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## Main Question or Discussion Point

The electric field of quasi-monochromatic, partially polarized light can be expressed by the following random process (Goodman, Statistical optics)

[tex]\bar{E}(t,\bar{x})=u_{x}(t,\bar{x})\bar{e}_{x}+u_{y}(t,\bar{y})\bar{e}_{y}[/tex]

[tex]u_{x}(t,\bar{x})=\Psi_{x} e^{i(\bar{k}\cdot\bar{x}-\omega t)}[/tex]

[tex]u_{y}(t,\bar{x})=\Psi_{y} e^{i(\bar{k}\cdot\bar{x}-\omega t)}[/tex]

where [itex]\Psi_{x}[/itex] and [itex]\Psi_{y}[/itex] are radom phasor sums which are circular complex Gaussian random variable. The joint statistics of [itex]u_{x}=a+bi[/itex] and [itex]u_{y}=c+di[/itex] describe the polarization state. Knowing that [itex]E(u_{x})=E(u_{y})=0[/itex], the covariance matrix of these two complex is given by

[tex]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}[/tex]

This matrix has [itex]6[/itex] free parameters. However, one often states that the polarization is determined by the coherency matrix

[tex]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}

[/tex]

which has only [itex]4[/itex] free parameters because two pairs of free parameters of [itex]C[/itex] are combined in two free parameters in [itex]J[/itex]. So we lost 2 degrees of freedom. Does this mean that [itex]E(ac)=E(bd)[/itex] and [itex]E(bc)=-E(ad)[/itex] or does this mean that the coherency matrix doesn't contain all information on the polarization state?

[tex]\bar{E}(t,\bar{x})=u_{x}(t,\bar{x})\bar{e}_{x}+u_{y}(t,\bar{y})\bar{e}_{y}[/tex]

[tex]u_{x}(t,\bar{x})=\Psi_{x} e^{i(\bar{k}\cdot\bar{x}-\omega t)}[/tex]

[tex]u_{y}(t,\bar{x})=\Psi_{y} e^{i(\bar{k}\cdot\bar{x}-\omega t)}[/tex]

where [itex]\Psi_{x}[/itex] and [itex]\Psi_{y}[/itex] are radom phasor sums which are circular complex Gaussian random variable. The joint statistics of [itex]u_{x}=a+bi[/itex] and [itex]u_{y}=c+di[/itex] describe the polarization state. Knowing that [itex]E(u_{x})=E(u_{y})=0[/itex], the covariance matrix of these two complex is given by

[tex]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}[/tex]

This matrix has [itex]6[/itex] free parameters. However, one often states that the polarization is determined by the coherency matrix

[tex]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}

[/tex]

which has only [itex]4[/itex] free parameters because two pairs of free parameters of [itex]C[/itex] are combined in two free parameters in [itex]J[/itex]. So we lost 2 degrees of freedom. Does this mean that [itex]E(ac)=E(bd)[/itex] and [itex]E(bc)=-E(ad)[/itex] or does this mean that the coherency matrix doesn't contain all information on the polarization state?