## Common functions expressed as Hypergeometric function

General Hypergeometric function can be expressed in power series

$\displaystyle {}_pF_q(a_1, a_2,... a_p ; b_1, b_2, ... b_q; z) = \sum_n \frac{(a_1)_n(a_2)_n ... (a_p)_n}{(b_1)_n (b_2)_n ... (b_q)_n} \frac{z^n}{n!}$

where $(a)_n$ is Pochhammar symbol,

$\displaystyle (a)_n = \frac{\Gamma(a+n)}{\Gamma(a)} = a(a+1)...(a+n-1)$

The General hypergeometric function satisfies the following differential equation,

$\displaystyle \frac{d}{dz}[(\theta_{b_1}-1)(\theta_{b_2}-1)... (\theta_{b_q}-1)]y = [\theta_{a_1}\theta_{a_2}...\theta_{a_p}] y$

where

$\displaystyle \theta_{a} = z\frac{d}{dz} + a$

For $p = q = 0$, the differential equation becomes

$\displaystyle \frac{d}{dz} y = y \Rightarrow y = {}_2F_1(;;z) = \exp(z)$

For $p = 0, q = 1$,

$\displaystyle \frac{d}{dz}\left( z\frac{d}{dz} + c -1 \right) y = y \Rightarrow \displaystyle z\frac{d^2y}{dz^2} + c \frac{dy}{dz} -y = 0$

For $p = 1, q = 0$

$\displaystyle \frac{d}{dz} y= \left(z\frac{d}{dz}+a \right)y \Rightarrow \displaystyle (z-1)\frac{d}{dz} y + ay = 0$

For $p = 1 = q$

$\displaystyle \frac{d}{dz}\left( z\frac{d}{dz} + c -1 \right) y = \left(z\frac{d}{dz}+a \right) y \Rightarrow \displaystyle z\frac{d^2y}{dz^2} + (c-z) \frac{dy}{dz} - ay = 0$

The Gauss Hypergeometric function is $p = 2, q = 1$,

$\displaystyle {}_2F_1(a,b;c;z) =\sum_n \frac{(a)_n(b)_n}{(c)_n} \frac{z^n}{n!}$

which satisfies,

$\displaystyle x(1-x) \frac{d^2y}{dx^2} + (c - (a+b+1)x)\frac{dy}{dx} - aby = 0$

There are some interesting expression for Pochhammar symbol

$\displaystyle (-n)_{k} = (-n)(-n+1)...(-n+k-1) \\ = (-1)^k (n)(n-1)...(n-k+1) \\ = (-1)^k \frac{n!}{(n-k)!}$

when $k = n$

$(-n)_n = (-1)^n n!$

when $k = n + r, r>0$

$(-n)_{n+r} = 0$

Here are list of common function into hypergeometric function

${}_0F_0(; ; z) = e^z$

${}_1F_0(-a; -z) = (1+z)^a$

$\displaystyle {}_0F_1\left(;\frac{1}{2}; -\frac{z^2}{4} \right) = \cos(z)$

$\displaystyle {}_0F_1\left(;\frac{3}{2}; -\frac{z^2}{4} \right) = \frac{1}{z} \sin(z)$

$\displaystyle {}_0F_1\left(;a+1; -\frac{z^2}{4} \right) = \frac{2^a}{z^a} \Gamma(a+1) J_a(z)$

where $J_a(z)$ is Bessel function of first kind, which satisfies

$\displaystyle z^2 \frac{d^2y}{dz^2} + z \frac{dy}{dz} + (z^2 - a^2)y = 0$

$\displaystyle {}_0F_1\left(; \frac{1}{2}; \frac{z^2}{4} \right) = \cosh(x)$

$\displaystyle {}_0F_1\left(;\frac{3}{2}; \frac{z^2}{4} \right) = \frac{1}{z} \sinh(z)$

