# Jones Vectors and Polarization

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1. Jan 19, 2016

### Yoni V

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

3. 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...

2. Jan 19, 2016

### blue_leaf77

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?

3. Jan 19, 2016

### 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.

4. Jan 19, 2016

### 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 $c_r$?

5. Jan 19, 2016

### 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...

6. Jan 19, 2016

### blue_leaf77

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

7. Jan 19, 2016

### 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?

8. Jan 19, 2016

### blue_leaf77

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 $E_0^2$.