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Jones Vectors and Polarization

  1. Jan 19, 2016 #1
    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. jcsd
  3. Jan 19, 2016 #2

    blue_leaf77

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    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?
     
  4. Jan 19, 2016 #3
    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.
     
  5. Jan 19, 2016 #4

    blue_leaf77

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    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##?
     
  6. Jan 19, 2016 #5
    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...
     
  7. Jan 19, 2016 #6

    blue_leaf77

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    The question only asks you to generate y-polarized output regardless of whatever phase delay it may have accumulated.
     
  8. Jan 19, 2016 #7
    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?
     
  9. Jan 19, 2016 #8

    blue_leaf77

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    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##.
     
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