From Special Relativity to Feynman Diagrams.pdf

Graphically by lines connecting two vertices by a

Info icon This preview shows pages 450–452. Sign up to view the full content.

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).
Image of page 450

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

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 |
Image of page 451
Image of page 452
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern