What are the observable effects of finite pieces of the loop corrections in QED? I'm lost amidst the calculation of regularization and renormalization process in QED. In addition to the divergent piece in the in the self-energy correction (similarly in vacuum polarization correction and vertex correction) there is also a finite correction:
$$\Sigma(p)=\frac{e^2}{8\pi^2\epsilon}(-\not p+4m)+\text{finite}$$
$$\Pi_{\mu\nu}(k)=\frac{e^2}{6\pi^2\epsilon}(k_\mu k_\nu-g_{\mu\nu}k^2)+\text{finite}$$
$$\Lambda^{(1)}_\mu(p)=\frac{e^2}{8\pi^2\epsilon}\gamma_\mu+\text{finite}$$
Are there any observable effect of the finite correction? It appears to me that both the finite and infinite parts of the corrections are absorbed into the definition of renormalized mass and coupling constant. 
 A: *

*The electron self-energy has no obvious/direct measurable consequence, because it is a correction to a fermionic object, and therefore it has, loosely speaking, a vanishing contribution to classical phenomena. Nevertheless, one may use this function to obtain several measurable predictions; for example, if this function has a non-vanishing imaginary part at $\not p=m$ it means that the particle is unstable (with a width $m\Gamma=\mathrm{Im}\Sigma(m)$).
The function is also indirectly related to many measurable effects, because the Ward-Takahashi identities relate $\Sigma$ to $\Lambda$ (see below). Also, the loop-corrected Dirac equation reads $$(\not p-m-\Sigma(\not p))\psi(p)=\bar\psi \Lambda\psi+\cdots$$ which means that $\Sigma$ is closely related to many atomic effects (e.g., Lamb shift, etc).

*The photon self-energy $\Pi_{\mu\nu}$ is related to the "screening of charge", that is, to the fact that for short distances the Coulomb law is modified into
$$
\frac{e^2}{\boldsymbol k^2}\to\frac{e^2}{\boldsymbol k^2}\left(1+\frac{\alpha}{15\pi}\frac{\boldsymbol k^2}{m^2}+\cdots\right)
$$

*The vertex function is related to the magnetic ($\mu_M$) and electric ($d_E$) moments of the lepton (e.g., to the electron anomalous magnetic moment - the Schwinger result - $a_e\sim\frac{\alpha}{2\pi}$). The explicit relation is obtained by expressing $\bar u\Lambda u$ in terms of the form factors,
$$
\bar u\Lambda u\sim \gamma^\mu F_1(q^2)+\sigma^{\mu\nu} F_2(q^2)+\gamma_5 F_3(q^2)
$$
(with some coefficients omitted) so that $d_E\sim F_3(0)$ and $\mu_M\sim 1+F_2(0)$.
