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Mandl & Shaw eqn (9.94)

  1. Jul 23, 2013 #1
    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:

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

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

    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 [itex]\lambda \rightarrow 0[/itex]" since none of the terms involve [itex]\lambda[/itex].
  2. jcsd
  3. Jul 24, 2013 #2
    Let us see only the p/ . p'/ term.
    γα/p'γμ/pγα=[-/p'γα+2p'αμα(-/p)+2pα),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'αγμ2pα,so you will get 4 in the numerator not 2.
  4. Jul 24, 2013 #3
    Thanks Andrien, that helps a lot. But it still leaves the middle term:

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

    Do you have any ideas about it?
  5. Jul 25, 2013 #4
    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.
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