Imagine a bead of mass m that slides frictionlessly around a circular wire ring of circumference L. (This is just like a free particle, except that Ψ(x + L) = Ψ(x).) Find the stationary states (with appropriate normalization) and the corresponding alIowed energies. Note that there are two independent solutions for each energy E_n -corresponding to clockwise and counter-clockwise circulation; call them (x) and (x). How do you account for this degeneracy, in view of the theorem in Problem 2.45 (why does the theorem fail, in this case)?

 

Problem 2.45

If two (or more) distinct⁴⁴ solutions to the (time-independent) Schrodinger equation have the same energy E, these states are said to be degenerate. For example, the free particle states are doubly degenerate-one solution representing motion to the right. and the other motion to the left. But we have never encountered normalizable degenerate solutions, and this is no accident. Prove the following theorem: In one dimension'Y' there are no degenerate bound states. Hint: Suppose there are t\-vo solutions, Ψ₁ and Ψ₂, with the same energy E. Multiply the Schrodinger equation for Ψ₁ by Ψ₂, and the Schrodinger equation for Ψ₂ by Ψ, and subtract, to show that (Ψ₂dΨ/dx –Ψ₁d₁Ψ₂/dx) is a constant. Use the fact that for normalizable solutions Ψ → 0 at +∞ to demonstrate that this constant is in fact zero. Conclude that Ψ₂ is a multiple of Ψ₁, and hence that the two solutions are not distinct.