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quantum scattering
In quantum mechanics, scattering is different than classical scattering because particle trajectory and velocity of particle do not have (well defined) meaning in quantum mechanics so quantities like impact parameter and relative velocity can’t be used to describe scattering. Quantum mechanics scattering is described by quantities like incoming current and scattered current and probabilities. Scattering in quantum mechanics is describe by incoming particles with state $${\Psi_i} $$and they interact with scattering potential and outgoing particle state is called scattering state. Usually $${\Psi_i} $$is background eigenstate that means the eigenstate of Hamiltonian without scattering potential. For finding cross section one needs to find the probability of scattered particle which is $${\Psi_s} $$so the main problem in quantum scattering is finding scattering state. So in quantum scattering we are looking for particle’s wave functions after scattering and then the $$|{\Psi_s}|^2 $$is probability of presence of scattered particles at $$\mathbf{r}$$. Cross section in quantum scattering is defined by
 * $$\sigma = \frac{j_s}{j_i}r^2, $$

Where j is quantum current
 * $$ j = \frac{\hbar}{2mi}\left(\Psi^* \mathbf \nabla \Psi - (\mathbf \nabla \Psi^*)\Psi \right), $$

Elastic scattering
In elastic scattering particle’s kinetic energy should be conserved after scattering. So the absolute value of incoming and outgoing wave number should be equal.
 * $$|\mathbf {k_i}|=|\mathbf {k_s}| $$

For a spinless particle and short range spherically symmetric scattering potential (the scattering potential is ignorable at large distance from potential) and $${r}>>\lambda $$ where $$\lambda $$ is particle de Broglie wavelength) finding scattering state is relatively easy. The incoming beam usually is written as plane wave $$e^{ikz} $$ and scattered wave usually is written in terms of spherical wave but the intensity of scattered wave is not the same in every direction. So in general the wave function far from scattering region is
 * $${\psi}= {\psi_i}+{\psi_s} =e^{ikz} + \frac{f(\theta,\phi)}{r} e^{ikr}$$

Where $$\psi_i $$  is incoming wave and $${f(\theta,\phi)}$$  is scattering amplitude that determines the intensity of scattered wave function. And if the scattering potential is spherically symmetric then scattering amplitude is just function of $$(\theta)$$. The differential cross section is
 * $$\sigma =|f(\theta)|^2$$

Far from scattering potential the particle can be considered free. The general wave function for free particle in the spherical coordinate with azimuthal symmetry is
 * $$ \psi (r, \theta)= \sum_{\ell=0}^\infty \ell ( a_{\ell } j_\ell ( k r ) + b_{\ell } n_\ell ( k r ) ) p _\ell ( { cos({\theta} ))} .$$

where $$j$$ and $$n$$ are  spherical Bessel functions

Because $${r}>>\lambda $$ so $${kr}>>1 $$  then spherical Bessel functions can be replaced by their asymptotic limits
 * $${j_l}=\sin \left( x-\frac{l\pi}{2} \right) $$
 * $${n_l}=-\cos \left( x-\frac{l\pi}{2} \right) $$

So the wave function turns to
 * $$\psi (x)=\sum_{\ell=0}^\infty C_{\ell} \frac{e^{i(kr+\frac{l\pi}{2}+\delta_\ell)}- e^{-i(kr+\frac{l\pi}{2}+\delta_\ell)}}{2ikr} p _\ell ( { cos({\theta} ))} .$$

Because these functions are complete sets so incoming wave function can be expanded in terms of these function


 * $$e^{ikz} = \sum_{\ell=0}^\infty i^\ell (2 \ell + 1) j_\ell(kr) P_\ell(\cos (\theta)) $$

Far from scattering region $$\psi$$ and $$\psi_i$$ should be equal so


 * $${C_\ell}=(2\ell+1) e^{i(\frac{l\pi}{2}+\delta_\ell)} $$




 * $${B_\ell}={C_\ell}{\sin(\delta_\ell)} $$

And scattering amplitude is :$$ f(\theta) = \frac{1}{k} \sum_{\ell=0}^\infty (2 \ell + 1) e^{2i\delta_\ell} \sin(\delta_\ell) P_\ell (\cos(\theta)). $$ And the differential cross section is
 * $$\frac{d\sigma(\vartheta)}{d \Omega} = \frac{1}{4k^2} \sum_{\ell=0}^\infty \sum_{\ell'=0}^\infty(2 \ell + 1)(2 \ell' + 1) e^{2i\delta_\ell-2i\delta_\ell'} \sin(\delta_\ell) \sin(\delta_\ell') P_\ell (\cos(\theta) P_{\ell'}(\cos (\theta)). $$

And total cross section is
 * $$\sigma = \frac{4 \pi}{k^2} \sum_{\ell = 0}^\infty (2 \ell + 1) \sin^2 \delta_\ell. $$

