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

+ 4 like - 0 dislike
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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?

asked May 17, 2014 in Theoretical Physics by coolcty (125 points) [ revision history ]
reshown May 23, 2014 by coolcty

1 Answer

+ 3 like - 0 dislike

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,

Screenshot from 2014 05 23 06 20 56

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

answered May 23, 2014 by JeffDror (650 points) [ revision history ]

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