$\displaystyle {}_0F_1\left(;a+1; \frac{z^2}{4} \right) = \frac{2^a}{z^a} \Gamma(a+1) I_a(z)$

where $I_a(z)$ is modified Bessel function of first kind, which satisfies

$\displaystyle z^2 \frac{d^2y}{dz^2} + z \frac{dy}{dz} - (z^2 + a^2)y = 0$

$\displaystyle {}_1F_1\left(\frac{1}{2}; \frac{3}{2}; -z^2 \right) = \frac{\sqrt{\pi}}{2z} Erf(z)$

where $Erf(z)$ is error function

$Erf(z) = \int_0^z \exp(-t^2) dt$

$\displaystyle {}_2F_1\left(-a,a; \frac{1}{2}; \sin^2(z) \right) = \cos(2az)$

$\displaystyle {}_2F_1\left(\frac{1}{2}+a, \frac{1}{2}-a; \frac{3}{2}; \sin^2(z) \right) = \frac{\sin(2az)}{2a \sin(z)}$

$\displaystyle {}_2F_1(1,1;2;-z) = \frac{1}{z} \log_e(z+1)$

$\displaystyle {}_2F_1(\frac{1}{2},-1;\frac{a}{2};z) = 1- \frac{z}{a}$

$\displaystyle {}_2F_1\left( \frac{1}{2}, 1; \frac{3}{2}; z^2 \right) = \frac{1}{2z} \log_e \left( \frac{1+z}{1-z} \right) = \frac{1}{z} \tanh^{-1}(z)$

$\displaystyle {}_2F_1 \left( \frac{1}{2}, 1; \frac{3}{2} ; -z^2 \right) = \frac{1}{z} \tan^{-1}(z)$

$\displaystyle {}_2F_1 \left( \frac{1}{2}, \frac{1}{2}; \frac{3}{2} ; z^2 \right) = \frac{1}{z}\sin^{-1}(z)$

$\displaystyle {}_2F_1 \left( \frac{1}{2}, \frac{1}{2}; \frac{3}{2} ; -z^2 \right) = \frac{1}{z}\sinh^{-1}(z)$

$\displaystyle {}_2F_1 \left( \frac{1}{2}, \frac{1}{2}; \frac{3}{2} ; \frac{1-z}{2} \right) = \frac{1}{\sqrt{2(1-z)}}\cos^{-1}(z)$

$\displaystyle {}_2F_1\left(-n, n+1; 1; \frac{1-z}{2} \right) = P_n(z)$

where $P_n(z)$ is Legendre function, which satisfies

$\displaystyle (1-z^2)\frac{d^2y}{dz^2} -2z \frac{dy}{dz} + n(n+1) y = 0$

$\displaystyle {}_2F_1\left(m-n,m+n+1; m+1; \frac{1-z}{2} \right) \\= (-1)^m\frac{(n-m)!m!2^m}{(n+m)!(1-x^2)^{\frac{m}{2}}} P_n^m(z), m\geq0$

where $P_n^m(z)$ is associate Legendre function, which satisfies

$\displaystyle (1-z^2)\frac{d^2y}{dz^2} -2z \frac{dy}{dz} + \left(n(n+1) -\frac{m^2}{1-z^2} \right)y = 0$

$\displaystyle {}_2F_1\left(\frac{1}{2}, \frac{1}{2}; 1; z^2 \right) = \frac{2}{\pi} K(z)$

where $K(z)$ is complete elliptic integral of 1st kind

$\displaystyle K(z) = \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-z^2 \sin^2(t)}} dt$

