Special Functions

Legendre Polynomials

Legendre Equation

Equation

\begin{align} (1-x^2) P''(x) - 2xP'(x) + \lambda P(x) = 0 \label{eq:leq} \end{align}

Boundary conditons

\[\mathrm{Nonsingularity at} x = \pm 1\]

Frobenius' Methods

尝试级数解

\begin{align} P(x) =& x^{s} (a_0 + a_1 x + a_2 x^2 + \cdots) \\ =& \sum_{j=0}^{\infty}a_j x^{s+j}, \quad a_0 \neq 0 \end{align}

代回原始方程 \eqref{eq:leq} , 得

\begin{align} \sum_{j=0}^{\infty}\left[ a_j (s+j) (s+j-1) x^{s+j-2} - a_j (s+j) (s+j-1) x^{s+j} - 2(s+j)a_j x^{s+j} + \lambda a_j x^{s+j} \right] = 0 \end{align}

$j=0$ 时, 得 $x$ 的最低阶即 $x-2$ ,对应的系数方程为

\begin{align} a_0 s (s-1) = 0 \end{align}

$j=1$ 时

\begin{align} a_1 (s+1)s = 0 \end{align}

并且 $a_0\neq 0$ , 所以

\begin{align} s(s-1) = 0 \quad \quad \mathrm{(indical\quad equation)} \end{align}

即 $s=0$ 或 $s=1$

$s=0$

此时得 recurrence relation 为

\begin{align} a_{j+2} = \frac{j(j+1)-\lambda}{(j+1)(j+2)}a_j \end{align}

假设 $a_1 = 0$ . 那么所有的奇数项都为 $0$ . 当 $j\to \infty$ 时, 级数解 $P(x = \pm 1)$ 不收敛, 不满足边界条件. 使其收敛, $\lambda$ 必须取

\begin{align} \lambda = l(l+1) \end{align}

其中 $l$ 是偶数. 这样的话, 级数解就在 $l$ 阶截断了.

$s=1$

此时得 recurrence relation 为

\begin{align} a_{j+2} = \frac{(j+1)(j+2) -\lambda}{(j+2)(j+3)} a_j \end{align}

而且必须有 $a_1=0$ 那么此时所有的奇数项都为 $0$ . 此时截断的条件为

\begin{align} \lambda = (l+1)(l+2) \end{align}

其中 $l$ 是偶数.

合并

$s=0, a_1 = 0$ 时, 对应的级数解, 都是 $x$ 的偶数次幂. $s = 1$ 时, 对应的级数解, 都是 $x$ 的奇数次幂, 其实对应于 $s=0, a_1 \neq 0$ 时解. 两种 $s$ 的取值, 对应的截断条件也可以合并为

\begin{align} \lambda = l(l+1) \end{align}

其中 $l$ 是整数. 合并的 recurrence relation 为

\begin{align} a_{j+2} = \frac{j(j+1)-\lambda}{(j+1)(j+2)}a_j \end{align}

有 $a_1, a_2$ 两个待定的系数, 这恰好是一个二阶微分方程所需要的!

Legendre Polynomials

下面是画前七阶 Legendre Polynomials 的图像

import numpy as np
from matplotlib import pyplot as plt
def P(N,x):
    l = N*(N+1)
    a = np.zeros(N+1)
    k = N//2

    if N > 0:
	if N%2 == 0:
	    a[0] = (-1)**k
	    a[1] = 0
	if N%2 == 1:
	    a[0] = 0
	    a[1] = (-1)**k

	for j in range(N-1):
	    a[j+2] = j*(j+1) - l 
	    a[j+2] = a[j+2] * a[j]
	    a[j+2] = a[j+2] / (j+1)
	    a[j+2] = a[j+2] / (j+2)
	a = a/a.sum()
	s = 0
	for i in range(N+1):
	    s = s + a[i]*x**i
    else:
	s =1 
    return s
n = 1000
nn = 8
x = np.linspace(-1,1,n)
y = np.zeros(n)
plt.figure(figsize=(6.0,10.0))    # must before plt.plot(x,y)
for i in range(nn):
    for j in range(n):
	y[j] = P(i,x[j])
    t = "$P_ { %s }(x)$"%i
    plt.plot(x,y,label=t)
plt.xlabel(r"$x$")
plt.ylabel(r"$P_ { l }(x)$")
plt.title("Legendre Polynomials")
plt.legend()
plt.savefig("./2019-06-03-专业笔记-PartialWaveExpansion/LegenderPolynomials.jpg"\
    ,writer='imageffmagick', dpi=100)

