Quantum Mechanical Scattering Theory

Incident Flux of particles with mass $m_1$ on a target of particles with mass $m_2$, measured at detector $\mathcal{D}$ in a solid angle $d\mathrm{\Omega }$.

$(1)+(2)\to (1)+(2)$ thus momentum and energy is exchanged.

Elastic Scattering: Internal states of particles $(1)$ and $(2)$ do not change on collision. I.e. an electron and an atom in the ground state keeps the atom in the ground state.

Start without considering any spin for the particles and that the target is thin, i.e. only a single scattering event. Neglect coherent effects (interference between scattered waves). The interactions are by central potentials, $V(|{\overrightarrow{r}}_1-{\overrightarrow{r}}_2|)=V(r)\Rightarrow \mu =\frac{m_1{m}_2}{m_1+m_2}$ scattering of relative particle.

$[{\mathrm{\nabla }}^2+{\kappa }^2-v(r)]{\mathrm{\Psi }}_{\overrightarrow{\kappa }}(\overrightarrow{r})=0$. ${\phi }_{\overrightarrow{\kappa }}\sim \exp (i\kappa z)+f_{\kappa }(\theta ,\phi )\frac{\exp (i\kappa r)}{r}$. $J(\overrightarrow{r})=-\frac{i\hslash }{2\mu }({\phi }_{\kappa }^{\ast }\mathrm{\nabla }{\phi }_{\kappa }-{\phi }_{\kappa }\mathrm{\nabla }{\phi }_{\kappa }^{\ast })=\frac{\hslash }{\mu }\mathrm{\Im }\{{\varphi }_k^{\ast }\mathrm{\nabla }{\varphi }_k\}$. ${\overrightarrow{J}}_i\Vert z:{\phi }_{\kappa }\sim \exp (i\kappa z),J_i=\frac{\hslash \kappa }{\mu }$.

${\overrightarrow{J}}_{SC}(\overrightarrow{r})\Rightarrow {\phi }_k\sim f_{\kappa }(\theta ,\phi )\frac{\exp (i\kappa r)}{r}$.

$$\begin{array}{rl}{\overrightarrow{J}}_{sc}\cdot \hat{r}& =-\frac{i\hslash }{2\mu }(\frac{\exp (i\kappa r)}{r}\frac{\partial }{\partial r}\left(\frac{\exp (i\kappa r)}{r}\right)-\frac{\exp (i\kappa r)}{r}\frac{\partial }{\partial r}\left(\frac{\exp (-i\kappa r)}{r}\right))\\ \text{(1)}& & =\frac{\hslash k}{\mu }\frac{1}{r^2}|f_{\kappa }(\theta ,\phi ){|}^2{\overrightarrow{J}}_{sc}\cdot \hat{\theta }& =-\frac{i\hslash }{2\mu }(\frac{\exp (i\kappa r)}{r}{f}_k^{\ast }\frac{1}{r}\frac{\exp (i\kappa r)}{r}\frac{\partial f_{\kappa }}{\partial \theta }-\frac{\exp (i\kappa r)}{r}\frac{1}{r}\frac{\exp (-i\kappa r)}{r}{f}_{\kappa }\frac{\partial f_{\kappa }^{\ast }}{\partial \theta })\\ \text{(2)}& & =\frac{\hslash }{\mu }\frac{1}{r^3}\mathrm{\Im }\left\{f_{\kappa }^{\ast }\frac{\partial f_{\kappa }}{\partial \theta }\right\}\text{(3)}& {\overrightarrow{J}}_{sc}\cdot \hat{\phi }& =\frac{\hslash }{\mu }\frac{1}{r^3\sin \theta }\mathrm{\Im }\left\{f_{\kappa }^{\ast }\frac{\partial f_{\kappa }}{\partial \phi }\right\}.\end{array}$$

Note that since we are far away, the radial component is the most important scattering current contribution.

$dn=F_i\varsigma d\mathrm{\Omega }={\overrightarrow{F}}_{SC}d\overrightarrow{S}$.

$F_i=\mathbb{C}|{\overrightarrow{J}}_i|=\mathbb{C}\frac{\hslash \kappa }{\mu }$.

