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I'm having some trouble deriving equation 6.44 on p.191 of Peskin and Schroeder (my book is the fifth printing, December 1997). The algebra comes close but I think that you need to do some arguing to actually get his answer--maybe this requires some Dirac equation reasoning that I'm not seeing. Here's the problem:

We're working on a Feynman integral for electrons and photons, so everything is sandwiched between u(p,s) electron state-functions. The precise step I'm on is converting the denominator of the integral into the integral ∫dx dy dz δ(1-x-y-z) 2/D^{3}. We are given that:

D = k^{2}+2k(yq-zp)+yq^{2}+zp^{2}-(x+y)m^{2}+iε

With the Dirac delta in the integral, we can always assume 1=x+y+z. He makes the definition l=k+yq-zp.

Then he saysSo I tried doing the algebra, and immediately it seems we need to use the Dirac equation to get another m After a bit of algebra we find that D simplifies to:

D=l^{2}-Δ+iε where Δ = -xyq^{2}+(1-z)^{2}m^{2}^{2}term.

Starting with the given D, I got

D=l^{2}-[z^{2}p^{2}]+2zpyq-y^{2}q^{2}+yq^{2}+[zp^{2}]-(1-z)m^{2}+iε

So after staring at that a while, I decided there's no way to get anything else into the coefficient of m^{2}without applying the Dirac equation. So I made the substitution p^{2}= m^{2}in the bracketed terms. [This is the only time I applied the Dirac equation. Is it valid?]

This gets us closer--it changes the coefficient of m^{2}into what we want. A few steps later I found:

D=l^{2}+2zpyq+yzq^{2}+yxq^{2}-(1-z)^{2}m^{2}+iε

=l^{2}-Δ+iε+yzq^{2}+2zpyq

=l^{2}-Δ+iε+yz[(p+q)^{2}-p^{2}]

Aside from that one application of the Dirac equation, I didn't use anything but algebra. Why are those pesky terms there? Should we be saying they cancel, somehow, or that q=0? (In Ryder's QFT, he mentions something like this on p.344, but I don't really get it.) If that's the case, why would q not drop out of the Δ term? Did I misapply the Dirac equation?

Thanks for any and all help!

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# Feynman Parameters-Peskin&Schroeder 6.44

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