From Special Relativity to Feynman Diagrams.pdf

Graphically by lines connecting two vertices by a

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graphically by lines connecting two vertices: by a line connecting x i and x j and oriented from x j to x i ; by a dashed, undirected line joining x i to x j . The reason why the latter line is not oriented is due to the symmetry of the photon propagator D F μν ( x i x j ) with respect to an exchange of x i and x j : D F μν ( x i x j ) = D F μν ( x j x i ). Each of these internal lines describes a virtual particle propagating between the two vertices. By virtual particle we mean a particle whose momentum does not satisfy the on-shell condition: p 2 m 2 = 0 for the electron, k 2 = 0 for the photon, k μ being the photon 4-momentum. Thenormalorderedterms,besidesthecontractions,willalsocontainun-contracted field operators, which we shall refer to as free . These operators will be represented by lines extending from the vertex in which they are computed to infinity: The line rep- resenting ψ( x ) will originate at infinity and end in x . In the matrix element between initial and final state, it will contribute only if the initial state contains a free electron or if the final state contains a free positron. In these two cases ψ( x ) will destroy the incoming electron or create the outgoing positron in x , respectively. In the first case, if the electron state is | p e , r = c ( p e , r ) | 0 , ψ( x ) would give the following non-vanishing contribution to the amplitude 15 : 0 | ψ( x ) | p e , r = d q 2 m 2 s = 1 u ( q , s ) 0 | c ( q , s ) c ( p e , r ) | 0 e iq · x = d q 2 m 2 s = 1 u ( q , s ) 0 |[ c ( q , s ), c ( p e , r ) ] | 0 e e iq · x = 2 mu ( p e , r ) e ip e · x . (12.132) Similarly, in the second case, if the final positron state is | p e + , r = d ( p e + , r ) | 0 , the field ψ( x ) will contribute the following non-vanishing quantity: p e + , r | ψ( x ) | 0 = 2 m v( p e + , r ) e ip e + · x . (12.133) By the same token we can show that a free ψ( x ) operator contributes only to those processeswithanincomingpositron,whichwillbedestroyedin x ,orwithanoutgoing electron, which will be created in x by the same operator. It will be represented by 15 Recall that, in the light of our comments below eq. (12.76), we have replaced everywhere the normalization volume V with 1/(2 E ), so that, for instance [ c ( p , r ), c ( q , s ) ] = ( 2 π) 3 2 E δ 3 ( p q ) . Note that, had we kept the normalization volumes, the calculation below would yield 0 | ψ( x ) | p , r = m E p V e ip · x , contributing a factor 1 / V to the amplitude, as anticipated in our discussion below eq. (12.76).

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470 12 Fields in Interaction a line originating in the vertex x and extending towards infinity. In these two cases ψ( x ) will therefore, contribute the following matrix elements to the amplitude: 0 | ψ( x ) | p e + , r = 2 m ¯ v( p e + , r ) e ip e + · x , p e , r | ψ( x ) | 0 = 2 m u ( p e , r ) e ip e · x . (12.134) We can also have a process with an electron in both the initial and final states. In this case the normal-ordered product : ψ α ( y β ( x ) : will contribute the following quantity to the amplitude: q e , r | : ψ α ( y β ( x ) : | p e , r = 0 | c ( q e , r ) ψ ( ( y β ( + ) ( x ) c ( p e , r ) | 0 = 0 | c ( q e , r ), ψ ( ( y ) × ψ β ( + ) ( x ), c ( p e , r ) | 0 = q e , r | ψ α ( y ) | 0 0 |
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• Fall '17
• Chris Odonovan

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