So,

$$\begin{array}{rl}dn& ={\overrightarrow{F}}_{SC}d\overrightarrow{S}\\ & =\mathbb{C}{\overrightarrow{J}}_{SC}\cdot \hat{r}(r^2\mathrm{\Omega })\\ & =\frac{F_i\mu }{\hslash k}{\overrightarrow{J}}_{SC}\cdot \hat{r}{r}^{2}d\mathrm{\Omega }\\ & =\frac{F_i\mu }{\hslash k}(\frac{\hslash k}{\mu }\frac{1}{r^2}|f_{\kappa }(\theta ,\phi ){|}^2)r^{2}d\mathrm{\Omega }\\ & =F_i|f_{\kappa }(\theta ,\phi ){|}^{2}d\mathrm{\Omega }.\\ \varsigma =\sigma (\theta ,\phi )& =|f_{\kappa }(\theta ,\phi ){|}^2.\end{array}$$

Then, $[{\mathrm{\nabla }}^2+{\kappa }^2-v(r)]{\mathrm{\Psi }}_{\overrightarrow{\kappa }}(\overrightarrow{r})=0$ can be solved with the Method of Green’s Function. C.f. ${\mathrm{\nabla }}_r^2\mathrm{\Phi }=\int {\mathrm{\nabla }}_r^{2}G(\overrightarrow{r}-{\overrightarrow{r}}^{\prime })\rho ({\overrightarrow{r}}^{\prime })d^3{\overrightarrow{r}}^{\prime }=-4\pi \rho$ solves ${\mathrm{\nabla }}^2\mathrm{\Phi }=-4\pi p$. We have $({\mathrm{\nabla }}^2+{\kappa }^2)G(\overrightarrow{r})={\delta }^{(3)}(\overrightarrow{r})$. So, ${\phi }_{\kappa }(\overrightarrow{r})={\phi }_0(\overrightarrow{r})+\int G(\overrightarrow{r}-{\overrightarrow{r}}^{\prime })v({\overrightarrow{r}}^{\prime }){\phi }_{\kappa }(r^{\prime })d^3{\overrightarrow{r}}^{\prime }$.

Using Fourier methods we would arrive at, ${\overrightarrow{G}}_{\pm }(\overrightarrow{r})=-\frac{1}{4\pi }\frac{\exp (\pm i\kappa r)}{r}$.

Since ${\phi }_{\overrightarrow{\kappa }}\sim \exp (i\kappa z)+f_{\kappa }(\theta ,\phi )\frac{\exp (i\kappa r)}{r}$, we choose $G_{+}(\overrightarrow{r})$ and find, ${\phi }_{\kappa }(\overrightarrow{r})=\exp (i\kappa z)-\int \frac{1}{4\pi }\frac{\exp (i\overrightarrow{\kappa }\cdot |\overrightarrow{r}-{\overrightarrow{r}}^{\prime }|)}{|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }|}v(r^{\prime })f_{\kappa }(\theta ,\phi )\frac{\exp (i\kappa r)}{r}{d}^3\overrightarrow{r}$. Note that $|{\overrightarrow{r}}^{\prime }|\ll |\overrightarrow{r}-{\overrightarrow{r}}^{\prime }|$ so $|\overrightarrow{r}-\overrightarrow{r}{|}^2=r^2+r^{\prime 2}-2r{r}^{\prime }\cos \alpha \approx r^2-2r{r}^{\prime }\cos \alpha =r^2(1-2\frac{r^{\prime }}{r}\cos \alpha )$. Then, $|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }|\approx r(1-\frac{r^{\prime }}{r}\cos \alpha )=r-r^{\prime }\cos \alpha =r-{\overrightarrow{r}}^{\prime }\cdot \hat{u}$

From the worksheet, we find $f_{\kappa }(\theta ,\phi )=-\frac{1}{4\pi }\int \exp (-i\kappa \hat{u}\cdot {\overrightarrow{r}}^{\prime })v({\overrightarrow{r}}^{\prime }){\phi }_k({\overrightarrow{r}}^{\prime })d^3{\overrightarrow{r}}^{\prime }$. Let $\overrightarrow{k}={\overrightarrow{k}}_d-{\overrightarrow{k}}_i$, ${\overrightarrow{k}}_d=\kappa \hat{u}$ is the detected wave. This is called the scattered/transferred wave vector.

