From Special Relativity to Feynman Diagrams.pdf

# 11127 and we note that there is no ambiguity when x y

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(11.127) and we note that there is no ambiguity when x 0 = y 0 since in this case ˆ φ( x ) and ˆ φ ( y ) commute. 13 Furthermore the fact that ˆ φ( x ) and ˆ φ ( y ) commute at space-like 13 For a hermitian (and thus neutral) field D F ( x y ) = c 0 | T ˆ φ( x ) ˆ φ( y ) | 0 . ( 11 . 128 )

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11.4 Invariant Commutation Rules and Causality 389 distances ensures that time ordering remains invariant under Lorentz transformations, and thus that the Feynman propagator is Lorentz invariant. To compute D F ( x y ) we note that if x 0 > y 0 , using ( 11.113 ) and the fact that the destruction operators annihilate the vacuum, we have 0 | T ˆ φ( x ) ˆ φ ( y ) | 0 = 0 | ˆ φ + ( x ) ˆ φ ( y ) | 0 = 0 | ˆ φ + ( x ), ˆ φ ( y ) | 0 = c D + ( x y ). Similarly for y 0 > x 0 we get 0 | T ˆ φ( x ) ˆ φ ( y ) | 0 = 0 |[ ˆ φ + ( y ), ˆ φ ( x ) ]| 0 = c D ( x y ). (11.129) We may then write the Feynman propagator in the following compact form: D F ( x y ) = θ( x 0 y 0 ) D + ( x y ) + θ( y 0 x 0 ) D ( x y ) (11.130) The physical meaning of the Feynman propagator is easily understood if we observe that for x 0 > y 0 , D F ( x y ) = 0 | ˆ φ + ( x ) ˆ φ ( y ) | 0 , that is D F ( x y ) measures the probability amplitude that a particle be created at y at the instant y 0 and then destroyed at x at the instant x 0 , while, if y 0 > x 0 , D F ( x y ) = 0 | ˆ φ + ( y ) ˆ φ ( x ) | 0 is the probability amplitude that an antiparticle be created in x at the time x 0 and then destroyed in y at the time y 0 . If we now use the explicit expression of D + ( x y ) and D ( x y ) we may write D F ( x y ) = θ( x 0 y 0 ) D + ( x y ) + θ( y 0 x 0 ) D ( x y ) = c d 3 p ( 2 π ) 3 1 2 E p θ( x 0 y 0 ) e i p · ( x y ) + θ( y 0 x 0 ) e i p · ( x y ) . (11.131) We are now going to prove that, using the Cauchy residue theorem, we can write the above expression for the Feynman propagator in terms of a an integral in the complex p 0 plane along the path C F in Fig. 11.1 : D F ( x y ) = i 2 d 3 p ( 2 π ) 3 C F dp 0 2 π e i p · ( x y ) p 2 m 2 c 2 = i 2 d 3 p ( 2 π ) 3 C F dp 0 2 π e i ( p 0 ( x 0 y 0 ) p · ( x y ) ) ( p 0 − ¯ p 0 ) ( p 0 + ¯ p 0 ) , (11.132) where, as usual, p · ( x y ) p 0 ( x 0 y 0 ) p · ( x y ) and ¯ p 0 | p | 2 + m 2 c 2 = E p / c > 0 . To show this we observe that if x 0 > y 0 we can close the contour C F in the lower p 0 half-plane, where the imaginary part of p 0 is negative, along a semi-circle
390 11 Quantization of Boson and Fermion Fields Fig.11.1 Integration in the complex p 0 plane Fig.11.2 The x 0 > y 0 case C ( ) of infinite radius, so that the integral along C ( ) be exponentially suppressed, see Fig. 11.2 . Indeed, denoting the integrand in ( 11.132 ) by f ( p 0 , p ) : f ( p 0 , p ) i 2 π 1 p 2 m 2 c 2 e i p · ( x y ) , the integral of f along C ( ) vanishes since lim Im ( p 0 ) →−∞ e 1 Im ( p 0 )( x 0 y 0 ) = 0 . We can then write: C F f ( p 0 , p ) dp 0 = C F + C ( ) f ( p 0 , p ) dp 0 = − 2 π i Res ¯ p 0 ( f ) = − 2 π i [ ( p 0 − ¯ p 0 ) f ( p 0 , p ) ] p 0 = ¯ p 0 = 2 ¯ p 0 e i p · ( x y ) p 0 = ¯ p 0 . Therefore, for x 0 > y 0 , the integral in ( 11.132 ) reads: d 3 p ( 2 π ) 3 C F dp 0 f ( p 0 , p ) = c d 3 p ( 2 π ) 3 e i p · ( x y ) 2 E p = D F ( x y ).

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