# 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