# Beta function in the Standard Model

In Srednicki's textbook "Quantum Field Theory", Problem 89.4 asks us to compute the leading terms in the beta function for each of the three gauge couplings of the Standard Model. These gauge couplings are $$g_{3}$$, $$g_{2}$$ and $$g_{1}$$, corresponding to the $$SU(3)$$, $$SU(2)$$, and $$U(1)$$ gauge groups respectively. The answer is given in section 97 of the same book:

With the usual fields of the Standard Model, we find from our results in sections 66 and 73 that the one-loop beta functions for the three gauge couplings are given by $$$$\mu \frac{d}{d\mu}g_{i} = \frac{b_{i}}{16\pi^{2}} g_{i}^{3} + O(g_{i}^{5}), \tag{97.26}$$$$ with $$$$b_{3} = -11 + \frac{4}{3}n, \tag{97.27}$$$$ $$$$b_{2} = -\frac{22}{3} + \frac{4}{3}n + \frac{1}{6}, \tag{97.28}$$$$ $$$$b_{1} = \frac{20}{9}n + \frac{1}{6}, \tag{97.29}$$$$ where $$n = 3$$ is the number of generations; the $$+\frac{1}{6}$$ contribution to $$b_{2}$$ and $$b_{1}$$ are from the $$\varphi$$ (Higgs) field.

My question is with $$b_{1}$$ [eq. (97.29)]. In section 66, the beta function for $$e$$ (coupling constant) in QED is derived to be $$$$\beta_{e}(e, \lambda) = \frac{1}{12\pi^{2}} (\Sigma_{\Psi} Q_{\Psi}^{2} + \frac{1}{4} \Sigma_{\varphi} Q_{\varphi}^{2}) e^{3} + ... \tag{66.29}$$$$ where $$Q_{\Psi}e$$ are the electric charges of the Dirac fields $$\Psi$$'s.

To compute $$b_{1}$$, I put $$Q_{\nu_{e}} = 0, Q_{e} = -1, Q_{\overline{e}} = +1, Q_{u} = \frac{2}{3}, Q_{d} = -\frac{1}{3}, Q_{\overline{u}} = - \frac{2}{3}, Q_{\overline{d}} = \frac{1}{3}$$, and $$Q_{\varphi_{2}} = \frac{1}{2}, Q_{\varphi_{4}} = \frac{1}{2}$$ into eq. (66.29). I also noticed that $$g_{1} = e/\cos\theta_{w}$$, and added an $$n$$ in the first term in the bracket in eq. (66.29) to account for three generations. However, I got $$$$\mu\frac{d}{d\mu}g_{1} = \frac{\cos^{2}\theta_{w}}{16\pi^{2}}(\frac{112}{27}n + \frac{1}{6})g_{1}^{3}.$$$$ This is not what it should be according to eqs. (97.26) and (97.29). What's wrong?

I also noticed that eq. (66.29) [which is supposed to lead to eq. (97.29) when applied to computing the beta function in the Standard Model] is derived from the Lagrangian in QED (section 66), which is different from the Lagrangian in the nonabelian gauge theory (section 73) from which eqs. (97.27) and (97.28) are stemmed. Is this the reason for the discrepancy between eq. (66.29) and eq. (97.29)? However, in Srednicki's book, the author did not give notice to this discrepancy but simply combined the results in sections 66 and 73 into a unified equation [eq. (97.26)], which implies that eq. (66.29) can be used to compute the beta function in the Standard Model. Is there something hidden and needs to be adjusted and clarified here?

You are misreading the book, which is consistent. The author has been putting you on "notice" relentlessly when he embeds QED into the SM. Without doing the problem for you, I can reassure you that your expression for the β-function of $$g_1$$ is completely nonsensical— you confused the hypercharge U(1) coupling evolution with that of the vector EM U(1), when the hypercharge interactions are magnificently chirally lopsided—Feynman's complaint to Weinberg that his model was so "cockeyed".
In fact, if you supplant the Weinberg angle and essentially write the height of the SM mixing triangle in the two alternate ways quantifying its area, $$e^2=\frac{g_1^2 g_2^2}{g_1^2+ g_2 ^2}~~,$$ (97.26,8,9) imply (66.29) without further ado.
Specifically, $$8\pi^2\frac {\partial g_2^2}{\partial \log \mu}= g_2^4 b_2 ~,\\ 8\pi^2\frac {\partial g_1^2}{\partial \log \mu}= g_1^4 b_1 ~.$$
You may then combine them simply to evaluate the above, $$8\pi^2\frac {\partial e^2}{\partial \log \mu}= 8\pi^2\frac {\partial \frac{g_1^2 g_2^2}{g_1^2+ g_2 ^2}}{\partial \log \mu} \\ = e^4 (b_1+b_2)=e^4 \Bigl ( \frac{1}{3}+ \frac{32}{9}n -\frac{22}{3}\Bigr ),$$ the first term being due to the higgs, the second to fermions, and the third to the Ws, the charged vectors missing from your (66.29) (and so are part of the ellipsis!): $$8\pi^2 \frac {\partial e^2}{\partial \log \mu} = \frac{4}{3} e^4 \Biggl (\frac{1}{4} + n \left (1+\frac{1}{3} + \frac{4}{3}\right ) +...\Biggr ).$$ Understand that your EM (66.29) involves vector couplings to fermions, Dirac particles now, so the electron suffices to also cover the positron: you do not need to double-count it!
• No, both the electron and the positron are packaged in the field $\Psi$ of (66.29). The same is true for (97), except the left and right modes are materially different and they are meant to be counted separately. So one does not count antiquarks on top of quarks, for example. Your text should be making that clear. – Cosmas Zachos Feb 18 at 2:04