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:

  1. 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)

  2. 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!

enter image description here

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

enter image description here

  • $\begingroup$ You must have a sign error in your reasoning. This is subtle enough that it’s worth writing out in mathematical detail. $\endgroup$
    – Gilbert
    May 21, 2022 at 11:55
  • $\begingroup$ I have added a mathematical description in the post. Could you perhaps take a look? $\endgroup$
    – ynyin
    May 21, 2022 at 12:59

2 Answers 2


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:

Simulation model and results

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:

Jones calculus equation

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:

With wrong fast axis

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 Shurcliff QWP matrices

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

The wikipedia Jones calculus table, with my annotation, is Wikipedia Jones calculus table

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.

  • 1
    $\begingroup$ Good job! It really answers the question which has puzzled me for a long time without a satisfying answer from people around. They all say that the problem of my reasoning is that the reflector should not change the handness of the circular polarization. Only you have pointed out correctly that it is because of the change in the angle of the fast axis of QWP which I did not realize. A big thanks for this detailed answer and for introducing this nice software you developed, which I will surely try in the near future. $\endgroup$
    – ynyin
    May 21, 2022 at 16:58
  • $\begingroup$ Dear Ed, when I tried to understand the Jones calculus you wrote above, I got a bit confused about the matrix for the QWP with its fast axis at +45d (the red one in your equation). I would expect there is a minus sign in front of #i# if I use the equation in Wikipedia(en.wikipedia.org/wiki/Jones_calculus), the equation for Quarter-wave plate with fast axis at angle \theta w.r.t the horizontal axis in the last table. I guess it could depend on the choice of coordinate system. Could you perhaps explain a bit more about this matrix? $\endgroup$
    – ynyin
    May 22, 2022 at 8:29
  • $\begingroup$ Thanks very much for the update; it clarifies the opposite sign in the matrix I would get if I use the Wikipedia equation. To be more clear, I assume the configuration in your software for the QWP is the 2nd case I drew here. That is why the x-polarized light becomes a left-handed circular one in your simulation. Is that right? $\endgroup$
    – ynyin
    May 22, 2022 at 13:09
  • $\begingroup$ From an old sim model I did about 30 years ago: chemistry.meta.stackexchange.com/a/5143/79678. I don’t draw the little pictures because there is no agreed upon standard way to do it. Kliger et al. note, on pages 24-26, that even the famous people, e.g., Jones, Maxwell, and Born, disagree about how those little diagrams should be drawn. So I just go with components and leave it at that. $\endgroup$
    – Ed V
    May 22, 2022 at 14:17
  • $\begingroup$ Sure, that makes sense! I was just thinking from an experimental point of view, that if I know the input polarization direction, how I should orientate the fast axis of my QWP w.r.t to the polarization direction so that I could get either right or left-handed light, and how to describe it mathematically. Now I think I am clear about it. Thanks so much! $\endgroup$
    – ynyin
    May 22, 2022 at 15:15

Your error is the reasoning with the flipping of the handness in the mirror. The handness may look like it was flipped, but the direction of the beam flipped also, meaning that the handness (which is also related to the beam direction) didn't flip.

Think about it in this way: if you put the mirror not perpendicular to the beam but with an angle, such that the reflected light would deviate with an angle of $1^\circ$ (almost no deviation), would you say the handness flipped? No, because the mirror doesn't flip handness, only changes direction.

  • $\begingroup$ Thanks for the answer and explanation. I am still not convinced that the handness will not flip after reflection. To my understanding, the rotation direction of the E field will not change when looking at it in a fixed lab frame, but its propagation direction is flipped. Since the handness is defined relative to its propagation direction, it means the handness is changed. Perhaps the question here has the same reasoning: physics.stackexchange.com/questions/122849/… $\endgroup$
    – ynyin
    May 21, 2022 at 15:51

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