泽尔尼克多项式

泽尔尼克多项式也可以表示为超几何函数

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

Noll 用一个J数字表示 [n,m]:如下表

由于

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

其中${\displaystyle k_{j}}$因j而异，

${\displaystyle k_{1}=1}$

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

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

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

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

必须先归一化

令${\displaystyle Z_{j}=Z_{j}/{\sqrt {(}}k_{j})}$

使得

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

归一化泽尔尼克多项式以Noll序列排列如下：

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

径向正交性

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

角度正交性

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

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

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

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

径向与角度正交性

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

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

${\displaystyle n-m}$ 与 ${\displaystyle n'-m'}$ 都是偶数.

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