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

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?

Answer 1

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 $|\Phi_x \rangle$.

Answer 2

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