solution6 - 20. If j1 j2 , then j =j1 j2 jX2 1 +j (2j + 1)...

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20. If j 1 j 2 , then j 1 + j 2 X j = j 1 j 2 (2 j +1)=2 j 1 + j 2 X j = j 1 j 2 j + j 1 + j 2 X j = j 1 j 2 1 . The second sum on the right is equal to the number of terms in the sum, which is j 1 + j 2 ( j 1 j 2 1) = 2 j 2 +1. The f rst sum on the right satis f es 2 j 1 + j 2 X j = j 1 j 2 j =( j 1 j 2 )+ ··· +( j 1 + j 2 ) +( j 1 + j 2 j 1 j 2 ) , whe rewehavereve r sedtheo rde rinthesecondl inere la t ivetothe f rst. Adding corresponding terms in each line we end up with 2 j 2 +1termsallequalto( j 1 j 2 )+( j 1 + j 2 )=2 j 1 . Thus, this sum is equal to (2 j 1 )(2 j 2 +1) . Adding both terms we end up with j 1 + j 2 X j = j 1 j 2 (2 j +1) = (2 j 1 )(2 j 2 +1)+(2 j 2 2 j 1 +1)(2 j 2 . 21. Consider three particles of spin 1 2 , and let ~ S 1 , ~ S 2 , and ~ S 3 denote the corresponding spin operators. In the combined spin space associatedw iththetota lsp inoperator ~ S = ~ S 1 + ~ S 2 + ~ S 3 , what irreducible invariant subspaces S ( s ) , occur, and how many spaces occur for each value of s . Solution: If we couple two of the spins, ~ S 1 and ~ S 2 ,say, together we obtain 2 subspaces, corresponding to s = s 12 =1= 1 2 + 1 2 , and s 12 =0= 1 2 1 2 : S 1 μ 1 2 S 2 μ 1 2 = S 12 (1) S 12 (0) These states are easily constructed: | 1 , 1 i 12 = | , i 12 | 1 , 0 i 12 = 1 2 £ | , i 12 + | , i 12 ¤ | 1 , 1 i 12 = | , i 12 | 0 , 0 i 12 = 1 2 £ | , i 12 | , i 12 ¤ When we couple the third spin to the s 12 = 0 subspace we get one subspace with spin s 123 =0+ 1 2 = 1 2 S 12 (0) S 3 μ 1 2 = S 0 μ 1 2 These states are also easily constructed: | 0 , 1 2 , 1 2 i = 1 2 £ | , i 12 | , i 12 ¤ | i 3 = 1 2 [ | , , i | , , i ] | 0 , 1 2 , 1 2 i = 1 2 £ | , i 12 | , i 12 ¤ | i 3 = 1 2 [ | , , i | , ,
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This note was uploaded on 12/19/2010 for the course PHYSICS ph 463 taught by Professor Paule. during the Fall '10 term at Missouri S&T.

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solution6 - 20. If j1 j2 , then j =j1 j2 jX2 1 +j (2j + 1)...

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