$\displaystyle {}_2F_1\left(-\frac{1}{2}, \frac{1}{2}; 1; z^2 \right) = \frac{2}{\pi} E(z)$

and $E(z)$ is complete elliptic integral of 2nd kind

$\displaystyle E(z) = \int_0^{\frac{\pi}{2}} \sqrt{1-z^2 \sin^2(t)} dt$

Reference

“Notes on hypergeometric functions” by John D. Cook (April 10, 2003)
“Generalized Hypergeometric Series” by W. N. Bailey, Cambridge (1935)
“Handbook of Mathematical Functions” by Abramowitz and Stegun (1964)
“The special functions and their approximations” by Yudell L. Luke v. 1 (1969)
“Concrete Mathematics” by Graham, Knuth, and Patashnik (1994)

In Wolfram research (http://functions.wolfram.com/functions.html), many functions are listed. We can click to a function, then we click “Representations through more general functions”, then “Through hypergeometric functions”, then we can see how the function looks like.

## Very short introduction to Partial-wave expansion of scattering wave function

In a scattering problem, the main objective is solving the Schrödinger equation

$H\psi=(K+V)\psi=E\psi$

where H is the total Hamiltonian of the scattering system in the center of momentum, K is the kinetic energy and V is the potential energy. We seek for a solution $\psi$,

$\displaystyle \psi_{k}^{+}(r)=e^{i\vec{k}\cdot \vec{r}}+f(\theta)\frac{e^{ikr}}{kr}$

The solution can be decomposed

$\displaystyle \psi_{k}^{+}(r)=R_{l}(k,r)Y_{lm}(\theta,\phi)=\frac{u_{l}(k,r)}{kr}Y_{lm}(\theta,\phi)$

The solution of $u_{l}(k,r)$ can be solve by Runge-Kutta method on the pdf

$\displaystyle \left(\frac{d^2}{d\rho^2} + 1 - \frac{l(l+1)}{\rho^2} \right)u_{l}(k,\rho)=U(\rho)u_{l}(k,\rho)$

where $\rho=kr, k=\sqrt{2\mu E}/\hbar, \mu=(m_1+m_2)/(m_1 m_2)$ and $U=V/E$.

For $U = 0$, the solution of $u_l$ is

$\displaystyle u_{l}(k,r)=\hat{j}_l(\rho) \xrightarrow{r\rightarrow \infty} \sin(r') = \frac{e^{ir'}-e^{-ir'}}{2i}$

where $r' = kr-l\pi/2$ and $\hat{j}_l$ is the Riccati-Bessel function. The free wave function is

$\displaystyle \phi_k(r)=e^{i\vec{k}\cdot\vec{r}}=\sum\limits_{l=0} P_l(\cos(\theta)) \frac{2l+1}{2ikr}i^l (e^{ir'}-e^{-ir'})$

where $P_l(x)$ is the Legendre polynomial.

Note that, if we have Coulomb potential, we need to use the Coulomb wave instead of free wave, because the range of coulomb force is infinity.

For $U\neq 0$, the solution of $u_l(r can be found by Runge-Kutta method, where R is a sufficiency large that the potential $V$ is effectively equal to 0.  The solution of $u_l(r>R)$ is shifted

$\displaystyle u_{l}(k,r>R)=\hat{j}_l(\rho)+\beta_l \hat{n}_l(\rho) \xrightarrow{r\rightarrow \infty} \frac{1}{2i}(S_l e^{ir'}-e^{-ir'})$

where $S_l$ is the scattering matrix element, it is obtained by solving the boundary condition at $r = R$. The scattered wave function is

$\displaystyle \psi_k(r)=\sum\limits_{l=0} P_l(\cos(\theta)) (2l+1) i^l \frac{u_l(r)}{kr}$

put the scattered wave function and the free wave function back to the seeking solution, we have the $f(\theta)$

$\displaystyle f(\theta) = \sum\limits_{l=0} P_l(\cos(\theta)) \frac{2l+1}{2ik} (S_l - 1)$

and the differential cross section

$\displaystyle \frac{d\sigma}{d\Omega}=|f(\theta)|^2$.

In this very brief introduction, we can see

• How the scattering matrix $S_l$ is obtained
• How the scattering amplitude $f(\theta)$ relates to the scattering matrix

But what is scattering matrix? Although the page did not explained very well, especially how to use it.

