In nuclear physics, the Optical Model means, we are treating the scattering problem is like optical wave problem. due to the incident beam can be treated as a wave-function. and this wave will be scattered by the target.

when the beam is far away from the target, the wave function of the incident beam should satisfy the Schrödinger equation in free space :

$\left( \frac {\hbar^2 } {2m} \nabla^2 + V(r) \right) \psi( \vec{r} ) = E \psi ( \vec{r} )$

and the plane wave solution is

$\psi ( \vec{r} ) \sim Exp ( \pm i \vec{k} \cdot \vec {r} )$

after the scattering, there will be some spherical wave come out. the spherical wave should also satisfy the free-space Schrödinger equation.

$\psi( \vec{r} ) \sim Y(\theta, \phi) \frac {Exp( \pm i \vec{k} \cdot \vec{r} ) }{r}$

Thus, the process of scattering can be think in this way:

$Exp( \pm i k z ) \rightarrow Exp( i k z ) + f ( \theta ) \frac { Exp ( i k r ) } {r}$

where f(θ) is a combination of spherical wave.

one consequence of using Optical Model is, we use complex potential to describe the nuclear potential terms in quantum mechanics.

when using a complex potential, the flux of the incident beam wave function can be non-zero. meanings that the particles in the beam are being absorbed or emitted. This corresponding to the inelastic scattering.

The reason for the “OPTICAL” is come form the permittivity and permeability of the EM field. for metallic matter, their permittivity or permeability may have a imaginary part. and this imaginary part corresponding to the absorption of the light. so, nuclear physics borrow the same idea.

the flux is defined as:

$J = \frac { \hbar }{ 2 i m} ( \psi^*(r) \nabla \psi(r) - \psi(r) \nabla \psi^* (r) )$

and the gradient of the flux, which is the absorption (sink) or emission ( source ) is:

$\nabla J = \frac {\hbar }{ 2 i m }( \psi^* \nabla^2 \psi - \psi \nabla^2 \psi^* )$

The Schrödinger equation gives the equation for the wave function:

$\nabla^2 \psi(r) = \frac { 2m} {\hbar^2} ( E - V(r)) \psi(r)$

when sub the Schrödinger equation in to the gradient of flux, we have:

$\nabla J = \frac {1} {i \hbar } ( V(r) - V^*(r) ) | \psi |^2 = \frac { 2} {\hbar } Im ( V) | \psi |^2$

we can see, if the source and the sink depend on the complex part of the potential. if the imaginary part is zero, the gradient of the flux is zero, and the wave function of the beam is conserved.