Recall the dyson series, now we can see, ${\phi }_{\kappa }({\overrightarrow{r}}^{\prime })=\exp (i{\overrightarrow{k}}_i\cdot {\overrightarrow{r}}^{\prime })+\int d^3{\overrightarrow{r}}^{″}{G}_{+}({\overrightarrow{r}}^{\prime }-{\overrightarrow{r}}^{″})v({\overrightarrow{r}}^{″}){\phi }_k({\overrightarrow{r}}^{″})$. Doing this insertion multiple times, like we did with the Dyson series,

$$\begin{array}{rl}{\phi }_{\kappa }(\overrightarrow{r})& =\exp (i{\overrightarrow{k}}_i\cdot \overrightarrow{r})+\int d^3{\overrightarrow{r}}^{\prime }{G}_{+}(\overrightarrow{r}-{\overrightarrow{r}}^{\prime })v({\overrightarrow{r}}^{\prime })(\exp (i{\overrightarrow{k}}_i\cdot {\overrightarrow{r}}^{\prime })+\int d^3{\overrightarrow{r}}^{″}{G}_{+}({\overrightarrow{r}}^{\prime }-{\overrightarrow{r}}^{″})v({\overrightarrow{r}}^{″}){\phi }_{\kappa }({\overrightarrow{r}}^{″}))\\ & =\exp (i{\overrightarrow{k}}_i\cdot \overrightarrow{r})+\int d^3{\overrightarrow{r}}^{\prime }{G}_{+}(\overrightarrow{r}-{\overrightarrow{r}}^{\prime })v({\overrightarrow{r}}^{\prime })(\exp (i{\overrightarrow{k}}_i\cdot {\overrightarrow{r}}^{\prime }))\\ \text{(4)}& & +\int d^3{\overrightarrow{r}}^{\prime }{G}_{+}(\overrightarrow{r}-{\overrightarrow{r}}^{\prime })v({\overrightarrow{r}}^{\prime })(\int d^3{\overrightarrow{r}}^{″}{G}_{+}({\overrightarrow{r}}^{\prime }-{\overrightarrow{r}}^{″})v({\overrightarrow{r}}^{″}){\phi }_{\kappa }({\overrightarrow{r}}^{″}))\end{array}$$

This is called the Born Expansion, when you stop with the first term. It is valid when future terms are smaller than the previous ones. Hence the series terms are monotomically decreasing after a certain point.

So,

$$\begin{array}{}\text{(5)}& {\phi }_{\kappa }(\overrightarrow{r})& \approx \exp (i{\overrightarrow{k}}_i\cdot \overrightarrow{r})+\int d^3{\overrightarrow{r}}^{\prime }{G}_{+}(\overrightarrow{r}-{\overrightarrow{r}}^{\prime })v({\overrightarrow{r}}^{\prime })(\exp (i{\overrightarrow{k}}_i\cdot {\overrightarrow{r}}^{\prime }))\end{array}$$

Then, $f_{\kappa }^{(B)}=-\frac{1}{4\pi }\int \exp (-i\kappa \hat{u}\cdot {\overrightarrow{r}}^{\prime })v({\overrightarrow{r}}^{\prime })\exp (i{\overrightarrow{k}}_i\cdot {\overrightarrow{r}}^{\prime })d^3{\overrightarrow{r}}^{\prime }=-\frac{1}{4\pi }\int \exp (-i\overrightarrow{k}\cdot {\overrightarrow{r}}^{\prime })v(\overrightarrow{r^{\prime }})d^3{\overrightarrow{r}}^{\prime }$ where $\overrightarrow{k}\equiv {\overrightarrow{k}}_d-{\overrightarrow{k}}_i$.

Then, ${\varsigma }^{(b)}\equiv {\sigma }^{(B)}(\theta ,\phi )=|f_{\kappa }(\theta ,\phi ){|}^2=\frac{{\mu }^2}{4{\pi }^2{\hslash }^2}{|\int d^3\overrightarrow{r}\exp (-i\overrightarrow{k}\cdot \overrightarrow{r})V(\overrightarrow{r})|}^2$.

