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Unformatted text preview: PHYS 507 Answers to Homework V 1. The assumption ~μ = k ~ S is not necessary. We only need to use the commutation relation [ S i ,μ j ] = i ¯ h X k ² ijk μ k , which expresses the fact that ~μ is a vector operator. d dt h S i i t = i ¯ h h [ H,S i ] i t = i ¯ h X j B j h [ μ j ,S i ] i t = i ¯ h X j B j h [ S i ,μ j ] i t = X jk B j ² ijk h μ k i t = ( ~ B × h ~μ i t ) i = +( h ~μ i t × ~ B ) i . 2. We have calculated the rotation operator for a single spin 1/2 particle before. It is D ( θ, ˆ x ) = exp i 2 θσ x ¶ = cos θ 2 i sin θ 2 σ x = • cos θ 2 i sin θ 2 i sin θ 2 cos θ 2 ‚ As a result we have the following D ( θ, ˆ x )  ↑i = cos θ 2  ↑i  i sin θ 2  ↓i , D ( θ, ˆ x )  ↓i = i sin θ 2  ↑i + cos θ 2  ↓i . (a) We will rotate the twoparticle state  , i by applying individual rotations to each particle using the relations above. Let us start with  ↑↓i . We have D ( θ, ˆ x )  ↑↓i = ( D 1 ( θ, ˆ x )  ↑i ) ⊗ ( D 2 ( θ, ˆ x )  ↓i ) = cos θ 2  ↑i  i sin θ 2  ↓i ¶ ⊗ i sin θ 2  ↑i + cos θ 2  ↓i ¶ = cos 2 θ 2  ↑↓i  i sin θ 2 cos θ 2 (  ↑↑i +  ↓↓i ) sin 2 θ 2  ↓↑i . Continue with  ↓↑i , D ( θ, ˆ x )  ↓↑i = ( D 1 ( θ, ˆ x )  ↓i ) ⊗ ( D 2 ( θ, ˆ x )  ↑i ) = i sin θ 2  ↑i + cos θ 2  ↓i ¶ ⊗ cos θ 2  ↑i  i sin θ 2  ↓i ¶ = cos 2 θ 2  ↓↑i  i sin θ 2 cos θ 2 (  ↑↑i +  ↓↓i ) sin 2 θ 2  ↑↓i . 1 For the state  , i we have D ( θ, ˆ x )  , i = 1 √ 2 cos 2 θ 2 + sin 2 θ 2 ¶ (  ↑↓i   ↓↑i ) =  , i ....
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 Spring '11
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