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Consider a Yang-Mills theory in $4D$ over a gauge group $G$

$$ \mathcal{L} = - \frac{1}{4} F^{a\mu\nu}F_{\mu\nu}^a + \bar \psi i D_\mu \gamma^\mu \psi + (D_\mu \phi)^\dagger D^\mu \phi $$

where $\psi$ is a Dirac spinor and $\phi$ a scalar, both in representations of $G$. (I'm omitting the Faddeev-Popov part as it's not important for this).

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

$$ Z_3 = 1+ \frac{g^2}{(4\pi)^2 \epsilon} \left( \frac{10}3 C_\text{ad} - \frac{8}{3} N_f T_f - \frac{2}{3} N_s T_s \right) $$

where $C_\text{ad}$ is the Casimir of the adjoint , $T_f$ and $T_s$ are the Dynkin indices of the reps of the spinor and scalar respectively (defined as $\text{Tr}(T^a T^b) = - T \delta^{ab} $), and $N_f$ $N_s$ are flavour numbers.

In terms of diagram, the fermion contribution comes from a single $\psi$ loop diagram, while the scalar contributes both a $\phi$ loop and a tadpole diagram, thanks to the quartic $AA\phi^\dagger\phi$ vertex.

Since the contributions of $\psi$ and $\phi$ to $Z_3$ are identical (mutatis mutandis), except for a factor of $4$, it's no surprise that the same holds for their contributions to the $\beta$ function, that is one is $4$ times the other. (EDIT: it's not obvious that this is true, but it's easily shown if you use the Slavnov-Taylor identity $g_0 = Z_1 Z_3^{-3/2} g = Z_{1\psi} Z_\psi^{-1} Z_3^{-1} g$ and noting $Z_{1\psi}$ and $Z_\psi$ receive no 1-loop contributions depending on $N_f$ nor $N_s$, so that the only dependence is from $Z_3$.)

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 $Z_3$ in the presence of only fermions is known, is it possible to prove that scalars would contribute $\frac{1}{4}$ as much without reviewing the calculation explicitly? I feel like there should be a simple principle at work here.

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    $\begingroup$ Well...the fermions have four components and the scalar has one... $\endgroup$ – ACuriousMind Jun 10 '16 at 9:29
  • $\begingroup$ that doesn't sound very convincing... it's not obvious why a spinor's component should contribute as much as a scalar. They make different Feynman diagrams. $\endgroup$ – Riccardo Antonelli Jun 10 '16 at 10:42
  • $\begingroup$ The "additional" diagram due to the quartic vertex (I wouldn't call it a tadpole) vanishes in dim reg if the scalar is massless, so the diagrams are effectively the same. $\endgroup$ – fqq Jun 10 '16 at 10:55
  • $\begingroup$ Oh right! I did not notice that. It doesn't seem too hard then to show the fermion loop diagram is 4 times the scalar one. $\endgroup$ – Riccardo Antonelli Jun 10 '16 at 12:08

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