What does it mean that Zernike's polynomials form a orthogonal basis on the unit disk?

Question

I am doing a project on adaptive optics and I would like to understand a little more about Zernike's polynomials. What does it mean that they form an orthogonal basis on the unit circle? What dimension is that basis seeing that we can have infinite Zernike's polynomials?

Answer 1

It means that any function (sensible) $f(\rho, \phi)$ defined on the unit circle can be expanded into a linear combination of Zernike polynomials:

$$ f(\rho, \phi) = \sum_{m,n}{a_{m,n}R_n^m(\rho)\cos{m\phi}+b_{m,n}R_n^{m}(\rho)\sin{m\phi}}$$

with suitable bounds on $m$ for all integer $n\ge 0$. That it works for any (sensible) function means the basis is complete.

When defining orthogonality over a shape, there may be some weighting factor for the Jacobian, and sometime one talks about orthogonality with respect to a weighting function. Generally one can say orthonormal means that if:

$$ f(\rho, \phi) = R_{n'}^{m'}(\rho)\cos{m'\phi} $$

then "normal" means:

$$ a_{n'}^{m'} = 1 $$

and "ortho" means all other coefficients are zero. That is:

$$ a_n^m = \delta_{nn'}\delta_{mm'} $$ $$ b_n^m = 0 $$

with similar result for $f(\rho, \phi)=R_{n'}^{m'}\sin{m\phi}$. (Just swap the rolls of $a$ and $b$).

Note the similarity to a Fourier series defined over a unit square, where the function is factored in series over each coordinate.

Because the domain of the function is the cylindrically symmetric unit circle, the problem is factored into a radial component and a cylindrical component. The cylindrical part is just a Fourier series on the unit ring. One can think of each term as a standing wave on a circle.

Likewise for the radial part. It's not a simple trig-function because the area element is $\rho d\rho d\phi$, as opposed to $dx dy$.

You can also compare them with the https://en.wikipedia.org/wiki/Solid_harmonics

Regarding the dimensionality of the space, for simplicity look at Fourier series of odd functions in 1D on an interval $[-\pi/2, +\pi/2]$. Any continuous (odd, just for simplicity) function can be represented as:

$$ g(x) = \sum_{n=1}^{\infty}a_n\times\big(\sqrt{\frac{\pi}2}\sin{nx}\big) $$

The basis states, $\hat e_n$, of this function space are:

$$ \hat e_n(x)\equiv \sqrt{\frac 2{\pi}}\sin{nx}$$

However, to make it look more like a vector space, we can use the notation:

$$\hat e_n(x) \rightarrow|n\rangle $$

e.g., $|3\rangle \propto \sin{3x}$

The coefficients are defined by the standard definition of the inner product:

$$ a_n = \langle g|n\rangle \equiv \int_{-\frac{\pi}2}^{+\frac{\pi}2} g(x)\hat e_n(x)dx = \sqrt{\frac 2{\pi}}\int_{-\frac{\pi}2}^{+\frac{\pi}2} g(x)\sin{nx}dx $$

That may bother you. In a typical inner product on $\mathbb{R}^N$, you sum the products over each dimension in the space:

$$ \vec u\cdot\vec v =u_1v_1+u_2v_2+\cdots u_Nv_N$$

while in our function space we have a countably infinite number of basis vectors $|1\rangle, |2\rangle,|3\rangle,\ldots$, and the integral is continuos.

To fix that you can define an inner product based solely on the norm:

$$\vec{u}\cdot\vec{v} \equiv \frac 1 4(||\vec{u}+\vec{v}||^2-||\vec{u}-\vec{v}||^2)$$

That works here, with the norm defined by:

$$||f(x)||^2 = \int_{-\frac{\pi}2}^{+\frac{\pi}2}f(x)^2dx$$

While it is not computationally efficient (at all), you can say:

$$a_n=\frac 1 4\big(||f(x)+\hat e_n(x)||^2- ||f(x)-\hat e_n(x)||^2\big)$$

Orthonormality is proven by showing:

$$ \langle n|m\rangle =\delta_{nm}$$

and then there's no worries.

The fact that $x\in[-\pi/2, \pi/2]$, is continuous with $\aleph^1$ cardinality, while $|n\rangle\,\forall n\in\mathbb{N}$ has $\aleph^0$ cardinality I believe is due to the requirement that $g(x)$ be "reasonable", but that's really a math question.

In 2 dimensions, over a square region, you would just define:

$$|n,m\rangle = \frac 2{\pi}\sin{nx}\sin{my}$$

with $(n,m)\in\mathbb{N} \otimes\mathbb{N}$

and be done.

With the unit circle factored in $\rho$ and $\phi$, representations of rotation groups come in to play (c.f. the spherical harmonics), so that for radial polynomial of degree $n$, only a finite number of angular functions ($m$) are possible.

Looking at the figure, you can see these are geometric constraints:

For $n=0$, $R_0^0=1$ is constant, so that $m=0$. In the 1st non-trivial polynomial, $n=1$, and the basis is:

$$ \rho\cos \phi = y $$ $$ \rho\sin \phi = x $$

AKA: the two possible dipoles.

At degree 2, you have the 2D quadrupole shapes:

$$ \rho^2\cos{2\phi} =y^2-x^2$$ $$ (2\rho^2-1) = 2(x^2+y^2) - 1$$ $$ \rho^2\sin{2\phi} =2xy$$

and so-on.