How can i mathematically describe two orthogonal states of polarized light? In the Poincaré sphere representation, any two diametrically opposite points are orthogonal polarized states of light.
To prove this i tried using the Stokes vector formalism and it's relation with the spherical coordinate system in the sphere, comparing two points that differ in the 2$\omega$ angle by $\pi$ the result is that the dot product between them vanishes, 
Is that a valid answer? 
Poincaré sphere and the spherical coordinates of the point P, picture from Polarized light, production and use by W. Shurcliff, page 16 : 

 A: From Wikipedia, the Stokes parameters of a fully-polarized state are
$$
S_0 = I, \\
S_1 = I\cos 2\psi\cos 2\chi,\\
S_2 = I\sin 2\psi\cos 2\chi,\\
S_3 = I\sin\chi.
$$


A polarization state with a diametrically opposite point on the same spherical surface must have Stokes parameters $S'_0 = S_0$, $S'_1 = -S_1$, $S'_2 = -S_2$, $S'_3 = -S_3$. Thus, you need to find new angular parameters $\psi'$ and $\chi'$ such that
$$
\cos 2\psi'\cos2\chi' = -\cos 2\psi\cos2\chi,\\
\sin2\psi'\cos2\chi' = -\sin2\psi\cos2\chi,\\
\sin2\chi' = -\sin2\chi.
$$
There are various solutions to this equation system, but we only need to show that one of these corresponds to an orthogonally polarized state. In this convention, $\psi$ is the angle between the semi-major axis and the x-coordinate and $\chi$ is such that the ratio between the semi-minor axis and the semi-major axis is given by $\tan\chi$; with the sign of $\chi$ denoting the handedness of the polarization state. Then, an orthogonally polarized state is found if we rotate the ellipse 90 degrees ($\psi' = \psi + \pi/2$) and change the handedness ($\chi' = -\chi$). It is readily verified that these relations satisfy the system of equations above and thus orthogonally polarized states have diametrically opposite points on the Poincaré sphere.
In terms of the angles of the Poincaré sphere ($2\psi$,$2\chi$), we may write
$$
2\psi' = 2\psi + \pi,\\
2\chi' = -2\chi.
$$
Then, our result is equivalent to the well-known result (see for example, Kittel - Introduction to Solid State Physics, Chapter 1) that a rotation by $\pi$ around an axis followed by a reflection in the plane normal to the rotation axis is equal to the inversion operation which maps any vector $\mathbf{r}$ to $-\mathbf{r}$.
A: Writing 
$$
\boldsymbol{P}=\frac{1}{2}\left(p_0\hat I +p_x\sigma_x+
p_y\sigma_y+p_z\sigma_z\right)
$$
the Stokes parameters are related to the Pauli matrices 
$$
p_x=\langle \sigma_x\rangle \, , \quad 
p_y=\langle \sigma_y\rangle \, ,\quad 
p_z=\langle \sigma_z\rangle \, .
$$
with $p_k=\hbox{Tr}(\boldsymbol{P}\sigma_k)$.  
If $\vert\psi_\pm\rangle$ are any to orthogonal polarization states, then you can write them in terms of eigenstates $\vert \uparrow\rangle$ and $\vert\downarrow\rangle$  as
\begin{align}
\vert \psi_+\rangle &= e^{-i\alpha/2} \cos\textstyle\frac{1}{2}\beta\vert \uparrow\rangle
+e^{i\alpha/2}\sin\textstyle\frac{1}{2}\beta\vert \downarrow\rangle\, ,\\
\vert \psi_-\rangle &= -e^{-i\alpha/2} \sin\textstyle\frac{1}{2}\beta\vert \uparrow\rangle
+e^{i\alpha/2}\cos\frac{1}{2}\beta\vert \downarrow\rangle\, .
\end{align}
If I understand your figure well, the azimuthal angle $\alpha=2\lambda$ in your notation; $0\le \beta\le \pi $ is the angle measured from the North Pole so $\beta=2\omega+\pi$ in your notation.
Assuming unit intensity, the polarization vectors associated with two orthogonal states $\vert \psi_\pm\rangle$ correspond to 
\begin{align}
(p_{+,x},p_{+,y},p_{+,z})&=(\cos\alpha\sin\beta,\sin\alpha\sin\beta,\cos\beta)\, ,\\
(p_{-,x},p_{-,y},p_{-,z})&=(-\cos\alpha\sin\beta-,\sin\alpha\sin\beta,-\cos\beta)
\end{align}
i.e. correspond to diametrically opposite vectors on the Poincare sphere since the second is just the negative of the first.
