# Rotation and Polarisation of Light using Jones Matrices

• Mr_Allod
In summary, the conversation discusses a problem with finding the value of theta in a rotation matrix. The individual converts the exponential into trigonometric components and takes the dot product to simplify the equation. The final equation is solved using trigonometric identities and the individual expresses the solution in terms of a quadratic of ##\tan\theta##. They then wonder if there is a simpler solution or assumption that can be made to find the value of ##\theta##. The expert summarizer notes that the final equation is correct and provides the final steps to solve for ##\theta##, while also pointing out a small error in the description of the problem.
Mr_Allod
Homework Statement
a) Rotate an arbitrary Jones matrix ##\vec J = \begin{bmatrix} A \\ Be^{i\delta} \end{bmatrix}## using ##R(-\theta)##
b) At a particular angle ##\theta## this rotation should result in an elliptically polarized wave of the form ##\begin{bmatrix} A' \\ iB' \end{bmatrix}## where the ##\hat x## and ##\hat y## components of this new Jones vector are orthogonal in the complex plane. Find this angle by computing the dot product of these components on an Argand Diagram and setting it equal to zero.
Relevant Equations
##R(-\theta) = \begin{bmatrix} \cos\theta && \sin\theta \\
-\sin\theta && \cos\theta \end{bmatrix}##
##e^{i\delta} = \cos\delta + i\sin\delta##
##\cos^2\theta - \sin^2\theta = \cos2\theta##
##\cos^2\theta + \sin^2\theta = 1##
##\sin\theta\cos\theta = \frac 1 2 \sin2\theta##
Hello there I am having trouble with part b) of this exercise. I can apply the rotation matrix easily enough and get:
$$R(-\theta) \vec J= \begin{bmatrix} A\cos\theta + B\sin{\theta}e^{i\delta} \\ -A\sin\theta + B\cos{\theta}e^{i\delta} \end{bmatrix}$$

I decided to convert the exponential into it's trigonometric components to make it easier to represent on an Argand Diagram:
$$R(-\theta) \vec J= \begin{bmatrix} (A\cos\theta + B\sin\theta\cos\delta) + iB\sin\theta\sin\delta \\ (-A\sin\theta +B\cos\theta\cos\delta) + iB\cos\theta\sin\delta \end{bmatrix}$$
Now I take the dot product of the real and imaginary components like so: $$(a\hat r +b\hat i) \cdot (c\hat r + d\hat i) = ac + db$$

Which gives me:
$$(A\cos\theta + B\sin\theta\cos\delta)(-A\sin\theta +B\cos\theta\cos\delta) + (B\sin\theta\sin\delta)(B\cos\theta\sin\delta)$$

And after multiplying out and using some trig. identities I can simplify it to:
$$-A^2\sin\theta\cos\theta + AB\cos2\theta\cos\delta + B^2\sin\theta\cos\theta$$
$$= \frac 1 2 \sin2\theta(B^2-A^2) + AB\cos2\theta\cos\delta = 0$$
From here the only thing I could thing of doing is expressing everything in terms of a quadratic of ##\tan\theta## and getting an answer for ##\theta## by taking the ##\tan^{-1}## of the roots. That of course gives two very complicated expressions in terms of ##A, B## and ##\cos\delta## and I have a suspicion the actual answer should be much neater. I wonder if there is some assumption I can make that would simplify it? Or is there some point earlier in the analysis where I can determine a value for ##\theta## just by visual inspection? If someone could help me with this I would appreciate it.

vanhees71 and PhDeezNutz
Mr_Allod said:
$$\frac 1 2 \sin2\theta(B^2-A^2) + AB\cos2\theta\cos\delta = 0$$

Assuming your final equation is correct, you've done the hard part. You just hit a 'blind spot' for the final simple steps:
$$\sin2\theta(B^2-A^2) = -2AB\cos2\theta\cos\delta$$$$\tan2\theta=\frac {2AB\cos\delta}{(A^2-B^2) }$$$$\theta=\frac 1 2 \left[ \tan^{-1}\frac {2AB\cos\delta}{(A^2-B^2) } \right]$$Being a bit picky, I'll also note that the question describes ##\vec J## as a 'Jones matrix, but this should be 'Jones vector'.

Last edited:
vanhees71

## 1. What is the Jones matrix method used for in the study of light?

The Jones matrix method is a mathematical tool used to analyze the polarization of light. It allows scientists to represent the polarization state of light using a 2x2 matrix, making it easier to understand and manipulate.

## 2. How does the Jones matrix represent the rotation of light?

The Jones matrix represents the rotation of light by using a complex number, known as the Jones vector, to describe the amplitude and phase of the electric field in a specific direction. The matrix then applies a rotation operation to this vector, resulting in a new vector that represents the rotated polarization state of light.

## 3. What is the difference between linear and circular polarization in terms of Jones matrices?

Linear polarization is represented by a Jones matrix with real values, while circular polarization is represented by a Jones matrix with complex values. This is because circularly polarized light has both horizontal and vertical components, while linearly polarized light only has one component.

## 4. How does the Jones matrix method help in the study of birefringent materials?

The Jones matrix method is useful in studying birefringent materials because it can accurately predict the behavior of polarized light as it passes through these materials. By using Jones matrices to represent the polarization state of light before and after passing through the material, scientists can calculate the amount of birefringence and the orientation of the material's optic axis.

## 5. Can the Jones matrix method be used for other types of electromagnetic waves?

Yes, the Jones matrix method can be applied to any type of electromagnetic wave, not just light. It is commonly used in the study of radio waves, microwaves, and other forms of electromagnetic radiation. The only difference is that the Jones matrix will have different values and dimensions depending on the specific wave being analyzed.

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