= Solution
The divergent part follows from the elementary integrals $\int_0^1x\,dx=1/2$ and $\int_0^1dx=1$:
$$
\Sigma_{\rm div}(\not p)=\boxed{\frac{e^2}{8\pi^2\epsilon}(\not p-4m).}
$$
Thus \b[a wave-function counterterm and a mass counterterm are required]. Write their contribution as
$$
\mathcal L_{\rm ct}=\delta Z_2\bar\psi i\not\partial\psi-\delta m_{\rm coeff}\bar\psi\psi.
$$
In the insertion convention of part (a), the inverse <Dirac propagator> receives $\delta Z_2\not p-\delta m_{\rm coeff}$. With $\kappa=e^2/(8\pi^2\epsilon)$, cancellation requires
$$
\boxed{\delta Z_2=-\kappa,\qquad\delta m_{\rm coeff}=-4\kappa m.}
$$
To distinguish the coefficient counterterm from the multiplicative mass renormalization, write $m_0=Z_m m$ and $\psi_0=Z_2^{1/2}\psi$. Then $\delta m_{\rm coeff}=m(\delta Z_2+\delta Z_m)$ at this order, giving $\delta Z_m=-3\kappa$. In <modified minimal subtraction>, replace $\kappa$ by $e^2\Delta_{\overline{\rm MS}}/(16\pi^2)$. No new derivative structure is needed for this two-point divergence; charge and photon <counterterms> are determined from other functions.
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