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 $s0$ 或 $s1$
$s=0$
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-physics-PartialWaveExpansion/LegenderPolynomials.jpg"\
,writer='imageffmagick', dpi=100)#+RESULTS: : None
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),"\\\\")#+RESULTS: : P_0(x)=& 1 \\ : P_1(x)=& x \\ : P_2(x)=& \frac{3 x^{2}}{2} - \frac{1}{2} \\ : P_3(x)=& \frac{5 x^{3}}{2} - \frac{3 x}{2} \\ : P_4(x)=& \frac{35 x^{4}}{8} - \frac{15 x^{2}}{4} + \frac{3}{8} \\ : P_5(x)=& \frac{63 x^{5}}{8} - \frac{35 x^{3}}{4} + \frac{15 x}{8} \\ : P_6(x)=& \frac{231 x^{6}}{16} - \frac{315 x^{4}}{16} + \frac{105 x^{2}}{16} - \frac{5}{16} \\ : P_7(x)=& \frac{429 x^{7}}{16} - \frac{693 x^{5}}{16} + \frac{315 x^{3}}{16} - \frac{35 x}{16} \\ 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 $s0$ 或 $s1$
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-physics-PartialWaveExpansion/BFF.jpg"\
,writer='imageffmagick', dpi=100)#+RESULTS: : time is 0.765683398
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-physics-PartialWaveExpansion/MBFF.jpg"\
,writer='imageffmagick', dpi=100)#+RESULTS: : time is 0.721994943
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
$$\begin{align} \int_0^{2\pi} \mathrm{d}\phi \int_0^{\pi}\sin \theta \mathrm{d}\theta \left[ Y_{l_1}^{m_1} (\theta,\phi) \right]^{*} Y_{l_2}^{m_2} (\theta,\phi) =\delta_{l_1l_2} \delta_{m_1m_2} \end{align}$$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