Why the terms in $\mathscr{L}_2$ are of second and higher order in e? in 11.1.9 of Weinberg's QFT book

Question

The Lagrangian density for electrons and photons is taken in the form of 11.1.1.

Introducing renormalized fields and charge and mass, the counterterms  can be written as

\(\mathscr{L}_2=-\frac{1}{4}(Z_3-1)F^{\mu\nu}F_{\mu\nu}-(Z_2-1)\bar\psi [\gamma_\mu\partial^\mu+m]\psi+Z_2\delta m\bar\psi\psi\\ -ie(Z_2-1)A_\mu\bar\psi\gamma^\mu\psi\)

The book states:

It will turn out that all of the terms in \(\mathscr L_2\) are of second order and higher order in e,...

Isn't the order of e in the expression just one or zero?

Answer 1

In the Lagrangian there is hidden $e$ dependence in the $1-Z_i$'s and $\delta m$. To see that each term must be order $e^2$ or higher remember how they are calculated. The $Z_i$ are wavefunction renormalization. Calculating them amounts to calculating diagrams with one particle coming in and another coming out. Since we are dealing with 3 point vertices the lowest order diagram will have one loop and have two interaction points. For example for the photon one needs to calculate,

The same goes for the other $Z_i$'s and $\delta m $.