- #1

HomogenousCow

- 736

- 210

I'm currently reading Schwartz's QFT text "Quantum Field Theory and the Standard Model" (although this question does not specifically pertain to this text) and I've got two questions:

1. At the beginning of chapter 6 he talks about how we may consider the system to be non-interacting in asymptotic times, ostensibly because "the particles are far apart". I find this very hand-wavy so my question is can this notion be made rigorous?

Take a massless scalar field with a ## \phi^3## interaction, what's stopping a final momentum eigenstate from splitting into two particles each with half the momenta?I don't see how one could find free asymptotic states in a theory such as this.

2. My second question relates to the first question. Take a massive scalar field with a ## \phi^3## interaction for example. Now even though the momentum operator is unaltered by this term, and hence the ##a_p## at fixed times are still creating/annihilating momentum eigenstates, the Hamiltonian is altered with a ## - \frac g {3!} \phi^3## term which means that the ##a^{\dagger}| 0 \rangle## states are no longer energy eigenstates. However the asymptotic states are said to be on-shell, so does that mean that we are altering the Lagrangian/Hamiltonian at asymptotic times? I'm very confused.

1. At the beginning of chapter 6 he talks about how we may consider the system to be non-interacting in asymptotic times, ostensibly because "the particles are far apart". I find this very hand-wavy so my question is can this notion be made rigorous?

Take a massless scalar field with a ## \phi^3## interaction, what's stopping a final momentum eigenstate from splitting into two particles each with half the momenta?I don't see how one could find free asymptotic states in a theory such as this.

2. My second question relates to the first question. Take a massive scalar field with a ## \phi^3## interaction for example. Now even though the momentum operator is unaltered by this term, and hence the ##a_p## at fixed times are still creating/annihilating momentum eigenstates, the Hamiltonian is altered with a ## - \frac g {3!} \phi^3## term which means that the ##a^{\dagger}| 0 \rangle## states are no longer energy eigenstates. However the asymptotic states are said to be on-shell, so does that mean that we are altering the Lagrangian/Hamiltonian at asymptotic times? I'm very confused.

Last edited: