泽尔尼克多项式

头15个泽尔尼克多项式

20个泽尔尼克多项式 以Noll序列表示

泽尔尼克多项式是一个以1953年获诺贝尔物理学奖荷兰物理学家弗里茨·泽尔尼克命名的正交多项式，分为奇、偶两类

奇多项式：

$Z_{n}^{m}(\rho ,\varphi )=R_{n}^{m}(\rho )\,\cos(m\,\varphi )\!$

偶多项式

$Z_{n}^{-m}(\rho ,\varphi )=R_{n}^{m}(\rho )\,\sin(m\,\varphi ),\!$

其中$n\geq m$ 为非负整数，

$\phi$为方位角

$0\leq \rho \leq 1$ 为径向距离

如果 n-m为偶数则

$R_{n}^{m}(\rho )=\sum _{k=0}^{\tfrac {n-m}{2}}{\frac {(-1)^{k}\,(n-k)!}{k!\left({\tfrac {n+m}{2}}-k\right)!\left({\tfrac {n-m}{2}}-k\right)!}}\;\rho ^{n-2\,k}$

如果n-m为奇数，则

$R_{n}^{m}(\rho )=0$

泽尔尼克多项式的超几何函数表示

泽尔尼克多项式也可以表示为超几何函数

${\begin{aligned}R_{n}^{m}(\rho )&={\binom {n}{\tfrac {n+m}{2}}}\rho ^{n}\ {}_{2}F_{1}\left(-{\tfrac {n+m}{2}},-{\tfrac {n-m}{2}};-n;\rho ^{-2}\right)\\&=(-1)^{\tfrac {n+m}{2}}{\binom {\tfrac {n+m}{2}}{\tfrac {n-m}{2}}}\rho ^{m}\ {}_{2}F_{1}\left(1+n,1-{\tfrac {n-m}{2}};1+{\tfrac {n+m}{2}};\rho ^{2}\right)\end{aligned}}$

Noll 序列

Noll 用一个J数字表示 [n,m]:如下表

n,m	0,0	1,1	1,−1	2,0	2,−2	2,2	3,−1	3,1	3,−3	3,3
j	1	2	3	4	5	6	7	8	9	10
n,m	4,0	4,2	4,−2	4,4	4,−4	5,1	5,−1	5,3	5,−3	5,5
j	11	12	13	14	15	16	17	18	19	20

泽尔尼克多项式

由于

$I_{j}=\int _{0}^{2\pi }\int _{0}^{1}Z_{j}^{2}\,\rho \,d\rho \,d\theta =k_{j}*\pi .$

其中$k_{j}$因j而异，

$k_{1}=1$

$k_{2}={\frac {1}{4}}$

$k_{3}={\frac {1}{4}}$

$k_{4}={\frac {1}{3}}$

$k_{5}={\frac {1}{6}}$

必须先归一化

令$Z_{j}=Z_{j}/{\sqrt {(}}k_{j})$

使得

$I_{j}=\int _{0}^{2\pi }\int _{0}^{1}Z_{j}^{2}\,\rho \,d\rho \,d\theta =\pi .$

归一化泽尔尼克多项式以Noll序列排列如下：

Noll index ($j$)	Radial degree ($n$)	Azimuthal degree ($m$)	$Z_{j}$	Classical name
1	0	0	$1$	Piston
2	1	1	$2\rho \cos \theta$	Tip (lateral position) (X-Tilt)
3	1	−1	$2\rho \sin \theta$	Tilt (lateral position) (Y-Tilt)
4	2	0	${\sqrt {3}}(2\rho ^{2}-1)$	Defocus (longitudinal position)
5	2	−2	${\sqrt {6}}\rho ^{2}\sin 2\theta$	Astigmatism
6	2	2	${\sqrt {6}}\rho ^{2}\cos 2\theta$	Astigmatism
7	3	−1	${\sqrt {8}}(3\rho ^{3}-2\rho )\sin \theta$	Coma
8	3	1	${\sqrt {8}}(3\rho ^{3}-2\rho )\cos \theta$	Coma
9	3	−3	${\sqrt {8}}\rho ^{3}\sin 3\theta$	Trefoil
10	3	3	${\sqrt {8}}\rho ^{3}\cos 3\theta$	Trefoil
11	4	0	${\sqrt {5}}(6\rho ^{4}-6\rho ^{2}+1)$	Third-order spherical
12	4	2	${\sqrt {10}}(4\rho ^{4}-3\rho ^{2})\cos 2\theta$	—
13	4	−2	${\sqrt {10}}(4\rho ^{4}-3\rho ^{2})\sin 2\theta$	—
14	4	4	${\sqrt {10}}\rho ^{4}\cos 4\theta$	—
15	4	−4	${\sqrt {10}}\rho ^{4}\sin 4\theta$	—

正交性

径向正交性

$\int _{0}^{1}\rho {\sqrt {2n+2}}R_{n}^{m}(\rho )\,{\sqrt {2n'+2}}R_{n'}^{m}(\rho )\,d\rho =\delta _{n,n'}.$

角度正交性

$\int _{0}^{2\pi }\cos(m\varphi )\cos(m'\varphi )\,d\varphi =\epsilon _{m}\pi \delta _{|m|,|m'|},$

$\int _{0}^{2\pi }\sin(m\varphi )\sin(m'\varphi )\,d\varphi =(-1)^{m+m'}\pi \delta _{|m|,|m'|};\quad m\neq 0,$

$\int _{0}^{2\pi }\cos(m\varphi )\sin(m'\varphi )\,d\varphi =0,$

其中 $\epsilon _{m}$ 称为Neumann因子，其数值为 2 如果满足 $m=0$ ，数值为 1，如果 $m\neq 0$.

径向与角度正交性

$\int Z_{n}^{m}(\rho ,\varphi )Z_{n'}^{m'}(\rho ,\varphi )\,d^{2}r={\frac {\epsilon _{m}\pi }{2n+2}}\delta _{n,n'}\delta _{m,m'},$

其中 $d^{2}r=\rho \,d\rho \,d\varphi$ 为 雅可比矩阵

$n-m$ 与 $n'-m'$ 都是偶数.

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