- #1

fluidistic

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However I have attempted to guess the matrix of a polarizer, say whose polarization axis is in the x-direction, without success. Let's say the incident light is ##(E_{0,x}, E_{0,y})##. I know that after the light passes through the polarizer, the E field becomes ##(E_{0,x}\cos (\theta), 0)##. The matrix ##P## representing the polarizer seems to be ##((\cos (\theta), 0), (0 ,0) )##. However this matrix does not satisfy the property of a projection operator, namely that applying the operator twice should yield the same operator. Here,##P^2 =((\cos ^2(\theta), 0), (0 ,0) ) ##, so it doesn't seem to be a projection. I tried several guesses of matrices that are projections, but I couldn't get the correct E field afterwards.

I have also noticed that if the matrix is allowed to depend on the incident field, then I can "escape" and indeed get correct results, but this seems like a cheat or something. It's as if a polarizer works differently according to its input, which is not the case for any other optical "instruments" or whatever they are called. In my example above, it would be to change ##\theta## to ##0## in the 2nd matrix, after the first matrix is applied with an arbitrary ##\theta##.

So, I am a bit lost. How should one think of a polarizer, and how to write it under matrix form?