下面是用 sympy 输出前七阶 Legendre Polynomials

from sympy import Rational as R
from sympy import Symbol
from sympy import factor
from sympy import latex

def P(N):
    l = N*(N+1)
    a = []
    k = N//2
    if N%2 == 0:
	a.append((-1)**k)
	a.append(0)
    if N%2 == 1:
	a.append(0)
	a.append((-1)**k) 
    for j in range(N-1):
	a.append(j*(j+1) - l )
	a[j+2] = a[j+2] * a[j]
	a[j+2] = R( a[j+2] , (j+1) )
	a[j+2] = R( a[j+2] , (j+2) )

    c = 0
    for i in range(N+1):
	c = c + a[i]

    P = 0
    x = Symbol('x')
    for i in range(N+1):
	a[i] = R(a[i], c) * x**i
	P = P + a[i]
    P = latex(P)
    return P

for i in range(8):
    print(r'P_%d(x)=&'%i,P(i),"\\\\")

结果

\begin{align*} : P_0(x)=& 1 \\ : P_1(x)=& x \\ : P_2(x)=& \frac{3 x^{2} - 1}{2} \\ : P_3(x)=& \frac{x \left(5 x^{2} - 3\right)}{2} \\ : P_4(x)=& \frac{35 x^{4} - 30 x^{2} + 3}{8} \\ : P_5(x)=& \frac{x \left(63 x^{4} - 70 x^{2} + 15\right)}{8} \\ : P_6(x)=& \frac{231 x^{6} - 315 x^{4} + 105 x^{2} - 5}{16} \\ : P_7(x)=& \frac{x \left(429 x^{6} - 693 x^{4} + 315 x^{2} - 35\right)}{16} \\ \end{align*}

其中 $a_1$ 和 $a_0$ r 取为, 满足 $P_l(1) = 1$ , 并且 $x$ 的最高次幂的系数为正的值.

Orthogonality

Legendre Polynomials 的正交归一性为

\begin{align} \int_{-1}^{1} P_l(x)P_m(x) \cdot \mathrm{d}x = \frac{2 \delta_{lm}}{2l+1} \end{align}

Associated Legendre Functions

Associated Legendre Equation

\begin{align} (1-x^2) P''(x) -2x P'(x) + \left[ \lambda - \frac{m^2}{1-x^2} \right]P(x) = 0 \end{align}

分母中的 $1-x^2$ 是个问题, 尝试通过做变换把它从分母中去掉. 令

\begin{align} P(x) = (1-x^2)^p \mathcal{P}(x) \end{align}

原式变为

\begin{align} \mathcal{P}(1-x^2)^{p-1}\left[ -2p(1+x^2) + 4p^2 x^2 +4px^2 +\lambda(1-x^2) -m^2 \right]\\ -2x \mathcal{P}'(1-x^2)^p(2p + 1) + \mathcal{P}''(1-x^2)^{p+1} = 0 \end{align}

这和原来的 Legendre equation 有些类似. $\mathcal{P}$ 的系数中有 $\lambda(1-x^2)$ 而且如果我们能提取出一个 $1-x^2$ 话, 它和 Legendre equation 就更像了. 我们还有 $p$ 是可调的, 可以令常数项和 $x^{2}$ 项前面的系数差一负号, 就可以提出啦! 即

\begin{align} -2p+4p^2+4p = -(-2p-m^2) \end{align}

得 $p = \frac{m}{2}$ .将它代回, 就可以得到

\begin{align} (1-x^2)\mathcal{P}''- 2x(m+1)\mathcal{P}' + \left[ \lambda - m(m+1) \right]\mathcal{P} = 0 \end{align}

