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 $m_0$?


1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.