Semicircular potential well

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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 0 to π where it cannot escape, because the potential from π to 2π is infinite. Instead there is total reflection, meaning the particle bounces back and forth between 0 to π. The Schrödinger equation for a free particle which is restricted to a semicircle (technically, whose configuration space is the circle S1) is

-\frac{\hbar^2}{2m}\nabla^2 \psi = E\psi \quad (1)

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[edit] Wave function

Using cylindrical coordinates on the 1 dimensional semicircle, the wave function depends only on the angular coordinate, and so

\nabla^2 = \frac{1}{s^2} \frac{\partial^2}{\partial \phi^2} \quad (2)

Substituting the Laplacian in cylindrical coordinates, the wave function is therefore expressed as

-\frac{\hbar^2}{2m s^2} \frac{d^2\psi}{d\phi^2} = E\psi \quad (3)

The moment of inertia for a semicircle, best expressed in cylindrical coordinates, is I \ \stackrel{\mathrm{def}}{=}\ \iiint_V r^2 \,\rho(r,\phi,z)\,r dr\,d\phi\,dz \!. Solving the integral, one finds that the moment of inertia of a semicircle is I = ms2, exactly the same for a hoop of the same radius. The wave function can now be expressed as -\frac{\hbar^2}{2I} \psi = E\psi , which is easily solvable.

Since the particle cannot escape the region from 0 to π, the general solution to this differential equation is

\ \psi (\phi) = A cos(m \phi) + B sin (m \phi) \quad (4)

Defining m=\sqrt {\frac{2 I E}{\hbar^2}} , we can calculate the energy as E= \frac{m^2 \hbar ^2}{2I} . We then apply the boundary conditions, where ψ and \frac{d\psi}{d\phi} are continuous and the wave function is normalizable:

\int_{0}^{\pi} \left| \psi ( \phi ) \right|^2 \, d\phi = 1\ \quad (5) .

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

\ \psi (0) = \psi (\pi) = 0 \quad (6) .

Since the wave function \ \psi(0) = 0 , the coefficient A must equal 0 because \ \cos (0) = 1 . The wave function also equals 0 at φ = π so we must apply this boundary condition. Discarding the trivial solution where B=0, the wave function \ \psi (\pi) = 0 = B sin (m \pi) only when m is an integer since sin(nπ) = 0. This boundary condition quantizes the energy where the energy equals E= \frac{m^2 \hbar ^2}{2I} where m is any integer. The condition m=0 is ruled out because ψ = 0 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 B= \sqrt{\frac{2}{\pi}} . The normalized wave function is

\ \psi (\phi) = \sqrt{\frac{2}{\pi}} sin (m \phi) \quad (7) .

The ground state energy of the system is E= \frac{\hbar ^2}{2I} . Like the particle in a box, there exists nodes in the excited states of the system where both \ \psi (\phi) and \ \psi (\phi) ^2 are both 0, which means that the probability of finding the particle at these nodes are 0.

[edit] Analysis

Since the wave function is only dependent on the azimuthal angle φ, the measurable quantities of the system are the angular position and angular momentum, expressed with the operators φ and Lz respectively.

Using cylindrical coordinates, the operators φ and Lz are expressed as φ and -i \hbar \frac{d}{d\phi} respectively, where these observables play a role similar to position and momentum for the particle in a box. The commutation and uncertainty relations for angular position and angular momentum are given as follows:

[\phi, L_z] = i \hbar \ \psi(\phi) \quad (8)
(\Delta \phi) (\Delta L_z) \geq \frac{\hbar}{2} where \Delta_{\psi} \phi = \sqrt{\langle {\phi}^2\rangle_\psi - \langle {\phi}\rangle_\psi ^2} and \Delta_{\psi} L_z = \sqrt{\langle {L_z}^2\rangle_\psi - \langle {L_z}\rangle_\psi ^2} \quad (9)

[edit] Boundary conditions

As with all quantum mechanics problems, if the boundary conditions are changed so does the wave function. If a particle is confined to the motion of an entire ring ranging from 0 to 2π, the particle is subject only to a periodic boundary condition (see particle in a ring). If a particle is confined to the motion of \frac{- \pi}{2} to \frac{\pi}{2} , the issue of even and odd parity becomes important.

The wave equation for such a potential is given as:

\ \psi_o (\phi) = \sqrt{\frac{2}{\pi}} cos (m \phi) \quad (10)
\ \psi_e (\phi) = \sqrt{\frac{2}{\pi}} sin (m \phi) \quad (11)

where \ \psi_o (\phi) and \ \psi_e (\phi) are for odd and even m respectively.

Similarly, if the semicircular potential well is a finite well, the solution will resemble that of the finite potential well where the angular operators φ and Lz replace the linear operators x and p.

[edit] See also