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I've been told that for any group of SM, the running of the corresponding coupling constant, $g$, is given by:

$$ \frac{dg}{d(\ln{Q})} = b·g^3/(16\pi^2) $$

Where $$ b = -\frac{11}{3}C_2(A) + \sum\Bigg[\frac{2}{3}T(R_f) + \frac{1}{3}T(R_s) \Bigg] $$ and

$$ C_2(A) = \begin{cases} N,\ {\rm for\ }SU(N)\\ 0,\ {\rm for\ }U(1) \end{cases}, \qquad T(R_f) = \begin{cases}\frac{1}{2},\ {\rm Weyl\ spinors\ in\ fundamental\ repr.\ of\ }SU(N)\\ N,\ {\rm Weyl\ spinors\ in\ adjoint\ repr.\ of\ }SU(N)\\ Y_f^2,\ {\rm for\ }U(1) \end{cases} $$

$Y_f$ is the hypercharge of the corresponding field. $T(R_s)$ takes the same values as $T(R_f)$ for each complex scalar in the corresponding representation.

I'm trying to obtain the correct $b$ terms for each group, achieving:

$$ b(SU(3)) = -(11/3)·3 + 6·(2/3)·(1/2) + (2/3)·(1/2)·3 + (2/3)·(1/2)·3 = -7 $$

In the RHS and reading from left to right we find the term corresponding to $C_2(A)$, followed by the term for QCD quark triplets, SU(2) lepton doublets, SU(2) right parts. This result is the same as Cheng and Li, so maybe it's correct. I'm not sure because I have used $T(R_f) = 1/2$ for right fields.

For SU(2):

$$ b(SU(2)) = -(11/3)·2 + 3·(2/3)·(1/2) + 3·(2/3)·(1/2) + 9·(2/3)·(1/2) + 3·6·(2/3)·(1/2) = 11/3 $$

Again, from left to right, $C_2(A)$ part, 3 lepton doublets, 3 right leptons, 3 quark lepton doublets with 3 colours each one, and 6 flavour with 3 colours each for right quarks.

For U(1):

$$ b(U(1)) = (2/3)·\{3[3(2·(1/6)^2 + (2/3)^2 + (1/3)^2)] + 2(1/2)^2 + 1\} + 2(1/2)^2(1/3) = 41/6 $$

Here, $\{···\}$ counts the 3 families with 3 colour copies for left and right quarks and the 3 families for left leptons and right charged leptons. The last contribution is the Higgs that counts as 2 complex scalars.

Checking with Cheng and Li, section 14.3, I know the 2nd and 3rd values are incorrect. Actually, for the 2nd one we should have a result less than zero and for the 3rd, $b(U(1)) = 4$. What am I doing wrong?

Find this formula here: https://en.wikipedia.org/wiki/Beta_function_(physics)#SU(N)_Non-Abelian_gauge_theory

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    $\begingroup$ You should get something proporional to $3\cdot\left[3\cdot\left(2\cdot\frac{1}{6^2}+\frac{2^2}{3^2}+\frac{1}{3^2}\right)+2\cdot\frac{1}{2^2}+1\right]+2\cdot\frac{1}{2^2}$ where the first factor of 3 comes from the repetition of families, the second one from the three colors of SU(3) and the 2's from the fact that doublets have two components. The $[\dots]$ term is the contribution of a family of quarks and leptons, and the last term comes from the Higgs. $\endgroup$
    – user178876
    Commented Apr 21, 2019 at 20:40
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    $\begingroup$ Please don't add useless comment like "edited" in the title or the body of the question. All of the posts on this site are version controlled, so just seemlessly integrate the new content into the post; interested parties can always see the changes made by looking at the edit history. $\endgroup$
    – Kyle Kanos
    Commented Apr 22, 2019 at 1:33
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    $\begingroup$ More imprtantly though you seem to be summing all the three gauge group representations for each gauge group. Also I suspect you should be considering left and right Weyl spinors $\endgroup$
    – innisfree
    Commented Apr 22, 2019 at 1:59
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    $\begingroup$ No, your SU(3) result is correct by accident. Only the quarks contribute - leptons and Higgs don’t interact under SU(3). The result is just -11/3 * 3 + 3.3.2 * 2/3 * 1/2 = -7 $\endgroup$
    – innisfree
    Commented Apr 22, 2019 at 2:52
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    $\begingroup$ 3 colour x 3 families x 2 quarks per family is the 3.3.2 $\endgroup$
    – innisfree
    Commented Apr 22, 2019 at 2:53

1 Answer 1

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You just read off your formulas, but properly.

For SU(3), you have N =3 , and 6 Weyl fermions, so $N_F=6$ (Dirac) flavors in the fundamental or antifundamental of SU(N), which count the same, as per your definitions, so, then, plain QCD: $$ b=-11 \cdot N/3 + 2 N_F /3 \to -11 +4 =-7. $$

For the EW SU(2), you have N =3 , and 4 Weyl weak doublets (1 for leptons and 3 color copies for quarks) per generation coupling to the W triplet. Call generations $N_G=3$, $$ b=-22/3 + 4 N_F/3 \to -22/3 + 4=-10/3 . $$

Cheng and Li take special pains in their digression at Fig 14.3 to explain this magnificent freak "situation", namely the equality of fermion contributions, 4, in both cases.

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  • $\begingroup$ Let me see if I understand it: in SU(3) you have triplets of the form $(q_R, q_G, q_B)$ for each quark flavour $q$, which is a total of 6. Since we have to count Weyl that are either left or right, you don't have 6 but 6·2 = 12 given by: $(q_R, q_G, q_B)_{L, R}$ $\endgroup$
    – Vicky
    Commented Apr 22, 2019 at 15:47
  • $\begingroup$ I have made an edition because I found that my value for U(1) differs from Cheng and Li, if you could show me the light I'd really appreciate it. The result is due to a previous calculus that @marmot let me in a comment $\endgroup$
    – Vicky
    Commented Apr 22, 2019 at 15:54
  • $\begingroup$ Fair enough. That's what all texts emphasize. $\endgroup$
    – Cosmas Zachos
    Commented Apr 22, 2019 at 15:55
  • $\begingroup$ I imagine that the normalization is the factor $\sqrt{3/5}$ for all hypercharges since $\alpha_1$ appears with an extra $5/3$ factor in page 441, Eq. (14.54). Including it you get $b(U(1)) = (41/6)·(3/5) = 41/10 = 4.1$. Is Cheng and Li's value for $b(U(1)) = 4$, $b_1$ in their notation, an approximation to just carry $4$ instead of $41/10$? $\endgroup$
    – Vicky
    Commented Apr 22, 2019 at 16:13
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    $\begingroup$ Well, they say they ignore the Higgs contribution, focussing on the fermions, only.... $\endgroup$
    – Cosmas Zachos
    Commented Apr 22, 2019 at 16:24

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