# Calculate Jones Vector of polarized light thorugh QWP

1. Jun 7, 2013

### mhsd91

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

Find the final polarization in terms of Jones vectors for vertical polarized, coherent light passing through a quarter waveplate rotated $\theta = \pi/4$. The Jones matrix for the rotated waveplate is given as,

$J_Q = \frac{1}{\sqrt{2}} \cdot \left( \begin{array}{cc} 1 + i\cos(2\theta) & i\sin(2\theta) \\ i\sin(2\theta) & 1-i\cos(2\theta) \end{array} \right)$

Which for the given angle results in:

$J_Q = \frac{1}{\sqrt{2}} \cdot \left( \begin{array}{cc} 1 & i \\ i & 1 \end{array} \right)$

3. The attempt at a solution

Now, this should be straight forward: $J' = J_Q J_{VLP}$:

$J' = \frac{1}{\sqrt{2}} \cdot \left( \begin{array}{cc} 1 & i \\ i & 1 \end{array} \right) \cdot \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = \frac{1}{\sqrt{2}}\left( \begin{array}{c} i \\ 1 \end{array} \right)$

The problem is that I am given the answer, which is not what i got, but rather:

$\frac{1}{\sqrt{2}}\left( \begin{array}{c} 1 \\ i \end{array} \right)$

So I wonder: One of three possible outcomes: (1) since Jones vectors gived us the phase difference are these two vectors equivalent!?

$\left( \begin{array}{c} 1 \\ i \end{array} \right) \equiv \left( \begin{array}{c} i \\ 1 \end{array} \right)$
.. overlooking the normalization constants.

(3) There is an error in the answer the problem.

Thank you very much!!:)

2. Jun 7, 2013

### MisterX

Perhaps your problem may come from some ambiguity as to what the plate was like before rotating. I get

$\frac{1}{\sqrt{2}}\left( \begin{array}{c} 1 \\ i \end{array} \right)$

When I consider the fast axis of the quarter wave plate to be in the vertical direction before rotation.

I get your answer when I consider the quarter wave plate before rotation to have the fast axis horizontal.

And no, they do not represent the same polarization. They do not differ by a phase.