Density matrix of a single qubit as a function of its Stokes Parameters $\newcommand{\bra}[1]{\left\langle#1\right|}
\newcommand{\ket}[1]{\left|#1\right\rangle}
\newcommand{\prom}[1]{\langle{#1}\rangle}
\newcommand{\matrixel}[3]{\bra{#1}{#2}\ket{#3}}$
How can I prove that the density matrix for a single qubit as a function of its Stokes parameters can be expressed by
$$\hat\rho = \frac{1}{2}\sum_{i=0}^{3} \frac{\mathcal{S}_i}{\mathcal{S}_0}\hat\sigma_i?$$
where $\mathcal{S_i}$ are the Stokes parameters defined as
\begin{align}
  \mathcal{S}_0 &\equiv  \mathcal{N}(\matrixel{R}{\hat\rho}{R} + \matrixel{L}{\hat\rho}{L}),\nonumber \\
  \mathcal{S}_1 &\equiv  \mathcal{N}(\matrixel{R}{\hat\rho}{L} + \matrixel{L}{\hat\rho}{R}),\nonumber \\
  \mathcal{S}_2 &\equiv  \mathcal{N}i(\matrixel{R}{\hat\rho}{L} - \matrixel{L}{\hat\rho}{R}),\nonumber \\
  \mathcal{S}_3 &\equiv  \mathcal{N}(\matrixel{R}{\hat\rho}{R} - \matrixel{L}{\hat\rho}{L}).
\end{align}
and where $\mathcal{N}$ is a constant dependent on the detector efficiency and light intensity.
I think it has something to do with the coherency matrix but I am not pretty sure if that is correct. If so, how can I prove it?
 A: Since this seems a homework exercise, here's a sketch: I'm not sure about the $\mathcal{N}$-part in the formular, but in general:
Note that $\sigma_i$ form an orthonormal basis of the Hermitian matrices according to the Hilbert-Schmidt inner-product ($\langle A,B\rangle:=\operatorname{tr} (A^{\dagger}B)\rangle$). This means that you can write 
$$ \rho=\sum_{i=0}^3 k_i \sigma_i $$
Now, you can take the expectation values $\operatorname{tr}(\sigma_i \rho)=\operatorname{tr}(\sigma_i \sum_j k_j\sigma_j)$ and compare the two sides. Note that you can write e.g. $\sigma_0=|R\rangle\langle R|+|L\rangle\langle L|$. Using the orthogonality on the right hand side you'll get $k_i=\mathcal{S}_i/\mathcal{N}$ (as I said, I don't quite get the $\mathcal{N}$-parameter, but this doesn't play any role anyway, as it gets divided out afterwards).
Finally, you have to make sure that $\rho$ is properly normalized, so you have to divide the right hand side by $\operatorname{tr}(\rho)=\operatorname{tr}(\sigma_0\rho)$. Putting everything together, you'll obtain the formula above.
EDIT: Let's have a look at coherency matrices and the Jones vector. The Jones vector (as in here) is a vector $e\in \mathbb{C}^2$ (more precisely, in the projective version thereof, since the global phase doesn't really matter). It is a pure state of polarization. By, definition, the coherency matrix is given by $ee^{\dagger}$, hence it is an operator in $\mathcal{B}(\mathbb{C}^2)=\mathbb{C}^{2\times 2}$. This is exactly the definition of the density matrix of a pure state. This gives you the argument: Since the general coherency matrix is a $2\times 2$ complex matrix (as a qubit-matrix) and behaves exactly like a mixture of pure polarization states, which are states in $P\mathbb{C}^2$ (as are quantum states), from a purely mathematical perspective, where the density matrix is a normalized mixture of pure quantum states, the polarization matrix is also a density matrix. 
This means, you can treat polarized light like a qubit with a qubit density matrix by identifying the Jones vector of a pure polarization state with a pure quantum state. You can actually also see this from the other direction: You can measure polarization in different ways: one basis is left-right, another is circular-anticircular a third one is 45°-polarization. If you have a look at the Jones calculus again (first table, exchange $|H\rangle$ by $|L\rangle$ and $|V\rangle$ by $|R\rangle$), you can immediately find the three spin-axes: If you fix left-right polarization, which is there called $|H\rangle,|V\rangle$ polarization, the measurement is then by the Pauli z, i.e. $\operatorname{tr}(\sigma_3 |L\rangle\langle L|)=1$ and $\operatorname{tr}(\sigma_3 |R\rangle\langle R|)=-1$ where 
$$ \sigma_3:=\begin{pmatrix}{} 1 & 0 \\ 0 & -1\end{pmatrix}$$
The other two bases of polarization are measured by $\sigma_1$ and $\sigma_2$ respectively. This gives a full equivalence between a general coherency matrix and a general spin-qubit matrix, hence the two pictures can be used interchangedly.
