# Stokes parameters

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

## 2 Definition

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

{\displaystyle {\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 ${\displaystyle P}$s are the Stokes parameters. ${\displaystyle P_{I}}$ is the total intensity, ${\displaystyle P_{Q}}$ is the polarization along the coordinate axes, ${\displaystyle P_{U}}$ is the polarization along the ${\displaystyle 45^{\circ }}$ line between the coordinate axes, and ${\displaystyle P_{V}}$ 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 ${\displaystyle ({\hat {x}},{\hat {y}})}$, the ${\displaystyle 45^{\circ }}$ rotated Cartesian basis ${\displaystyle ({\hat {a}},{\hat {b}})}$, and the circular basis ${\displaystyle ({\hat {r}},{\hat {l}})}$, which are defined as follows:

{\displaystyle {\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

{\displaystyle {\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 ${\displaystyle {\hat {x}}}$, ${\displaystyle {\hat {y}}}$, ${\displaystyle {\hat {a}}}$, ${\displaystyle {\hat {b}}}$, ${\displaystyle {\hat {l}}}$, and ${\displaystyle {\hat {r}}}$ are all unit vectors. From these definitions, we can solve for ${\displaystyle E_{a}}$, ${\displaystyle E_{b}}$, ${\displaystyle E_{l}}$, and ${\displaystyle E_{r}}$ in terms of ${\displaystyle E_{x}}$ and ${\displaystyle E_{y}}$:

{\displaystyle {\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

{\displaystyle {\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_{l}^{2}\rangle -\langle E_{r}^{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 ${\displaystyle E_{x}=a+ib}$ and ${\displaystyle E_{y}=c+id}$, where ${\displaystyle a}$, ${\displaystyle b}$, ${\displaystyle c}$, ${\displaystyle d}$ ${\displaystyle \in {\mathcal {R}}}$. Using these definitions, calculate the Stokes parameters- you’ll see that you only end up with real values.

## 4 Proof

Here we explicitly do out the math (in case you don’t believe the above and are too lazy to do it yourself.)

${\displaystyle E_{a}={\frac {E_{x}+E_{y}}{\sqrt {2}}},\,E_{b}={\frac {E_{x}-E_{y}}{\sqrt {2}}}\,\!}$

Calculating ${\displaystyle P_{U}}$ by substitution:

{\displaystyle {\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!!!

${\displaystyle E_{l}={\frac {E_{x}+iE_{y}}{\sqrt {2}}},\,E_{r}={\frac {E_{x}-iE_{y}}{\sqrt {2}}}\,\!}$

Calculating ${\displaystyle P_{V}}$ by substitution:

{\displaystyle {\begin{aligned}P_{V}&=\langle E_{l}^{2}\rangle -\langle E_{r}^{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}}\,\!}