Thursday, 4 June 2015

quantum mechanics - How to prove the equivalence of two different definitions of $S$-operator?


I read there are two definitions about $S$-operator:


The first one (e.g (8.49) in Greiner's Field Quantization) is: $$S_{fi}\equiv \langle \Psi_p^{-}| \Psi_k^{+}\rangle$$ where $|\Psi_p^{-}\rangle$ is a state in Heisenberg picture which is $| p \rangle$ at $t=+\infty$ when you calculate the $|\Psi_p^{-}\rangle$ in Schrodinger picture , called out state. $| \Psi_k^{+}\rangle$ is a state in Heisenberg picture which is $| k \rangle$ at $t=-\infty$, called in state.


So$$S_{fi}\equiv \langle \Psi_p^{-}| \Psi_k^{+}\rangle= \langle p|(\Omega_-)^\dagger\Omega_+|k \rangle$$


In this case the S-operator $\hat S=(\Omega_-)^\dagger\Omega_+$, where Møller operator $$\Omega_+ = \lim_{t\rightarrow -\infty} U^\dagger (t) U_0(t)$$ $$\Omega_- = \lim_{t\rightarrow +\infty} U^\dagger (t) U_0(t)$$ So $$S=U_I(\infty,-\infty)$$


Another definition (e.g (9.14) (9.17) (9.99) in Greiner's Field Quantization) is : $$S_{fi}\equiv \langle \Psi_p^{-}| \Psi_k^{+}\rangle\equiv\langle \Psi_p^{-}| \hat S ^\prime |\Psi_k^{-}\rangle=\langle \Psi_p^{+}| \hat S ^\prime |\Psi_k^{+}\rangle$$ where S-operator $\hat S ^\prime |\Psi_p^{-}\rangle =|\Psi_p^{+}\rangle$ that is $\hat S^\prime = \Omega_+(\Omega_-)^\dagger$.



It seems that these two definitions are differnt, but many textbook can derive the same dyson formula for these two S-operators. https://en.wikipedia.org/wiki/S-matrix#The_S-matrix


How to prove: $$\Omega_+(\Omega_-)^\dagger= e^{i \alpha}(\Omega_-)^\dagger\Omega_+$$


related to this question: There are two definitions of S operator (or S matrix) in quantum field theory. Are they equivalent?




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...