Example: $V(r)=\frac{A}{r^2}$. Then,

$$\begin{array}{rl}\int d^3\overrightarrow{r}\frac{\exp (-ikr\cos \theta )}{r^2}& =\int dr{r}^{2}d\mathrm{\Omega }\frac{\exp (-ikr\cos \theta )}{r^2}\\ & =\int drd\mathrm{\Omega }\exp (-ikr\cos \theta )\\ & =\int drd(\cos \theta )d\phi \exp (-ikr\cos \theta )\\ & =2\pi \int drd(\cos \theta )\exp (-ikr\cos \theta )\\ & =2\pi \int dr\frac{\exp (-ikr)-\exp (ikr)}{ikr}\\ & =-4\pi {\int }_0^{\mathrm{\infty }}dr\frac{\sin kr}{ikr}\\ f_{\kappa }^{(B)}& =-\frac{\pi \mu A}{k{\hslash }^2}\\ & =-\frac{\pi \mu A}{\sqrt{k_i^2+k_d^2-2k_i{k}_d\cos \theta )}{\hslash }^2}\\ & =-\frac{\pi \mu A}{\sqrt{2k^2(1-\cos \theta )}{\hslash }^2}\\ \text{(6)}& & =-\frac{\pi \mu A}{2k\sin \frac{\theta }{2}{\hslash }^2}.\text{(7)}& {\varsigma }^{(B)}& =\frac{{\pi }^2{\mu }^2{A}^2}{4{\hslash }^4{k}^2{\sin }^2\frac{\theta }{2}}.\end{array}$$

This approximation then works better for higher energy since the perturbative series would be smaller, and thus more like a perturbation. Also, ’small’ $A$.

From before, we had the CSCO: $\{H,L^2,L_z\}$.

For a central potential, $H=H_0+V(r)$, $H|k\ell m⟩=E|k\ell m⟩$, $L^2|k\ell m⟩={\hslash }^2\ell (\ell +1)|k\ell m⟩$, $L_z|k\ell m⟩=\hslash m|k\ell m⟩$.

For waves, we have $E=\frac{{\hslash }^2{k}^2}{2\mu }$. Our wavefunction is then ${\mathrm{\Psi }}_{k,\ell m}=\frac{u_{k,\ell }}{r}{Y}_{\ell }^m$. The radial wave function is then, $(-\frac{{\hslash }^2}{2\mu }\frac{d^2}{d{r}^2}+\frac{{\hslash }^2\ell (\ell +1)}{2\mu r^2}+V(r))u_{k,\ell }=\frac{{\hslash }^2{k}^2}{2\mu }$, $(-\frac{{\hslash }^2}{2\mu }\frac{d^2}{d{r}^2}+V_{eff}(r))u_{k,\ell }=\frac{{\hslash }^2{k}^2}{2\mu }$.

Far away, we throw away $V_{eff}$ as it is well localized for $V(r)\sim \frac{1}{r^n}$ with $n>1$. This gives the waves, $u_{k,\ell }(r)=A_{\pm }\exp (\pm ikr)$. Note that $A_{-}$ is the incident and $A_{+}$ is the scattered wave. Since we don’t lose any particles, no transmission or absorption, $|A_{+}|=|A_{-}|$. So, $u_{k,\ell }\sim |A|(\exp (ikr)\exp (i{\phi }_R)+\exp (-ikr)\exp (i{\phi }_I))=\mathbb{C}\sin (kr-\beta \ell )$ with ${\beta }_{\ell }=\frac{\pi }{2}-\frac{{\phi }_R-{\phi }_I}{2}$.

Then our asymptotic waves are then ${\mathrm{\Psi }}_{k,\ell ,m}\sim \mathbb{C}\frac{\sin (kr-{\beta }_{\ell })}{r}{Y}_{\ell }^m(\theta ,\phi )$.

What if $V=0$. Ansatz: Our phase is going to be exactly opposite, ${\varphi }^{\prime }=\pi -\varphi$, which is the case for slow particles. Our solution from the spherical box, without the boundary conditions, ${\mathrm{\Psi }}_{k\ell m}^{(0)}(r)=\sqrt{\frac{2}{\pi }}k{j}_{\ell }(kr)Y_{\ell }^m(\theta ,\phi )$. The asymptotic limit gives ${\mathrm{\Psi }}_{k\ell m}^{(0)}(r)=\sqrt{\frac{2}{\pi }}\frac{\sin (kr-\frac{\ell \pi }{2})}{r}{Y}_{\ell }^m(\theta ,\phi )$. Thus, ${\beta }_{\ell }=\frac{\ell \pi }{2}$. Compare with ${\beta }_{\ell }=\frac{\pi }{2}-\frac{{\phi }_R-{\phi }_I}{2}$ we see that the phase shift is then, ${\phi }_I-{\phi }_R=(\ell -1)\pi$. Set ${\beta }_{\ell }=\frac{\ell \pi }{2}+{\delta }_{\ell }$, where the ${\delta }_{\ell }$ is the phase shift of the partial wave.

