# Relation between bare and full scalar booson propagator.

1. Jun 8, 2013

### center o bass

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. Jun 9, 2013

### andrien

It is probably DF'=ZDF,where DF' is the modified propagator.