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

[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?

 

[geoduck]

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?

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.

 

[DrDu]

How do we then conclude that this pole is actually the mass of the particle and not the 'mass' of the bound state?

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.

 

[andrien]

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