Jones Vectors and Polarization

[Yoni V]

Homework Statement

Linearly polarized light in the x direction with wave number $k_0$ travels in the z direction. It enters a medium such that a RHCP component of the wave and a LHCP component each accumulate a phase of $n_Rk_0z$ and $n_Lk_0z$ respectively, where z is the distance traveled inside the medium.
a. What is the polarization of the wave as a function of z?
b. For which values of z do we get a linear polarization in the y direction?

Homework Equations

The Attempt at a Solution

It's kind of a warm-up question and should be pretty easy, but I guess I'm missing something elementary...
I suppose I should write the wave in terms of $\hat r$ and $\hat l$, apply the given transformation for each component, and then rewrite the answer in terms of $\hat x, \hat y$. But I'm stuck on step 1, which is writing my initial x-polarized wave in terms of $\hat r, \hat l$. Any advice, an analogous example or a more general treatment?

Thanks...

Edit: latex fixes...

 

[blue_leaf77]

I suppose I should write the wave in terms of $\hat{r}$ and $\hat{l}$,

Yes that's right. Now it's a matter of finding the correct coefficients for $\hat{r}$ and $\hat{l}$.
Do you know the Jones vectors for linearly and circularly polarized light?

 

[Yoni V]

Yes, the vectors are $\frac{1}{\sqrt 2}(1,-i), \frac{1}{\sqrt 2}(1,i)$ for circularly polarized light, and (a,b) for some linear polarization.
But I'm not sure how the initial linear input should be written.

I thought that if $E_0 = E_{x0} \hat x$ then
$$\frac{1}{\sqrt 2}E_{x0} (exp(in_Rk_0z)+exp(in_Lk_0z))\hat x+\frac{i}{\sqrt 2}E_{x0} (exp(in_Lk_0z)-exp(in_Rk_0z))\hat y$$
but that doesn't seem right, given the fact that I'm left with an imaginary y component, and then for the second part of the question, I need the x component to be 0. But then I get $z= \frac{2\pi m -\pi}{k_0(n_L-n_R)}$, which leaves the y component with imaginary amplitude.

 

[blue_leaf77]

We want to write $E_0 \hat{x} = E_0 \left( c_l \hat{l} + c_r \hat{r} \right)$. Using the property
$$
\begin{aligned}
&\hat{l}^\dagger \hat{r} = 0 \\
&\hat{l}^\dagger \hat{l} = 1 \\
&\hat{r}^\dagger \hat{r} = 1 \\
\end{aligned}
$$
, how can you determine $c_l$ and $c_r$?

 

[Yoni V]

Ok, we can solve for them by
$$
\begin{pmatrix}1\\
0
\end{pmatrix}=\frac{c_{l}}{\sqrt{2}}\begin{pmatrix}1\\
i
\end{pmatrix}+\frac{c_{r}}{\sqrt{2}}\begin{pmatrix}1\\
-i
\end{pmatrix}\Rightarrow c_{l}=c_{r}=\frac{1}{\sqrt{2}}$$
and now
$$
\mathbf{E} = E_{0}\left(\frac{1}{\sqrt{2}}e^{in_{R}k_{0}z}\hat{r}+\frac{1}{\sqrt{2}}e^{in_{L}k_{0}z}\hat{l}\right)
= E_{0}\left(\frac{1}{2}e^{in_{R}k_{0}z}\left(\hat{x}-i\hat{y}\right)+\frac{1}{2}e^{in_{L}k_{0}z}\left(\hat{x}+i\hat{y}\right)\right)
= E_{0}\left(\frac{1}{2}\left(e^{in_{R}k_{0}z}+e^{in_{L}k_{0}z}\right)\hat{x}+\frac{i}{2}\left(e^{in_{L}k_{0}z}-e^{in_{R}k_{0}z}\right)\hat{y}\right)
$$
which is different from what I had before only by a factor of $1/\sqrt{2}$. The problem with y-polarization remains, hinting that this is still incorrect...

 

[blue_leaf77]

The problem with y-polarization remains, hinting that this is still incorrect...

The question only asks you to generate y-polarized output regardless of whatever phase delay it may have accumulated.

 

[Yoni V]

Ok, got it, it indeed doesn't matter- we can decompose the expression for the y component and take the real part.
One last thing, I didn't understand your discussion of the dot products. Is there anything wrong with how I calculated the coefficients?

 

[blue_leaf77]

Ok, got it, it indeed doesn't matter- we can decompose the expression for the y component and take the real part.
One last thing, I didn't understand your discussion of the dot products. Is there anything wrong with how I calculated the coefficients?

I don't know how you calculated the coefficients which lead to how it looks like in post #3, but the form you derived there bears violation of energy conservation. You can check that the form in post #2 carries different energy from the incident one, which is $E_0^2$.