Troubleshooting Mandl & Shaw QFT: Deriving eqn (9.94) on pg 192 of 2nd edition

[Vic Sandler]

I tried to derive eqn (9.94) on page 192 of the second edition of Mandl and Shaw QFT and failed. Can someone help me see what I am doing wrong?

Ignoring factors that do not change from eqn (9.92) to eqn (9.94), noting that f(k) has been set to 1, and dropping terms linear in k and k squared as described in the text, I get:

\frac{\gamma^{\alpha}(\not{p'}+m)\gamma^{\mu}(\not{p}+m)\gamma_{\alpha}}{((p'-k)^2 - m^2)((p-k)^2 - m^2)}

= \gamma^{\mu}\frac{(-2p'p)}{(-2p'k)(-2pk)} + \frac{4m(p' + p)^{\mu}}{(-2p'k)(-2pk)} + \gamma^{\mu}\frac{(-2m^2)}{(-2p'k)(-2pk)}

The first term on the right hand side is the same as in the book, but divided by -2. Perhaps the other two terms combine in some way to fix it up, but I don't see it. I also don't see what terms are meant by the author when he says "the dots indicate terms which are finite in the limit \lambda \rightarrow 0" since none of the terms involve \lambda.

 

[andrien]

Let us see only the p/ . p'/ term.
γ^α/p'γ_μ/pγ_α=[-/p'γ^α+2p'^α]γ_μ[γ_α(-/p)+2p_α),now /p and /p' operating on free particles states will give m which will cancel with the m already written in (/p+m) and (/p'+m) terms.You are left with
2p'^αγ_μ2p_α,so you will get 4 in the numerator not 2.

 

[Vic Sandler]

Thanks Andrien, that helps a lot. But it still leaves the middle term:

\frac{4m(p' + p)^{\mu}}{(-2p'k)(-2pk)}

Do you have any ideas about it?

 

[andrien]

There are no such terms because as I have written earlier owing to -/p(-/p') acting on free particle spinor you get a factor of -m which will cancel with m in (/p -/k +m) and also dropping linear terms and quadratic terms in k,you are left only with the first term already written in the book.