How to "think" of a polarizer in matrix representation?

  • #1
fluidistic
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From what I remember of my optics course, any element such as a lens (be it thick or thin), can be represented by a matrix. So they are sort of operators, and it is then easy to see how they transform an incident ray, since we can apply the matrix to the electric field vector and see how it gets modified. So far so good.

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?
 

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  • #2
vanhees71
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If you have the polarizer in ##x##-direction (i.e., letting an em. plane-wave mode with ##\vec{k}=k \vec{e}_3## through if it is polarized in ##x## direction and blocking it if it's polarized in ##y##-direction) the polarizer is described by the projection operator
$$\hat{P}_{\varphi=0}=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}.$$
If it's rotated to an arbitrary angle ##\varphi## you need the rotation matrix
$$\hat{D}(\varphi)=\begin{pmatrix} \cos \varphi & -\sin \varphi \\ \sin \varphi & \cos \varphi \end{pmatrix}$$
to get a polarization-filter matrix for orientation ##\varphi## relative to the ##x## axis,
$$\hat{P}(\varphi)=\hat{D}(\varphi) \hat{P}(\varphi=0) \hat{D}^{-1}(\varphi)= \begin{pmatrix}\cos^2 \varphi & \cos \varphi \sin \varphi \\ \cos \varphi \sin \varphi & \sin^2 \varphi \end{pmatrix},$$
which is also a projection matrix of course, i.e., ##\hat{P}^2(\varphi)=\hat{P}(\varphi)##.

The meaning of the matrix is clear: Entering with a ##x##-polarized beam, after the filter the polarization is in direction ##(\cos \varphi,\sin \varphi)## and the magnitude is smaller by a factor ##|\cos \varphi|##, which reflects Maluss' Law. For a ##y##-polarized beam after the filter the polarization is also in direction ##(\cos \varphi,\sin \varphi)##, and the magnitude is smaller by a factor ##|\sin \varphi|## (again reflecting Malus's) Law.

For more info on this "Jones calculus", see

https://en.wikipedia.org/wiki/Jones_calculus
 
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  • #3
fluidistic
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Thank you very much, I could follow everything except the fact that P(theta) equals D(theta) P(theta equals 0) inverse of D(theta). It looks like a change of basis to me rather than a rotation. I would have expected the last term not to be there. Could you please shed some light on this? (Sorry for using theta instead of phi).
 
  • #4
vanhees71
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The idea with the rotation matrix is that you know the Jones matrix for the polarization filter for the special case that it's directed in ##1## direction. Now everything is covariant under rotations in electromagnetism (because rigid rotations are a subgroup of the Lorentz group). So you can get the Jones matrix for the polarization filter in an arbitrary direction by first changing to a reference frame, where the filter is oriented in ##x## direction and then back to the original frame, where it is rotated by an angle ##\varphi## wrt. the ##1##-axis. That's where the typical construction ##\hat{P}(\varphi)=\hat{D}(\varphi) \hat{P}(0) \hat{D}^{-1}(\varphi)## comes from.
 

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