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Unformatted text preview: PHYS651: Problem Set 6 Solutions 1 . [ S ,S ] = i 4 2 [ , ] [ , ] [ , ] + [ , ] We can use that = 2 g 1 and the fact that 1 commutes with everything. Therefore = i 4 2 2 [ , ] ( ) Evaluate = 2 g  = 2 g  2 g + = 2 g  2 g + 2 g  = 2 g  2 g + 2 g  2 g + so the first 4 terms is the commutator that we need. Plugging it in yields [ S ,S ] = i 4 i g  g + g  g  g + g  g + g = ig S  ig S  ig S + ig S Now, you should by now know pretty well what is the algebra of the lorentz group, lets just hope you dont have any nightmares about it and that this is the last question about it. 2 . ( a ) Using the definitions given in the problem we obtain L 1 = J 2 , 3 , L 2 = J 3 , 1 , L 3 = J 1 , 2 and K 1 = J , 1 , K 2 = J , 2 , K 3 = J , 3 , Together with the commutation relation for J from problem 1 we get for example L 1 ,L 2 = J 23 ,J 31 = i ( g 33 J 21 g 23 J 31 g 31 J 23 + g 21 J 33 ) = iJ 21 = iL 3 K 1 ,K 2 = J 01 ,J 02 = i ( g 10 J 02 g 00 J 12 g 12 J 00 + g 02 J 10 ) = iJ 12 = iL 3 L i ,K j = 1 2 ilm J lm ,J j = i 1 2 ilm ( g m J lj g l J mj g mj J l + g lj J m ) = i ilm J m = i ilm K m or again in generalized notation L i ,L j = i ijk L k K i ,K j = i...
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 Fall '03
 TYE,HENRY
 Quantum Field Theory

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