From Special Relativity to Feynman Diagrams.pdf

# Φ x d 3 p 2 π 3 v 2 e p a p e i p x b p e i p x

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φ( x ) = d 3 p ( 2 π ) 3 V 2 E p a ( p ) e i p · x + b ( p ) e i p · x , (11.20) ˆ π( x ) = t ˆ φ = i d 3 p ( 2 π ) 3 V E p 2 a ( p ) e i p · x b ( p ) e i p · x , (11.21) ˆ π( x ) = t ˆ φ = − i d 3 p ( 2 π ) 3 V E p 2 a ( p ) e i p · x b ( p ) e i p · x . (11.22) 2 Actually we shall only consider the quantum description of the interaction between the electro- magnetic field and a Dirac field.

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11.2 Quantization of the Klein–Gordon Field 365 Note that had we considered an hermitian field ˆ φ ( x ) = ˆ φ( x ), corresponding to a real classical field, hermiticity would have identified b a . For a non-hermitian field instead the a and b operators are independent. 3 Equations( 11.19 ) and/or ( 11.20 ) can be inverted to compute the operators a ( p ), b ( p ) and their hermitian conjugates in terms of ˆ φ and ˆ φ . To this end let us define the following function: f p = 1 2 E p V e i p · x , and prove that a ( q ) = i d 3 x f q ( x )∂ t ˆ φ( x ) ˆ φ( x )∂ t f q ( x ) = i d 3 x f q ( x ) ˆ π ( x ) ˆ φ( x )∂ t f q ( x ) , (11.23) a ( q ) = − i d 3 x f q ( x )∂ t ˆ φ ( x ) ˆ φ ( x )∂ t f q ( x ) (11.24) = − i d 3 x f q ( x ) ˆ π( x ) ˆ φ ( x )∂ t f q ( x ) . (11.25) Let us recall here some useful properties which we shall extensively use in the following: d 3 x e ± i p · x = ( 2 π ) 3 δ 3 ( p ) ; f ( p ) = d 3 q δ 3 ( p q ) f ( q ), E p m 2 c 4 − | p | 2 c 2 = E p . (11.26) Consider the first of ( 11.23 ) and let us rewrite the first term on the right hand side using for ˆ φ the expansion ( 11.19 ): i d 3 x f q ( x )∂ t ˆ φ( x ) = d 3 x d 3 p ( 2 π ) 3 E p E q a ( p ) 2 e i ( p q ) · x b ( p ) 2 e i ( p + q ) · x = a ( q ) 2 b ( q ) 2 e 2 i E q t , (11.27) where we have used the fact that p = q implies E p = E q . By the same token we prove that: i d 3 x ˆ φ( x )∂ t f q ( x ) = a ( q ) 2 + b ( q ) 2 e 2 i E q t . (11.28) 3 We have used a similar argument after ( 10.30 ), in the classical case.
366 11 Quantization of Boson and Fermion Fields Summing ( 11.27 ) and ( 11.28 ) the terms with b ( q ) drop out and we find the first of ( 11.23 ). We can prove similar formulas for b and b : b ( q ) = i d 3 x f q ( x )∂ t ˆ φ ( x ) ˆ φ ( x )∂ t f q ( x ) = i d 3 x f q ( x ) ˆ π( x ) ˆ φ ( x )∂ t f q ( x ) , (11.29) b ( q ) = − i d 3 x f q ( x )∂ t ˆ φ( x ) ˆ φ( x )∂ t f q ( x ) = − i d 3 x f q ( x ) ˆ π ( x ) ˆ φ( x )∂ t f q ( x ) , (11.30) by showing that the following properties hold: i d 3 x f q ( x )∂ t ˆ φ ( x ) = − a ( q ) 2 e 2 i E q t + b ( q ) 2 , i d 3 x ˆ φ ( x )∂ t f q ( x ) = a ( q ) 2 e 2 i E q t + b ( q ) 2 . (11.31) From ( 11.23 ) and ( 11.29 ), and from the commutation relations ( 11.6 ), we may now compute the commutators among a ( q ), a ( q ), b ( q ), b ( q ) a ( p ), a ( q ) = d 3 x d 3 y f p ( x ) ˆ π ( x ) ˆ φ( x )∂ t f p ( x ), f q ( y ) ˆ π( y ) ˆ φ ( y )∂ t f q ( y ) = d 3 x d 3 y f p ( x )∂ t f q ( y ) ˆ φ ( y ), ˆ π ( x ) t f p ( x ) f q ( y ) ˆ φ( x ), ˆ π( y ) = i d 3 x d 3 y f p ( x )∂ t f q ( y ) t f p ( x ) f q ( y ) δ 3 ( x y ) = i d 3 x f p ( x )∂ t f q ( x ) t f p ( x ) f q ( x ) = i 2 V i E q E p i E p E q ( 2 π ) 3 δ 3 ( p q ) = ( 2 π ) 3 V δ 3 ( p q ).

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