## 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.