hypergeometric | Nuclear Physics 101

Asymptotic normalization coefficient

May 2, 2021

GoLuckyRyan Basic ANC, Coulomb, Hankel, hypergeometric, Whittaker Leave a comment

The Asymptotic normalization coefficient or the ANC is thought to be an alternative to the spectroscopic factor. As the name suggested, this is the coefficient for asymptote of the radial wave function.

For a single nucleon adding/removal reaction, the reaction probability is

$\displaystyle T^2 = \left|\left< \phi_b \Psi_B| V | \phi_a \Psi_A \right>\right|^2$

Assume the interaction $V$ only acts on the nucleon that being add or remove, thus

$\displaystyle \left< \Psi_B | \Psi_A \right> = \phi(r)$

which is the bound state wave function. Due to the nuclear interaction, the bound state wave function most probably not normalized. In other point of view, suppose the nucleus B = A + 1, the wave function of nucleus B could be

$\displaystyle \left| \Psi_B \right> = \beta_{00} [\phi_{0} \times \Psi_A(0) ]_{J_B} + ... + \beta_{ij} [\phi_{i} \times \Psi_A(j) ]_{J_B}$

where $\phi_i$ is an orbital, $\Psi_A(j)$ is the $j$ -th excited state of nucleus A, and the square bracket is the angular coupling, antisymmetric, and normalization operator. As the wave function of nucleus B must be normalized, Thus,

$\displaystyle \sum_{ij} \beta_{ij}^2 = 1$ .

And the bound state can be approximated

$\displaystyle \phi(r) \approx \beta_{00} \phi_{0}$ .

The spectroscopic factor of orbital $\phi_0$ in this A + 1 = B reaction is

$\displaystyle \beta_{00}^2 = \int r^2 \phi^2(r) dr$

At far away distance, the nucleus potential is very weak or effectively zero. The Schrodinger equation becomes

$\displaystyle \left( - \frac{\hbar^2}{2m} \nabla^2 + \frac{Q_1Q_2e^2}{r} \right) \Phi = E \Phi$

where $e^2 = 1.44~\textrm{MeV.fm}$ . The Coulomb potential is still here because it is a long range force. Separate the radial and angular part,

$\displaystyle \left( - \frac{\hbar^2}{2m} \frac{1}{r^2} \left( \frac{d}{dr} r^2 \frac{d}{dr} \right) + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2} + \frac{Q_1Q_2 e^2}{r} \right) R(r) = E R(r)$

This radial equation was solved before in this post. There are two solutions for $u(r) = R(r)/r$ , one is bound (in mathematics) and one is unbound. I state the mathematically-bounded solution in here

$\displaystyle F_l(r, \eta) = \frac{2^l e^{-\pi \eta /2} \left| \Gamma(l+1+i\eta) \right|}{(2l+1)!} (kr)^{l+1} e^{ikr} {_1F_1(l+1+i\eta, 2l+2, -2ikr)},$

$\displaystyle k^2 = \frac{2mE}{\hbar^2}$

$\displaystyle \eta = \frac{k}{2E} Q_1 Q_2 e^2 = \frac{1}{\hbar} \sqrt{\frac{m}{2E}} Q_1 Q_2 e^2$

There is another Coulomb wave function $G_l(r)$ , although there is no simple form of the function, the two Coulomb wave function, one behave like a sine wave ( $F_l(r)$ ) another one behave like a cosine wave ( $G_l(r)$ ), they can be combined into a complex function, the Coulomb Hankel function, (it is different from the Hankel function of the first and second kind )

$\displaystyle H_l^{\pm}(x, \eta) = D_l^{\pm} x^{l+1} e^{\pm i x} U(l+1\pm \eta, 2l+2, \mp 2ix)$

$\displaystyle D_l^{\pm} = (\mp 2i) ^{2l+1} \frac{\Gamma(l+1\pm i \eta)} {C_l }$

$\displaystyle C_l = \frac{2^l}{2^{\eta \pi /2}} \sqrt{\Gamma(l+1+i\eta) \Gamma(l+1-i\eta)}$

where $U(a,b,z)$ is the confluent hypergeometric function of the second kind. In Mathematica, the function is given is built-in

HypergeometrixU[l+1+i eta, 2l+2, -2 i r]

And

$\displaystyle F_l(r, \eta) = \frac{1}{2i} ( H_l^+(kr, \eta) - H_l^-(kr, \eta) )$

$\displaystyle G_l(r, \eta) = \frac{1}{2} ( H_l^+(kr, \eta) + H_l^-(kr, \eta) )$

We can see, the analogy

$\displaystyle F_l(x, \eta) \sim \sin(x) = \frac{1}{2i} ( e^{ix} - e^{-ix} ) \\ G_l(x, \eta) \sim \cos(x) = \frac{1}{2} ( e^{ix} + e^{-ix} )$

The long range (unbound, scattering) behaviour of the Coulomb wave function is

$\displaystyle F_l(r\rightarrow \infty, \eta) = \sin \left( k r - l \frac{\pi}{2} - \eta \log(2kr) + \sigma_l \right)$

$\displaystyle G_l(r\rightarrow \infty, \eta) = \cos \left( k r - l \frac{\pi}{2} - \eta \log(2kr) + \sigma_l \right)$

where $\sigma_l = \arg(\Gamma(l+1+i\eta))$ is the Coulomb phase shift.

