In quantum mechanics, the case of a particle in a one-dimensional ring is similar to the particle in a box. The particle follows the path of a semicircle from to where it cannot escape, because the potential from to is infinite. Instead there is total reflection, meaning the particle bounces back and forth between to . The Schrödinger equation for a free particle which is restricted to a semicircle (technically, whose configuration space is the circle ) is
Using cylindrical coordinates on the 1-dimensional semicircle, the wave function depends only on the angular coordinate, and so
Substituting the Laplacian in cylindrical coordinates, the wave function is therefore expressed as
The moment of inertia for a semicircle, best expressed in cylindrical coordinates, is . Solving the integral, one finds that the moment of inertia of a semicircle is , exactly the same for a hoop of the same radius. The wave function can now be expressed as , which is easily solvable.
Since the particle cannot escape the region from to , the general solution to this differential equation is
Defining , we can calculate the energy as . We then apply the boundary conditions, where and are continuous and the wave function is normalizable:
Like the infinite square well, the first boundary condition demands that the wave function equals 0 at both and . Basically
Since the wave function , the coefficient A must equal 0 because . The wave function also equals 0 at so we must apply this boundary condition. Discarding the trivial solution where B=0, the wave function only when m is an integer since . This boundary condition quantizes the energy where the energy equals where m is any integer. The condition m=0 is ruled out because everywhere, meaning that the particle is not in the potential at all. Negative integers are also ruled out since they can easily be absorbed in the normalization condition.
We then normalize the wave function, yielding a result where . The normalized wave function is
The ground state energy of the system is .
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