# Representing a polarization vector for light as a 'manifold of two state'

Context: I was reading a paper (Phys. Rev. A 68, 052307) which involved mesoscopic coherent states of light. There, in order to calculate the uncertainty of a single PoincarĂ© angle(see polarization parametrization, for both polarization elipse and PoincarĂ© sphere) http://en.wikipedia.org/wiki/Polarization_%28waves%29#Parameterization) due to shot noise (by calculating $|\langle \Psi_k|\Psi_a\rangle|^2$ where $\Psi_k$ and $\Psi_a$ are two mesoscopic coherent pulses with the same amplitude and on a same great circle on the poincarĂ© sphere)

I got found the following representation presented as a "manifold of two state in cartesian $(x,y)$ at $45^o$ from horizontal", where $(\Theta_k, \Phi_k)$ are PoincarĂ© angles and $\gamma, \delta$ are projections on axes $x,y$, and $\alpha$ is the coherent amplitude:

$|\Psi(\Theta_k, \Phi_k)\rangle=|\alpha \gamma(\Theta_k, \Phi_k) \rangle \otimes |\alpha \delta(\Theta_k, \Phi_k) \rangle$

$\gamma = (1-i)e^{i \frac{\Phi_k}{2}} cos\frac{\Theta}{2} +(1+i)e^{i \frac{-\Phi_k}{2}}sin\frac{\Theta}{2}$

$\delta = (1+i)e^{i \frac{\Phi_k}{2}} cos\frac{\Theta}{2} +(1-i)e^{i \frac{-\Phi_k}{2}}sin\frac{\Theta}{2}$

And I don't understand. Thinking 'classicaly', writing with the polarization elipse (parameters $\Psi$ and $\chi$) I got

$\vec{E(t)}=|E|e^{i\omega t}[cos\chi,\pm i sin\chi]R_\Psi$

Where $R_\Psi$ is a rotation of $\Psi$ and the $\pm i$ account for clockwise and counter clockwise, but its different. I've put this here just for a base of what I expect the answer to cover. I've had many ideas of 'mappings' but none of them were good.

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Could you put a little more effort into asking a clear question? What is the context of your question? What are you trying to represent? Where did you find this representation? – user1504 Nov 28 '12 at 16:37
I added a lot of context. Let me know if its not enough. – galmeida Nov 28 '12 at 16:57
Maybe you could post a reference to the paper you were reading, this might help. – au700 Nov 29 '12 at 0:10
Done.LMK if its not enough. – galmeida Nov 29 '12 at 1:07