The ANC $C$ is the coefficient between the bound state wave function and the Coulomb wave function at $r \rightarrow \infty$ . Since it is the bound state long range behaviour, $k = i \kappa$ , we have to use the Coulomb Hankel function.

For neutron, $\eta = 0$

$\displaystyle H_0^\pm(x, 0) = e^{\pm i x}$

We can see that, when $x = i \kappa r$ , the $latex H_0^\pm (i \kappa r, 0 ) = e^{-\kappa r} is a bounded and real solution.

This is the same for higher $l$ that the decay factor $e^{-\kappa r }$ appeared in the Hankel function.

In the following, we use the neutron and simplify the Coulomb wave function for few $l$ . When no charge, $\eta = 0$ . And since it is a bound state,

In fact, the solution for $\eta = 0$ is the Spherical Bessel function J. As the $k = i \kappa$ is pure imaginary, and we know that the Spherical Bessel function J and Y are unbound for pure imaginary position.

The solution is the Spherical Hankel function $h_l^\pm$ . This is the Coulomb Hankel function divided by the radius. In Mathematica, the built-in function is

SphericalHankelH1[l, i kappa r]

In fact,

$\displaystyle h_l^+(x) = \frac{-1}{x} H_l^+(x, 0) \\ h_l^-(x) = \frac{(-1)^l}{x} H_l^-(x,0)$

In this paper N.K. Timofeyuk, PRC 88, 044315 (2013), in equation 3, the Whittaker function should be the Whittaker W-function. In Mathematica

WhittakerW[ - i eta, l + 1/2, 2 kappa r ]

The difference between the Whittaker W-function and the Hankel function with complex argument is the normalization factor

$\displaystyle W_{-i \eta, l+1/2}(2 \kappa r ) = (2 \kappa r)^{l+1} e^{-\kappa r} U(l+1+i \eta, 2l+2, 2 \kappa r)$

Since the ANC is the proportional factor between the bound state wave function and the Whittaker W-function (or the Coulomb wave function, or the spherical Hankel function ), so the normalization factor is important.

As we know that the neutron should behave as Spherical Hankel function, thus, we checked that

$\displaystyle W_{0, l+1/2}(2 \kappa r) = (-1) i^l ( \kappa r ) h_{l}^+(i \kappa r) = \sqrt{\frac{2 \kappa r}{\pi}} K_{l+1/2}(\kappa r)$

where $K_{\alpha}$ is the modified Bessel function of the 2nd kind. Here are a list of the Whittaker W-function for $\eta = 0$

$\displaystyle W_{0, 1/2}( 2 x) = e^{-x}$

$\displaystyle W_{0, 3/2}( 2 x) = \frac{e^{-x}}{x} (1+x)$

$\displaystyle W_{0, 5/2}( 2 x) = \frac{e^{-x}}{x^2} (3 + 3x+x^2)$

$\displaystyle W_{0, 7/2}( 2 x) = \frac{e^{-x}}{x^3} (15 + 15x+ 6x^2 + x^3)$

$\displaystyle W_{0, 9/2}( 2 x) = \frac{e^{-x}}{x^4} (105 + 105x+ 45x^2 + 10x^3 + x^4)$

$\displaystyle W_{0, 11/2}( 2 x) = \frac{e^{-x}}{x^5} (945 + 945x+ 420x^2 + 105x^3 + 15x^4+x^5)$

For me, the problem of the ANC is that, the Coulomb Hankel function or the Whittaker W-function cannot be normalized. And without a proper normalization, how can we compare the magnitude of two wave functions??

For example, I calculated the 1s1/2 bound state wave function from a Woods-Saxon potential for neutron. The parameters are $V_0 = -50.4, R_0 = 3.1246, a_0 = 0.67, V_{SO} = 19.6, R_{SO} = 3.0238, a_{SO} = 0.66, m_\mu = 884.297$ , the energy is -2.227 MeV and $\kappa = 0.3181$ . And here is the comparison

$\displaystyle \phi(r) = C \frac{W_{0, 1/2}(2 \kappa r )}{\kappa r}$

The Woods-Saxon bound state is supposed to be a pure wave function that has SF = 1 or ANC = 1. But since the Whittaker cannot be normalized, and depends on the “definition”, where the $\kappa$ appears in the denominator or not, the ANC can be different. And for this example the ANC = 0.7, for a pure state.

