Supplement 3.2: Polarisation of Electromagnetic Waves: Jones Vectors and Jones Matrices (1/2)

Polarisation may be depicted easier and more clearly if the y and z components of the electric field are described by vector elements. Interaction of light and matter changes for example its intensity and polarisation due to reflection or refraction. These changes can be easily described through a matrix which interacts with the vector and changes its values. This procedure has been developed by R. Clark Jones, thus the field vectors and the matrices are named Jones Vectors and Jones Matrices.

Field components as column matrix: Jones Vectors

For simplicity and clarity, a description of the wave through a complex exponential function instead of a sine function is helpful:

\vec{E} (x, t) = {\vec{E}}_{o} e^{i (k x - ω t)}

Complex exponential functions ↓

The complex exponential function is an example of complex functions which consist of a real part and an imaginary part: $z = ℜ (z) + i ℑ (z)$ . $ℜ$ denotes the real part, $ℑ$ the imaginary part and $i = \sqrt{- 1}$ is the imaginary unit.

For the complex exponential function, the following Euler formulas are valid:

e^{\pm i α} = cos α \pm i sin α

The advantage of the complex exponential function in comparison to the sine and cosine function grounds on the fact that they do not change when differentiated or integrated, what facilitates the calculations in the event of solving differential equations for example.

Several relations can be derived from the Euler formulas. Those are used on the following pages:

e^{i π} = - 1 e^{i π / 2} = i e^{- i π / 2} = - i

This is a plane wave propagating in x direction, though the following depiction can also be used for waves with a radial propagation and others. We write the electric field as a column vector with the components in the y,z plane. This is the Jones Vector:

\vec{E} = (\begin{matrix} E_{y} \\ E_{z} \end{matrix}) = (\begin{matrix} E_{y, o} e^{i (k x - ω t)} \\ E_{z, o} e^{i (k x - ω t + φ)} \end{matrix})

Often only the amplitudes and phases of the wave components are of interest instead of the spectral properties of the light when examining polarisation. In that case, one may leave out the exponential function which describes the monochromatic wave and write instead:

(\begin{matrix} E_{y, o} e^{i φ_{y}} \\ E_{z, o} e^{i φ_{z}} \end{matrix})

One may simplify even more and equate the intensity with 1 if only the type of polarisation is of interest. The intensity though is proportional to the second power of the field strength as it was shown in the Section about electromagnetic waves. The result of $E_{y}^{2} + E_{z}^{2}$ thus shall be 1. This explains the expressions in the following table of Jones Vectors for various polarisations.

Jones Vector	Jones Vector for intensity=1	Type of polarisation
$(\begin{matrix} E_{y, o} e^{i φ_{y}} \\ 0 \end{matrix})$	$(\begin{matrix} 1 \\ 0 \end{matrix})$	linear along the y axis
$(\begin{matrix} 0 \\ E_{z, o} e^{i φ_{z}} \end{matrix})$	$(\begin{matrix} 0 \\ 1 \end{matrix})$	linear along the z axis
$E_{y, o} e^{i φ_{y}} (\begin{matrix} 1 \\ 1 \end{matrix})$	$\frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ 1 \end{matrix})$	linear diagonally in the first and third quadrant of the y,z plane
$E_{y, o} e^{i φ_{y}} (\begin{matrix} 1 \\ - 1 \end{matrix})$	$\frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ - 1 \end{matrix})$	linear diagonally in the second and fourth quadrant of the y,z plane
$E_{y, o} e^{i φ_{y}} (\begin{matrix} 1 \\ - i \end{matrix})$	$\frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ - i \end{matrix})$	right circular
$E_{y, o} e^{i φ_{y}} (\begin{matrix} 1 \\ i \end{matrix})$	$\frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ i \end{matrix})$	left circular

Example: right circular polarisation ↓

At this kind of polarisation, the z wave component is at 90° resp. $π / 2$ ahead to the y wave component. Hence with $E_{y, o} = E_{z, o}$ it becomes:

\begin{array}{l} (\begin{matrix} E_{y, o} e^{i φ_{y}} \\ E_{z, o} e^{i φ_{z}} \end{matrix}) = (\begin{matrix} E_{y, o} e^{i φ_{y}} \\ E_{y, o} e^{i (φ_{y} - \frac{π}{2})} \end{matrix}) \\ = E_{y, o} e^{i φ_{y}} (\begin{matrix} 1 \\ e^{- i \frac{π}{2}} \end{matrix}) = E_{y, o} e^{i φ_{y}} (\begin{matrix} 1 \\ - i \end{matrix}) \end{array}

This method uses monochromatic waves and is therefore only applicable to completely polarised light.

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Supplement 3.2: Polarisation of Electromagnetic Waves: Jones Vectors and Jones Matrices (1/2)

Field components as column matrix: Jones Vectors