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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 $pi$ 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 ## Wave function

## 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.## 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. ## See also

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

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

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

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

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

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

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 $frac\{dpsi\}\{dphi\}$ are continuous and the wave function is normalizable:

- $int\_\{0\}^\{pi\}\; left|\; psi\; (phi\; )\; right|^2\; ,\; dphi\; =\; 1\; quad\; (5)$.

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.

Using cylindrical coordinates, the operators $phi$ and $L\_z$ are expressed as $phi$ and $-i\; hbar\; frac\{d\}\{dphi\}$ 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 $Delta\_\{psi\}\; phi\; =\; sqrt\{langle\; \{phi\}^2rangle\_psi\; -\; langle\; \{phi\}rangle\_psi\; ^2\}$ and $Delta\_\{psi\}\; L\_z\; =\; sqrt\{langle\; \{L\_z\}^2rangle\_psi\; -\; langle\; \{L\_z\}rangle\_psi\; ^2\}\; quad\; (9)$

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.

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Last updated on Thursday November 22, 2007 at 15:18:25 PST (GMT -0800)

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