Sunday, 12 May 2019

quantum field theory - Scattering, Perturbation and asymptotic states in LSZ reduction formula


I was following Schwarz's book on quantum field theory. There he defines the asymptotic momentum eigenstates $|i\rangle\equiv |k_1 k_2\rangle$ and $|f\rangle\equiv |k_3 k_4\rangle$ in the S-matrix element $\langle f|S|i\rangle$ as the eigenstates of the full Hamiltonian i.e., $H=H_0+H_{int}$. Therefore, the states $|i\rangle=|k_1 k_2\rangle$ is defined as



$$|k_1 k_2\rangle=a_{k_1}^{\dagger}(-\infty) a_{k_2}^{\dagger}(-\infty)|\Omega\rangle$$


where $|\Omega\rangle$ is the vacuum of the full interacting theory. Then the LSZ reduction formula connects the S-matrix element $\langle f|S|i\rangle$ to the Green' functions of the interaction theory defined as $$G^{(n)}(x_1,x_2,...x_n)=\langle \Omega|T[\phi(x_1)\phi(x_2)...\phi(x_n)]|\Omega\rangle.$$ Here are a few doubts.


Doubt 1 When the particles are far away, the interaction can be considered to be adiabatically switched off. Therefore, at $t=\pm\infty$ the states are really free particle states and should have been written as


$$|k_1 k_2\rangle=a_{k_1}^{\dagger}(-\infty) a_{k_2}^{\dagger}(-\infty)|0\rangle$$


and $$|k_3 k_4\rangle=a_{k_3}^{\dagger}(+\infty) a_{k_4}^{\dagger}(+\infty)|0\rangle$$ where $|0\rangle$ is the vacuum of the free thoery. I do not understand why these the states $|i\rangle$ and $|f\rangle$ are derived from $|\Omega\rangle$ instead of $|0\rangle$.


Doubt 2 The initial and final states were derived from the vacuum of the interacting theory $|\Omega\rangle$. According to my understanding, this suggests that the states $|i\rangle\equiv |k_1 k_2\rangle$ and $|f\rangle\equiv |k_3 k_4\rangle$ are eigenstates of the full Hamiltonian $H$. Since then there is no perturbation, there should not be any scattering or transition at all.




More references Even Peskin and Schroeder, Bjorken and Drell, Srednicki take the same approach as Schwartz; they too define the external momentum eigenstates to the eigenstate of the full Hamiltonian $H$. However, if the system was initially in a stationary state why should it undergo a transition in absence of any perturbation?




No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...