Jones vector

In wave optics, the Jones vector is an important tool for representing the polarisation of light. We explicitly write the phases of the field components in a plane harmonic wave as:

$$\vec{E}(\vec{r},t)=\left(|{\vec{E}}_X|e^{i{\phi }_x}\hat{x}+|{\vec{E}}_Y|e^{i{\phi }_y}\hat{y}\right)\exp \left[i\left(\vec{k}\cdot \vec{r}-\omega t+{\phi }_o\right)\right]$$

Then, we define the Jones vector as a 2D vector $\vec{J}$, where each component $J_x,J_y\in \mathbb{C}$. Note that the components need not be in the $\hat{x},\hat{y}$ direction. They can be defined in any two basis vectors orthogonal to the direction of propagation and to each other.

Basics

Some key remarks. First is that the Jones vector need not be unique.

$$\left[\begin{array}{c}{E}_{ox}{e}^{i{\phi }_x}\\ E_{oy}{e}^{i{\phi }_y}\end{array}\right]$$

and

$$\left[\begin{array}{c}{E}_{ox}\\ E_{oy}{e}^{i({\phi }_y-{\phi }_x)}\end{array}\right]$$

represent the same polarisation state. In fact, $\vec{J}$ and $\alpha \vec{J}$ do as well, where $\alpha \in \mathbb{C}$.

Next, typically $\vec{J}$ is normalised to 1, i.e.,

$$|\vec{J}|={\vec{J}}^{\ast }\cdot \vec{J}=⟨J|J⟩=1$$

Then, the Jones vectors are orthogonal to each other, i.e.,

$${\vec{J}}_1^{\ast }\cdot {\vec{J}}_2=⟨J_1|J_2⟩=0$$

Interpretation

The Jones vectors for a:

• Linear polarisation state have purely real components or has no phase difference.
  • We should specify the angle they’re polarised at. $\alpha =\arctan \frac{J_y}{J_x}$
• Circular polarisation has the vector components have equal magnitudes, with a 90 degree phase difference. Note that circular/elliptical always have some complex component.
• Elliptical for everything else.

To determine whether a circular/elliptical polarisation is right/left-handed, we follow a similar pattern as regular polarisation states with the phase difference $\Delta \phi ={\phi }_y-{\phi }_x$.

• If positive, it is left-hand polarised, like $[1i{]}^T$.
• If negative, it is right-hand polarised, like $[1-i{]}^T$.