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 $\mathbf{\Psi }\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{L}\mathbf{)}\mathbf{=}\mathbf{\Psi }\left(x\right)$find the stationary states (with appropriate normalization) and the corresponding allowed energies. Note that there are two independent solutions for each energy En-corresponding to clockwise and counter-clockwise circulation; call them${\mathbf{\Psi }}_{\mathbf{n}}^{\mathbf{+}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ and${\mathbf{\Psi }}_{\mathbf{n}}^{-}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ How do you account for this degeneracy, in view of the theorem in Problem 2.45 (why does the theorem fail, in this case)?

The equation for the time-dependent Schrödinger equation is,

$\mathit{i}\sout{\mathbf{h}}\frac{\mathbf{d}}{\mathbf{d}\mathbf{t}}\mathbf{|}\mathit{\Psi }\left(t\right)\mathbf{>}\mathbf{=}\hat{\mathbf{H}}\mathbf{|}\mathit{\Psi }\left(t\right)\mathbf{>}$

One coordinate, x, is all that is required to describe a bead's location on a ring because the ring is in a two-dimensional plane. The next step is to solve the Schrödinger equation for the wave function in one dimension Ψ(x, t).

The value of x is determined around the circumference.

Here given equation is:

$$\begin{gathered} -\frac{{\sout{h}}^2}{2m}\frac{d^2\psi }{d{x}^2}=E\psi \\ \frac{d^2\psi }{d{x}^2}=-k^2\psi , \end{gathered}$$

where k =$\frac{\sqrt{2mE}}{\sout{h}}$

$$\begin{gathered} \psi \left(x\right)=A{e}^{ikx}+B{e}^{-ikx} \\ A{e}^{ikx}{e}^{iKL}+B{e}^{-ikx}{e}^{iKL}=A{e}^{ikx}+B{e}^{-ikx} \\ A{e}^{ikL}+B{e}^{-ikL}=A+B......\left(1\right) \end{gathered}$$

Now,

$$\begin{gathered} A{e}^{i\pi /2}{e}^{ikL}+B{e}^{-i\pi /2}{e}^{-ikL}=A{e}^{i\pi /2}+B{e}^{-i\pi /2} \\ iA{e}^{ikL}-iB{e}^{-ikL}=iA-iB \\ A{e}^{ikL}-iB{e}^{-ikL}=A-B.....\left(2\right) \end{gathered}$$

Add equations (1) and (2), and we get,

$2A{e}^{ikL}=2A$

When A= 0 ,$e^{ikL}=1,$$kL=2n\pi$, n is an integer.

If A = 0, then $B{e}^{-ikL}=\mathrm{B}$ .

This leads to the same conclusions.

Now, every positive n has two solutions ${\psi }_n^{+}\left(x\right)=A{e}^{i\left(2n\pi x/L\right)}$ and ${\psi }_n^{-}\left(x\right)=A{e}^{i\left(2n\pi x/L\right)}$.

For n = 0 , there is just one solution:

${\int }_0^L{\left|{\psi }_{\pm }\right|}^{2}dx=1A=B=1/\sqrt{L}$

Any other solution will be a linear combination of these as:

$$\begin{gathered} {\psi }_n^{\pm }\left(x\right)=\frac{1}{\sqrt{L}}{e}^{\pm i\left(2n\pi x/L\right)} \\ {E}_n=\frac{2n^2{\pi }^2{\sout{h}}^2}{m{L}^2},n=0,1,2,3,4,... \end{gathered}$$

Hence, we conclude that this theorem fails because:

(1)$\psi$ does not go to zero at infinity

(2) x is restricted to a finite range

(3) we are unable to determine the constant K