Frobenius' Methods

这个变换后的方程就可以由 Frobenius' Methods 得到级数解

尝试级数解

\begin{align} P(x) =& x^{s} (a_0 + a_1 x + a_2 x^2 + \cdots) \\ =& \sum_{j=0}^{\infty}a_j x^{s+j}, \quad a_0 \neq 0 \end{align}

代回原始方程 \eqref{eq:leq} , 得

\begin{align} \sum_{j=0}^{\infty} \left[ (s+j)(s+j-1)a_jx^{s+j-2} -(s+j)(s+j-1)a_jx^{s+j} \\ -2a_j(m+1)(s+j)x^{s+j} + a_j(\lambda -m(m+1))x^{s+j} \right] = 0 \end{align}

$j=0$ 时, 得 $x$ 的最低阶即 $x-2$ ,对应的系数方程为

\begin{align} a_0 s (s-1) = 0 \end{align}

$j=1$ 时

\begin{align} a_1 (s+1)s = 0 \end{align}

并且 $a_0\neq 0$ , 所以得到了和 Legendre equation 一样的 indical equation

\begin{align} s(s-1) = 0 \quad \quad \mathrm{(indical\quad equation)} \end{align}

即 $s=0$ 或 $s=1$

当 $s=0$时, recurrence relation 为

\begin{align} a_{j+2} = \frac{(j+m)(j+m+1)-\lambda}{(j+1)(j+2)}a_j \end{align}

假设 $a_1 = 0$ . 那么所有的奇数项都为 $0$ . 当 $j\to \infty$ 时, 级数解 $P(x = \pm 1)$ 不收敛, 不满足边界条件. 使其收敛, $\lambda$ 必须取

\begin{align} \lambda = (j+m)(j+m+1) \end{align}

其中 $l$ 是偶数. 这样的话, 级数解就在 $l$ 阶截断了. 与 Legendre equation 类似, $s=1$ 时得到的结果是对应的奇数项, 所以 $l$ 取整数就是最后的结果.

$\lambda = l(l+1)$

在实际问题中, 一般 $\lambda = l(l+1)$ , 此时截断为
\begin{align} j = l - m \end{align}
也就是说只有在 $m\le l$ 时才有收敛的解.

Bessel Functions

Bessel ODE

\begin{align} x^2 J_{\nu}'' +x J_{\nu}' + (x^2 - \nu^2)J_{\nu} = 0 \end{align}

Frobenius' Methods and Bessel Functions of the First Kind

尝试级数解

\begin{align} J_{\nu}(x) =& x^{s} (a_0 + a_1 x + a_2 x^2 + \cdots) \\ =& \sum_{j=0}^{\infty}a_j x^{s+j}, \quad a_0 \neq 0 \end{align}

代回 Bessel ODE , 得

\begin{align} \sum_{j = 0}^{\infty} \left[ (s+j)(s+j-1) + (s+j) -\nu^2 \right] a_j x^j + a_{j}x^{j+2} = 0 \end{align}

$j=0$ 时, 得 $x$ 的最低阶即 $x$ ,对应的系数方程为

\begin{align} ( s^2 - \nu^2 ) a_0 = 0 \end{align}

并且 $a_0\neq 0$ , 所以

\begin{align} (s+\nu)(s-\nu) = 0 \quad \quad \mathrm{(indical\quad equation)} \end{align}

得

\begin{align} s = \pm \nu \end{align}

$j=1$ 时

\begin{align} (s+1+\nu)(s+1-\nu)a_1 = 0 \end{align}

由于 $s = \pm \nu$ , 所以 $a_1 = 0$

以下先讨论 $\nu = n$ 是整数的情况

$s = n$

此时得 recurrence relation 为

\begin{align} a_{j+2} = - \frac{1}{(j+2)(2n +j + 2)} a_j \end{align}

任意系数为

\begin{align} a_{2p} = (-1)^p \frac{a_0 n!}{2^{2p}p! (n+p)!} \end{align}

Bessel ODE 的解为

\begin{align} J_n(x) = \sum_{s=0}^{\infty} \frac{(-1)^s}{s!(n+s)!} \left( \frac{x}{2} \right)^{n+2s} \end{align}

上式中按习惯取 $a_0 = \frac{1}{2^n n!}$ .

