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Irreducible graphs and srednicki's book

  1. Feb 7, 2014 #1
    Srednicki's QFT book uses scalar [tex]\phi^3 [/tex] as an example of a QFT. To make a calculation to all orders, Srednicki claims (chpt 19) you calculate all 1P1 graphs for 2 external lines, giving the self-energy. He then says you calculate all the 1p1 graphs for 3 external lines, giving the 3-point vertex function. Then you calculate n>3 vertex functions by drawing 1p1 graphs, but using the 3-point vertex for vertices and exact propagators for propagators. Then any amplitude can be calculated by tree level graphs using these n-point vertices and exact propagators.

    I think this is mostly correct, but doesn't Srednicki need to say that you have to calculate the 3-point vertex function using the exact propagator? If not, you'd be missing a lot of diagrams that can contribute to a scattering process.
     
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  3. Feb 7, 2014 #2

    Avodyne

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    No, because the sum of all 1PI diagrams with 3 external lines includes the loop corrections to the internal propagators.
     
  4. Feb 8, 2014 #3
    I see. So in ø4, the sum of 1PI with 4 external lines already includes loop corrections to internal propagators. Or more generally, in øn, saying that the n-point vertex is the sum of 1PI diagrams already takes into account loop corrections to internal propagators.

    I'm a little confused about the meaning of the vertex function and 1PI with regards to the 2-point function however. The self-energy is the 1PI diagram, but the 2-point vertex function is actually defined as the full 2-point Green's function, divided by the external lines (each represented by a full 2-point Green's function), so the renormalized 2-point vertex is:

    1/ΔR ΔR 1/ΔR=1/ΔR =p2-mR2-∑R(p2R)

    whereas if you just take

    1/ΔRRRRΔR)1/ΔR=p2-mR2+∑R(p2R)

    If you keep careful track of i's, then you get the same result, that the two expressions differ in the sign of the renormalized self energy ∑R.

    It would seem to me that the former is not 1PI, but the latter is.

    So is it safe to say that the 2-point vertex (the former expression) is not 1PI? Only the 3-point vertex you can say is constructed from 1PI? I mean, you took ΔR, which is not 1PI (only ∑R is), and you cut two of the ends, so everything in between remains the same, and everything in between wasn't 1PI.
     
  5. Feb 8, 2014 #4

    Avodyne

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    Your 2nd expression has no meaning.

    The self-energy ∑ is the sum of 1PI diagrams with two external lines (and no propagators for those lines). The exact propagator is then given by a geometric series that sums up to Δ as given by your first expression.
     
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