# 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’
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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