$s = -n$

此时得 recurrence relation 为

\begin{align} a_{j+2} = \frac{1}{(j+2)(2n -j - 2)} a_j \end{align}

当 $j = 2n-2$ 时上式发散, 并且不满足级数是从 $x^{-n}$ 开始的假设. 因此 $s = -n$ 不给出解.

$\nu = n$ 是整数的情况给出了 Bessel 的一个解, 称为 Bessel function of first kink. 且有以下性质

\begin{align} J_{-n}(x) = (-1)^n J_n(x) \end{align}

code

下面是画前3阶 Bessel functions of the first kind 的图像

import numpy as np
from matplotlib import pyplot as plt
import time

start = time.process_time()

def J(n,x,up=10):
    J = 0
    for s in range(up):
	js = (-1)**s
	js = js/(np.math.factorial(s))
	js = js/(np.math.factorial(s+n))
	js = js*(x/2)**(n+2*s)
	J = J + js
    return J


N = 1000
x = np.linspace(1e-8,30,N)
up =100
def pltj(J,n,x,N):
    j = np.zeros(N)
    for i in range(N):
	j[i] = J(n,x[i],up = up)
    return j

for i in range(4):
    l = r"$J _ { %d } (x)$" %i
    plt.plot( x, pltj(J,i,x,N), label = l)

plt.plot(x, 1/np.sqrt(x),label=r"$x^{-1/2}$") 

plt.ylim(-.5,1.)
plt.xlabel("x")
plt.ylabel(r"$J_n(x)$")
plt.legend()
plt.title("Bessel Functions")

end = time.process_time()
print('time is',end-start)
plt.savefig("./2019-06-03-专业笔记-PartialWaveExpansion/BFF.jpg"\
    ,writer='imageffmagick', dpi=100)

可以看出它在 $x=0$ 处没有发散.

the Bessel Functions of the Second Kind

Definition

Bessel ODE 是一个二阶微分方程, 它应该有两个独立的解. 上面 $\nu$ 取整数的时候只给出了一个解. 它的另一个解由 Neumann functions 给出, 定义如下

\begin{align} Y_{\nu}(x) = \frac{\cos (\nu\pi) J_{\nu}(x)-J_{-\nu(x)}}{\sin(\nu\pi)} \end{align}

当 $\nu$ 不整数的时候, 上式就是 Bessel functions of the first kind 的线性组合. 当 $\nu = n$ 取整数的时候, 它应取极限

\begin{align} Y_n(x) = \lim_{\nu\to n}Y_{\nu}(x) \end{align}

它在 $x = 0$ 处是发散的. 它就是 Bessel Functions 的第二个解, Bessel functions of the second kind . 这里不详细讨论它.

Hankel Functions

Hankel Functions 曾经也常被叫做 Bessel functions of the third kind . 它由 Bessel functions of the first kind and the second kind 的线性组合定义

\begin{align} H_{\nu}^{(1)} = J_{\nu}(x) + \mathrm{i}Y_{\nu}(x) \\ H_{\nu}^{(2)} = J_{\nu}(x) - \mathrm{i}Y_{\nu}(x) \end{align}

Modified Bessel Functions

MOdified Bessel equation

\begin{align} \rho^2 \frac{\mathrm{d}^2}{\mathrm{d}\rho^2}P_{\nu}(k\rho) + \rho \frac{\mathrm{d}}{\mathrm{d}\rho}P_{\nu}(k\rho) - (k^2\rho^2 + \nu^2) P_{\nu} P_{\nu}(k\rho) = 0 \end{align}

它与 Bessel ODE

\begin{align} x^2 J_{\nu}'' +x J_{\nu}' + (x^2 - \nu^2)J_{\nu} = 0 \end{align}

相比, 在于 $k^2\rho^2$ 前面的符号变了.

