From Special Relativity to Feynman Diagrams.pdf

Γ μ n μ p ν 4 γ 5 2 γ μ γ ν 2 ? μν n μ p

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γ μ ) n μ p ν = − 4 γ 5 ( 2 γ μ γ ν 2 η μν ) n μ p ν = − 2 γ 5 γ μ n μ p . (10.199) where the property n · p = 0 has been used. In the rest frame p = 0 , n μ W μ becomes: ( n · W )( p = 0 ) = 2 γ 5 ( n i γ i ) p 0 γ 0 = − 2 mc γ 5 γ 0 γ i n i = − 2 mc γ 5 α i n i = − mc · n , (10.200) where we have used the property i = 2 γ 5 α i , (10.201) which can be verified using ( 10.194 ), ( 10.68 ) and ( 10.105 ). Thus we have found a Lorentz scalar quantity that in the rest frame reduces to n · : O n ≡ − 1 mc n μ W μ p = 0 −→ O ( 0 ) n = n · . (10.202) In the particular case of n pointing along the z -axis, n = n z = ( 0 , 0 , 0 , 1 ) , from ( 10.105 ) we find O ( 0 ) n z = − 1 mc n μ W μ p = 0 = 2 σ 3 0 0 2 σ 3 = 3 . (10.203) Clearly, using the transformation property ( 10.196 ) of W μ and the Lorentz invariance of the expression of O n , in a generic frame S we find O n = − 1 mc n μ W μ = S ( p ) O ( 0 ) n S ( p ) 1 , (10.204) that is if u ( 0 , r ), v( 0 , r ) are eigenvectors on · n , u ( p , r ), v( p , r ) are eigenvectors on O n corresponding to the same eigenvalues, which is the content of ( 10.190 ) and ( 10.191 ). We can define projectors P r on eigenstates of O n corresponding to the eigenvalues ε r / 2 = ± / 2: P r 1 2 1 + ε r 2 O n = 1 2 1 + ε r 1 mc γ 5 n p . (10.205)
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348 10 Relativistic Wave Equations In the rest frame the above projector reads: P ( 0 ) r 1 2 1 + ε r γ 5 n i α i = 1 2 + ε r n · σ 0 0 1 2 + ε r n · σ . (10.206) The matrices P r project on both positive and negative energy solutions with the same spin component along n . Let us now define two operators + , r , , r projecting on positive and negative solutions with a given spin component r , respectively: + , r u ( p , s ) = δ rs u ( p , s ) ; + , r v( p , s ) = 0 , , r u ( p , s ) = 0 ; , r v( p , s ) = δ rs v( p , s ). (10.207) They have the following general form: ( + , r ) α β = u α ( p , r ) ¯ u β ( p , r ) ; ( , r ) α β = − v α ( p , r ) ¯ v β ( p , r ), (10.208) as it follows from the orthogonality properties ( 10.168 ) and ( 10.169 ). To find the explicit expression of these matrices in terms of p and n , we notice that they are obtained by multiplying to the right and to the left the projectors P r on the spin state r by the projectors ± on the positive and negative energy states: ± , r = ± P r ± = ± 1 2 ( 1 ± ε r γ 5 n ) = ± 1 4 mc ( p ± mc )( 1 ± ε r γ 5 n ), where we have used the property: 1 + ε r 1 mc γ 5 np ( p ± mc ) = ( p ± mc )( 1 ± ε r γ 5 n ), (10.209) which can be easily verified using the fact that p and n anticommute: np = − pn . 10.7 Dirac Equation in an External Electromagnetic Field We shall now study the coupling of the Dirac field to the electromagnetic field A μ . To this end, as we did for the complex scalar field in Sect.10.2.1 , we apply the minimal coupling prescription, namely we substitute in the free Dirac equation p μ p μ + e c A μ , (10.210) that is, in terms of the quantum operator i μ i μ + e c A μ . (10.211) In the convention which we adopt throughout the book, the electron has charge e = −| e | < 0 . The coupled Dirac equation takes the following form:
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10.7 Dirac Equation in an External Electromagnetic Field 349 ( i μ + e c A μ μ mc ψ( x ) = 0 . (10.212) Using the covariant derivative introduced in ( 10.36 ), ( 10.212 ) takes the form i γ μ D μ mc ψ( x ) = 0 , (10.213) Just as in the case of the complex scalar field, the resulting equation is not invariant
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