# 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?

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 $$|\Phi_x \rangle$$.

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 $\alpha~=~-i[\beta_-,~\beta_+]$.

• How do you know $[\beta_-,\beta_+]$ is a number? Feb 11, 2019 at 20:07
• I doubt it is, furthermore I don't think BCH is applicable here because I don' think $\beta$ has to be small in any sense. Mar 24, 2021 at 21:31