Modified Bessel Functions of the First Kind

它的解只须做代换 $k\to \mathrm{i}k$ 即可得到, 即 modified Bessel functions of the first kind

\begin{align} I_{\nu}(x) = \mathrm{i}^{-\nu}J_{\nu}(\mathrm{i}x) = \sum_{s=0}^{\infty} \frac{1}{s!\Gamma(s+\nu+1)} \left( \frac{x}{2} \right)^{\nu+2s} \end{align}

它的系数都是正的, 没有振荡. 它有

\begin{align} I_n(x) = I_{-n}(x) \end{align}

Code

下面是画前3阶 modified Bessel functions of the first kind 的图像

import numpy as np
from matplotlib import pyplot as plt
import time

start = time.process_time()

def J(n,x,up=10):
    J = 0
    for s in range(up):
	js = 1/(np.math.factorial(s))
	js = js/(np.math.factorial(s+n))
	js = js*(x/2)**(n+2*s)
	J = J + js
    return J


N = 1000
x = np.linspace(1e-8,3,N)
up =100
def pltj(J,n,x,N):
    j = np.zeros(N)
    for i in range(N):
	j[i] = J(n,x[i],up = up)
    return j

for i in range(4):
    l = r"$I _ { %d } (x)$" %i
    plt.plot( x, pltj(J,i,x,N), label = l)


plt.xlabel("x")
plt.ylabel(r"$I_n(x)$")
plt.legend()
plt.title("Modified Bessel Functions")

end = time.process_time()
print('time is',end-start)
plt.savefig("./2019-06-03-专业笔记-PartialWaveExpansion/MBFF.jpg"\
    ,writer='imageffmagick', dpi=100)

Modified Bessel Functions of the Second Kind

modified Bessel functions of the second kind 有时也叫做 Whittaker functions , 定义如下

\begin{align} K_{\nu}(x) = \frac{\pi}{2}\frac{I_{-\nu}(x)-I_{\nu}(x)}{\sin(\nu\pi)} \end{align}

moddified Bessel function 对于 Bessel function 的关系, 就像 $\sinh$ 对于 $\sin$ 的关系, 所以有时它们也叫做 hyperbolic Bessel functions .

Spherical Bessel Functions

对于球对称的问题, 经常会遇到径向方程

\begin{align} r^2 \frac{\mathrm{d}^2 R}{\mathrm{d}r^2} + 2r \frac{\mathrm{d}R}{\mathrm{d}r} +[k^2r^2 - l(l + 1)]R = 0 \end{align}

这个方程和 Bessel ODE 很像, 区别在于 $l(l+1)$ , 而不是一个数的平方. 但是只要做如下变量代换

\begin{align} R(kr) = \frac{Z(kr)}{\sqrt{kr}} \end{align}

就可以化为 Bessel ODE 的形式

\begin{align} r^2 \frac{\mathrm{d}^2 Z}{\mathrm{d}r^2} + r \frac{\mathrm{d}Z}{\mathrm{d}r} +\left[ k^2r^2 -\left( l + \frac{1}{2} \right) \right]Z = 0 \end{align}

那么它的解就可以写出

\begin{align} R(kr) = \frac{C}{\sqrt{kr}}J_{l+1/2}(kr) \end{align}

它就是 Spherical Functions . $C$ 是一个任意常数, 因为方程是齐次的

Definition

取定常数 $C$ , Spherical Functions 定义如下

\begin{align} j_n(x) =& \sqrt{\frac{\pi}{2x}}J_{n+1/2}(x) \\ y_n(x) =& \sqrt{\frac{\pi}{2x}}Y_{n+1/2}(x) \\ h_n^{(1)}(x) =& j_n(x) +\mathrm{i}y_n(x) \\ h_n^{(2)}(x) =& j_n(x) -\mathrm{i}y_n(x) \\ \end{align}

