ller}~{\rm operators}.\nonumber\end{aligned}$$

We see that $$\begin{aligned}
e^{i H \tau} \Omega_\pm &=& \lim_{T_\mp \rightarrow \mp \infty} e^{i H
  T_\mp } e^{-i H_0 (T_\mp~-~\tau)} , \\
&=& \Omega_\pm e^{i H_0 \tau}\end{aligned}$$ and $$|\psi; {\rm out} \rangle = \Omega_- \Omega_+^\dagger |\psi; {\rm in}\rangle$$ If the incoming state is $|k_1, k_2, \cdots k_N; {\rm F}\rangle$, it follows that $$|k_1, k_2, \cdots k_N; {\rm in}\rangle = \Omega_+ |k_1, k_2, \cdots
 k_N; {\rm F}\rangle$$ has eigenvalue $\sum k_{i0}$ for the [*total*]{} Hamiltonian $H$. A similar result is true for $$|k_1, k_2, \cdots k_N; {\rm out}\rangle = \Omega_- |k_1, k_2, \cdots
 k_N; {\rm F}\rangle .$$ We note that the scattering amplitude is $$\langle \xi; {\rm out} | \psi; {\rm in}\rangle = \langle \xi; {\rm
  in}|\Omega_+ \Omega_-^\dagger |\psi; {\rm in}\rangle.$$ In other words, the LSZ $S$-matrix is $$S = \Omega_+ \Omega_-^\dagger, \quad |\psi; {\rm out} \rangle =
S^\dagger |\psi; {\rm in}\rangle.$$ Between the “free” states, the $S$-operator is different: $$\langle \xi; {\rm out} | \psi; {\rm in}\rangle = \langle \xi; {\rm
  F}|\Omega_-^\dagger  \Omega_+ |\psi; {\rm F}\rangle.$$

The LSZ formalism works exclusively with the in- and out-states, as Haag’s theorem shows that $\Omega_\pm$ do not exist in quantum field theories. The creation-annihilation operators $c_k^{\rm in (out)
\dagger}, c_k^{\rm in 