## 3-D spherical infinite well

the potential is

$V(r,\theta,\phi) = \begin{pmatrix} 0 & |r|

The Laplacian in spherical coordinate is:

$\nabla^2 = \frac{d^2}{dr^2} + \frac{2}{r}\frac{d}{dr} - \frac{L^2}{r^2}$

since the $L$ is the reduced angular momentum operator, if we set the solution be:

$\psi(r,\theta,\phi) = R(r) Y_{lm}(\theta,\phi)$

Then the angular part was solved and the radial part becomes:

$L^2 Y_{lm} = l(l+1) Y_{ml}$

$\left(r^2 \frac{d^2}{dr^2} + 2 r \frac{d}{dr}+(k^2 r^2 - l(l+1))\right)R(r) = 0$

$k^2 = 2 m E/ \hbar^2$

The radial equation is the spherical Bessel function.

The solution was common written as:

$R(r) = j_l( k r) = \left( - \frac{r}{k} \right)^l \left(\frac{1}{ r} \frac{d}{dr}\right)^l \frac{sin(k r)}{kr}$

The Boundary condition fixed the k and then the energy,

$j_l ( k_{nl} a ) = 0$

the all possible root are notated as n. thus the quantum numbers for this system are:

• n , the order of root
• l , the angular momentum

We can see in here, the different between Coulomb potential and spherical infinite well:

• there is no restriction on n and l, therefore, there will be 1s, 1p, 1d, 1f orbit.
• the energy level also depend on angular momentum, since it determined the order of spherical Bessel function.

we can realized the energy level by the graph of Bessel function. we set some constants be 1, the root are :

$k_{nl} a = \frac{1}{\hbar} \sqrt{ 2 m a^2} \sqrt{E_{nl}} = \pi \sqrt{E_{nl}}$

Thus, we plot

$j_l( \pi \sqrt{E_{nl}})$

Since $k_{nl}$ is a scaling factor to “force” the function to be zero at the boundary. Interestingly, the spherical Bessel function is not normalizable or orthogonal with $r^2$, i.e.

$\int_{0}^{\infty} j_l(r) j_{l'}(r) r^2 dr$

is diverged. Of course, the spherical Bessel function is a “spherical wave” that propagating in space, same as plane wave, which is also not normalizable or orthogonal. However, in the infinite spherical well, with the scaling factor $k$ for same $l$, they are orthogonal! And for difference $l$, the spherical part are already orthogonal. Thus, the wave function for difference energy are orthogonal. i.e. the eigen states are orthogonal! Power of math! In the following plots, the left is $l = 0$ and the right is $l = 1$.

In fact, even for Woods-Saxon potential, the numerical solution of the eigen wave-function are also orthogonal for same angular momentum.

## Scattering phase shift

for a central potential, the angular momentum is a conserved quantity. Thus, we can expand the wave function by the angular momentum wave function:

$\sum a_l Y_{l , m=0} R_l(k, r)$

the m=0 is because the spherical symmetry. the R is the radial part of the wave function. and a is a constant. k is the linear momentum and r is the radial distance.

$R_l(k,r) \rightarrow J_{Bessel} (l, kr )$

which is reasonable when r is infinite and the nuclear potential is very short distance. when r goes to infinity,

$J_{Bessel} (l,kr) \rightarrow \frac {1}{kr} sin( k r - \frac{1}{2} l \pi )$

for elastic scattering, the probability of the current density is conserved in each angular wave function, thus,

the effect of the nuclear potential can only change the phase inside the sin function:

$\frac{1}{kr} sin( k r - \frac {1}{2} l \pi +\delta_l )$

with further treatment, the total cross section is proportional to $sin^2(\delta_l)$.

thus, by knowing the scattering phase shift, we can know the properties of the nuclear potential.

for more detail : check this website