I measured some wavefront aberrations using a Shack-Hartmann sensor. The output is a finite set of Zernike coefficients $\{c_{ij}\}$. My wavefront is $$W(\rho, \theta) = \sum_{ij}c_{ij}Z_i^j$$ with $Z_i^j$ the Zernike polynomials.

The definition of the RMS error (in the above link, and also in general) is $$\sigma^2 = \int d\theta \int d\rho \rho (W(\rho, \theta) - \bar{W}(\rho, \theta))^2 = \langle W^2 \rangle - \langle W \rangle^2$$ where $$\bar{W} = \langle W \rangle = \int_0^{2\pi} d\theta \int_0^{1} d\rho \rho (W(\rho, \theta)$$

The link then says that this evaluates to $$\sigma^2 = \sum_{ij}|c_{ij}|^2$$

I see that the first term evaluates to that due to orthogonality relations. But I think $\langle W \rangle \neq 0$, since $$Z_n^m(\rho, \theta) = R_n^m(\rho, \theta)\cos{(m\theta)}$$ (or for a different $m$ with sine instead of cosine) and I know that $$\int_0^{2\pi} cos^2 = \int sin^2 = \pi$$

Where am I wrong?


1 Answer 1


The perfect wavefront should be flat i.e. $W_{\text{perfect}}(\rho, \theta) = 0 $. However, your flat wavefront can be tilted and shifted with respect to your measurement plane, which is expressed by the first three Zernike polynomials: $Z_0^0$, $Z_1^{-1}$, $Z_1^1$. For this reason, the authors of the paper you have linked are suggesting that: $\sigma^2 = \sum_{j=3}C^2_j$ (notice that summation starts from 3). The "$\bar{W}$" in the integral that they've presented stands for "mean wavefront optical path difference" in a given point which accounts for a "displacement" of your wavefront from a flat surface of your aperture/measurement plane. Numerically you could implement $\bar{W}(\rho, \theta)$ as a "z" value of a surface fitted to your measured wavefront.

If we assume that the measured (or already compensated) wavefront is "flat" in terms of its total tilt or displacement, then you would calculate RMS error in the following way:

$$\sigma^2 = \int_\text{unit disk} (W(\rho, \theta)-W_{\text{perfect}})^2 \rho \text{d}\rho\text{d}\theta = \int_\text{unit disk} (W(\rho, \theta))^2 \rho \text{d}\rho\text{d}\theta$$.

Please notice here, that I've rescaled the radius of the aperture to 1 as the Zernike polynomials are orthogonal in those boundaries. Your decomposition should take that into account. Also please remember that the Zernike polynomials are not orthonormal, i.e. $\langle Z_i^j , Z_k^l\rangle = C_{ijkl} \delta_{ik}\delta_{jl}$, where constant can be different from 1 for different polynomials. This fact should also be included in your decomposition (for details see Wikipedia article). If you used a commercially available device your coefficients are probably correctly processed. If so, they should be prompted with a correct unit, for example $\mu$m.

Finally: Assuming that all the normalization conditions are met, the integral above can be seen as an inner product of a vector of your decomposed coefficients with itself so:

$$\sigma^2 = \int_\text{unit disk} (W(\rho, \theta))^2 \rho \text{d}\rho\text{d}\theta = \langle W, W \rangle = \sum C_j^2$$.


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