# Inconsistency between Jones vectors and Stokes parameters

Plane wave equation : $$$$\vec{E}(z,t)=\left| \begin{array}{c} E_{0}^{x}e^{i\delta_{x}} \\ E_{0}^{y}e^{i\delta_{y}} \\ 0 \\ \end{array}\right|e^{i(\omega t -kz)}\quad,\quad E_0^{x,y} \in\mathbb{R}$$$$ which can be rewritten at $$z=0$$ and changing the time origin : $$$$\vec{E}(0,t)=\left| \begin{array}{c} E_{0}^{x} \\ E_{0}^{y}e^{i\delta} \\ 0 \\ \end{array}\right|e^{i(\omega t)}\quad,\quad E_0^{x,y} \in\mathbb{R}$$$$ with $$\delta = \delta_y - \delta_x \in [0,2\pi [$$.

The $$z$$ axis is in the direction of the Poynting vector of the plane wave (same direction as $$\vec{k}$$), as on following picture.

The Jones vectors are a representation of degenerate light polarization states :

• horizontal : the $$x$$ axis, in the sequel $$E_H = E_x$$

$$$$| H \rangle = (1,0)^T$$$$

• vertical : the $$y$$ axis, in the sequel $$E_V = E_y$$

$$$$| V \rangle = (0,1)^T$$$$

• left circular (clockwise) $$$$| L \rangle = \frac{1}{\sqrt{2}} (1,i)^T \qquad (1)$$$$

• right circular (counter clockwise) $$$$| R \rangle = \frac{1}{\sqrt{2}} (1,-i)^T \qquad (2)$$$$

For left and right circular polarizations, I follow the convention of increasing phase ($$\omega t - kz$$) with time.

That is to note horizontal and vertical polarizations can be expressed with left and right circular polarizations : $$$$|H\rangle = \frac{1}{\sqrt{2}} (|R\rangle + |L\rangle)$$$$ and $$$$|V\rangle = \frac{i}{\sqrt{2}} (|R\rangle - |L\rangle)$$$$ In latter equation, the phase factor $$i$$ could be omitted, but I prefer to keep it, otherwise there is no more consistency between Jones vectors.

With $$E_H$$ and $$E_V$$, one can derive Stokes parameters : $$$$\begin{split} S_{0} &= (E_{0}^{H})^{2} + (E_{0}^{V})^{2} \newline S_{1} &= (E_{0}^{H})^{2} - (E_{0}^{V})^{2} \newline S_{2} &= 2E_{0}^{H}E_{0}^{V}\cos(\delta) \newline S_{3} &= 2E_{0}^{H}E_{0}^{V}\sin(\delta) \end{split}$$$$

and $$\delta = \delta_y - \delta_x = \delta_V - \delta_H$$

For circular polarizations, I consider $$E_0^x = E_0^y = \frac{E_0}{\sqrt{2}}$$, where $$(E_0)^2$$ can be interpreted as the total light intensity.

Left circular polarization is characterized by $$\delta = \frac{\pi}{2}$$ and right one by $$\delta = - \frac{\pi}{2}$$. Then, for left circular polarization, $$S_3 = (E_0)^2$$ and for right circular polarization, $$S_3 = - (E_0)^2$$. This explains the position of left and right circular polarizations one the Poincaré sphere.

But $$S_3$$ can also be expressed as : $$$$S_3 = i(E_{H}E_{V}^{\star}- E_{H}^{\star}E_{V})$$$$

If I replace $$E_H$$ and $$E_V$$ by their decomposition (Eqs. 1 and 2) in the $$(E_L, E_R)$$ basis (I implicitely identify the complex electric field $$E$$ with Jones vectors), one get : $$$$\begin{split} S_3 &=i(E_{H}E_{V}^{\star}- E_{H}^{\star}E_{V})\newline &= \frac{i}{2}[(E_R + E_L)(-i)(E_R^\star - E_L^\star)] - (E_R^\star + E_L^\star)(i)(E_R - E_L)]\newline &=\frac{1}{2}[(E_R + E_L)(E_R^\star - E_L^\star)] + (E_R^\star + E_L^\star)(E_R - E_L)]\newline &=|E_R|^2 - |E_L|^2 \end{split}$$$$

However, this would mean that $$S_3= (E_0)^2$$ for right circular polarization and $$S_3 = - (E_0)^2$$ for left one, which contradicts the previous results and the mapping on the Poincaré sphere !

I must be missing something, where is my error ?

• the question seem pretty complete, but why do $E_x$, $E_y$ even show up? They just appear...maybe start with the plane wave whence the came? That could also be used to introduce $\delta_{(x, y, H, V}}$, which aren't defined in equations. But why even both with (x, y), and write the plane wave on $E_H\hat x$, $E_V\hat y$ ...it would be clearer.
– JEB
Commented Apr 21 at 17:39
• I updated my question by adding the definition of the plane wave and the fact that I identify horizontal polarization with $x$-axis and vertical with $y$ axis. It should be clearer now. Commented Apr 21 at 19:38
• In your 2nd Eq. you seem to be meaing $\vec{E}(0,t)$ instead of $\vec{E}(z,t),$ because how it's written now, it has a $z$-dependent LHS while the RHS is $z$-independent. Commented Apr 22 at 8:48
• Yes, I corrected this. The position on the $z$-axis is arbitrary. Both LHS and RHS are independent of the position on $z$-axis, it does not change computations. Commented Apr 22 at 9:54
• So you have $E_{(x, y)}$ real, and $E_{(H, V)} = E_{(x, y)}$, and then are taking complex conjugates of $E_{(H, V)}$...what am I missing?
– JEB
Commented Apr 22 at 14:31

As you explain, your definitions for $$|H\rangle, |V\rangle, |R\rangle$$ and $$|L\rangle$$ are related by: \begin{align} |H\rangle &= \frac{1}{\sqrt{2}} (|R\rangle + |L\rangle) \\ |V\rangle &= \frac{i}{\sqrt{2}} (|R\rangle - |L\rangle) \tag{1} \end{align} But that means that for the coefficients used with these basis vectors you have: \begin{align} E_H &= \frac{1}{\sqrt{2}} (E_R + E_L) \\ E_V &= \frac{1}{i\sqrt{2}} (E_R - E_L) \quad \color{red}{\bf \Leftarrow\text{!!}} \tag{2} \end{align} where the second line has $$1/i$$ instead of the factor $$i$$ in $$(1)$$. (If the basis has covariant vectors then the coefficients are contravariant! Or you could call it the difference between an active and a passive transformation.) You forgot to use that in evaluating your second definition of $$S_3$$ (below the 2nd image) where you erroneously use $$E_V=\tfrac{i}{\sqrt{2}}(E_R - E_L)$$.
To see that $$(1)$$ really leads to $$(2)$$ with the surprising $$1/i$$ factor, just solve for: $$E = E_H \binom{1}{0}+E_V \binom{0}{1} = E_L\ \frac1{\sqrt{2}} \binom{1}{i}+E_R\ \frac1{\sqrt{2}} \binom{1}{-i}$$ which will unambiguously give you $$E_H$$ and $$E_V$$ expressed in $$E_R$$ and $$E_L$$