Recall from before, ${\phi }_k\sim \exp (ikz)+f_k(\theta ,\phi )\frac{\exp (ikr)}{r}$. The first term is the free wave, ${\mathrm{\Psi }}_{k\ell m}^{(0)}$. The whole thing is related to ${\mathrm{\Psi }}_{k\ell m}$.

Then, we want to find $f_k(\theta ,\phi )$ in terms of ${\delta }_{\ell }$ to get ${\sigma }_{tot}$.

Note, $\exp (ikz)=\sum _{\ell =0}^{\mathrm{\infty }}{i}^{\ell }\sqrt{4\pi (2\ell +1)}{j}_{\ell }(kr)y_{\ell }^0(\theta ,\phi )=\sum _{\ell }{c}_{\ell }{\psi }_{k\ell 0}^{(0)}(\overrightarrow{r})$. Also, ${\phi }_k\sim \sum _{\ell }{\tilde{c}}_{\ell }{\psi }_{k,\ell ,0}(\overrightarrow{r})$.

Then, we have ${\mathrm{\Psi }}_{k,\ell ,m}\sim \tilde{c}\frac{\sin (kr-\frac{\ell \pi }{2}+{\delta }_{\ell })}{r}{y}_{\ell }^m$ and ${\mathrm{\Psi }}_{k,\ell ,m}\sim c\frac{\sin (kr-\ell \pi /2)}{r}{y}_{\ell }^m$ note that $\tilde{c}=c\exp (i{\delta }_{\ell })$.

Starting with

$$\begin{array}{rl}{\phi }_k\sim \sum _{\ell }{\tilde{c}}_{\ell }{\psi }_{k\ell 0}(\overrightarrow{r})& {=}^{r\to \mathrm{\infty }}\sum _{\ell }{i}^{\ell }\sqrt{4\pi (2\ell +1)}{y}_{\ell }^0\exp (i{\delta }_{\ell })\frac{\sin (kr-\ell \pi /2+{\delta }_{\ell })}{kr}\\ & =\sum _{\ell }{i}^{\ell }\sqrt{4\pi (2\ell +1)}{Y}_{\ell }^0(\theta )\exp (i{\delta }_{\ell })\frac{\exp (i(kr-\ell \pi /2))\exp (i{\delta }_{\ell })-\exp (-i(kr-\ell \pi /2))\exp (-i{\delta }_{\ell })}{2irk}\\ & =\sum _{\ell }{i}^{\ell }\sqrt{4\pi (2\ell +1)}{Y}_{\ell }^0(\theta )\frac{\exp (i(kr-\ell \pi /2))\exp (2i{\delta }_{\ell })-\exp (-i(kr-\ell \pi /2))}{2irk}\\ & =\sum _{\ell }{i}^{\ell }\sqrt{4\pi (2\ell +1)}{Y}_{\ell }^0(\theta )\frac{\exp (i(kr-\ell \pi /2))(1+2i\exp (i{\delta }_{\ell })\sin {\delta }_{\ell })-\exp (-i(kr-\ell \pi /2))}{2irk}\\ & =\sum _{\ell }{i}^{\ell }\sqrt{4\pi (2\ell +1)}{Y}_{\ell }^0(\theta )(\frac{\exp (i(kr-\ell \pi /2))-\exp (-i(kr-\ell \pi /2))}{2irk}\\ & +2i\exp (i{\delta }_{\ell })\sin {\delta }_{\ell }\frac{\exp (i(kr-\ell \pi /2))-\exp (-i(kr-\ell \pi /2))}{2irk})\\ & =\sum _{\ell }{i}^{\ell }\sqrt{4\pi (2\ell +1)}{Y}_{\ell }^0(\theta )(\frac{\sin (kr-\ell \pi /2)}{kr}\\ & +2i\exp (i{\delta }_{\ell })\sin {\delta }_{\ell }\frac{\exp (i(kr-\ell \pi /2))-\exp (-i(kr-\ell \pi /2))}{2irk})\\ & =\exp (ikz)+\sum _{\ell }{i}^{\ell }\sqrt{4\pi (2\ell +1)}{Y}_{\ell }^0(\theta )(2i\exp (i{\delta }_{\ell })\sin {\delta }_{\ell }\frac{\exp (i(kr-\ell \pi /2))-\exp (-i(kr-\ell \pi /2))}{2irk})\\ & =\exp (ikz)+\sum _{\ell }{i}^{\ell }\sqrt{4\pi (2\ell +1)}{Y}_{\ell }^0(\theta )\frac{(-i{)}^{\ell }}{k}\exp (i{\delta }_{\ell })\sin {\delta }_{\ell }\\ \text{(8)}& & =\exp (ikz)+\sum _{\ell }\sqrt{4\pi (2\ell +1)}{Y}_{\ell }^0(\theta )\frac{1}{k}\exp (i{\delta }_{\ell })\sin {\delta }_{\ell },\text{(9)}& & =\exp (ikz)+\frac{\exp (ikr)}{r}{f}_k(\theta ).\end{array}$$

