Recent content by UnreliableObserver

Finally getting back around to this :) So it seems that was indeed it. The problem was that I was writing down all these momentum eigenstates without paying attention to which momentum operators they are actually eigenstates of! The operators ##(a_p^{\pm\infty})^\dagger## don't just "create...

Hi Demystifier. I agree with everything you've said, however the states ##S|i\rangle## and ##|f\rangle## you refer to are the states ##U(t_f,t_i)|k_1\cdots k_m\rangle## and ##|p_1\cdots p_n\rangle## I wrote down in the Schrodinger-picture expression ##(1)##. These are different to the "messy"...

Thanks for the reply vanhees. I don't think this is true. Using ##a_p(t)=U^\dagger (t, t_0)a_p U(t, t_0)##, ##(3)## can be written as $$\bra{\Omega}U^\dagger (t_f, t_0)a_{p_1} U(t_f, t_0)\cdots U^\dagger (t_f, t_0)a_{p_n}\underbrace{U(t_f, t_0) U^\dagger (t_i, t_0)}_{U(t_f, t_i)}a_{k_1}^\dagger...

In scattering theory, the quantity of interest is the amplitude for the system—initially prepared as a collection of (approximate) momentum eigenstates—to evolve into some other collection of momentum eigenstates. For example, for ##m\to n## scattering, the amplitude we're interested in is...