'Off-shell' particle in an external field....

[asimov42]

TL;DR

Questioning why particles are said to be 'off-shell' in an external field...

Silly question but could someone explain why a real, 'observable' particle is said to be 'off-shell' in an external field? @A. Neumaier 's excellent FAQ notes that the mass shell constraints ceases to have meaning in this case. I'm just not fully clear on why (probably obvious) given that energy conservation still holds.

 

[asimov42]

Or is the use of 'off shell' in this case simply to indicate that the field may influence the energy of the particle?

 

[A. Neumaier]

could someone explain why a real, 'observable' particle is said to be 'off-shell' in an external field?

Or is the use of 'off shell' in this case simply to indicate that the field may influence the energy of the particle?

Off shell just means that the standard asymptotic (or free field) relation $p^2=m^2$ (in units where $c=1$), which defines the on shell condition, is not satisfied.

 

[asimov42]

Thanks @A. Neumaier!