# Confused about the scattering Operator in LSZ reduction formula

In Greiner's Field quantization book, Chapter 9 on the LSZ reduction formalism, he states

$$S_{fi}=\langle q_1,...,q_m;\text{out}| p_1,...,p_m;\text{in}\rangle\tag{9.10}$$

where $$S$$ is the scattering matrix.

My confusion is this doesn't seem to agree with the usual definition in non-relativistic quantum mechanics. I think if written in this way, there is no time evolution on the in state, and we are simply doing an inner product between the out and the in state. I think the transition amplitude in nonrelativistic quantum mechanics is given by $$|\langle\text{out state}|S|\text{in state}\rangle|^2.$$

Greiner then goes on to say 9.10 implies $$\hat{\phi}_{\text{out}}=\hat{S}^{-1}\hat{\phi}_{\text{in}}(x)\hat{S}.\tag{9.11}$$ How does this work?

Even more confusingly, Greiner claims $$|p_1,...,p_n;\text{in}\rangle=\hat{S}|p_1,...,p_n;\text{out}\rangle\tag{9.14}$$

Whereas I clearly remember $$S|\text{in state}\rangle=|\text{out state}\rangle$$ in the quantum mechanics.

In Greiner's Field quantization book, Chapter 9 on the LSZ reduction formalism, he states $$S_{fi}=\langle q_1,...,q_m;\text{out}| p_1,...,p_m;\text{in}\rangle\tag{9.10}$$
Greiner then goes on to say 9.10 implies $$\hat{\phi}_{\text{out}}=\hat{S}^{-1}\hat{\phi}_{\text{in}}(x)\hat{S}.\tag{9.11}$$ How does this work?
Eq.9.10 is a consequence of how operators evolve in Heisenberg's scheme. And Greiner might have used the property $$\hat{S}^{-1}=S^{\dagger}$$
• In tha Heisenberg picture the evolution of the field operator is given by solving the evolution equation $i\partial_t\phi=[\phi,H]$. Which has the solution $\phi(x,t)=U^\dagger\phi(x)U$, in the context of quantum field theory the $U$ evolution operator is called the $S$ operator ($S$-matrix). May 27, 2022 at 14:23