ps08soln - PHYS651: Problem Set 8 Solutions 1. In this...

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PHYS651: Problem Set 8 Solutions 1 . In this problem set, I use the notation x P = ( x 0 , - ~x ) and x T = ( - x 0 ,~x ). ( a ) The complex KG fields are φ ( x ) = Z d 3 ~ p (2 π ) 3 1 p E ~ p ± a ~ p e - ip · x + b ~ p e ip · x ² φ ( x ) * = Z d 3 ~ p (2 π ) 3 1 p E ~ p ± b ~ p e - ip · x + a ~ p e ip · x ² Under parity we have that U ( P ) a ~ p U - 1 ( P ) = a - ~ p , U ( P ) b ~ p U - 1 ( P ) = b - ~ p So, φ ( x ) P Z d 3 ~ p (2 π ) 3 1 p E ~ p ± a - ~ p e - ip · x + b - ~ p e ip · x ² φ ( x ) ~ p →- ~ p = Z d 3 ~ p (2 π ) 3 1 p E ~ p ± a - ~ p e - i ˜ p · x + b - ~ p e i ˜ p · x ² = φ ( x P ) where ˜ p · x = p · x P . Under charge conjugation U ( C ) a ~ p U - 1 ( C ) = b ~ p U ( C ) b ~ p U - 1 ( C ) = a ~ p φ φ * obviously. Finally inder time reversal we have U ( T ) a ~ p U - 1 ( T ) = a - ~ p U ( T ) b ~ p U - 1 ( T ) = b - ~ p This is just like parity but T has the extra feature that it acts on c- numbers (commuting number): U ( T )(c-number) = (c-number * ) U ( T ).
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using this we find φ ( x ) T Z d 3 ~ p (2 π ) 3 1 p E ~ p ± a ~ p e ip · x + b ~ p e - ip · x ² φ ( x ) ~ p →- ~ p = Z d 3 ~ p (2 π ) 3 1 p E ~ p ± a ~ p e i ˜ p · x + b ~ p e - i ˜ p · x ² = φ ( x T ) where ˜ p · x = - p · x T . The only difference is the sign in the exponentials. ( b ) I used the notation ( - 1) μ is 1 for μ = 0 and -1 otherwise. j μ ( x ) = i [ φ * μ φ - ( μ φ * ) φ ] j μ x P ( - 1) μ j μ ( x P ) ( - 1) μ is from μ C - j μ ( x ) just interchange φ φ * T ( - 1)[ - ( - 1) μ ] j μ ( x T ) from the ‘i’ and μ Under CPT we get: j μ x CPT = - j μ ( - x ) 2 . see P.S., page 65-70. The only difference from the previous calculation is that we now need to include spins and spinors in the picture. 3 . The last line in the table on p. 71 in P.S. shows that for any fermion bilinear B μ 1 ··· μ n CPT ( - 1) n B μ 1 ··· μ n ( - x ) In other words, every Lorentz index gives a minus sign under CPT. We know from question 1 that this is also true for scalar bilinear φ * φ . Finally, we know that μ → - μ under CPT. Lorentz invariance and hermicity of the lagrangian require to have only power of bilinear ( ¯ ψψ or φ * φ ). Since all lorentz indices must be contracted ( L is a scalar) either among themselves or using a derivative, we conclude that: L ( x ) CPT = L ( - x ) ** I will add here the proof that ¯ ψiγ μ μ ψ is CPT invariant since a lot of people didn’t do this properly and it is the only tricky one because of the ’i’
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factor. Using table on p. 71, you can easily see that this term is C and P
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This note was uploaded on 09/28/2008 for the course PHYS 651 taught by Professor Tye,henry during the Fall '03 term at Cornell University (Engineering School).

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ps08soln - PHYS651: Problem Set 8 Solutions 1. In this...

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