at the point of scattering ( t = 0 ), the wave function and the incoming and out-going wave function can be related as:

$\left| \psi_{in} \right> \rightarrow \left| \psi \right> \leftarrow \left| \psi_{out}\right>$

where the incoming and out-going wavefunction is very far away from the field of the scatter, and thus, they are free and we say they are asymptotic.

Let a time propagator with a full Hamiltonian be U(t), and a time propagator with free Hamiltonian be $U^0(t)$. thus the time behavior of the scattering wave functions can be related as:

$U(t) \left| \psi \right> \rightarrow U^0(t) \left| \psi_{in} \right>$ for $t \rightarrow - \infty$

$U(t) \left| \psi \right> \rightarrow U^0(t) \left| \psi_{out} \right>$ for $t \rightarrow + \infty$

in equation:

$\left| \psi \right> = lim_{t \rightarrow - \infty} U^\dagger (t) U^0(t) \left| \psi_{in} \right> = \Omega_+ \left| \psi_{in} \right>$

$\left| \psi \right> = lim_{t \rightarrow + \infty} U^\dagger (t) U^0(t) \left| \psi_{out} \right> = \Omega_- \left| \psi_{out} \right>$

Thus, we have the scattering matrix or the S-matrix.

$\left| \psi_{out} \right> = \Omega_-^\dagger \Omega_+ \left|\psi_{in} \right> = S\left| \psi_{in} \right>$

Now assume for a particular state generated by an accelerator is Φ, and a particular out-going asymptotic state is χ. we have:

$\left| \phi \right> = \Omega_+ \left| \phi_+ \right>$

$\left| \chi \right> = \Omega_+ \left| \chi_- \right>$

thus, the probability amplitude for the scattering between these 2 states is:

$\omega( \chi \leftarrow \phi) = | \left<\chi_- | \phi_+ \right> |^2 =| \left<\chi|\Omega_-^\dagger \Omega_+| \phi \right> |^2 = |\left< \chi | S | \phi \right>|^2$

if we expand a wave function  in momentum basis:

$\left| \psi \right> = \int d^3p \left| p \right> \left< p | \psi \right>$

$\psi_{out}(p) = \int d^3p' \left

\psi_{in}(p')$

now, we are going to show the energy conservation of the scattering operator or matrix, by showing that the scattering operator S commutes with the Hamiltonian. from

$Exp(\frac{i}{\hbar} H \tau) \Omega_\pm =\Omega_\pm Exp( \frac{i}{\hbar} H^0 \tau )$

differential it then we have :

$H \Omega_\pm = \Omega_\pm H^0$

thus,

$H^0 = \Omega_\pm^\dagger H \Omega_\pm$

$S H^0 = H^0 S$

together with the wavefunction:

$0 = \left = (E_{p'} -E_p ) \left< p'|S|p\right>$

thus implies,

$\left = \delta(E_{p'}-E_p) g( p' \leftarrow p)$

since, at the forward direction,  the change of momentum is zero, we can write S = 1 + R, then,

$\left =\delta(p-p') - 2 \pi i \delta(E_{p'}-E_p) t( p' \leftarrow p)$

the $t(p' \leftarrow p)$ is called on-shell T-matrix. since the energy must be equal, required by the delta function, thus, the momentum magnitude must be equal, therefore, the 2 momentums s on a shell. The T-matrix also related to the scattering amplitude by:

$f(p' \leftarrow p) = - (2 \pi)^2 m t(p' \leftarrow p)$

then the S-matrix becomes,

$\left = \delta(p - p') + \frac{i}{2\pi m} \delta(E_{p'} - E ) f(p' \leftarrow p)$