Some Examples

\begin{align} j_0(x) =& \quad\frac{1}{x}\sin x \\ y_0(x) = & - \frac{1}{x}\cos x \end{align} \begin{align} j_1(x) =&\quad \frac{1}{x^2}\sin x - \frac{1}{x}\cos x \\ y_1(x) =&-\frac{1}{x^2}\cos x -\frac{1}{x}\sin x \end{align} \begin{align} j_2(x) = & \quad\left( \frac{3}{x^3} - \frac{1}{x} \right)\sin x -\frac{3}{x^2}\cos x \\ y_2(x) = & -\left( \frac{3}{x^3} -\frac{1}{x} \right)\cos x - \frac{3}{x^2}\sin x \end{align} \begin{align} j_3(x) = &\quad \left( \frac{15}{x^4} - \frac{6}{x^2} \right)\sin x +\left( - \frac{15}{x^3} + \frac{1}{x} \right) \cos x \\ y_3(x) = & -\left( \frac{15}{x^4} -\frac{6}{x^2} \right) \cos x +\left( -\frac{15}{x^3} + \frac{1}{x} \right) \sin x \end{align}

Asymptotic Forms

as $x \to \infty$

\begin{align} j_n(x) =& \frac{1}{x}\sin (x - \frac{n\pi}{2}) \\ y_n(x) =& - \frac{1}{x}\cos (x - \frac{n\pi}{2}) \end{align}

Closure Relation

Closure Relation for spherical Bessel functions

\begin{align} \int_0^{\infty} r^2 j_l(k_1r)j_l(k_2r) \mathrm{d}r = \frac{\pi}{2k_1^2}\delta(k_1-k_2) \end{align}

Spherical Harmonics

Associated Legendre Functions

\begin{align} P_l^m(x) = (-1)^m \frac{\mathrm{d}^m}{\mathrm{d}x^m}P_l(x) \end{align}

Spherical Harmonics

\begin{align} Y_l^m (\theta,\phi) \equiv \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}} P_l^m(\cos\theta)e^{\mathrm{i}m\phi} \end{align}

它们在整个球面上是正交归一的

\begin{align} \int_0^{2\pi} \mathrm{d}\phi \int_0^{\pi}\sin \theta \mathrm{d}\theta \left[ Y_{l_1}^{m_1} (\theta,\phi) \right]^{*} Y_{l_2}^{m_2} (\theta,\phi) =\delta_{l_1l_2} \delta_{m_1m_2} \end{align}

Summary

Legendre equation and its orthogonality

\begin{align} (1-x^2) P''(x) - 2xP'(x) + \lambda P(x) = 0 \end{align} \begin{align} \int_{-1}^{1} P_l(x)P_m(x) \cdot \mathrm{d}x = \frac{2 \delta_{lm}}{2l+1} \end{align}

associated Legendre equation

\begin{align} (1-x^2) P''(x) -2x P'(x) + \left[ \lambda - \frac{m^2}{1-x^2} \right]P(x) = 0 \end{align}

解为

\begin{align} P_l^m(x) = (-1)^m \frac{\mathrm{d}^m}{\mathrm{d}x^m}P_l(x) \end{align}

Spherical Harmonics

\begin{align} Y_l^m (\theta,\phi) \equiv \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}} P_l^m(\cos\theta)e^{\mathrm{i}m\phi} \end{align}

orthogonality

Bessel ODE

\begin{align} x^2 J_{\nu}'' +x J_{\nu}' + (x^2 - \nu^2)J_{\nu} = 0 \end{align}

the Bessel functions of the first kind

\begin{align} J_{-n}(x) = (-1)^n J_n(x) \end{align}

Neumann functions ( the Bessel functions of the second kind )

\begin{align} Y_{\nu}(x) = \frac{\cos (\nu\pi) J_{\nu}(x)-J_{-\nu(x)}}{\sin(\nu\pi)} \end{align} \begin{align} Y_n(x) = \lim_{\nu\to n}Y_{\nu}(x) \end{align}