Where $$\delta_{\ell} $$is phase shift that depends on l and scattering potential and is positive for attractive potential and negative for repulsive potential. The phase shift can be determine with solving the Schrodinger equation in the scattering region where the potential is not ignorable and connect it with the large distance solution with suitable boundary conditions.

hard sphere
For the hard sphere problem the potential is infinity inside the sphere and zero outside. So the boundary condition here is
 * $$ \psi(d,\theta)=0. $$

The general solution for Schrodinger equation free particle is
 * $$ \psi (r, \theta)= \sum_{\ell=0}^\infty ({C_\ell}{\cos(\delta_\ell)} j_\ell ( k r ) + {C_\ell}{\sin(\delta_\ell)}n_\ell ( k r ) ) p _\ell ( { cos({\theta} ))} .$$

Applying boundary condition to equation gives
 * $$ \tan(\delta_\ell)= \frac{j_\ell ( k r )}{ n_\ell ( k r ) }$$

In low energy limits the l=0 has the most contribution in the wave function in that case the total cross section will be For $$ \ell=0$$   $$ \delta_0={kd}$$ The total cross section is
 * $$\sigma = \frac{4 \pi}{k^2} \sin^2 kd $$

This is a remarkable result -- it says that the cross section is not the area of the sphere presented to the oncoming particles, as it is in classical mechanics, but rather {\it four times} this amount -- the entire surface area of the sphere. Of course, at energies approaching zero, the de Broglie wavelength is larger than the dimension of the object (see discussion below), so perhaps it is to be expected that the classical result is not recovered. But at higher energies, it is possible to include all the partial waves, with the result that
 * $$\lim_{k \rightarrow \infty} \sigma = 2 \pi d^2,$$

still twice the classical result.

Born approximation
Using green function method for solving Schrodinger equation one can find
 * $$ \Psi= \Psi_i-\frac{2{m}}{\hbar}\int\frac{\Psi(\mathbf{r '})e^{i({k}|\mathbf r -\mathbf r '|)}}{4\pi |\mathbf r -\mathbf r '|}{v}(\mathbf{r '})d^3\mathbf{r '} \,\!$$

Far away from scattering region $$|\mathbf r -\mathbf r '| $$can be approximated by $$r-\mathbf{k}\cdot\mathbf{r}$$ Now the wave function can be written as
 * $$ \Psi(\mathbf{r},t) = \Psi_i-\frac{2{m}}{\hbar} e^{i{k}{r}}\int\frac{\Psi(\mathbf{r '})e^{i-\mathbf{k}\cdot\mathbf{r}}}{4\pi |\mathbf r -\mathbf r '|}{v}(\mathbf{r '})d^3\mathbf{r '} \,\!$$

This equation is not use full because the wave function is inside the integral. For solving that one can approximate the wave function inside the integral with incoming wave function this is called Born approximation. With using born approximation
 * $$f({\theta}) = - \frac{2{m}}{\hbar^2 } \int\frac{\Psi(\mathbf{r '})e^{i({k}{r}\cos(\theta ' ))}}{4\pi}{v}(\mathbf{r '})d^3\mathbf{r '}=- \frac{2 \pi \hbar^2\sin (\frac{\theta}{2})} \int\Psi({r '})\sin (k\sin (\frac{\theta}{2})r ') {v}({r '})d{r '}$$

Yukawa scattering
For Yukawa scattering the potential is
 * $$V_\text{Yukawa}(r)= -g^2\frac{e^{-\mu r}}{r},$$

So the scattering amplitude will be
 * $$f({\theta}) = - \frac{{m} g^2}{2 {k}\pi \hbar^2\sin (\frac{\theta}{2})} \int\Psi({r '})\sin (k\sin (\frac{\theta}{2})r ') \frac{e^{-\mu r}}{r} d{r '}=-\frac{{m} g^2}{\hbar^2 ({k} \sin (\frac{\theta}{2}) )^2+\mu^2)}$$

rutherford scattering
rutherford scattering is Coulomb potential witch is Yukawa potential when $$\mu=0$$
 * $$ V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}, \, $$

so for rutherford scattering the cross section is
 * $$f({\theta})=-\frac{2{m}{q_1}{q_2} }{4\pi\epsilon_0\hbar^2 ({k} \sin (\frac{\theta}{2}) )^2)}$$

In Rutherford scattering the result from first born approximation is the same as classical scattering.

Inelastic scattering
If the incoming particles or target particles or both have internal structure energy exchange is between incoming and target particles with excitation of internal states so in this case if the energy of incoming particles is enough to excite internal structure then the energy of incoming particle and scattered particle is not the same any more. Neutron scattering from molecular systems is an example for inelastic scattering.