I Spectrum energy of a particle moving on a circumference

[Lola1]

If you consider the one-dimensional case of a particle constrained to move on the edge of a circumference, the energy spectrum is continuous and two times degenerate. Why the fact that the particle can move in clockwise and counterclockwise implies that the spectrum is degenerate twice?
In any case thanks to the help

 

[mfb]

Moving clockwise and anticlockwise at the same absolute momentum has the same energy as it is symmetric. Two states with the same energy.

 

[Lola1]

Mathematically double degeneration is known by 2 that multiplies the exponents of the exponential solution? (Which we get by imposing the boundary conditions):
ψ (x) = α \exp (2i (nπ / L ) x) + β \exp (-2i (nπ / L) x)

 

[mfb]

Mathematically double degeneration is known by 2 that multiplies the exponents of the exponential solution?

I don't understand that sentence.

You can write down the general solution like that, but you can also let n be positive or negative, and of course you can always have superpositions of those solutions.

 

[Lola1]

I meant that the two who multiplies the exponents indicates that the energy spectrum is degenerate twice...is it true? Furthermore if the energy spectrum is degenerate twice means that his energy levels are equally spaced (but with spacing that is En=(2n^2π^2)⋅(ħ^2)/mL^2 ,unlike the energy levels of the particle in a box where the levels are En=(n^2π^2)⋅(ħ^2)/mL^2 ) or are one double the previous one?

 

[mfb]

The 2 in the exponent is part of a factor (2pi), the circumference of a circle. It has nothing to do with multiplicities.

The difference between energy levels is independent of the multiplicities. It arises from the detailed structure of the potential, while the multiplicity is based on the symmetry. Every symmetric potential will have this multiplicity, but it can have different steps between the energy levels.