From Special Relativity to Feynman Diagrams.pdf

# So that at each vertex there are two fermion and one

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, so that at each vertex there are two fermion and one boson lines. Let us denote F i , F e and B i , B e the number of internal and external lines. Since an external line is connected to one vertex and an internal line connects two vertices, it is easy to see that 27 2 F i + F e = 2 V , 2 B i + B e = V . (12.258) If we solve the above equations in F i and B i and substitute the result into ( 12.257 ) we find D G = 4 3 2 F e B e . (12.259) We conclude that the degree of divergence does not depend on the number of vertices and internal lines, but only on the number of external lines. In particular we see that D G > 0 only for a limited number of diagrams. In general when this happens we say that the theory is renormalizable . Therefore QED is renormalizable. On the other hand the renormalizability property is related to the physical dimen- sion of the coupling constant. Let us first recall from Sect.12.3.5 that, using nat- ural units [ c ] , [ ] , the fermions have dimension 3 2 in mass units and the bosons dimension 1. Furthermore the action of a theory is dimensionless, so that the Lagrangian density has dimension (in mass) [ M 4 ] . On the other hand, if from each vertex f fermionic and b bosonic lines originate, respectively, we must have that the dimension of the coupling constant λ in front of the interaction Lagrangian density is [ λ ] = [ M 4 ν ] = [ M d λ ] , 27 To derive these relations, one can cut each internal fermion line of a diagram into two parts. The total number of lines so obtained should be twice the number of vertices. In this counting however, each internal line contributes two units (i.e. a total of 2 F i units) and each external ones a single unit (i.e. a total of F e units). A similar argument applies to the boson lines.

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12.8 A Pedagogical Introduction to Renormalization 513 where ν = 3 2 f + b . In particular for quantum electrodynamics we find ν = 3 2 f + b = 4, so that the coupling constant e 4 π , ( e 4 π c in the usual units ) is indeed dimen- sionless. For a general theory we can generalize ( 12.258 ) as follows: 2 F i + F e = fV . 2 B i + B e = bV . (12.260) Substituting the values of F i and B i in ( 12.257 ) we find D G = bV B e + 3 2 ( f V F e ) 4 V + 4 = V b + 3 2 f 4 + 4 3 2 F e B e = − d λ V 3 2 F e B e + 4 , (12.261) where we have denoted by d λ = b + 3 2 f 4 the physical dimension of λ . Therefore: If b + 3 2 f < 4, that is if d λ > 0, as the perturbative order V increases, D G decreases and amplitudes are finite. We say in this case that the theory is super-renormalizable . If b + 3 2 f = 4, so that d λ = 0, the coupling constant is dimensionless. In this case D G is independent of V and the theory is renormalizable . If b + 3 2 f > 4 , d λ < 0, the theory is not-renormalizable since, by increasing the order of the diagram, that is the number of the vertices, the degree of divergence also increases and we are left with an infinite number of divergent diagrams.
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• Fall '17
• Chris Odonovan

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