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

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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?

reshown May 23, 2014

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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 $\delta m$.

answered May 23, 2014 by (650 points)

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