According to the wikipedia entry on the fine-structure constant:
In fact, α is one of the about 20 empirical parameters in the Standard Model of particle physics, whose value is not determined within the Standard Model.
but, the wikipedia list of parameters does not mention α:
me Electron mass 511 keV
mμ Muon mass 105.7 MeV
mτ Tau mass 1.78 GeV
mu Up quark mass μMS = 2 GeV 1.9 MeV
md Down quark mass μMS = 2 GeV 4.4 MeV
ms Strange quark mass μMS = 2 GeV 87 MeV
mc Charm quark mass μMS = mc 1.32 GeV
mb Bottom quark mass μMS = mb 4.24 GeV
mt Top quark mass On-shell scheme 172.7 GeV
θ12 CKM 12-mixing angle 13.1°
θ23 CKM 23-mixing angle 2.4°
θ13 CKM 13-mixing angle 0.2°
δ CKM CP-violating Phase 0.995
g1 or g' U(1) gauge coupling μMS = mZ 0.357
g2 or g SU(2) gauge coupling μMS = mZ 0.652
g3 or gs SU(3) gauge coupling μMS = mZ 1.221
θQCD QCD vacuum angle ~0
v Higgs vacuum expectation value 246 GeV
mH Higgs mass ~ 125 GeV (tentative)
Is α one of the basic parameters of the Standard Model?
If not, then is there a simple formula for α in terms of these other parameters?
(My guess is that α can be derived from g1/g2/g3. However, I have been unable to find an explicit formula so far.)
Best Answer
Your guess is correct. After electroweak symmetry breaking, the coupling constant for the residual $U(1)_\textrm{EM}$ gauge group can be written as a function of the couplings of the broken $SU(2)_L \times U(1)_\textrm{Y}$ gauge groups: $$ \alpha = \frac{1}{4\pi}\frac{g^2 g\prime^2}{g^2+g\prime^2} = \frac{e^2}{4\pi} $$
These couplings, however, are running parameters, defined at a particular energy scale. In your table, the energy scale is $\mu=M_Z$. If you plugged in the numbers from your table, you would calculate $\alpha$ at $\mu=M_Z$, which is $\alpha(\mu=M_Z) \approx 1/128$.
The fine-structure constant is usually considered to be the IR fixed-point of $\alpha$, which is $\alpha(\mu = m_e) = 1/137$, i.e., $\alpha$ at low energy. To calculate this from your table, you would have to run $e$ to a lower energy scale, with the $\beta$-function: $$ \frac{\partial e(\mu)}{\partial \log \mu} \equiv \beta(e) = \frac{e(\mu)^3}{12\pi^2} $$ With this, and your knowledge of $\alpha(\mu=M_Z)$, you could recover $\alpha(\mu = m_e)= 1/137$.