Semicircular potential well

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

:$-frac\left\{hbar^2\right\}\left\{2m\right\} abla^2 psi = Epsi quad \left(1\right)$

Wave function

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

:$abla^2 = frac\left\{1\right\}\left\{s^2\right\} frac\left\{partial^2\right\}\left\{partial phi^2\right\} quad \left(2\right)$

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

:$-frac\left\{hbar^2\right\}\left\{2m s^2\right\} frac\left\{d^2psi\right\}\left\{dphi^2\right\} = Epsi quad \left(3\right)$

The moment of inertia for a semicircle, best expressed in cylindrical coordinates, is $I stackrel\left\{mathrm\left\{def\left\{=\right\} iiint_V r^2 , ho\left(r,phi,z\right),r dr,dphi,dz !$. 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 $-frac\left\{hbar^2\right\}\left\{2I\right\} psi = Epsi$, which is easily solvable.

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

:$psi \left(phi\right) = A cos\left(m phi\right) + B sin \left(m phi\right) quad \left(4\right)$

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

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

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

:$psi \left(0\right) = psi \left(pi\right) = 0 quad \left(6\right)$.

Since the wave function $psi\left(0\right) = 0$, the coefficient A must equal 0 because $cos \left(0\right) = 1$. 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 $psi \left(pi\right) = 0 = B sin \left(m pi\right)$ only when m is an integer since $sin \left(n pi\right) = 0$. This boundary condition quantizes the energy where the energy equals $E= frac\left\{m^2 hbar ^2\right\}\left\{2I\right\}$ 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 $B= sqrt\left\{frac\left\{2\right\}\left\{pi$. The normalized wave function is

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

The ground state energy of the system is $E= frac\left\{hbar ^2\right\}\left\{2I\right\}$. Like the particle in a box, there exists nodes in the excited states of the system where both $psi \left(phi\right)$ and $psi \left(phi\right) ^2$ 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 $-i hbar frac\left\{d\right\}\left\{dphi\right\}$ 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:

:$\left[phi, L_z\right] = i hbar psi\left(phi\right) quad \left(8\right)$

:$\left(Delta phi\right) \left(Delta L_z\right) geq frac\left\{hbar\right\}\left\{2\right\}$ where $Delta_\left\{psi\right\} phi = sqrt\left\{langle \left\{phi\right\}^2 angle_psi - langle \left\{phi\right\} angle_psi ^2\right\}$ and $Delta_\left\{psi\right\} L_z = sqrt\left\{langle \left\{L_z\right\}^2 angle_psi - langle \left\{L_z\right\} angle_psi ^2\right\} quad \left(9\right)$

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 $frac\left\{- pi\right\}\left\{2\right\}$ to $frac\left\{pi\right\}\left\{2\right\}$, the issue of even and odd parity becomes important.

The wave equation for such a potential is given as:

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

where $psi_o \left(phi\right)$ and $psi_e \left(phi\right)$ 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.

ee 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
* Delta potential well (QM)

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