Coulomb wave function (II)

April 17, 2020

GoLuckyRyan Basic Coulomb, Gamma Function, hypergeometric, phase shift, scattering 5 Comments

In the previous post, we tried to derived to Coulomb wave function, and the regular Coulomb wave function $F_l(x)$ is derived, except for the normalization constant, and the mysterious Coulomb phase shift.

From this arxiv article, the Coulomb “Hankel” function is

$\displaystyle H_L^{pm}(x) = D_L^{\pm} x^{L+1} e^{\pm i x} U(L+1 \pm i \eta, 2L+2, - \pm 2 i x)$

$\displaystyle D_L^{\pm} = (-\pm 2i)^{2L+1} \frac{\Gamma(L+1\pm i \eta)}{ C_L \Gamma(2L+2)}$

$\displaystyle C_L = 2^L \frac{\sqrt{\Gamma(L+1+ i \eta) \Gamma(L+1- i \eta)}}{ e^{\eta \pi/2}\Gamma(2L+2)}$

where $U(a,b,z)$ is the confluent hypergeometric function of the second kind.

The regular Coulomb wave function is

$\displaystyle F_L(x) = C_L x^{L+1} e^{\pm i x} ~_1F_1(L+1 \pm i \eta, 2L+2, - \pm 2 i x) = \frac{1}{2i} \left( H_L^+ - H_L^- \right)$

I am fail to prove the last equality. The irregular Coulomb wave function is

$\displaystyle G_L(x) = \frac{1}{2} \left( H_L^+ + H_L^- \right)$

The form of the coefficience $C_L$ is different from the form give in this post, but they are the same.

$\displaystyle \frac{\sqrt{\Gamma(L+1+ i \eta) \Gamma(L+1- i \eta)}}{ \Gamma(2L+2)} = \frac{\left| \Gamma(L+1+i \eta) \right|}{(2L+1)!}$

In Mathematica 11.2, the $U(a,b,z)$ is fail to evaluate when $\eta > 2$ for $l = 0$ , and $\eta > 0.1$ for $l = 5$ .

Update 20200420, settign WorkingPrecision to 45 can solve the problem. Here are some plots for the Coulomb wave functions with $\eta = 10$ .

Annotation 2020-04-20 223816.png

Annotation 2020-04-20 223849.png

Annotation 2020-04-20 224003.png

When $\eta$ getting larger, the wave function pushed further.

Annotation 2020-04-20 224017.png

Coulomb wave function

April 17, 2020

GoLuckyRyan Basic Coulomb, Gamma Function, hypergeometric, Legendre polynomial, phase shift, Riccati-Bessel, scattering Leave a comment

The Coulomb wave function is the solution of a REPULSIVE Coulomb potential. The Schrödinger equation is

$\displaystyle \left( -\frac{\hbar^2}{2m} \nabla^2 + \frac{Q_1Q_2 e^2}{r} \right) \Phi = E \Phi$

where $e^2 = 1.44~\textrm{MeV.fm}$ , separate the radial and angular part as usual,

$\displaystyle \left( -\frac{\hbar^2}{2m} \frac{1}{r^2} \left( \frac{d}{dr} r^2 \frac{d}{dr} \right) + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2} + \frac{Q_1Q_2 e^2}{r} \right) R(r) = E R(r)$

setting

$\displaystyle k^2 = \frac{2mE}{\hbar^2}, \eta =\frac{k}{2E} Q_1Q_2 e^2$

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

using the usual substitution $R = u / r$

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

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

Setting $x = kr$

$\displaystyle \frac{d^2u(x)}{dx^2} + \left( 1 - \frac{l(l+1)}{x^2} - \frac{2\eta}{x} \right) u(x) =0$

The short range behaviour is approximated as

$\displaystyle \frac{d^2u(x)}{dx^2} = \frac{l(l+1)}{x^2}u(x) \Rightarrow u(x) \approx x^{l+1}~\textrm{or}~ x^{-l}$

and for long range behaviour, it should approach Riccati–Bessel functions $\hat{j}_l, \hat{n}_l$ with phase shift.

