# What are the asymptotic momentum eigenstates? Dressed quanta or quanta of free theory?

Suppose I consider an interacting theory, say QED (with electrons and photons). Let, free electrons, I mean the quanta of the free Dirac Lagrangian. The dressed electron differs from the free electron by a renormalized mass and renormalized charge.

When we consider scattering, do we consider scattering between free Dirac electrons or renormalized electrons? In other words, I want to understand whether the asymptotic momentum eigenstates in the LSZ formalism, are electrons of the free theory or dressed electrons.

• @ Solenodon Paradoxus- I apologize but again I'm throughly confused. What are asymptotic momentum eigenstates or dressed electrons? Are they eigenstates of the full Hamiltonian $H=H_0+H_{int}$ of the interacting theory or that of $H_0$ with bare mass replaced by physical mass? If they are eigenstates of the full Hamiltonian, they'll be "stationary states", and hence there would not be any transition/scattering. – SRS Dec 3 '16 at 10:09
• @ Solenodon Paradoxus- In Matthew Schwartz's book on quantum field theory, it derives the in (and out) state from the interacting vacuum $|i\rangle\sim a_{p_1}^\dagger(-\infty) a_{p_1}^\dagger(-\infty)|\Omega\rangle$ which confuses me. Because if in states and out states are derived from $|\Omega\rangle$ don't they implicitly become eigenstate of the full Hamiltonian? This is what troubles me. The same derivation of LSZ reduction is also followed in Mark Srednicki's book. But he used $|0\rangle$ for the interacting vacuum and $|\slashed{O}\rangle$ for the free theory vacuum. – SRS Dec 3 '16 at 10:45
• @SRS Does the full interacting Hamiltonian commute with $a_{p_1}^{\dagger}(-\infty)$? – Prof. Legolasov Dec 3 '16 at 10:55
• @SRS you take an eigenstate $\left| \Omega \right>$ of the full Hamiltonian and apply an operator to it. Why do you think that the resulting state is also an eigenstate? This only happens when that operator commutes with the Hamiltonian, and in your case it doesn't. – Prof. Legolasov Dec 3 '16 at 11:54