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.


Its dressed electrons.

Remember how we are only considering one-particle-irreducible diagrams when evaluating the S-matrix element between two asymptotic states? Well, this is exactly because the asymptotic states are dressed. Intuitively, the diagrams with self-energy corrections are already accounted for because the states are dressed and we have to "amputate the legs" to calculate the amplitude between dressed states.

| cite | improve this answer | |
  • $\begingroup$ @ 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. $\endgroup$ – SRS Dec 3 '16 at 10:09
  • 1
    $\begingroup$ @SRS They couldn't be eigenstates of the full Hamiltonian because there won't be any scattering this way, as you correctly noticed. They are equivalent to free theory states with mass replaced with physical mass though, but what we are interested in is not the structure of Fock space (which is trivial), but how incoming and outgoing states relate through quantum dynamics of the full theory. $\endgroup$ – Prof. Legolasov Dec 3 '16 at 10:21
  • $\begingroup$ @ 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. $\endgroup$ – SRS Dec 3 '16 at 10:45
  • $\begingroup$ @SRS Does the full interacting Hamiltonian commute with $a_{p_1}^{\dagger}(-\infty)$? $\endgroup$ – Prof. Legolasov Dec 3 '16 at 10:55
  • 1
    $\begingroup$ @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. $\endgroup$ – Prof. Legolasov Dec 3 '16 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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