- #1
kof9595995
- 679
- 2
In page 110(section 3.1), after giving the relation between in and out states and free particle states,
[itex]\Psi_{\alpha}^{\pm}=\Omega(\mp\infty)\Phi_{\alpha}[/itex]...(3.1.13)
Then he says in page 111:
[itex]\Psi_{\alpha}^{\pm}=\Omega(\mp\infty)\Phi_{\alpha}[/itex]...(3.1.13)
Then he says in page 111:
But I don't see why it's necessary.However, it should be kept in mind that [itex]\Omega(\mp\infty)[/itex] in Eq.(3.1.13) gives meaningful results only when acting on a smooth superposition of energy eigenstates.