# QMB09sol_2 - Physics 139B Solutions to Homework Set 2 Fall...

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Unformatted text preview: Physics 139B Solutions to Homework Set 2 Fall 2009 1. Liboff, problem 9.32 on page 395. (a) The key equation is given in Table 9.4 on p. 379 of Liboff: L ± | ℓ , m ) = planckover2pi1 [( ℓ ∓ m )( ℓ ± m + 1)] 1 / 2 | ℓ , m ± 1 ) . (1) We will also need to use: vector L 2 | ℓ , m ) = planckover2pi1 2 ℓ ( ℓ + 1) | ℓ , m ) . (2) When coupling two p states, there are nine possible states in the product basis: | 1 , m ℓ 1 ) 1 ⊗ | 1 , m ℓ 2 ) 2 , where m ℓ 1 , m ℓ 2 = +1 , , − 1. Using vector L 2 = vector L 2 1 + vector L 2 2 + 2 L 1 z L 2 z + L 1+ L 2 − + L 1 − L 2+ , we can operate with vector L 2 on the nine states of the product basis: vector L 2 | 1 , 1 ) 1 ⊗ | 1 , 1 ) 2 = 6 planckover2pi1 2 | 1 , 1 ) 1 ⊗ | 1 , 1 ) 2 , vector L 2 | 1 , 1 ) 1 ⊗ | 1 , ) 2 = 4 planckover2pi1 2 | 1 , 1 ) 1 ⊗ | 1 , ) 2 + 2 planckover2pi1 2 | 1 , ) 1 ⊗ | 1 , 1 ) 2 , vector L 2 | 1 , 1 ) 1 ⊗ | 1 , − 1 ) 2 = 2 planckover2pi1 2 | 1 , 1 ) 1 ⊗ | 1 , − 1 ) 2 + 2 planckover2pi1 2 | 1 , ) 1 ⊗ | 1 , ) 2 , vector L 2 | 1 , ) 1 ⊗ | 1 , 1 ) 2 = 4 planckover2pi1 2 | 1 , ) 1 ⊗ | 1 , 1 ) 2 + 2 planckover2pi1 2 | 1 , 1 ) 1 ⊗ | 1 , ) 2 , vector L 2 | 1 , ) 1 ⊗ | 1 , ) 2 = 4 planckover2pi1 2 | 1 , ) 1 ⊗ | 1 , ) 2 + 2 planckover2pi1 2 | 1 , 1 ) 1 ⊗ | 1 , − 1 ) 2 + 2 planckover2pi1 2 | 1 , − 1 ) 1 ⊗ | 1 , 1 ) 2 , vector L 2 | 1 , ) 1 ⊗ | 1 , − 1 ) 2 = 4 planckover2pi1 2 | 1 , ) 1 ⊗ | 1 , − 1 ) 2 + 2 planckover2pi1 2 | 1 , − 1 ) 1 ⊗ | 1 , ) 2 , vector L 2 | 1 , − 1 ) 1 ⊗ | 1 , 1 ) 2 = 2 planckover2pi1 2 | 1 , − 1 ) 1 ⊗ | 1 , 1 ) 2 + 2 planckover2pi1 2 | 1 , ) 1 ⊗ | 1 , ) 2 , vector L 2 | 1 , − 1 ) 1 ⊗ | 1 , ) 2 = 4 planckover2pi1 2 | 1 , − 1 ) 1 ⊗ | 1 , ) 2 + 2 planckover2pi1 2 | 1 , ) 1 ⊗ | 1 , − 1 ) 2 , vector L 2 | 1 , − 1 ) 1 ⊗ | 1 , − 1 ) 2 = 6 planckover2pi1 2 | 1 , − 1 ) 1 ⊗ | 1 , − 1 ) 2 . In evaluating the above, I replaced vector L 2 with vector L 2 1 + vector L 2 2 +2 L 1 z L 2 z + L 1+ L 2 − + L 1 − L 2+ , and then used eqs. (1) and (2) repeatedly. Likewise, we can operate with vector L 2 on the nine states of the total angular mo- mentum bases, | ℓ , m ; 1 1 ) , vector L 2 | 2 , m ; 1 1 ) = 6 planckover2pi1 2 | 2 , m ; 1 1 ) , m = 2 , 1 , , − 1 , − 2 , vector L 2 | 1 , m ; 1 1 ) = 2 planckover2pi1 2 | 2 , m ; 1 1 ) , m = 1 , , − 1 , vector L 2 | , 0 ; 1 1 ) = 0 , It is now straightforward to verify Table 9.5 on p. 394 of Liboff. First, by operating with L z = L z 1 + L z 2 , it follows that in the expansion | ℓ m ; ℓ 1 ℓ 2 ) = summationdisplay m 1 ,m 2 C m 1 m 2 | ℓ 1 m 1 ) 1 ⊗ | ℓ 2 m 2 ) 2 , only terms with m = m 1 + m 2 appear on the right hand side above. Second, by operating with vector L 2 and using the results above, we can verify the entries of Table 9.5. We give two examples. To check that | 2 , 1 ; 1 , 1 ) = radicalBig 1 2 bracketleftbigg | 1 , 1 ) 1 ⊗ | 1 , ) 2 + | 1 , ) 1 ⊗ | 1 , 1 ) 2 bracketrightbigg , we note that vector...
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QMB09sol_2 - Physics 139B Solutions to Homework Set 2 Fall...

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