If I send two elliptically polarised beams into the outputs of a (say 50:50) beam splitter which is at some angle $\theta$ to the $x$-axis, what sort of beam emerges? As far as I'm aware they do not simply add.


  • What if the two beams are not orthogonal in their elliptical polarisations?
  • What if the two beams have passed through different optical elements so may not be coherent?
  • What if the beam splitter is a polarising beam splitter?

Currently, I am using the fact that antipodal points on the Poincare sphere represent orthogonal polarisations of light. Since a PBS run 'forward' produces orthogonal polarisations of light, I'd imagine any two orthogonal beams of light, regardless of their polarisation, would have the same effect. For this reason, I think the wave resulting from passing two beams of light $\textbf{$E_{1}$}$ and $\textbf{$E_{1}$}$ would be $(\textbf{$E_{1}$} + \textbf{$E_{1}$})$cos$(2\Psi)$cos$(2\chi)$ with $\Psi, \chi$ defined in the image below (if one input wave is along $S_{1}$ and the other at the marked point.)

Example Poincare sphere

(Image taken from Wikipedia).

  • $\begingroup$ I know the name Poincare should have an accent, I'm just not sure how to add it since using the LaTeX version doesn't seem to work. $\endgroup$ Feb 20, 2019 at 8:32

1 Answer 1


It is important to realise that any polarisation of light can be split into a linear superposition of any two orthogonal basis polarisations. Two orthogonal linear polarisations are simply an example of this. Therefore, a wave which is incident on a PBS will be split into two orthogonal basis polarisations, which are chosen by the orientation of the PBS. One of these basis vectors will be reflected, and the other will be transmitted through the PBS. Adding these resultant waves will give the input field. This is the principle of reciprocity.

If two elliptical waves are sent into the two inputs of a PBS, this decomposition occurs to both waves and the resultant waves are formed from, for example, the x-axis polarisation of the first input wave and the y-axis polarisation of the second input wave. This is therefore generally elliptical too. A great picture representing this can be found in this manufacturers guide (figure 1).

For a 50:50 beam splitter, the splitting does not happen by polarisation, but by intensity. The resulting outputs from this beam splitter is therefore two beams which are a half of the addition of the input fields each.


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