Semicircular Potential Well - Wave Function

Wave Function

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:

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Like the infinite square well, the first boundary condition demands that the wave function equals 0 at both and . Basically

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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.

We then normalize the wave function, yielding a result where . The normalized wave function is

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The ground state energy of the system is . Like the particle in a box, there exists nodes in the excited states of the system where both and are both 0, which means that the probability of finding the particle at these nodes are 0.