,s in α s in β ,co s 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 e -I β = u . We may rewrite this in the form 2 v s in α 2 cos α 2 e - i β = 2 u s in 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

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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 s in α e - i β s in α e i β - cos α U = 10 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 + ) m n = 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 0 and
S y = i 2 ( S - - S + ) = h 2 0 - i 3 0 0 i 3 0 - 2 i 0 0 2 i 0 - i 3 0 0 i 3 0 The eigenstates in the above representation are very simple: χ 3/2 = 1 0 0 0 ; χ 1 /2 = 0 1 0 0 ; χ - 1/2 = 0 0 1 0 ; χ - 3/2 = 0 0 0 1 5.

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• Spring '05
• mokhtari
• Trigraph, Spin multiplicity, Singlet state, Doublet state, triplet state

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