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.

  • 1
    $\begingroup$ No there are no observable effect of the finite correction. All that matter are p (momentum) dependent terms because physical observables are always $\mathcal{O}(p1)-\mathcal{O}(p2)$. Rest you can absorb in your definitions. ($\mathcal{O}(p)$ can be any of the quantities you have defined in your question) $\endgroup$ Commented Oct 17, 2016 at 10:32
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    $\begingroup$ @seeking_infinity What do you think is the correction to the Coulomb Potential due to? Isn't it due to the finite part of the corrections? $\endgroup$
    – SRS
    Commented Oct 17, 2016 at 11:16
  • $\begingroup$ You can be more specific. Which term are you talking about? If you mean correction to QED vertex, correction to that term is $\propto \frac{\sigma^{\mu \nu}}{2m} p_{\nu}$. $\endgroup$ Commented Oct 18, 2016 at 5:44
  • $\begingroup$ I'm talking about the "finite" parts of $\Sigma(p),\Pi^{\mu\nu}$ and $\Lambda_\mu$. $\endgroup$
    – SRS
    Commented Oct 18, 2016 at 6:57
  • $\begingroup$ you said "correction to the coulomb potential". I am not sure what exactly you mean by this. Remember, in electrodynamics as well, we have learnt that potentials are not physical objects, its potential difference that we can observe. And when we calculate the difference, finite parts cancel. $\endgroup$ Commented Oct 18, 2016 at 7:30

1 Answer 1

  1. 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).

  2. 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 $$ \frac{e^2}{\boldsymbol k^2}\to\frac{e^2}{\boldsymbol k^2}\left(1+\frac{\alpha}{15\pi}\frac{\boldsymbol k^2}{m^2}+\cdots\right) $$

  3. 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 $\mu_M\sim 1+F_2(0)$.


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