# What degenerates mode in a fiber have in common? Do they have the same eigenvalues?

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)?

• Who says that modes can't share eigenvalues? – Emilio Pisanty Sep 4 '17 at 0:20
• 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. – dmckee Sep 4 '17 at 0:27
• 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... – ZeroTheHero Sep 4 '17 at 0:48
• @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. – WetSavannaAnimal Sep 4 '17 at 2:13

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.