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Question about the jones matrix for a quarter wave plate

  1. Apr 15, 2013 #1
    1. The problem statement, all variables and given/known data
    So I need to identify the optical filter with jones matrix

    [tex]A=\frac{1}{sqrt{2}}e^{-i\pi /4}\left[\begin{matrix}1&-i \\ -i&1 \end{matrix}\right][/tex]

    2. Relevant equations

    See attempt

    3. The attempt at a solution

    So I have applied the rotation matrix like this : [itex]RAR^{-1}[/itex][tex]R(\theta)=\left[\begin{matrix}\cos\theta&-\sin\theta \\ \sin\theta&\cos\theta \end{matrix}\right][/tex]I wind up with [tex]R(\pi /4)AR^{-1}(\pi /4)=e^{i\pi /2}\left[\begin{matrix}1&0 \\ 0&-1 \end{matrix}\right][/tex]Now my problem is what does the [itex]e^{i\pi /2}[/itex] mean? what makes it different from a quarter wave plate?
    Last edited: Apr 15, 2013
  2. jcsd
  3. Apr 26, 2014 #2
    not sure, but from Euler's formula which is exp(i*alpha)=cos(alpha)+i*sin(alpha) I know that exp(i*pi/2)=cos(pi/2)+i*sin(pi/2)=i. Then substituting i instead of exp(i*pi/2) in your last formula I ended up with matrix {{i,0},{0,-i}}
    Your matrix looks like a half wave plate with extra term of exp(i*pi/2). It might be also a HWP with fast axis at pi/4 angle w.r.t. horizontal axis: {{cos 2*pi/4, sin 2*pi/4},{sin 2*pi/4, -cos 2*pi/4}}={{i,0},{0,-i}}
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