Question about the jones matrix for a quarter wave plate

[richyw]

Homework Statement

So I need to identify the optical filter with jones matrix

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

Homework Equations

See attempt

The Attempt at a Solution

So I have applied the rotation matrix like this : $RAR^{-1}$$$R(\theta)=\left[\begin{matrix}\cos\theta&-\sin\theta \\ \sin\theta&\cos\theta \end{matrix}\right]$$I wind up with $$R(\pi /4)AR^{-1}(\pi /4)=e^{i\pi /2}\left[\begin{matrix}1&0 \\ 0&-1 \end{matrix}\right]$$Now my problem is what does the $e^{i\pi /2}$ mean? what makes it different from a quarter wave plate?

 

[vredina97]

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}}