# Can the 'mass' of bound states show up full propagator?

1. Jun 12, 2013

### center o bass

The result of the Kallen-Lehmann spectral representation is that the two point correlation (and thus also the full propagator) has a pole in the physical mass of the particle. In Peskin and Schroeder it is also argued that multiparticle states show up as a cut, but bound states can also show up as poles.

Let's then say that we calculate the full propagator to a certain order and we find that we have a pole for a certain mometum - how do we then conclude that this pole is actually the mass of the particle and not the 'mass' of the bound state?

2. Jun 19, 2013

### geoduck

The mass of the particle is what you start with in the Lagrangian. If a bound state appears then it'll look like something else.

3. Jun 19, 2013

### DrDu

I don't think there is a difference. You could for example consider a proton moving in an electron gas with chemical potential mu. When interaction is switched on, the ground state of the interacting proton will evolve into the ground state of a bound hydrogen atom.

4. Jun 19, 2013

### andrien

for a multiparticle state it is a bound state,for a single particle it is the physical mass.