Do Zernike polynomials form an orthogonal basis of $L^2 ( \mathbb{D} )$

polynomialsreference-requestspecial functions

The family of Zernike polynomials is defined as follow over the unit disc $\mathbb{D}= \{ x \in \mathbb{R}^2, \ \lVert x \rVert \leq 1 \}$.
For $n \geq 0$ and $0 \leq m \leq n$ such that $n-m$ is even, we define the Zernike polynomials for $r \in [0,1]$ and $\theta \in [0,2\pi)$ as
$$Z_n^m (r, \theta) = P_n^m (r) \cos(m \theta)\quad \text{ and } \quad Z_{n}^{-m} (r,\theta) = P_n^m (r) \sin (m \theta),$$
where $$P_n^m (r) = \sum_{k=0}^{(n-m)/2} \frac{(-1)^k (n-k)!}{k!(\frac{n+m}{2}-k)!(\frac{n-m}{2}-k)!} r^{n – 2k}.$$
The Zernike polynomials are known to be orthogonal. Up to normalization, do they form an orthonormal basis of square integrable functions over $\mathbb{D}$, i.e. is the system complete?

Best Answer

The (complex) Zernike polynomials $V_{n}^{l}(x,y)$, of total degree $n$, with $|l|\leq n,~n-|l|\text{ even}$, are such that $$ V_{n}^{l}(x,y)=R_{n}^{l}(\rho)e^{il\varphi},\quad \text{with }x=\rho\cos\varphi,~y=\rho\sin\varphi, $$ where the radial polynomials $R_{n}^{l}(\rho)$, of degree $n$, satisfy the orthogonality relations $$ \int_{0}^{1}R_{n}^{l}(\rho)R_{m}^{l}(\rho)\rho d\rho=c_{n}^{l}\delta_{n,m}, $$ for some nonzero constant $c_{n}^{l}$. There are precisely $(n+1)(n+2)/2$ linearly independent polynomials $V_{n}^{l}$ of degree $\leq n$, and thus every monomial $x^{i}y^{j}$ is a linear combination of a finite number of $V_{n}^{l}$. From the Weierstrass theorem, it follows that the set is complete in $L^{2}(\mathbb{D})$.

Some details about the Zernike polynomials can be found in Appendix VII of

M. Born, E. Wolf, Principles of optics : electromagnetic theory of propagation, interference, and diffraction of light, 7th ed. (2019) Cambridge University Press

Related Question