For elastic scattering, the scattered state’s energy should equal the incident state’s energy. However, the equation for them are different.

$H_0 \left| \phi \right> = E \left| \phi \right>$

$H_0 + V \left| \psi \right> = E \left| \psi \right>$

the Lippmann-Schwinger equation proposed an equation ( or solution )

$\left| \psi ^{\pm} \right> = \frac{1} { E - H_0 \pm i \epsilon } V \left|\psi^{\pm} \right> + \left| \phi \right>$

the complex energy is the key, without it, the operator will be singular. to compute the wave function, we apply the position bra on the right:

$\left< x | \psi ^{\pm} \right> = \int {dx'^3} \left< x| \frac {1}{E-H_0 \pm \epsilon } | x' \right> \left< x' | V | \psi^{\pm} \right> + \left< x | \phi \right>$

to compute the

$< x | \frac {1} {E-H_0 \pm \epsilon } |x'>$

we have to use the momentum ket, since it is commute with the Hamiltonian. then we have

$< x | \frac {1} {E-H_0 \pm \epsilon } |x'> = - \frac { 2 m } { 4 \pi \hbar^2 } \frac {1} {|x-x'| } Exp ( \pm i k |x-x'| )$

which is the Green function. We can also see this by compare the equation with the scatter equation.thus , the solution is

$\left< x | \psi ^{\pm} \right> =- \frac { 2 m } { \hbar^2 } \int {dx'^3} \frac{1}{4 \pi |x-x'| }Exp( \pm i k |x-x'|) \left< x' | V | \psi^{\pm} \right> + \left< x | \phi \right>$

for the potential is local. which means it is confined in a small space. or

$\left < x' | V | x'' \right> = V(x') \delta(x'-x'')$

and for the detector is far, plus using plane wave as incident wave the solution is

$\psi^{pm}(x) = \frac{1}{\sqrt {2 \pi \hbar}^3 } \left( Exp( \pm i k z ) + \frac { Exp( \pm i k \cdot r ) } {r} f(k, k') \right)$

$f(k,k') = - \frac { (2 \pi )^3 } { 4 \pi } \frac { 2 m } {\hbar ^2 } < k' |V | \psi ^{pm} >$

$k = p / \hbar$

since the scattered wave can be a composition of  different wave-number k. thus, the matrix

$T = < k'| V | \psi^ {\pm} >$

also called the transition matrix, which mean, it is the transition through potential V to a wave-number k.

we can see that the equation is very similar to the scatter equation, since basically, they are the solution of the same Schrödinger equation.

if we further apply the Born approximation, we will get the same result that the scattered wave amplitude is a Fourier transform of the nuclear potential.