Semicircular potential well

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 $\pi$ where it cannot escape, because the potential from $\pi$ to $2 \pi$ is infinite. Instead there is total reflection, meaning the particle bounces back and forth between $0$ to $\pi$. The Schrödinger equation for a free particle which is restricted to a semicircle (technically, whose configuration space is the circle $S^1$) is

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

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 $I=m s^2$, exactly the same for a hoop of the same radius. The wave function can now be expressed as $-\frac{\hbar^2}{2I} \frac{d^2\psi}{d\phi^2} = E\psi$, which is easily solvable.

Since the particle cannot escape the region from $0$ to $\pi$, 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 $\psi$ 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 $\phi = 0$ and $\phi = \pi$. 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 $\phi= \pi$ 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 $\sin (n \pi) = 0$. 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 $\psi = 0$ 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.

Analysis

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

Using cylindrical coordinates, the operators $\phi$ and $L_z$ are expressed as $\phi$ 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 $\Delta_{\psi} L_z = \sqrt{\langle {L_z}^2\rangle_\psi - \langle {L_z}\rangle_\psi ^2} \quad (9)$

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 $2 \pi$, 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 $\phi$ and $L_z$ replace the linear operators x and p.

See also

• particle in a ring
• particle in a box
• finite potential well
• Delta function potential
• gas in a box
• Particle in a spherically symmetric potential