Mass of the asymptotic fields: physical or bare?

Question

If I understand it correct, then the physical mass $m$ of a particle, is the mass in presence of the interaction (i.e., the mass of the dressed particle) where as the bare mass $m_0$ is the mass in absence of interaction. However, in the derivation of LSZ reduction formula, as given in Bjorken and Drell, it is said in Eqn. 16.6 that the 'in' state $\phi_{in}(x)$ at the asymptotic past $t\rightarrow -\infty$ (and similarly, the 'out' state $\phi_{out}(x)$ at the asymptotic future $t\rightarrow +\infty$) obeys free Klein-Gordon equation with the physical mass m.

But since the 'in state' is a free-particle state, shouldn't the KG equation be written in terms of th bare mass $m_0$?

Answer 1

The physical mass can be defined as the pole of the propagator $$\int d^4x e^{-iq\cdot x} \cdot\langle \Omega|T\{\Psi_l(x)\Psi^\dagger_{l^\prime}(0)\}|\Omega\rangle$$ where $\Psi_l$ are renormalized fields.

The LSZ reduction formula says that if a one-particle state with momentum squared $m^2$ has non-vanishing matrix elements with the states $\Psi^\dagger_l|\Omega\rangle$, then $m^2$ is the pole of the propagator. So the square of the physical mass of a field equals to the momentum squared of the in state (or out state), which is just the mass squared present in the asymptotic K-G equation.

Note that the in state is not a free state (they are energy eigenstates belonging to two different systems, which have a one-to-one correspondence, but it doesn't mean they have the same mass) and that the mass can only be measured via interaction.