Set $u(x) = x^{l+1} \exp(i x) y(x)$ , the equation of $u(x)$ becomes,

$\displaystyle x \frac{d^2y}{dx^2} + (2L+2 + 2ix) \frac{dy}{dx} + (2i(L+1) - 2\eta) y = 0$

Change of variable $z = - 2i x$ ,

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

This is our friend Laguerre polynomial again!!! with $\alpha = 2L+1, n = -(L+1+i\eta)$ . Since the $n$ is not an integer anymore, we go to a more general case, that is the Kummer’s equation,

$\displaystyle z \frac{d^2 w}{dz} + (b-z) \frac{dw}{dz} - a w = 0$

The solution of Kummer’s equation is the confluent hypergeometric function

$w(z) = _1F_1(a, b, z)$

Thus, the solution for the radial function is

$\displaystyle u_l(x) =A x^{l+1} e^{ix} _1F_1(L+1+i \eta, 2L+2, -2 i x )$

where $A$ is a normalization constant by compare the long range behaviour with Riccati-Bessel function. The full solution is,

$\displaystyle u_l(x) = \\ F_l(x) = \frac{2^l e^{-\pi \eta/2} |\Gamma(l+1+i\eta)| }{(2l+1)!} x^{l+1} e^{ix}~_1F_1(L+1+i \eta, 2L+2, -2 i x )$

At long range,

$\displaystyle F_l(x \rightarrow \infty) = \sin \left( x - l \frac{\pi}{2} - \eta \log(2x) + \sigma_l \right)$

where $\sigma_l = \arg( \Gamma(l+1+i\eta) )$ is the Coulomb phase shift.

Using Kummer’s transform

$\displaystyle _1F_1(a,b,z) = e^z ~_1F_1(b-a, b, -z)$

The solution can be written as

$\displaystyle u_l(x) = \\ F_l(x) = \frac{2^l e^{-\pi \eta/2} |\Gamma(l+1+i\eta)| }{(2l+1)!} x^{l+1} e^{-ix}~_1F_1(L+1-i \eta, 2L+2, 2 i x )$

Another solution should behave like $\cos$ and unbound at $x = 0$ .

Set $u(x) = x^{-l} \exp(i x) y(x)$ , the equation of $u(x)$ becomes,

$\displaystyle x \frac{d^2y}{dx^2} + (-2L + 2ix) \frac{dy}{dx} + (-2i L - 2\eta) y = 0$

Change of variable $z = -2ix$ ,

$\displaystyle z \frac{d^2y}{dz^2} + (-2L -z ) \frac{dy}{dx} - ( - L + i \eta) y = 0$

$\displaystyle u_l(x) =A x^{-l} e^{ix}~ _1F_1(-L+i \eta, -2L, -2 i x )$

However, $_1F_1(a, b, z)$ does not exist for non-position b.

We can also transform into Whittaker’s equation by change of variable $z = 2ix$

$\displaystyle -4 \frac{d^2u(x)}{dz^2} + \left( 1 - \frac{-4l(l+1)}{z^2} - \frac{4i\eta}{z} \right) u(x) =0$

$\displaystyle \frac{d^2u(x)}{dz^2} + \left( -\frac{1}{4} +\frac{i\eta}{z}- \frac{l(l+1)}{z^2} \right) u(x) =0$

using $\displaystyle l(l+1) = (l+1/2)^2 - 1/4$

$\displaystyle \frac{d^2u(x)}{dz^2} + \left( -\frac{1}{4} +\frac{i\eta}{z} + \frac{ 1/4 - (l+1/2)^2}{z^2} \right) u(x) =0$

This is the Whittaker’s equation with $\kappa = i \eta, \mu = l+1/2$

The solutions are

$u_l(x) = e^{-ix} (2ix)^{l+1} ~_1F_1(l+1-i \eta, 2l+2, 2ix)$

$u_l(x) = e^{-ix} (2ix)^{l+1} U(l+1-i \eta, 2l+2, 2ix)$

I still cannot get the second solution, the $G_l$ . According to Wolfram, https://mathworld.wolfram.com/CoulombWaveFunction.html

$\displaystyle G_l(x) = \frac{2}{\eta C_0^2(\eta)} F_l(x) \left( \log(2x) + \frac{q_l(\eta)}{p_l(\eta)} \right) + \frac{x^{-l}}{(2l+1) C_l(\eta)} \sum_{K=-l}^\infty a_k^l(\eta) x^{K+l}$ ,

where $q_l, p_l, a_k^l$ are defined inAbramowitz, M. and Stegun, I. A. (Eds.). “Coulomb Wave Functions.” Ch. 14 in Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 537-544, 1972.

Common functions expressed as Hypergeometric function

August 30, 2017

GoLuckyRyan Basic Bessel function, elliptic integral, Gamma Function, hypergeometric, Legendre polynomial Leave a comment

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.

Nuclear Physics 101

Asymptotic normalization coefficient

Coulomb wave function (II)

Coulomb wave function

Common functions expressed as Hypergeometric function

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