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Given the Stokes Vector

$$S_i = I_0\begin{pmatrix} 1 \\ 0.6 \\ -0.2 \\ 0.07 \end{pmatrix}$$

how do I quantify the polarization? Is there a way to say $S_i$ is $x$% right- or left-hand-circular, $y$% right- or left-slant, and $z$% vertical or horizontal? I can easily calculate the degree of polarization,

$$p = \frac{\sqrt{s_1^2+s_2^2+s_3^2}}{s_0}=0.64$$

So the light is 64% polarized, 36% unpolarized. Can I quantify how much of that polarization could be categorized as each of the different types of polarization? From the simple cases, I realize this matches horizontal linear polarization the closest. If I setup the system of equations

$$\begin{pmatrix} 1 \\ 0.6 \\ -0.2 \\ 0.07 \end{pmatrix} = a \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} + b \begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \end{pmatrix} + c \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} + d \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix} + e \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} + f \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix}$$

I only have 4 equations and 6 unknowns.

As an application of this, say I have radiation that should have linear vertical polarization. I want to quantify how much of the received radiation is actually linear vertical (the co-pol level), and how much of the received radiation is linear horizontal (the cross-pol level). I could potentially treat the problem as if $s_2=s_3=0$, letting $c = d = e = f =0$, ignoring the slant and circular polarization. This would leave me with 2 equations and 2 unknowns but this seems wrong as $s_2\neq s_1\neq 0$.

Hecht's optics books (E4) discusses Stokes vectors in Section 8.13.1 but I did not find a description of how to decompose a Stokes vector into the different polarization states.

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  • $\begingroup$ What do you mean by "quantify the polarization", and what do you mean by "co-polarization" and "cross-polarization"? Presumably those are with respect to some external frame of reference, but if you don't specify what that is, it's mostly impossible to answer how you quantify them. $\endgroup$ – Emilio Pisanty Oct 10 '18 at 14:35
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how do I quantify the polarization?

Using the Stokes parameters - that is precisely what they are.

Is there a way to say $S_i$ is $x$% right- or left-hand-circular, $y$% right- or left-slant, and $z$% vertical or horizontal?

Depending on what you mean, this is precisely what they are.

  • The $Q=S_1$ parameter gives you the balance between fully horizontal ($Q=1$) through to fully vertical ($Q=-1$) polarization.
  • The $U=S_2$ parameter gives you the balance between fully diagonal ($U=1$) through to fully antidiagonal ($U=-1$) polarization.
  • The $V=S_3$ parameter gives you the balance between fully left-handed circular ($V=1$) through to fully right-handed circular ($V=-1$) polarization.

However, it seems that this isn't quite what you want to do - it seems that you want a unique decomposition of your light as a convex combination of fully-polarized states. This is provably impossible, and that is the reason why your equations are coming out with more unknowns than equations.

As an illustrative example, consider the limiting case of fully de-polarized light, whose Stokes vector $$ \begin{pmatrix}1\\0\\0\\0\end{pmatrix} = \frac12\begin{pmatrix}1\\1\\0\\0\end{pmatrix} +\frac12\begin{pmatrix}1\\-1\\0\\0\end{pmatrix} = \frac12\begin{pmatrix}1\\0\\1\\0\end{pmatrix} +\frac12\begin{pmatrix}1\\0\\-1\\0\end{pmatrix} = \frac12\begin{pmatrix}1\\0\\0\\1\end{pmatrix} +\frac12\begin{pmatrix}1\\0\\0\\-1\end{pmatrix} $$ can be seen equally happily as an evenly-weighted incoherent mixture of horizontal+vertical, or diagonal+antidiagonal, or left-circular+right-circular polarizations. The process of incoherent mixture entails an inherent loss of information - you can make unpolarized light using any of those combinations, but you can't recover which combination was used if all you have is the unpolarized light.


What you can do, so long as $p>0$ and your light is not fully polarized, is to split up your light into a polarized and a depolarized components, i.e. as $$ \begin{pmatrix} 1 \\ 0.6 \\ -0.2 \\ 0.07 \end{pmatrix} = (1-p)\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + p \begin{pmatrix} 1 \\ 0.94 \\ -0.31 \\ 0.11 \end{pmatrix}, $$ where that last vector is simply $\frac1p (p,S_1,S_2,S_3)$. That last component is now fully polarized, and it is this polarization state which will govern all of the polarization-dependent interactions that you'll be able to extract from your light source (except, of course, the actual degree of polarization).

If you now want to understand the polarization described by this state, you can use any of the multiple descriptions found in the Wikipedia article as well as elsewhere. For your specific case, unless I'm messing up quite badly, for the fully-polarized component you have $Q=0.94$, $U=-0.31$, $V=0.11$, and using $$ L=Q+iU $$ you can get the major and minor axes of the polarization ellipse as \begin{align} A&=\sqrt{\frac12(1-|L|)} = 0.998 B&=\sqrt{\frac12(1-|L|)} = 0.055, \end{align} giving an ellipticity of $\varepsilon =B/A= 0.055$. (This is small enough that I'd start to worry about numerical precision, but you don't state anything about that in your question so I'm not going to touch that. In any case, your initial state has $V>0$ so you're required to have a nonzero ellipticity.) Finally, the ellipse angle can be obtained as $$ \theta = \frac12 \arg(L) = -9.2°. $$

In other words: your light is about $36\%$ unpolarized light, added to $64\%$ of an ever-so-slightly elliptical polarization set at an angle just below the $x$ axis.

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