According to the definition degenerates modes in a optical fiber "have the same propagation constants". These modes can be combined to from the linearly polarized modes.

This "constant" should be the $\beta$ constant that appear in the Bessel equation for fields (see e.g. "Optical Fiber Communications - Principles and Practice" - John M. Senior). The eigenvalues that are associated to a mode in a optical fiber in the core and in the cladding are $$\sqrt{n_{core}k^2-\beta^2}\,\,\,,\,\,\,\,\, \sqrt{n_{cladding}k^2-\beta^2}$$ Where $k=\frac{2\pi \nu}{ c}$.

So if $\beta$ is the same for these "degenerates modes" does it mean that these modes also have the same eigenvalues (since they have the same frequency $\nu$)?

This doesn't seem reasonable, since one eigenvalue should correspond to a particular mode and not to two or more modes that can be combined.

So what does the degenerates modes have in common (which constant or parameter)?

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    $\begingroup$ Who says that modes can't share eigenvalues? $\endgroup$ – Emilio Pisanty Sep 4 '17 at 0:20
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    $\begingroup$ For instance the idealized (amagnetic) hydrogen atom has degenerate modes for all non-s orbitals. It is common for absolute introductory treatments of the eigen-problem to ignore degenerate modes until the basic terminology and phenomenology is developed, but it's an important topic. $\endgroup$ – dmckee Sep 4 '17 at 0:27
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    $\begingroup$ moreover if the two modes did not have the same eigenvalue it would not be terribly meaningful to combine them to form linearly polarized modes... $\endgroup$ – ZeroTheHero Sep 4 '17 at 0:48
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    $\begingroup$ @ZeroTheHero In practice they generally don't - the degeneracy is broken by manufacturing imperfevctions. That's why fibers mess with polarization state in the most maddening, environment-dependent ways. $\endgroup$ – WetSavannaAnimal Sep 4 '17 at 2:13

Emilio Pisanty answers this most succinctly with:

"Who says that modes can't share eigenvalues?"

So yes, degenerate modes share the same eigenvalue. This is the fundamental definition of "degeneracy".

Degenerate modes share eigenvalues of the linear operator $\mathscr{L} = \nabla^2 + k^2 n^2$, which is the operator whose eigenvalue problem we solve for weak guidance, scalar waveguide theory. Any linear superposition of such modes is also a mode of the waveguide. We change this operator to the exact one when we do vector waveguide theory, but the idea is the same of course.

The simplest example is the two orthogonal polarizations of the fundamental mode of a perfectly round fiber. They span a two dimensional vector space of modes, which comprises any rotation of one of the polarization states as well as any complex superposition of these states, i.e. it includes circular and elliptical polarizations in the fundamental mode. The two complex co-efficients of the Jones vector for the polarization state are the superposition weights for the orthogonal modes. In general even round fibers are non degenerate owing to manufacturing imperfections, which means that their Jones matrices are nonscalar, i.e. they transform polarization states in ways that can be maddeningly dependent on environment (i.e. the propagation constants and Jones matrix drift with temperature changes and vibrations).

The notion of "eigenvector" is a special case of the more general idea of an "invariant subspace" of an operator, the former being the one dimensional case of the latter. And it is quite possible for the transformation wrought on the invariant subspace by the operator to be a uniform scaling, i.e. in which case every vector within the invariant subspace is an eigenvector with the same eigenvalue.


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