mhsd91
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Homework Statement
Find the final polarization in terms of Jones vectors for vertical polarized, coherent light passing through a quarter waveplate rotated [itex]\theta = \pi/4[/itex]. The Jones matrix for the rotated waveplate is given as,
[itex]J_Q = \frac{1}{\sqrt{2}} \cdot <br /> \left( \begin{array}{cc}<br /> 1 + i\cos(2\theta) & i\sin(2\theta) \\<br /> i\sin(2\theta) & 1-i\cos(2\theta) \end{array} \right)<br /> [/itex]
Which for the given angle results in:
[itex] <br /> J_Q = \frac{1}{\sqrt{2}} \cdot <br /> \left( \begin{array}{cc}<br /> 1 & i \\<br /> i & 1 \end{array} \right)[/itex]
The Attempt at a Solution
Now, this should be straight forward: [itex]J' = J_Q J_{VLP}[/itex]:
[itex] <br /> J' = \frac{1}{\sqrt{2}} \cdot <br /> \left( \begin{array}{cc}<br /> 1 & i \\<br /> i & 1 \end{array} \right) \cdot<br /> <br /> \left( \begin{array}{c}<br /> 0 \\<br /> 1 \end{array} \right)<br /> <br /> <br /> = \frac{1}{\sqrt{2}}\left( \begin{array}{c}<br /> i \\<br /> 1 \end{array} \right)<br /> <br /> [/itex]
The problem is that I am given the answer, which is not what i got, but rather:
[itex] \frac{1}{\sqrt{2}}\left( \begin{array}{c}<br /> 1 \\<br /> i \end{array} \right)[/itex]
So I wonder: One of three possible outcomes: (1) since Jones vectors gived us the phase difference are these two vectors equivalent!?
[itex] \left( \begin{array}{c}<br /> 1 \\<br /> i \end{array} \right)<br /> <br /> \equiv<br /> <br /> \left( \begin{array}{c}<br /> i \\<br /> 1 \end{array} \right)<br /> [/itex]
.. overlooking the normalization constants.
(2) I have calculated wrong: please help!
(3) There is an error in the answer the problem.
Thank you very much!:)