Jones Matrix

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Jones Matrix
관련코스 현대광학
소분류 물리
선행 키워드 편광
연관 키워드 Mueller Matrix

In Jones Matrix representation, the polarization state of light is represented as a 2x1 matrix, usually in normalized form.

When the light is assumed to be propagating in z-direction, the two components in the column matrix represent the electric field amplitude of light in x- and y-direction, respectively.

[math]\textbf{E}_0=E_{0x}\textbf{i}+E_{0y}\textbf{j} [/math]

Polarization State

Polarization State Jones vector
Linearly polarized in x-direction [math]\left[ \begin{array}{c} 1 \\ 0 \end{array} \right][/math]
Linearly polarized in y-direction [math]\left[ \begin{array}{c} 0 \\ 1 \end{array} \right][/math]
Linearly polarized in 45 degrees from x-direction [math]\frac1{\sqrt{2}}\left[ \begin{array}{c} 1 \\ 1 \end{array} \right][/math]
Left circularly polarized [math]\frac1{\sqrt{2}}\left[ \begin{array}{c} 1 \\ i \end{array} \right][/math]
Right circularly polarized [math]\frac1{\sqrt{2}}\left[ \begin{array}{c} 1 \\ -i \end{array} \right][/math]

Polarizing Optical Elements

Linear polarizer

Horizontal polarizer [math]\left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right][/math]
Vertical polarizer [math]\left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right][/math]
Polarizer in 45 degree from x-axis [math]{1\over 2}\left[ \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right][/math]
Polarizer in -45 degree from x-axis [math]{1\over 2}\left[ \begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array} \right][/math]

Phase retarder

Note that an error has been fixed for QWP with slow axis at [math]\pm45[/math] degree cases.

Quarter-wave plate, vertical slow axis [math]\left[ \begin{array}{cc} 1 & 0 \\ 0 & i \end{array} \right][/math]
Quarter-wave plate, horizontal slow axis [math]\left[ \begin{array}{cc} 1 & 0 \\ 0 & -i \end{array} \right][/math]
Quarter-wave plate, slow axis at -45 degrees from x-axis [math]\frac{1}{\sqrt{2}}\left[ \begin{array}{cc} 1 & -i \\ -i & 1 \end{array} \right][/math]
Quarter-wave plate, slow axis at 45 degrees from x-axis [math]\frac{1}{\sqrt{2}}\left[ \begin{array}{cc} 1 & i \\ i & 1 \end{array} \right][/math]
Half-wave plate, horizontal or vertical slow axis [math]\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right][/math]

Reflection and Transmission

Reflection Matrix [math]\left[ \begin{array}{cc} -r_\parallel & 0 \\ 0 & r_\perp \end{array} \right][/math]
Transmission Matrix [math]\left[ \begin{array}{cc} t_\parallel & 0 \\ 0 & t_\perp \end{array} \right][/math]

Coordinate Transformation

CoordRot.png

Even in cases where a new x-y coordinate is used, the physics shouldn't change.

What if we use a new coordinate system which is rotated by an angle [math]\alpha[/math]?

Then both Jones vector [math]\mathbf{J}[/math] and Jones matrix [math]\mathbb{A}[/math] transforms accordingly:

[math]\mathbf{J}'=\mathbb{R}(\alpha) \mathbf{J}[/math]
[math]\mathbb{A}'=\mathbb{R}(\alpha) \mathbb{A} \mathbb{R}(-\alpha)[/math]

where rotation matrix is defined as below:

[math]\mathbb{R}(\alpha)=\left[ \begin{array}{cc} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{array} \right][/math]

Now let's deal with some examples below.

Waveplate with fast axis at 45 degree from x-direction

Let's see if we can derive a Jones matrix for QWP with fast axis at +45 degree, by coordinate rotation.

We'll start with the Jones matrix for QWP with fast axis vertical, and apply the rotation transformation.

[math]\mathbb{A}'=\mathbb{R}(\frac{\pi}{4}) \mathbb{A} \mathbb{R}(-\frac{\pi}{4}) \\ =\left[\begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array}\right] \left[\begin{array}{cc} 1 & 0 \\ 0 & -i \end{array}\right] \left[\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}\right] = \left[\begin{array}{cc} 1-i & -1-i \\ -1-i & 1-i \end{array}\right] = (1-i) \left[\begin{array}{cc} 1 & -i \\ -i & 1 \end{array}\right][/math]

where we have neglected multiplicative constant for normalization throughout.

This result is consistent with the one listed in the table, and when this matrix operates on a horizontal polarization, it generates right-handed circular polarization as expected.

[math]\left[\begin{array}{cc} 1 & -i \\ -i & 1 \end{array}\right] \left[\begin{array}{c} 1 \\ 0 \end{array}\right] = \left[\begin{array}{c} 1 \\ -i \end{array}\right][/math]