When considering a Jones matrix

$$J=\ \left( \begin{array}{ccc} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \\ \end{array} \right) $$

I understand that the effect of a device described by this Jones matrix on a linearly polarized light is rotation by angle $\phi$. I identified the corresponding device as a Faraday rotator. I found eigenvalues to be

$$\lambda_1=e^{i\phi} \quad \text{and} \quad \lambda_2=e^{-i\phi}$$

and corresponding eigenvectors as

$$\vec{v}_1=\frac{1}{\sqrt{2}} \left( \begin{array}{ccc} 1 \\ i \\ \end{array} \right) \quad \text{and} \quad \vec{v}_2=\frac{1}{\sqrt{2}}\ \left(\begin{array}{ccc} 1 \\ -i \\ \end{array} \right)\ $$

which have the form of left and right circular polarization states. I found Jones matrix in its own diagonal frame to be

$$J'=\ \left( \begin{array}{ccc} e^{i\phi} & 0 \\ 0 & e^{-i\phi} \\ \end{array} \right)\ $$

When asked to explain rotational effect of a device described by the first Jones matrix by considering an incident polarised wave to be a superposition of eigenpolarizations, how can I approach this?


The vectors are $v_1,v_2$ are circularly polarised light, in which the polarisation vector rotates over time. So the matrix $J'$ describes changing the phase relations between $v_1$ and $v_2$. One way to explain the relation would be just to work out the original vector as a superposition of $v_1$ and $v_2$ and then apply $J'$.

Another way to think about it is that $v_1$ and $v_2$ have a polarisation that rotates over time. Linear polarisation in some direction is like the rotation of $v_1$ and $v_2$ cancelling out at a particular angle. And the matrix $J'$ just makes the rotation cancel out at a different angle.

  • $\begingroup$ the quarter wave plate induces a phase change of 90 degrees, which is absent in the OP. $\endgroup$ – Peter Diehr Mar 12 '16 at 15:59

Since the matrix shown is a rotation matrix, which rotates an input vector by the angle $\phi $, it should be the Jones matrix for a half wave plate.

Click for a list of devices and their corresponding Jones matrix representations.

The half wave plate "flips" the normally incident linear polarization across the fast axis; this means that if the input polarization makes an angle $\theta $ with the fast axis, the total rotation will be $2\theta $.

A detailed analysis is found at Newport's polarization page.

  • $\begingroup$ Would you be able to explain this in more detail, please? I did consider a half wave plate at first but because all sources (including the one your provided) describe half wave plate with an additional factor of 1/2, i.e. having Jones matrix $J=\ \left( \begin{array}{ccc} e^{i\frac{\phi}{2}} & 0 \\ 0 & e^{-i\frac{\phi}{2}} \\ \end{array} \right)\ $ it didn't seem to be the answer? At the same time, the main use of half-wave plate is to rotate linear polarization by arbitrary angle. It's all a bit confusing. $\endgroup$ – reynevan Mar 12 '16 at 15:40
  • $\begingroup$ @reynevan: I've added to the answer; for the rotation matrix in the OP -- and the one in the reference provided -- the incident plane polarization would be described by the half angle. Choice of representation doesn't change the physics; in this case it is a simple change of variable. $\endgroup$ – Peter Diehr Mar 12 '16 at 15:58

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