Physics, asked by rudrapsb887, 1 year ago

In QED/Yang Mills, why do fermions contribute 4 times as much as scalars to vacuum polarization?

Answers

Answered by Rohitdas800
0
Consider a Yang-Mills theory in 4D4D over a gauge group GG

L=−14FaμνFaμν+ψ¯iDμγμψ+(Dμϕ)†DμϕL=−14FaμνFμνa+ψ¯iDμγμψ+(Dμϕ)†Dμϕ

where ψψ is a Dirac spinor and ϕϕ a scalar, both in representations of GG. (I'm omitting the Faddeev-Popov part as it's not important for this).

Using Lorenz-Feynman gauge with ξ=1ξ=1, in dim reg and in the MS scheme, the renormalization for the gluon wavefunction is to one loop

Z3=1+g2(4π)2ϵ(103Cad−83NfTf−23NsTs)Z3=1+g2(4π)2ϵ(103Cad−83NfTf−23NsTs)

where CadCad is the Casimir of the adjoint , TfTf and TsTsare the Dynkin indices of the reps of the spinor and scalar respectively (defined as Tr(TaTb)=−TδabTr(TaTb)=−Tδab), and NfNf NsNs are flavour numbers.

In terms of diagram, the fermion contribution comes from a single ψψ loop diagram, while the scalar contributes both a ϕϕ loop and a tadpole diagram, thanks to the quartic AAϕ†ϕAAϕ†ϕ vertex.

Since the contributions of ψψ and ϕϕ to Z3Z3 are identical (mutatis mutandis), except for a factor of 44, it's no surprise that the same holds for their contributions to the ββ function, that is one is 44times the other. (EDIT: it's not obvious that this is true, but it's easily shown if you use the Slavnov-Taylor identity g0=Z1Z−3/23g=Z1ψZ−1ψZ−13gg0=Z1Z3−3/2g=Z1ψZψ−1Z3−1gand noting Z1ψZ1ψ and ZψZψ receive no 1-loop contributions depending on NfNf nor NsNs, so that the only dependence is from Z3Z3.)

Now, I can go through the three diagrams explicitly and show this, but is it possible to derive this factor through a simpler argument? In other words, assuming the result for Z3Z3 in the presence of only fermions is known, is it possible to prove that scalars would contribute 1414 as much without reviewing the calculation explicitly? I feel like there should be a simple principle at work here.

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