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There are well-known formulas for spin sums that you can apply after squaring the matrix element. If you have the textbook by Peskin and Schroder, I'm sure they are in there (look for spin sum in the index). Otherwise, any other textbook on QFT should have them. For example $$ \sum_s u_s(p)\bar u_s(p)=m-p_\mu\gamma^\mu $$


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The system of equations belong to the ESU, and presumably gaussian, where charge^2 = force*length^2. Putting this in, you get $F^2 * L^4 / F * L * L^3$ gives F = E/L. Hint, the correcting factor for SI is to replace $e^2$ with $e^2/4\pi$. If you really are going to read a lot of older physics, it is best to be versed in the gaussian units: their units ...


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Without having access to the book, the best guess I can come up with is that the formula is written in a natural unit system where $c=1$, $\hbar=1$, and $e=1$. In this system, charge is unitless, and energy has the same unit as inverse length. So, using $\equiv$ to mean "congruent to" (i.e. having the same units), ...


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OK, I figured out the answer. Prahar is right that it it's due to a difference in the definition of the amplitude. But it was really not obvious to me why the amplitudes would differ by a kinematic factor, so I'll write out the full explanation in case it's ever useful to anyone else. The amplitude that should be used in the second equation is not the ...



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