From before, ${\phi }_{k\ell m}\sim C\exp (i{\delta }_{\ell })\frac{\sin (kr-\ell \pi /2+{\delta }_{\ell })}{r}{Y}_{\ell }^m(\theta ,\phi )$. We found, $f(\theta ,\phi )=\frac{1}{k}\sum _{\ell }\sqrt{4\pi (2\ell +1)}{Y}_{\ell }^0(\theta )\exp (i{\delta }_{\ell })\sin {\delta }_{\ell }$.

For $r>r_0$,

$$\begin{array}{rl}(\frac{d^2}{d{r}^2}+k^2-\frac{\ell (\ell +1)}{r^2})u_{k,\ell }& =0\\ {\phi }_{k,\ell ,m}& =\frac{u_{k,\ell }(r)}{r}{Y}_{\ell }^m(\theta ,\phi ).\end{array}$$

We have,

$$\begin{array}{}\text{(10)}& u_{k,\ell }(r)=kr({\tilde{C}}_{1\ell }{j}_{\ell }(kr)+{\tilde{C}}_{2\ell }{n}_{\ell }(kr)).\end{array}$$

Our boundary condition now is,

$$\begin{array}{}\text{(11)}& {\phi }_{k\ell m}(r_0)=0.\end{array}$$

Then, $\frac{{\tilde{C}}_{1\ell }}{{\tilde{C}}_{2\ell }}=-\frac{n_{\ell }(k{r}_0)}{j_{\ell }(k{r}_0)}$.

At $r\to \mathrm{\infty }$, $u_{k,\ell }\sim {\tilde{C}}_{1\ell }\sin (kr-\ell \pi /2)+{\tilde{C}}_{2\ell }\cos (kr-\ell \pi /2)={\tilde{C}}_{1\ell }\sin (kr-\ell \pi /2)-{\tilde{C}}_{1\ell }\frac{j_{\ell }(k{r}_0)}{n_{\ell }(k{r}_0)}\cos (kr-\ell \pi /2)$

Note, $\sin (\alpha +\delta )=\sin \alpha \cos \delta +\cos \alpha \sin \delta$.

Then, $u_{k,\ell }\sim \sqrt{{\tilde{C}}_{1\ell }+{\tilde{C}}_{2\ell }}\sin (kr-\ell \pi /2+{\delta }_{\ell })$ with $\tan {\delta }_{\ell }=-\frac{{\tilde{C}}_{2\ell }}{{\tilde{C}}_{1\ell }}=\frac{j_{\ell }(k{r}_0)}{n_{\ell }(k{r}_0)}$.

For $\ell =0$, ${\delta }_0={\tan }^{-1}\frac{j_0(k{r}_0)}{n_0(k{r}_0)}={\tan }^{-1}\frac{\sin (k{r}_0)/k{r}_0}{-\cos (k{r}_0)/k{r}_0}=-k{r}_0$.

