How do you convert a Stokes vector into a Jones vector? I am only concerned about fully polarised light, and I need to convert the Stokes parameters (or the azimuth and ellipticity angles) as measured from a polarimeter into Jones vectors for further analysis.
A trivial example, for right hand circular polarisation. $$\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \\ \end{pmatrix} \to \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -i \end{pmatrix} $$
A Jones vector is transformed to its Stokes vector by the Hopf Map:
$$(z_1,\,z_2)\mapsto \left(2\,z_1\,z_2^\ast,\,|z_1|^2-|z_2|^2\right)$$
i.e.
$$s_x = 2\,\mathrm{Re}(z_1\,z_2^\ast),\quad s_y = 2\,\mathrm{Im}(z_1\,z_2^\ast),\quad s_z=|z_1|^2-|z_2|^2$$
The three Stokes parameters are the real and imaginary parts of the complex value $2\,z_1\,z_2^\ast$ and the real value $|z_1|^2-|z_2|^2$. The forth Stokes parameter is not relevant here as Jones vectors describe pure states, so the forth parameter $|z_1|^2 + |z_2|^2$ is always constant and equal to unity.
The Hopf map is, as you will be aware, a many to one map, as described in MikeA99's Answer. So when we invert it, we don't get a single answer, we get an equivalence class of Jones vectors equivalent modulo a phase $e^{i\,\varphi}$ common to both Jones vector elements. Moreover, we can only invert the Stokes parameters of pure polarization states, i.e. those with $s_x^2+s_y^2+s_z^2=1$, as opposed to partially depolarized quantum mixtures, which have $s_x^2+s_y^2+s_z^2<1$. We can still, however, invert the polarized part of these mixtures though.
The equivalence classes are Fibers of the Hopf Fibration. One representative of the equivalence class that corresponds to $(s_x,\,s_y,\,s_z)$ is given by the map:
$$\begin{equation} \label{InverseHopfMap} (s_x,\,s_y,\,s_z)\mapsto\left\{ \begin{array}{ll} \displaystyle{\left(0+\frac{s_x}{\sqrt{2}\sqrt{1+s_z}}\,i,\,\frac{s_y}{\sqrt{2}\sqrt{1+s_z}}+\sqrt{\frac{1+s_z}{2}}\,i\right)}&\displaystyle{s_z\neq -1}\\\\ \displaystyle{(i,\,0)}&\displaystyle{s_z = -1} \end{array} \right. \end{equation}$$
i.e.
$$z_1 = \frac{s_x}{\sqrt{2}\sqrt{1+s_z}}\,i,\quad z_2 = \frac{s_y+(1+s_z)\,i}{\sqrt{2}\,\sqrt{1+s_z}}\quad\text{if}\quad s_z\neq-1;\quad z_1=i,\;z_2=0 \quad\text{if}\quad s_z=-1$$
whence all other class members can be found by multiplying both $z_1,\,z_2$ by a common phase factor.
