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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 variable. 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}<br /> E(aa)&E(ac)&E(ab)&E(ad)\\<br /> E(ca)&E(cc)&E(cb)&E(cd)\\<br /> E(ba)&E(bc)&E(bb)&E(bd)\\<br /> E(da)&E(dc)&E(db)&E(dd)<br /> \end{bmatrix}=\begin{bmatrix}<br /> E(aa)&E(ac)&0&E(ad)\\<br /> E(ac)&E(cc)&E(bc)&0\\<br /> 0&E(bc)&E(aa)&E(bd)\\<br /> E(ad)&0&E(bd)&E(cc)<br /> \end{bmatrix}
This matrix has 6 free parameters. However, one often states that the polarization is determined by the coherency matrix
J=\begin{bmatrix}<br /> E(u_{x}u_{x}^{\ast})&E(u_{x}u_{y}^{\ast})\\<br /> E(u_{y}u_{x}^{\ast})&E(u_{y}u_{y}^{\ast})<br /> \end{bmatrix}=\begin{bmatrix}<br /> 2E(aa)&E(ac)+E(bd)+i(E(bc)-E(ad))\\<br /> E(ac)+E(bd)-i(E(bc)-E(ad))&2E(cc)<br /> \end{bmatrix}<br />
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?
\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 variable. 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}<br /> E(aa)&E(ac)&E(ab)&E(ad)\\<br /> E(ca)&E(cc)&E(cb)&E(cd)\\<br /> E(ba)&E(bc)&E(bb)&E(bd)\\<br /> E(da)&E(dc)&E(db)&E(dd)<br /> \end{bmatrix}=\begin{bmatrix}<br /> E(aa)&E(ac)&0&E(ad)\\<br /> E(ac)&E(cc)&E(bc)&0\\<br /> 0&E(bc)&E(aa)&E(bd)\\<br /> E(ad)&0&E(bd)&E(cc)<br /> \end{bmatrix}
This matrix has 6 free parameters. However, one often states that the polarization is determined by the coherency matrix
J=\begin{bmatrix}<br /> E(u_{x}u_{x}^{\ast})&E(u_{x}u_{y}^{\ast})\\<br /> E(u_{y}u_{x}^{\ast})&E(u_{y}u_{y}^{\ast})<br /> \end{bmatrix}=\begin{bmatrix}<br /> 2E(aa)&E(ac)+E(bd)+i(E(bc)-E(ad))\\<br /> E(ac)+E(bd)-i(E(bc)-E(ad))&2E(cc)<br /> \end{bmatrix}<br />
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?