Hankel Functions, 曾经也常被叫做 Bessel functions of the third kind . 它由 Bessel functions of the first kind and the second kind 的线性组合定义

\begin{align} H_{\nu}^{(1)} = J_{\nu}(x) + \mathrm{i}Y_{\nu}(x) \\ H_{\nu}^{(2)} = J_{\nu}(x) - \mathrm{i}Y_{\nu}(x) \end{align}

MOdified Bessel equation

\begin{align} \rho^2 \frac{\mathrm{d}^2}{\mathrm{d}\rho^2}P_{\nu}(k\rho) + \rho \frac{\mathrm{d}}{\mathrm{d}\rho}P_{\nu}(k\rho) - (k^2\rho^2 + \nu^2) P_{\nu} P_{\nu}(k\rho) = 0 \end{align}

$k^2\rho^2$ 前面的符号变了 Modified Bessel Functions of the First Kind

\begin{align} I_n(x) = I_{-n}(x) \end{align}

它没有振荡. 它的另一个解为 Modified Bessel Functions of the Second Kind

\begin{align} K_{\nu}(x) = \frac{\pi}{2}\frac{I_{-\nu}(x)-I_{\nu}(x)}{\sin(\nu\pi)} \end{align}

Spherical Bessel Functions

\begin{align} r^2 \frac{\mathrm{d}^2 R}{\mathrm{d}r^2} + 2r \frac{\mathrm{d}R}{\mathrm{d}r} +[k^2r^2 - l(l + 1)]R = 0 \end{align} \begin{align} j_n(x) =& \sqrt{\frac{\pi}{2x}}J_{n+1/2}(x) \\ y_n(x) =& \sqrt{\frac{\pi}{2x}}Y_{n+1/2}(x) \\ h_n^{(1)}(x) =& j_n(x) +\mathrm{i}y_n(x) \\ h_n^{(2)}(x) =& j_n(x) -\mathrm{i}y_n(x) \\ \end{align}

asymptotic as $x\to\infty$

\begin{align} j_n(x) =& \frac{1}{x}\sin (x - \frac{n\pi}{2}) \\ y_n(x) =& - \frac{1}{x}\cos (x - \frac{n\pi}{2}) \end{align}

Closure Relation for spherical Bessel functions

\begin{align} \int_0^{\infty} r^2 j_l(k_1r)j_l(k_2r) \mathrm{d}r = \frac{\pi}{2k_1^2}\delta(k_1-k_2) \end{align}

Reference

梁昆淼编, 刘法,缪国庆修订, 数学物理方法(第四版)

11.5 球贝塞尔方程

Arfken, Weber, Harris, Mathematical Methods for Physicists 7ed:

Chap 7.5 Series Solutions-Frobenius' Methods
Example 8.3.1 Legendre Equation
Chap 14 Bessel Functions
Chap 15.1 Legendre Polynomials

J. J Sakurai, Jim Napolitano, Modern Quantum Mechanics 2ed:

Chap 6.4 Phase Shifts and Partial Waves

R. Mehrem, arXiv:0909.0494v4: https://arxiv.org/pdf/0909.0494.pdf

Special Functions

Legendre Polynomials

Legendre Equation

Frobenius' Methods

\(s=0\)

\(s=1\)

合并

Legendre Polynomials

Orthogonality

Associated Legendre Functions

Associated Legendre Equation

Frobenius' Methods

Bessel Functions

Bessel ODE

Frobenius' Methods and Bessel Functions of the First Kind

\(s = n\)

\(s = -n\)

code

the Bessel Functions of the Second Kind

Definition

Hankel Functions

Modified Bessel Functions

MOdified Bessel equation

Modified Bessel Functions of the First Kind

Code

Modified Bessel Functions of the Second Kind

Spherical Bessel Functions

Definition

Some Examples

Asymptotic Forms

Closure Relation

Spherical Harmonics

Associated Legendre Functions

Spherical Harmonics

Summary

Reference