# Photon propagator counterterm in QED

The lagrangian for QED including counterterms is

$$\mathcal{L} = -\frac{1}{4}F_{\mu \nu}^{2}+i\bar{\psi}\gamma^{\mu}{\partial_{\mu}} \psi-m_R \bar{\psi}\psi-e_R \bar{\psi}\gamma^{\mu}A_{\mu}\psi- \frac{1}{4}\delta_{3}F_{\mu \nu}^{2} + ...$$

where ... means the rest of the counterterms. I have a problem obtaining the amplitude of the feynman diagram corresponding to the photon propagator. In the book Quantum Field Theory and the standard model of M. Schwartz this amplitude is stated to be $-i\delta_{3}(p^{2}g^{\mu\nu}-p^{\mu}p^{\nu})$. I have not been able to obtain the same result.

My procedure is the following:

$$- \frac{1}{4}\delta_{3}F_{\mu \nu}^{2}=-\frac{1}{2}\delta_3 (\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}) =\frac{1}{2}\delta_{3}A_{\nu}(\partial_{\mu}\partial^{\mu}A^{\nu}-\partial_{\mu}\partial^{\nu}A^{\mu})= \frac{1}{2}\delta_{3}A_{\nu}(\partial^2g^{\mu \nu}-\partial^{\mu}\partial^{\nu})A_{\mu}$$

From this the amplitude can be deduced considering that $p^{\mu} = i \partial^{\mu}$ and multiplying the amplitude by i. Then the amplitude is

$$-\frac{1}{2}i\delta_3 (p^2 g^{\mu \nu}-p^{\mu}p^{\nu})$$

As you can see I obtained a factor of 1/2 that doesn't appear in the result given by Schwartz. Where did I commit a mistake?

An arbitrary interaction of the form $$\mathcal L_\mathrm{int}=\frac{g}{n_1!\ n_2!\ \cdots \ n_j!}\phi_1^{n_1}\phi_2^{n_2}\cdots\phi_j^{n_j}$$ where all the $\phi_i$ are different, has an associated Feynman factor $$ig$$ because the factors $n_1!\ n_2!\ \cdots \ n_j!$ cancel off with the numerical factors coming from the equivalent permutations in the Wick-Dyson expansion. For example, the interaction $\mathcal L_\mathrm{int}=\frac{g}{4!}\phi^4$ has a Feynman factor $ig$.
In your case, $\phi\to A$ and $n=2$, which means that $$\mathcal L_\mathrm{int}\sim \delta \frac{1}{2}A\partial A$$ and therefore the Feynman factor is $i\delta_3 p$
• Thanks! but then why for other terms this rule of permutations doesn't apply? (or maybe I'm nott understanding correctly). For example, the counterterm for the electron propagator is $$i \delta_2 \bar{\psi}\gamma^{\mu} p_{\mu} \psi -(\delta_m + \delta_2)m_R \bar{\psi}\psi$$ and the amplitude is $i(\gamma^{\mu} p_{\mu} -(\delta_m + \delta_2)m_R)$ – Pam Mar 24 '17 at 17:21
• @Pam Recall that $\psi$ and $\bar \psi$ are to be regarded as different, independent operators. Therefore, in this case $\phi_1\to\psi$ and $\phi_2\to \bar\psi$, and both operators have $n_1=n_2=1$. – AccidentalFourierTransform Mar 24 '17 at 17:24