ב"ה Is there any way to build a beam splitter that will split two incoming beams differently? Say, beam coming from the left (x axes) will transfer 70% and reflect 30% while beam coming from bottom (y axes) will transfer 40% and reflect 60%? I would like to create a interferometer like this: interferometer similar (but different!!) to mach-zender BS1 splits 50/50 (R/T) but BS2 will split path A 70/30 and path B 40/60. Now assuming their is a phase shift between path A and B we will get the following results when we send single photons:

  1. if only path A is open then detector Dx will count 70% of the photons and Dy 30%.
  2. if path B is open then Dx will count 60% of the photons and Dy 40%.
  3. if both paths are open then we will get interference between the two paths but in proportion to the different R/T ratios.

is their any way to construct a interferometer like that?

  • 2
    $\begingroup$ Can you arrange to have different polarizations for the two beams? $\endgroup$
    – The Photon
    Dec 5, 2022 at 21:08
  • $\begingroup$ Thanks. I explained a little better what I want to achieve $\endgroup$
    – Gilad
    Dec 6, 2022 at 19:28

1 Answer 1


It is reasonably easy to show, see Montgomery Section 9.10 that in a 4-port directional coupler (port numbers $k=1,2,3,4$) if ports are to be matched that is all $S_{kk}=0$ and say ports $1,3$ are isolated, that is $S_{13}=0$ then if the junction is also lossless $\mathbf {S} \cdot \tilde {\mathbf{S}^*} = \mathbf I$ and reciprocal $\mathbf {S} = \tilde {\mathbf{S}}$ then you must also have

$S_{12}=S_{34}$, $S_{14}=S_{23}$ and $|S_{12}|^2+|S_{24}|^2=1$ Therefore the type of coupler you postulate cannot be lossless, or matched or properly isolated.

Having different polarizations, per @ThePhoton question, of course does not contradict this analysis, for then those beams really propagate through different circuits.

  • $\begingroup$ Thanks. I edited my question and explained better what I'm trying to achieve. I understand from your answer that it is not possible. Am I correct? $\endgroup$
    – Gilad
    Dec 6, 2022 at 19:30
  • $\begingroup$ You are still asking if BS2 can split Beam A and Beam B differently, I believe my answer above is still relevant and it is no, you cannot have that if the splitter is to be passive, reciprocal and lossless. I have not had time to look at it what scattering matrix you might need but I am quite sure that it is possible to synthesize such if you drop at least one of the three requirements; it is not difficult to work it out and is a good exercise to do it at least once if you have never done it. $\endgroup$
    – hyportnex
    Dec 6, 2022 at 20:06
  • $\begingroup$ Thanks alot. Could you please explain what is a non passive-reciprocal-lossless BS? how does it work? I searched but couldn't find any simple explanation, Could you please give me some references that may explain how to figure out the scattering matrix I need? Also, assuming I could use a polarizing BS with BS1 and get two polarized beams coming into BS2, would that help? Thanks Alot!! $\endgroup$
    – Gilad
    Dec 11, 2022 at 18:21
  • $\begingroup$ Given the scattering matrix $\mathbf S$ of a system, then it is called reciprocal if $\mathbf S = \tilde {\mathbf S}$ (tilde denotes transpose), passive if $\mathbf I-\mathbf S \tilde {\mathbf S^*}$ positive definite; reactive (passive lossless) if $\mathbf I=\mathbf S \tilde {\mathbf S^*}$ . All amplifiers are non-reciprocal but they are also active (non-passive), the most common non-reciprocal elements are based on magnetic effects such ferrite, these are passive and to some extent lossy, too. My practical knowledge is below <100GHz, not in photonics; for real advice ask somebody in that. $\endgroup$
    – hyportnex
    Dec 11, 2022 at 19:54

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