For slow particles, $k{r}_0\ll 1$. For $\rho \to 0$, $j_{\ell }\to \frac{{\rho }^{\ell }}{(2\ell +1)!!},n_{\ell }\to \frac{(2\ell +1)!!}{{\rho }^{\ell +1}}$.

Then for slow particles, ${\delta }_{\ell }={\tan }^{-1}(-\frac{(k{r}_0{)}^{2\ell +1}}{((2\ell +1)!!{)}^2})$. Hence, the $\ell =0$ term dominates. So, ${\sigma }_{tot}\approx \frac{4\pi }{k^2}{\sin }^2{\delta }_0\approx 4\pi r_0^2$. Thus, the area of the sphere behaves as the cross section. Hence, the particle ’sees’ the entire area of the sphere. Also, we see that the cross section is dependent on the energy. Higher energy will see less cross section.

Classically, we would expect a cross section of $\pi r^2$.

Colliding two identical particles. Detector 1 is at an angle $\theta$ and detector 2 is at an angle $\pi -\theta$.

We then need to symmetrize/ antisymmetrize our wavefunctions. Then, ${\phi }_k\sim \exp (ikz)=f_k(\theta )\exp (ikr)/r$ and $z=|z_1-z_2|$ and $k=\overrightarrow{k}\cdot |{\overrightarrow{x}}_1-{\overrightarrow{x}}_2|$. So, ${\mathrm{\Psi }}_{S,A}\sim \exp (ikz)\pm \exp (-ikz)+(f_k(\theta )\pm f(\pi -\theta ))\frac{\exp (ikr)}{r}$.

Then the differential cross section is ${\left(\frac{d\sigma }{d\mathrm{\Omega }}\right)}_{boson,fermion}=|f(\theta ){|}^2+|f(\pi -\theta ){|}^2\pm 2\mathrm{\Re }[f^{\ast }(\theta )f(\pi -\theta )]$. The last term is the interference term between the identical particles. Classically we would have the first two terms.

For a spin-1/2 fermion, $\left(\frac{d\sigma }{d\mathrm{\Omega }}\right)=\frac{3}{4}{\left(\frac{d\sigma }{d\mathrm{\Omega }}\right)}_A+\frac{1}{4}{\left(\frac{d\sigma }{d\mathrm{\Omega }}\right)}_S=\frac{3}{4}|f(\theta )-f(\pi -\theta ){|}^2+\frac{1}{4}|f(\theta )+f(\pi -\theta ){|}^2=|f(\theta ){|}^2+|f(\pi -\theta ){|}^2-\mathrm{\Re }[f^{\ast }(\theta )f(\pi -\theta )]$.

For $\theta =\frac{\pi }{2}$, ${\left(\frac{d\sigma }{d\mathrm{\Omega }}\right)}_{Classical}=2|f\left(\frac{\pi }{2}\right){|}^2$. For $\theta =\frac{\pi }{2}$, ${\left(\frac{d\sigma }{d\mathrm{\Omega }}\right)}_{Spinlessboson}=2|f\left(\frac{\pi }{2}\right){|}^2+2\text{Re}[f^{\ast }(\theta )f(\pi -\theta )|=4|f\left(\frac{\pi }{2}\right)|$. For $\theta =\frac{\pi }{2}$, ${\left(\frac{d\sigma }{d\mathrm{\Omega }}\right)}_{Spin-1/2}=2|f\left(\frac{\pi }{2}\right){|}^2-\text{Re}[f^{\ast }(\theta )f(\pi -\theta )|=|f\left(\frac{\pi }{2}\right){|}^2$.

For a spin-1 boson, $\left(\frac{d\sigma }{d\mathrm{\Omega }}\right)=\frac{1}{4}{\left(\frac{d\sigma }{d\mathrm{\Omega }}\right)}_A+\frac{3}{4}{\left(\frac{d\sigma }{d\mathrm{\Omega }}\right)}_S=\frac{1}{4}|f(\theta )-f(\pi -\theta ){|}^2-\frac{3}{4}|f(\theta )+f(\pi -\theta ){|}^2=|f(\theta ){|}^2+|f(\pi -\theta ){|}^2+\mathrm{\Re }[f^{\ast }(\theta )f(\pi -\theta )]$