- #1

Mr_Allod

- 42

- 16

- 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.

$$

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.