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STOKES PARAMETERS

FIXME: seems to be different from wikipedia. check on that. (i.e. right vs left polarization, a vs b).

## 1 A Note on Notation

In the following, remember that $E_{x}$ and $E_{y}$ are COMPLEX numbers!!! We will use the “physicist notation" when talking about complex numbers. Recall that for some complex number $z=x+iy$, where $x,y\in {\mathcal {R}}$, the complex conjugate is defined as $z^{*}=x-iy$ and the squared magnitude is given by $z^{2}=zz^{*}=x^{2}+y^{2}$.

## 2 Definition

Stokes parameters: used to describe polarization state of EM radiation.

${\begin{aligned}P_{I}&=\langle E_{x}^{2}\rangle +\langle E_{y}^{2}\rangle \\P_{Q}&=\langle E_{x}^{2}\rangle -\langle E_{y}^{2}\rangle \\P_{U}&=\langle E_{x}E_{y}^{*}\rangle +\langle E_{y}E_{x}^{*}\rangle \\P_{V}&=i(\langle E_{x}E_{y}^{*}\rangle -\langle E_{y}E_{x}^{*}\rangle )\end{aligned}}\,\!$
These 4 $P$s are the Stokes parameters. $P_{I}$ is the total intensity, $P_{Q}$ is the polarization along the coordinate axes, $P_{U}$ is the polarization along the $45^{\circ }$ line between the coordinate axes, and $P_{V}$ is circular polarization.

## 3 Changing Bases

We could just leave it at that, and take these definitions as given. Every definition has a story, so let’s understand it. However, a physical picture is not really obvious to me from these definitions. This is not very obvious (in my opinion) just from inspection of these formulas. Let’s break this down further, and switch bases to make this a little more immediate. Denote the Cartesian basis $({\hat {x}},{\hat {y}})$, the $45^{\circ }$ rotated Cartesian basis $({\hat {a}},{\hat {b}})$, and the circular basis $({\hat {r}},{\hat {l}})$, which are defined as follows:

${\begin{aligned}({\hat {a}},{\hat {b}})&={\Big (}{\frac {{\hat {x}}+{\hat {y}}}{\sqrt {2}}},{\frac {{\hat {x}}-{\hat {y}}}{\sqrt {2}}}{\Big )}\\({\hat {r}},{\hat {l}})&={\Big (}{\frac {{\hat {x}}-i{\hat {y}}}{\sqrt {2}}},{\frac {{\hat {x}}+i{\hat {y}}}{\sqrt {2}}}{\Big )}\end{aligned}}\,\!$
and

${\begin{aligned}E&=E_{x}{\hat {x}}+E_{y}{\hat {y}}\\&=E_{a}{\hat {a}}+E_{b}{\hat {b}}\\&=E_{l}{\hat {l}}+E_{r}{\hat {r}}\end{aligned}}\,\!$
Note that ${\hat {x}}$, ${\hat {y}}$, ${\hat {a}}$, ${\hat {b}}$, ${\hat {l}}$, and ${\hat {r}}$ are all unit vectors. From these definitions, we can solve for $E_{a}$, $E_{b}$, $E_{l}$, and $E_{r}$ in terms of $E_{x}$ and $E_{y}$:

${\begin{aligned}E_{a}&={\frac {E_{x}-E_{y}}{\sqrt {2}}}\\E_{b}&={\frac {E_{x}+E_{y}}{\sqrt {2}}}\\E_{l}&={\frac {E_{x}-iE_{y}}{\sqrt {2}}}\\E_{r}&={\frac {E_{x}+iE_{y}}{\sqrt {2}}}\end{aligned}}\,\!$
It can now be shown that

${\begin{aligned}P_{I}&=\langle E_{x}^{2}\rangle +\langle E_{y}^{2}\rangle \\&=\langle E_{a}^{2}\rangle +\langle E_{b}^{2}\rangle \\&=\langle E_{l}^{2}\rangle +\langle E_{r}^{2}\rangle \\P_{Q}&=\langle E_{x}^{2}\rangle -\langle E_{y}^{2}\rangle \\P_{U}&=\langle E_{a}^{2}\rangle -\langle E_{b}^{2}\rangle \\P_{V}&=\langle E_{r}^{2}\rangle -\langle E_{l}^{2}\rangle \end{aligned}}\,\!$
Things to notice:

- All the Stokes parameters are
*real* numbers. It might not be obvious because of all the complex conjugation, but it is easy to prove it to yourself. It is left as a simple exercise for the reader. For example, to do this, define $E_{x}=a+ib$ and $E_{y}=c+id$, where $a$, $b$, $c$, $d$ $\in {\mathcal {R}}$. Using these definitions, calculate the Stokes parameters- you’ll see that you only end up with real values.

## 4 Actually doing the math...

$E_{a}={\frac {E_{x}+E_{y}}{\sqrt {2}}},\,E_{b}={\frac {E_{x}-E_{y}}{\sqrt {2}}}\,\!$
Calculating $P_{U}$ by substitution:

${\begin{aligned}P_{U}&=\langle E_{a}^{2}\rangle -\langle E_{b}^{2}\rangle \\&={\Big \langle }{\Big (}{\frac {E_{x}+E_{y}}{\sqrt {2}}}{\Big )}{\Big (}{\frac {E_{x}^{*}+E_{y}^{*}}{\sqrt {2}}}{\Big )}{\Big \rangle }-{\Big \langle }{\Big (}{\frac {E_{x}-E_{y}}{\sqrt {2}}}{\Big )}{\Big (}{\frac {E_{x}^{*}-E_{y}^{*}}{\sqrt {2}}}{\Big )}{\Big \rangle }\\&={\frac {1}{2}}\langle E_{x}^{2}+E_{x}E_{y}^{*}+E_{x}^{*}E_{y}+E_{y}^{2}\rangle -{\frac {1}{2}}\langle E_{x}^{2}-E_{x}E_{y}^{*}-E_{x}^{*}E_{y}+E_{y}^{2}\rangle \\&=\langle E_{x}E_{y}^{*}\rangle +\langle E_{x}^{*}E_{y}\rangle \\\end{aligned}}\,\!$
where we use the linearity of expectation in the last line. And lo and behold- it matches the definition from earlier!!!

$E_{r}={\frac {E_{x}+iE_{y}}{\sqrt {2}}},\,E_{l}={\frac {E_{x}-iE_{y}}{\sqrt {2}}}\,\!$
Calculating $P_{V}$ by substitution:

${\begin{aligned}P_{V}&=\langle E_{r}^{2}\rangle -\langle E_{l}^{2}\rangle \\&={\Big \langle }{\Big (}{\frac {E_{x}-iE_{y}}{\sqrt {2}}}{\Big )}{\Big (}{\frac {E_{x}^{*}+iE_{y}^{*}}{\sqrt {2}}}{\Big )}{\Big \rangle }-{\Big \langle }{\Big (}{\frac {E_{x}+iE_{y}}{\sqrt {2}}}{\Big )}{\Big (}{\frac {E_{x}^{*}-iE_{y}^{*}}{\sqrt {2}}}{\Big )}{\Big \rangle }\\&={\frac {1}{2}}\langle E_{x}^{2}-iE_{x}^{*}E_{y}+iE_{x}E_{y}^{*}+E_{y}^{2}\rangle -{\frac {1}{2}}\langle E_{x}^{2}+iE_{x}^{*}E_{y}-iE_{x}E_{y}^{*}+E_{y}^{2}\rangle \\&=i(\langle E_{x}E_{y}^{*}\rangle -\langle E_{x}^{*}E_{y}\rangle )\end{aligned}}\,\!$
## 5 Examples