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?

  • $\begingroup$ hint: Orthogonal polynomials and second hint: study Functional analysis which treat functions "similar" like vectors in linear algebra and define operations on functions such as dot product, construct basis in functional space (similar to basis in linear algebra...) $\endgroup$ Commented May 19, 2021 at 16:40
  • $\begingroup$ I was confused by this question until I looked up the Zernike polynomials and discovered they are defined on the (two-dimensional) unit disk instead of the (one-dimensional) unit circle. $\endgroup$
    – d_b
    Commented May 19, 2021 at 17:43
  • $\begingroup$ Thank you @KamilKiełczewski, I'll look up functional analysis $\endgroup$ Commented May 19, 2021 at 18:58
  • $\begingroup$ Thanks @d_b, I edited the question accordingly $\endgroup$ Commented May 19, 2021 at 18:58

1 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:

enter image description here

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.

  • $\begingroup$ Thank you for your answer! I'm familiar with the concept of orthonormality, it's just that I've mostly worked with cartesian coordinates. The thing that got me vexed here is that if we have an infinite number of Zernike polynomials and they form an orthogonal base, then is the dimension infinite too? $\endgroup$ Commented May 19, 2021 at 18:55
  • $\begingroup$ @schrodingal Linear algebra works on finite spaces (usually) - but functional analysis "works" on functional spaces which usually have infinite dimensions (which is quite typical situation there)... (e.g. orthonal basis there has usually infinite number of elements (functions)... ) $\endgroup$ Commented May 19, 2021 at 20:19
  • $\begingroup$ @schrodingal I'll edit. $\endgroup$
    – JEB
    Commented May 19, 2021 at 22:05

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