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

Physics Asked by 346699 on December 17, 2020

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

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

Another definition (e.g (9.14) (9.17) (9.99) in Greiner’s Field Quantization) is :
$$S_{fi}equiv langle Psi_p^{-}| Psi_k^{+}rangleequivlangle Psi_p^{-}| hat S ^prime |Psi_k^{-}rangle=langle Psi_p^{+}| hat S ^prime |Psi_k^{+}rangle$$
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: $$Omega_+(Omega_-)^dagger= e^{i alpha}(Omega_-)^daggerOmega_+$$

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

One Answer

I think this is a Baker-Campbell-Hausdorff (BCH) rule type of result. I will define the operators $$ Omega_pm~=~e^{ibeta_pm}, $$ so that $$ (Omega_-)^daggerOmega_+~=~e^{-ibeta_-}e^{ibeta_+} $$ $$ =~left(1~-~ibeta_-~-~frac{1}{2}beta_-^2right)left(1~+~ibeta_+~-~frac{1}{2}beta_+^2right)~+~O(beta^3). $$ A similar expression is derived from $Omega_+(Omega_-)^dagger$. We may then easily see that $$ (Omega_-)^daggerOmega_+~=~Omega_+(Omega_-)^dagger~+~[beta_-,~beta_+]. $$ By BCH allows us to write this as $$ (Omega_-)^daggerOmega_+~=~e^{[beta_-,~beta_+]}Omega_+(Omega_-)^dagger. $$ From there it is a matter of defining $alpha~=~-i[beta_-,~beta_+]$.

Answered by Lawrence B. Crowell on December 17, 2020

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