Coherency matrix of partially polarized light doesn't contain all information on polarization state?

Question

The electric field of quasi-monochromatic, partially polarized light can be expressed by the following random process (Goodman, Statistical optics) $$\bar{E}(t,\bar{x})=u_{x}(t,\bar{x})\bar{e}_{x}+u_{y}(t,\bar{y})\bar{e}_{y}$$ $$u_{x}(t,\bar{x})=\Psi_{x} e^{i(\bar{k}\cdot\bar{x}-\omega t)}$$ $$u_{y}(t,\bar{x})=\Psi_{y} e^{i(\bar{k}\cdot\bar{x}-\omega t)}$$ where $\Psi_{x}$ and $\Psi_{y}$ are radom phasor sums (which are circular complex Gaussian random variables). The joint statistics of $u_{x}=a+bi$ and $u_{y}=c+di$ describe the polarization state. Knowing that $E(u_{x})=E(u_{y})=0$, the covariance matrix of these two complex is given by $$C=\begin{bmatrix} E(aa)&E(ac)&E(ab)&E(ad)\\ E(ca)&E(cc)&E(cb)&E(cd)\\ E(ba)&E(bc)&E(bb)&E(bd)\\ E(da)&E(dc)&E(db)&E(dd) \end{bmatrix}=\begin{bmatrix} E(aa)&E(ac)&0&E(ad)\\ E(ac)&E(cc)&E(bc)&0\\ 0&E(bc)&E(aa)&E(bd)\\ E(ad)&0&E(bd)&E(cc) \end{bmatrix}$$ This matrix has 6 free parameters. However, one often states that the polarization is determined by the coherency matrix $$J=\begin{bmatrix} E(u_{x}u_{x}^{\ast})&E(u_{x}u_{y}^{\ast})\\ E(u_{y}u_{x}^{\ast})&E(u_{y}u_{y}^{\ast}) \end{bmatrix}=\begin{bmatrix} 2E(aa)&E(ac)+E(bd)+i(E(bc)-E(ad))\\ E(ac)+E(bd)-i(E(bc)-E(ad))&2E(cc) \end{bmatrix} $$ which has only $4$ free parameters because two pairs of free parameters of $C$ are combined in two free parameters in $J$. So we lost 2 degrees of freedom. Does this mean that $E(ac)=E(bd)$ and $E(bc)=-E(ad)$ or does this mean that the coherency matrix doesn't contain all information on the polarization state?

Answer 1

Jones Matrices cannot express partial polarization.

For that you need Stokes Vectors and the related Mueller Matrices (for transforming one Stokes vector into another).

Interestingly a Stokes Vector can be trivially separated into a fully polarized vector and a fully unpolarized vector.