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When we bring two waveguides close together we can see some coupling between the modes of the two waveguides and verify energy transfer. Can the same transfer happen in modes within a single waveguide? In a square or rectangular waveguide we have a degeneracy in the polarization of TE modes. Even with dimensions of single mode, we end up with two polarization modes (TE00x and TE00y). If TE00x is excited in the waveguide, will there be, after some propagation length, any power in the TE00y mode? If not, how is it different from the two waveguide scenario?

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  • $\begingroup$ I am not familiar with this notation, what waveguide mode is $\text{TEoox}$ or $\text{TEooy}$? $\endgroup$
    – hyportnex
    Apr 7, 2023 at 21:05
  • $\begingroup$ I use that notation to represent the fundamental modes (not necessarily degenerate) such that TE00x is polarized in the x direction and TE00y in the y direction $\endgroup$
    – Bidon
    Apr 11, 2023 at 3:57

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In a perfect waveguide, the different modes are orthogonal and do not couple. However, waveguides are never perfect. Therefore, some coupling does occur. To modes that, one must have some knowledge of the imperfections or else, use some empirical information, such as the average coupling per unit length.

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  • $\begingroup$ I have heard that rectangular waveguides maintain polarization, while square ones may lead to scrambling of polarization. Could you comment on that? $\endgroup$
    – Bidon
    Apr 5, 2023 at 15:42
  • $\begingroup$ Due to the discrete rotation symmetry the maintenance of polarization is less robust in square waveguides. The two modes are degenerate. Therefore, slight imperfections are more likely to cause coupling. $\endgroup$ Apr 6, 2023 at 2:41
  • $\begingroup$ Could you point me to a reference that explains it? $\endgroup$
    – Bidon
    Apr 11, 2023 at 3:55

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