## Optical Model

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.

## Differential Cross Section III

a single diagram can illustrate everything — the relation between the yield ( the number of particle detected per second in the detector ) and the differential cross section.

the upper drawing is a big view, and the lower one is a zoom-in. the red-circles on the upper drawing are same as the lower one.

From the upper one, the number of particles that scatted by the target and go to direction (θ,φ) is :

$\frac {N_0}{S} \times ( \Delta \sigma \times N) = n$

the left hand side can be interpolated at follow:

$\frac{N_0}{S}$ is the number of particle per second per unit area, or the flux.

$\Delta \sigma \times N$ is the total area of  the cross section that deflect or scatter particle to direction (θ,φ).

The left hand side can also be viewed as :

$N_0 \times \frac { \Delta \sigma \times N } {S}$

where the fraction after multiplication is the chance of getting scattered to the direction (θ,φ). ( the requirement of the flux $N_0$ is in HERE. )

and the right hand side is the number of particle detected at direction (θ,φ).

since both $n$ and $\Delta \sigma$ depend on (θ,φ). thus we can differential it and get the angle dependent of these 2.

$\frac { N_0 N}{S} \frac {d \sigma}{d \Omega} = \frac {d n}{d \Omega}$

if we set the detector moving as radius R. thus the detector area is:

$D_A = R^2 d \Omega$

Therefore, the number of particles will be detected per second on the detector with some area (= Yield ) is:

$Y = \frac {d n}{d \Omega} R^2 d \Omega = \frac { N_o N}{S} R^2 \frac{ d \sigma}{d \Omega} d \Omega$

finish. Oh, the unit of the differential cross section is barn = $10^{-28} m$, recall that a nucleus radius is about $10^{-14} m$

the relation between the differential cross section to the nuclear potential was discussed on HERE.

## density, flux & luminosity

density is a consideration factor for scattering experiment. In low density, both for target and the beam, then the probability of collision will be small and experiment will be time consuming and uncertainly increase. Remember that the size of nuclear is 1000 times less then the atom. the cross section area of it will be 1000 x 1000 times lesser. the chance for a nucleus-nucleus collision is very small. for example, if there is only 1 particle in the area on 1 atom, the chance for hitting the nucleus is $1 / ( \pi 10^6)$ = 0.000003, 3 in 1 million. it is just more than nothing. thus, in order to have a hit, we have to send more then 3 million particles for 1 atom. in some case, the beam density is small, say, 0.3 million particles per second on an area of 1 atom. then we have to wait 10 second for 1 hit.

density is measured in particle per area for target .

for beam, since particle is moving in it, time is included in the unit. there are 2 units – flux and luminosity. flux is particle per second, and luminosity is energy per second per area. since energy of the beam is solely by the number of particle. so, density of beam is particle per second per area. but in particle physics, the energy of particle was stated. thus, the luminosity is equally understood as density of beam, and their units are the same as particle per second per area.

In daily life, density is measured by mass per volume. although the unit are different, they are the same thing – ” how dense is it? ”

in solid, the density is highest compare to other state of matter. from wiki, we can check the density. and coveted it in to the unit we want. for example, copper has density $8.94 g / cm^3$. its molar mass is $29 g / mole$. thus, it has $0.31 N_A$ copper atom in $1 cm^3$. and $N_A = 6.022 \times 10^{23}$, which is a huge number, so, the number of atom on $1 cm^2$ is $3.25 \times 10^{15}$ .

how about gas? the density depends on temperature and pressure, at $0^o C$ and 1 atm pressure, helium has density  $1.79 \times 10^{-4} g / cm^3$ and the molar mass is 2. thus, the number of Helium atom in $1 cm^2$ is $1.42 \times 10^{13}$. when the temperature go to -100 degree, the density will increase.

beside of the number of atom per area. we have to consider the thickness of the target. think about a target is a layer structure, each layer has certain number of atom per area. if the particle from the beams miss the 1st layer, there will be another layer and other chance for it to hit. thus. more the thickness, more chance to hit.

For a light beam, the power $P$ and the wavelength $\lambda$ determine the flux of photon. power is energy per second. and energy of single photon is inversely proportional to its wavelength. the density of a light beam is given by :

$L = n/area = P \frac { \lambda} { h c} = P \lambda \times 5 \times 10^{15} [W^{-1}][nm^{-1}][s^{-1}][m^{-2}]$

where, $L$ is the luminosity and $n$ is the flux. for typical green class 4 laser, which has power more then 0.5 W and wavelength is about 500nm. the flux is about  $n=1.3 \times 10^18 [s^{-1}]$ photons per second per unit area. if the laser spot light is about 5 mm in diameter. thus, the density of the beam is $L=1.7^{18} [s^{-1}][cm^2]$.

for laser pointer in office, which is class 1 laser. the power is less then0.4 mW, say, 0.1 mW. for same spot size, the luminosity  is still as high as $6.6^{14} [s^{1}][cm^2]$.

on LHC, the beam flux can be $10^{34} [s^{-1}][cm^2]$. by compare the the density of solid copper. it is much denser. thus, a collision in LHC is just like smashing 2 solid head to head and see what is going on.