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I came across this difficulty comparing several versions of the Feynman rules for QED. I traced the problem back to a statement of Weinberg's in "The Quantum Theory of Fields (Vol 1)", Chapter 5, page 195. Weinberg is busy setting up for deriving free field expressions for different particle types. He has got to the point where he is deriving the field operators as an integral expression. Specifically, he derives two equations:

##\displaystyle \psi_l^+(x) = \sum_{\sigma , n} (2 \pi )^{3/2} \int d^3p u_l(\textbf{p}, \sigma, n) e^{i p \cdot x} a(\textbf{p}, \sigma, n)##

and

##\displaystyle \psi_l^-(x) = \sum_{\sigma , n} (2 \pi )^{3/2} \int d^3p v_l(\textbf{p}, \sigma, n) e^{-i p \cdot x} a^{\dagger} (\textbf{p}, \sigma, n)##

Now, I understand the overall features here. But Weinberg continues on to say

The factors ##(2 \pi )^{3/2}## could be absorbed into the definition of ##u_l## and ##v_l##, but it is conventional to show them explicitly in these Fourier integrals.

Now, it's a Fourier integral, so I do see the value in showing the ##2 \pi##s. But by the time the Feynman rules are derived, practically everyone has redefined the field operators so that the ##(2 \pi )^{3/2}## factor disappears.

And when we get to the amplitude, Weinberg's insistence on keeping the ##2 \pi##s alters the amplitude compared to other sources. As an example, see here, equations 2.1 and 2.2. Weinberg's set of Feynman rules gives an overall factor of ##(2 \pi)^{-6}## to these amplitudes.

How do we get rid of those extra ##2 \pi##s?

Thanks!

-Dan

Addendum: Problem solved! Weinberg's extra ##(2 \pi)^{-3/2}## factor in his version of the Feynman rules is there to get rid of that extra ##(2 \pi)^{3/2}## factor in the Fourier integral. I saw it just as I posted this.