To visualize these maps, one can imagine the Jones vectors as unit quaternions, so that $(z_1,\,z_2) \cong \mathrm{Re}(z_1) + \mathrm{Im}(z_1)\,\mathbf{i} + \mathrm{Re}(z_2)\,\mathbf{j} + \mathrm{Im}(z_2)\,\mathbf{k}$. In this representation, multiplication of both Jones vector elements by a common phase factor corresponds to after-multiplication (recall quaternion multiplication is non-commutative) by the unit quaternion $\exp(\theta\,\mathbf{k}),\,\theta\in\mathbb{R}$. By doing so, any unit quaterion can be translated to the so-called equatorial sphere $\mathbb{E}$ of pure quaternions with no real part. Indeed, there are two phases that will do this; one puts the quaternion on the Northern hemisphere, the other on the antipodean point on the Southern hemisphere. We chose one hemisphere consistently, say the North. Nextly as in the diagram below, we project the Northern hemisphere point $Q$ to the point $Q^\prime$ on the 2-sphere $\left\{q_x\,\mathbf{i}+q_y\,\mathbf{j}+q_z\,\mathbf{k}|\;\left(q_x-\frac{1}{2}\right)^2+q_y^2+q_z^2=\frac{1}{2};\;q_x \geq0\right\}$ of radius $\frac{1}{2}$ centered at $\frac{1}{2}\,\mathbf{i}$ with the center $C$ of $\mathbb{E}$ as the projection point. It is easy to show that $\angle Q\,C\,N$ is half of $\angle Q^\prime\,C_2\,N$. Witness that antipodal points on $\mathbb{E}$ map to the same point we simply multiply an antipode by the phase factor $-1$ and that the whole equator of $\mathbb{E}$ maps to the South pole of the smaller 2-sphere in the figure. The inverse map to from the Stokes parameters on the Poincaré sphere to the equatorial sphere $\mathbb{E}$ is readily visualized by the inverse projection, and then after-multiplication by the phase factor $\exp(\theta\,\mathbf{k}),\,\theta\in\mathbb{R}$ reaches every Jones vector that condenses under the Hopf map to the point in question on the Poincaré sphere.
Posting to an old link, but I thought I'd leave this for future searchers.
Yes, a Jones vector can only represent the perfectly polarized part of the Stokes vector. And yes, because there is an arbitrary phase, there are an infinity of Jones' vectors that fit a given Stokes vector representation. However, that doesn't prevent us from constructing a Jones vector from a Stokes vector. It just means we have to assign an arbitrary phase, and be careful not to use it for anything that would trip on that ambiguity (such as interference)
Here is a little Matlab program to construct a Jones vector representing the perfectly polarized part of a Stokes vector and also spit out the degree of polarization in case you would like to handle the unpolarized part separately. I have chosen the phase so that the horizontal component of the Jones vector is always purely real.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Stokes2Jones
%
% Takes a stokes vector as an input and creates a matching Jones vector.
% The Jones vector only has the polarized part of the intensity, of course.
% Also, since the Jones vector can differ by an arbitrary phase, the phase
% is chosen which makes the horizontal component purely real
%
% Inputs:
% S = a stokes vector entered as a 4 component row vector
%
% Outputs:
% Jv = a 2 component complex column vector which is the Jones vector
% p = the degree of polarization from the Stokes vector so you can do
% something with the unpolarized part separately if you like
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [Jv,p] = Stokes2Jones(S)
% Calculate the degree of polarization
p=sqrt(S(2)^2+S(3)^2+S(4)^2)/S(1);
% Normalize the Stokes parameters (first one will be 1, of course)
Q=S(2)/(S(1)*p);
U=S(3)/(S(1)*p);
V=S(4)/(S(1)*p);
% And construct the 2 Jones components
A=sqrt((1+Q)/2);
if A == 0;
B=1;
else
B = U/(2*A)-i*V/(2*A);
end;
% put them together in a vector with the amplitude of the polarized part
Jv = sqrt(S(1)*p)*[A;B];
return
Stokes parameters are written in terms of intensities; Jones vectors use the electric field components. It's easy to go from Jones to Stokes, but the other way does not have a general method.
See < http://en.wikipedia.org/wiki/Mueller_calculus#Mueller_vs._Jones_calculi>
You can find tables of equivalents in the literature; but these are mostly for special cases. Check an optics textbook like Fowles or Hecht, and look in their chapter on polarized light, and the references.
Shurcliff's monograph on Polarized Light certainly discusses this, but is long out of print.
Let's assume a fully polarized Stokes vector, i.e. $I=p=1$. The last three components of the Stokes vector, $(S_1,S_2,S_3)$, are then parametrized according to \begin{equation} \vec S=\pmatrix{ \cos 2\psi\cos2\chi\\ \sin 2\psi\cos2\chi\\ \sin 2\chi\tag{1} } \end{equation} In the $HV$ basis where $H=(1,0)^T, D=\tfrac 1{\sqrt{2}}(1,1)^T$ and $R=\tfrac 1 {\sqrt 2}(1, -i)^T$, a Jones Vector that reproduces this Stokes vector is given by \begin{align} J=\pmatrix{ \cos\psi\cos\chi+i\sin\psi\sin\chi\\ \sin\psi\cos\chi-i\cos\psi\sin\chi\tag{2} } \end{align} This Jones vector is of course up to a global phase factor. It seems that this expression could be simplified further but I could not find a nicer expression.
