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Feynman Parameters-Peskin&Schroeder 6.44

  1. May 11, 2013 #1
    Feynman Parameters--Peskin&Schroeder 6.44

    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/D3. We are given that:
    D = k2+2k(yq-zp)+yq2+zp2-(x+y)m2+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 says
    So I tried doing the algebra, and immediately it seems we need to use the Dirac equation to get another m2 term.

    Starting with the given D, I got
    D=l2-[z2p2]+2zpyq-y2q2+yq2+[zp2]-(1-z)m2+iε

    So after staring at that a while, I decided there's no way to get anything else into the coefficient of m2 without applying the Dirac equation. So I made the substitution p2 = m2 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 m2 into what we want. A few steps later I found:

    D=l2+2zpyq+yzq2+yxq2-(1-z)2m2+iε

    =l2-Δ+iε+yzq2+2zpyq

    =l2-Δ+iε+yz[(p+q)2-p2]
    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!
     
    Last edited: May 12, 2013
  2. jcsd
  3. May 13, 2013 #2
    that term is zero.just see the feynman diagram of vertex correction where
    p'=p-k+k',now he defines k'=k+q so p'=p+q,now p' and p are outgoing and ingoing states so use the onshell condition p'2=p2=m2
    p'2=p2+q2+2pq.so q2+2pq=0 which is needed.
     
  4. May 13, 2013 #3
    Ah! Thanks very much andrien. That's really helpful. That answers both of my questions: why the terms cancel and why q2 is not 0 in the Δ term.

    What had me really confused was this snippet from Ryder:
    (Lewis H. Ryder, Quantum Field Theory, Second edition, p.344)

    I guess based on what you say, that quote from Ryder is wrong?
     
  5. May 13, 2013 #4

    Hepth

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    Gold Member

    In PS they're letting q^2 be off-shell, as they mention that in a scattering process q^2 < 0.

    So they don't use the on-shell condition q^2 = 0 for photons.

    In Ryder, are they just computing a diagram where the photon is real? I don't have a copy in front of me, but if they are then they use that condition.
     
  6. May 13, 2013 #5
    The diagram is the same as the one in this post: https://www.physicsforums.com/showthread.php?t=690885 [Broken] . That quote comes from the "Anomalous magnetic moment of the electron" section of Ryder, so I'm assuming they're calculating the same diagram and using the same method.

    About the notation for the diagrams: In both Ryder and Peskin&Schroeder, q is the incoming photon and p is the incoming electron. (There is small difference in some of the other definitions but I don't think they matter). The integration variable (the free variable inside the e-e-photon vertex) is k.

    If we are allowed to apply the Dirac equation to the incoming/outgoing electrons, then is it possible that Ryder is saying we can apply the Klein Gordon equation to the incoming photon? But then it is on-shell, right, so they must be doing something different from P&S? Does that mean there's a similar but slightly different way to evaluate that diagram? Or is it just a typo in Ryder?
     
    Last edited by a moderator: May 6, 2017
  7. May 15, 2013 #6
    what actually Ryder has done is not wrong but it may confuse you because he has used some simplification rather earlier.When you write the vertex part then after some manipulation you try to express it with two form factors(there would be four if electromagnetism does not conserve parity) which are multiplied by terms γμ and σμv respectively.Now the physical interpretation of those factors is given by the amount of charge(F1(0)) and amount of magnetic moment [1+F2(0)](ryder uses λ type thing for F's).The interpretation of the form factor really comes while taking the limit q2→0 which is what ryder has done but rather early.This F2(0) really gives α/2∏ correction which is schwinger term.P&S rather do it after expressing everything with q2 and then take the limit q2→0 in last for getting schwinger term.I hope it will clear you confusion.
     
  8. May 15, 2013 #7
    Hmm that is interesting. Why would P/S not do the same thing (usually removing unnecessary variables is an obvious way to make things easier)?

    Anyway, thanks for all your help!
     
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