Question

(a) Prove that there are no degenerate bound states in an infinite (−∞ < x < ∞) one-dimensional space. That is, if ψ1(x) and ψ2(x) are two bound-state solutions of − (h^2/2m) (d^2ψ dx^2) + V (x)ψ = Eψ for the same energy E, it will necessarily follow that ψ2 = Cψ1, where C is just a constant (linear dependence). Bound-state solutions should of course vanish at x → ±∞.

(b) Imagine now that our particle is restricted to move along a circular ring of radius R. It would still be a 1D system as we need only one coordinate x (as measured along the ring) to specify the position. It is, however, different from case (a) as x is limited to the circumference of the ring.

We want to find all stationary states (that is, eigenfunctions, including normalization factors) and energy values for the particle on this ring in the absence of an external potential (V (x) = 0 ). The main consideration here is that the wave function is single-valued – you are thus faced with what is called “periodic boundary conditions”: after going around the circle, you get to the same point. In addition to finding stationary states, you are asked to assess the degeneracy of all energy levels and relate your observations to the result of case (a). For better clarity, you are suggested to use both standing-wave (like sin(kx)) and running-wave (like exp(ikx)) solutions in your analysis. Any commentary and interpretations are welcome, including comparisons to states in the infinite square well potential.

Answer 1

Note that in the case of ring, we obtain degenerate eigenvalues, which correspond to clockwise and counterclockwise traveling waves.