Question 5.11: The Lowest-Order Terms of the β Function Show that (5) in Ex......
The Lowest-Order Terms of the \beta Function
Show that (5) in Example 4.3 or (4.156)
\begin{aligned} \beta(g)= & -\frac{g^3}{(4 \pi)^2}\left(11-\frac{2}{3} N_{\mathrm{F}}\right)-\frac{g^5}{(4 \pi)^4}\left(102-\frac{38}{3} N_{\mathrm{F}}\right)-\frac{g^7}{(4 \pi)^6}\left(\frac{2857}{2}-\frac{5033}{18} N_{\mathrm{F}}+\frac{325}{54} N_{\mathrm{F}}^2\right) . \end{aligned} (4.156)
indeed yields the lowest-order contribution to the \beta function:
\alpha_{S}\left(-q^{2}\right)=\frac{\alpha_{s}\left(\mu^{2}\right)}{1+\frac{11-2 N_{\mathrm{f}}\left(-q^{2}\right) / 3}{4 \pi} \alpha_{s}\left(\mu^{2}\right) \ln \left(-\frac{q^{2}}{\mu^{2}}\right)}. (1)
Here \mu^{2} denotes the renormalization point where \alpha_{s}\left(-q^{2}\right) assumes the renormalized value \alpha_{s}\left(\mu^{2}\right) . N_{\mathrm{f}}\left(-q^{2}\right) is the number of light quarks with masses smaller than -q^{2}.
Equation (1) yields for g\left(-q^{2}\right)
g\left(-q^{2}\right)=\frac{g\left(\mu^{2}\right)}{\sqrt{1+\frac{11-2 N_{\mathrm{f}}\left(-q^{2}\right) / 3}{(4 \pi)^{2}} g^{2}\left(\mu^{2}\right) \ln \left(-\frac{q^{2}}{\mu^{2}}\right)}} (2)
with \alpha_{s}=g^{2} / 4 \pi. We multiply (2) by the demominator on the right-hand side and differentiate the resulting equation with respect to \mu^{2} :
\begin{aligned} g^{\prime}\left(\mu^{2}\right)= & g\left(-q^{2}\right) \frac{\partial}{\partial \mu^{2}} \sqrt{1+\frac{11-\frac{2}{3} N_{\mathrm{f}}\left(-q^{2}\right)}{(4 \pi)^{2}} g^{2}\left(\mu^{2}\right) \ln \left(-\frac{q^{2}}{\mu^{2}}\right)} \\ = & g\left(-q^{2}\right) \frac{\frac{1}{2} \frac{11-2 N_{\mathrm{f}}\left(-q^{2}\right) / 3}{(4 \pi)^{2}}}{\sqrt{1+\frac{11-2 N_{\mathrm{f}}\left(-q^{2}\right) / 3}{(4 \pi)^{2}} g^{2}\left(\mu^{2}\right) \ln \left(-\frac{q^{2}}{\mu^{2}}\right)}} \\ & \times\left[2 g\left(\mu^{2}\right) g^{\prime}\left(\mu^{2}\right) \ln \left(-\frac{q^{2}}{\mu^{2}}\right)-\frac{1}{\mu^{2}} g^{2}\left(\mu^{2}\right)\right] . & (3) \end{aligned}
Once more we insert (2), into this expression and collect the terms containing g^{\prime} :
g^{\prime}\left(\mu^{2}\right)\left[1+\frac{11-\frac{2}{3} N_{\mathrm{f}}\left(-q^{2}\right)}{(4 \pi)^{2}} g^{2}\left(\mu^{2}\right) \ln \left(-\frac{q^{2}}{\mu^{2}}\right)\right]
\begin{aligned} & \quad=\frac{11-\frac{2}{3} N_{\mathrm{f}}\left(-q^{2}\right)}{(4 \pi)^{2}}\left[g^{2}\left(\mu^{2}\right) g^{\prime}\left(\mu^{2}\right) \ln \left(-\frac{g^{2}}{\mu^{2}}\right)-\frac{g^{3}\left(\mu^{2}\right)}{2 \mu^{2}}\right], & (4) \\ & g^{\prime}\left(\mu^{2}\right)=-\frac{1}{2 \mu^{2}} \frac{11-\frac{2}{3} N_{\mathrm{f}}\left(-q^{2}\right)}{(4 \pi)^{2}} g^{3}\left(\mu^{2}\right) .& (5) \end{aligned}
By means of (5) the \beta function now assumes the form
\beta=\mu \frac{\partial g(\mu)}{\partial \mu}=2 \mu^{2} \frac{\partial g\left(\mu^{2}\right)}{\partial \mu^{2}}=-\left[11-\frac{2}{3} N_{\mathrm{f}}\left(-q^{2}\right)\right] \frac{g^{3}\left(\mu^{2}\right)}{(4 \pi)^{2}}. (6)
In fact this is identical with the first term in (5.182).
\beta(g)=-\left(11-\frac{2}{3} N_{\mathrm{f}}\right) \frac{g^3}{(4 \pi)^2}-\left(102-\frac{38}{3} N_{\mathrm{f}}\right) \frac{g^5}{(4 \pi)^4}+O\left(g^7\right), (5.182)
We can see immediately where the higher terms come from. From Example 4.3, (1) is derived from vacuum polarization graphs. Consequently we obtain the higher-order contributions to the \beta function if higher graphs are taken into account in this calculation.
Finally, we summarize the basic ideas of this last calculation. The renormalization is carried out for a given value \mu^{2} of the momentum transfer, i.e., all divergencies are subtracted in that renormalization scheme by relating to the measured value of the coupling constant at the point, i.e. to \alpha_{s}\left(\mu^{2}\right). This procedure yields a renormalized coupling constant \alpha_{s}\left(\mu^{2}\right) which has to be identified with the physical coupling constant. However, we must decide on what we mean by “physical QCD coupling” and then we must express the effective (running) coupling \alpha_{s}\left(Q^{2}\right) in terms of it. In QED the effective coupling is expressed in terms of the fine structure constant \alpha \sim 1 / 137 which is defined as the effective coupling in the limit -q^{2} \rightarrow 0. In QCD \alpha_{s}\left(-q^{2}\right) diverges as -q^{2} \rightarrow 0. Hence we cannot define \alpha_{s}\left(-q^{2}\right) in terms of its value at -q^{2}=0. Instead we choose some value of -q^{2}, say -q^{2}=\mu^{2}, and define the experimental QCD coupling to be \alpha_{s} \equiv \alpha_{s}\left(\mu^{2}\right). The renormalization point \mu^{2} is, of course, arbitrary. Had we instead chosen the point \bar{\mu}^{2} then the two couplings would be related by
\alpha_{s}\left(\bar{\mu}^{2}\right)=\frac{\alpha_{s}\left(\mu^{2}\right)}{1+\left(\frac{11-2 N_{\mathrm{f}} / 3}{4 \pi}\right) \ln \left(\frac{\bar{\mu}^{2}}{\mu^{2}}\right)} .