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

Question

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?

Quantumwhisp · Answer

I'll offer a derivation, although I may gloss over possible subleties with different hilbert spaces. In short, the two definitions yield DIFFERENT Operators, that are only unitary equivalent:
We defined the s-matrix elements as
begin{align}
S_{pk} = langle Psi_p^-| Psi_k^+ rangle
end{align}
You showed yourself how to derive the first identity about the S Operator, and it is also the one that is given by Weinberg (3.2.4):
begin{align}
S_{pk} = langle Phi_p | S | Phi_k rangle
end{align}
With $|Phi_k>$ states of a free theory that are related to $Psi_k$.
What you ask for now is wether the same operator $S$ gives the same Matrix-elements as well between the -in or -out states. This is not the case.
Instead, define an operator $tilde{S}$ that maps -out states to -in states of the same label:
begin{align}
tilde{S} |Psi^-_{k} rangle =  |Psi^+_{k}  rangle
end{align}
Then
begin{align}
langle Psi^-_{p} |tilde{S}|Psi^-_{k} rangle = langle Psi^-_{p} |Psi^+_{k} rangle = S_{pk}
end{align}
So this operator $tilde{S}$ is not the same operator, but it has the same matrix elements for another choice of a basis. It bugged me as well until I found out, because other authors (for example Peskin / Schroeder) or Schwartz) use this definition.
That the operators are different can be seen when writing them down as linear combinations of $|Psi_x rangle$ and $|Phi_x rangle$.

Lawrence B. Crowell · 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 $alpha~=~-i[beta_-,~beta_+]$.

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

2 Answers

Add your own answers!

Ask a Question