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Relation between bare and full scalar booson propagator.

  1. Jun 8, 2013 #1
    One can show that at around ##p \approx m## where m is the physical mass the full propagator ##D_F## is something like

    $$D_F = \frac{Z}{p^2 - m^2}.$$

    Where ##Z = (1 - \Sigma '(m))^{-1}##, ##\Sigma## is the self energy and m is the physical mass of the particle. If i were now to write a relation between ##D_F## and the bare propagator ##D_F^0## my first guess would be that to a given order ##D_F^0 = D_F/Z##. But that is not true and of course I do see the full mass is still there.. So I have read that the relation is in fact the opposite, namely ##D_F^0 = Z D_F##, but I have problems proving it. How is that proof done?
     
  2. jcsd
  3. Jun 9, 2013 #2
    It is probably DF'=ZDF,where DF' is the modified propagator.
     
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