Beta function for QED What is the beta function for quantum electrodynamics?
I can't find it anywhere except on Wikipedia article on beta functions, where the one-loop beta function is given by:
$$\beta(\alpha) = \frac{\alpha^2}{3\pi}$$
However, using the definition of the beta function
$$\beta(\alpha) = \frac{d\alpha}{d\ln(\mu)}$$
one gets:
$$\frac{d\alpha}{d\ln(\mu)} = \frac{\alpha^2}{3\pi}$$
After rearranging
$$3\pi\frac{d\alpha}{\alpha^2} = d\ln(\mu)$$
and integrating
$$-\frac{3\pi}{\alpha} = \ln(\mu) + C = \ln(\mu) - \ln(\mu_0) = \ln(\mu/\mu_0),$$
where $C = -\ln(\mu_0)$ is an integration constant, the result is:
$$\alpha(\mu) = -\frac{3\pi}{\ln(\mu/\mu_0)} = \frac{3\pi}{\ln(\mu_0/\mu)}$$
This gives non-negative monotonically increasing values for $\alpha$ for $\mu < \mu_0$ with a singularity at $\mu_0$ (I guess that would be the Landau pole).
However, for any finite non-zero value of $\mu_0$, $\alpha(0)$ seems to be
$$\alpha(0) = \lim_{\mu\to0^+}\frac{3\pi}{\ln(\mu_0/\mu)} = 
\lim_{\ln(\mu)\to-\infty}\frac{3\pi}{\ln(\mu_0)-\ln(\mu)} = 0,$$
while on the other hand $\alpha \approx 1/137$ should appear on low energies instead of zero.
Where is the catch?
Is there a term missing from the beta function above? For example, replacing $\alpha^2$ with $(\alpha - \alpha_0)^2$ should give the expected result, but it doesn't seem to be written like that anywhere. Is that replacement somehow silently implied nevertheless?
 A: First, a small correction to the algebra.
I follow you up to here.
\begin{equation}
3\pi \frac{{\rm d} \alpha}{\alpha^2} = d \ln \mu
\end{equation}
Then I think we need to proceed carefully. The way I would go would be to integrate from a reference scale $\mu_0$ to an energy scale $q$. Then
\begin{equation}
3\pi \int_{\mu_0}^q \frac{{\rm d} \alpha}{\alpha^2} =  \ln \frac{q}{\mu_0}
\end{equation}
Rearranging this yields
\begin{equation}
\frac{1}{\alpha(q)} - \frac{1}{\alpha(\mu_0)}  = -\frac{1}{3\pi} \ln \frac{q}{\mu_0}
\end{equation}
If you like, you can rearrange this to obtain
\begin{equation}
\alpha(q) = \frac{\alpha(\mu_0)}{1-\frac{1}{3\pi}\alpha(\mu_0) \ln \frac{q}{\mu_0}}
\end{equation}
In deriving this, we have implicitly assumed we can use a one loop approximation to the beta function. This requires that so-called "large-logs" do not appear. In practice, this means choosing a reference scale $\mu_0$ that is not too different from the energy scale of the processes you are interested in, $q$.
By taking $q \ll \mu_0$, you are breaking this assumption, and would need to resume the loarge logs in the beta function.
Or, you can choose a reference scale $\mu_0$ which is at the energies of the experiments you are interested in. At low energies, $\alpha(\mu_0)$ is small, and so $\alpha(q) \approx \alpha(\mu_0)$.
If that seems unsatisfying (what do I mean by large energies?), just note that $\alpha(q)$ will increase slightly at increasing $q$. If you keep matching $\mu_0$ to the energy scale $q$, eventually you will reach a point where the $\alpha(\mu_0)$ term in the denominator can't be ignored for an appreciable range of $q$.
