# How to prove the equivalence of two different difinitions 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?

## 1 Answer

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? – MannyC Feb 11 at 20:07