TITLE: How to understand the polarization rotator using a PBS, a quarter-wave plate, and a mirror?
QUESTION [1 upvotes]: In optics experiments, I often see the following optics configuration to rotate the polarization of an incident linearly-polarized laser beam. The final reflected beam has its polarization rotated by 90 degrees. My question is:

Between the quarter plate and the mirror( reflecting surface), the following figure indicates the handness of the circular polarization does not change when it is reflected back. But from what I learned, the polarization should change its handness while being reflected by a mirror. (see, e.g. this question: https://physics.stackexchange.com/q...se-polarization-of-circularly-polarised-light)

If the circular polarization changes its handness, then after the quaterplate it should become the same linear polarization as the incident laser beam, meaning that it should pass through the PBS again and not be reflected away.


For the mathematical description of the process, I have included the Jones calculus below.
Where could I be wrong in understanding its principle? Thanks!

Here is how I use Jones calculus to describe the process:

REPLY [0 votes]: The problem is that you need to be careful with the fast axis of the quarter wave plate (QWP). Here is how your example works, using the same optical calculus software as I used in another answer here.
The Jones calculus model for your example is here:

The output Jones vectors are shown in the four dialog boxes under the simulation model. Note that the right handed coordinate system is such that the first QWP has its fast axis at +45 degrees, while the second QWP, which is simply a convenient way of dealing with the reflected ray from the mirror, has its fast axis at -45 degrees. This is because the reflected ray “sees” the QWP’s fast axis rotated 45 degrees the other way. The output is linearly polarized, as expected, and orthogonal to the incident ray’s linear polarization.
The Jones calculus equation is this:

Comparison with the simulation results in the first figure shows that the simulation results differ only due to round off error.
If the fast axis has been kept at +45 degrees, the output Jones vector would have been this:

In this case, the output ray would be linearly x polarized, just like the input light.
Added to address the OP’s query. The Jones calculus matrices I used in my software back in 1990-1992, and in the answer above, are from the following books:
W.A. Shurcliff, Polarized Light, Harvard University Press, Cambridge, MA, 1962, Appendix 2.
Kliger, D. S.,Lewis, J. W., Randall, C. E., Polarized Light in Optics and Spectroscopy, 1st ed., Academic Press, Boston, 1990, Appendix B II.
The QWP matrices, from p. 168 in Shurcliff, are

and those from Kliger et al., from p. 282, are

The wikipedia Jones calculus table, with my annotation, is

Using the general expression below the table, and substituting as shown, yields the QWP matrix. Comparing with what Kliger et al. give, the QWP matrix for $\theta = +45$ degrees corresponds to their matrix for $\rho = -45$ degrees fast axis orientation.
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