To go the other way around, start with an arbitrary Jones vector \begin{align} J=\pmatrix{ \cos\theta e^{i\alpha}\\ \sin\theta\tag{3} } \end{align} Then the corresponding Stokes vector is given by \begin{align} \vec S=\pmatrix{ \cos 2\theta\\ \sin2\theta \cos\alpha\\ \sin2\theta \cos\alpha }\tag{4} \end{align}
Finally, we can express these quantities using the density operator, which allows us to connect the degree of polarization $p$ more closely to the Jones vector. The density matrix for a pure state is given by \begin{align} \rho=|\psi\rangle\langle \psi |\tag{5} \end{align} while for a mixed state it is given by \begin{align} \rho=\sum_i p_i|\psi_i\rangle\langle \psi_i |\tag{6} \end{align} where $\sum_i p_i=1$ and $\psi_i$ form an orthonormal basis.
For a general Jones vector, this matrix is given by \begin{align} \rho=\left( \begin{array}{cc} \cos ^2(\theta ) & e^{i \alpha } \sin (\theta ) \cos (\theta ) \\ e^{-i \alpha } \sin (\theta ) \cos (\theta ) & \sin ^2(\theta ) \\ \end{array} \right)\tag{7} \end{align} To find the density matrix for the Stokes vector, we have to create a mixed and we have to define an orthonormal basis first. We can easiliy construct a vector that is orthogonal to (2) using \begin{align} J_\perp=\pmatrix{-J_2^*\\J_1^*}=\pmatrix{ -\sin\psi\cos\chi-i\cos\psi\sin\chi\\ \cos\psi\cos\chi-i\sin\psi\sin\chi } \tag{8} \end{align} Finally, the density matrix of the Stokes vector is given by \begin{align} \rho=(\tfrac 1 2+\tfrac{p}{2})|J\rangle\langle J|+(\tfrac 1 2-\tfrac{p}{2})|J_\perp\rangle\langle J\perp| \end{align}
\begin{align} \rho=\left( \begin{array}{cc} \frac{1}{2} (1+p \cos (2 \xi ) \cos (2 \psi )) & \frac{1}{2} p (\cos (2 \xi ) \sin (2 \psi )+i \sin (2 \xi )) \\ \frac{1}{2} p (\cos (2 \xi ) \sin (2 \psi )-i \sin (2 \xi )) & \frac{1}{2} (1-p \cos (2 \xi ) \cos (2 \psi )) \\ \end{array} \right) \tag{9} \end{align}
To reconstruct the Stokes vector, we can use$^1$: \begin{align} S_i=S_0\mathrm{Tr}\left( \rho \,\hat\sigma_i \right) \end{align} where \begin{align} &\hat\sigma_0=1\!\!1, &&\hat\sigma_1=|R\rangle\langle L|+|L\rangle\langle R|, \\ &\hat\sigma_2=i(|R\rangle\langle L|-|L\rangle\langle R|), &&\hat\sigma_3=|R\rangle\langle R|-|L\rangle\langle L| \end{align} Note that $\hat \sigma_2$ has a minus sign compared to the source$^1$. This is because the authors use an alternative definition of $D$.
Alternatively, we could construct a density matrix straight from the Stokes parameters using \begin{align} \rho=\frac 1 2\sum_{i=0}^3 \frac{S_i}{S_0}\hat\sigma_i \end{align}
Disclaimer: I derived a lot of these formulas myself. Although I checked them with Mathematica, if you want to use these for anything serious you might want to check them yourself.
