Talk:Jones calculus

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Do we know who Mr. Jones who invented this was? I ask because I suspect it was Isaac Newton's contemporary Sir William Jones and we might as well link to that if so. -- Paul Drye

I'm not sure. --Georg Muntingh

I very much doubt it. Try this:

Jones, R. C., ``New Calculus for the Treatment of Optical Systems I. Description and Discussion of the Calculus, Journal of the Optical Society of America, 31, 488-493 (1941)

I think we should expand this section to include elliptical jones vectors and how to get their magnitudes.

The polarizer with azimuth 0:

P(0)={\begin{pmatrix}1&0\\0&0\\\end{pmatrix}}

The rotation by $\varphi$ :

R(\varphi )={\begin{pmatrix}\cos \varphi &-\sin \varphi \\\sin \varphi &\cos \varphi \\\end{pmatrix}}

The polarizer with azimuth $\varphi$ :

P(\varphi )=R(\varphi )P(0)R(-\varphi )={\begin{pmatrix}\cos ^{2}\varphi &\cos \varphi \sin \varphi \\\sin \varphi \cos \varphi &\sin ^{2}\varphi \\\end{pmatrix}}

--HarpyHumming 22:01, 28 Feb 2004 (UTC)

conflicting definitions for right- and left-hand circular

In a couple of places (e.g. "The Depolarisation of Electromagnetic Waves", Beckmann 1968) I've seen a conflicting definition of Jones vectors for right- and left-hand polarisations, where right circular is (1, i) - not the (1, -i) that is quoted here. In this Wiki article, I assume the wave propagates along +z, so E has components in x and y directions. Beckmann assumes the wave propagates along +x, so the corresponding E vectors are y and z (if we stick with a right-handed coordinate system!), thus a right circular wave screws clockwise if viewed in the direction of propagation. A right circular wave, by his definition, has Ey = 1 and Ez = +i, which correspond to Ex = 1 and Ey = +i in this article, hence my confusion. Any ideas?

Rob Granger. —The preceding unsigned comment was added by 213.123.216.147 (talk) 10:39, 11 April 2007 (UTC).[reply]