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

This preview shows pages 1–2. Sign up to view the full content.

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 | , ,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online