Section 13.5: Particle on a Ring

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For an infinite square well potential, a particle is confined to a box of length L by two infinitely high potential energy barriers:

         V(x) = ∞, for x ≤ 0 ,           V(x) = 0 for 0 < x < L ,      and    V(x) = ∞ for x ≥ L .             (13.15)

For a particle on a ring (or a hoop) of radius R, the particle only has one degree of freedom, although the ring exists in two dimensions (on the xy plane). We begin by using the Schrödinger equation in cylindrical coordinates:

−ħ²/2m [(1/ρ)(∂/∂ρ)(ρ(∂/∂ρ)) + (1/ρ²)∂²/∂θ² + ∂²/∂z²]ψ(r) + V(r)ψ(r) = Eψ(r) .            (13.16)

Since the hoop is on the xy plane, ρ = R, z = 0, which are both constant. We therefore find that Eq. 913.16) reduces to

−ħ²/2m [ (1/R²) d²/dθ² ] ψ(θ) = Eψ(θ) ,                    (13.17)

since V = 0.  Eq.(13.17) can be written in a more standard form, d²/dθ²ψ(θ) + (2mR²E/ħ²)ψ(θ) = 0, which yields the equation

[d²/dx² + k²R²]ψ(θ) = Eψ(θ) ,                    (13.18)

upon making the substitution, k² = 2mE/ħ².  The general solution is therefore ψ(θ) = Ae^ikRθ, where we allow k to take either positive or negative values.  We next require that such solutions are valid wave functions: they must be single valued over the extent of the ring.  This is an issue since the ring repeats every 2π.  We require solutions that obey the periodic boundary condition, ψ(θ) = ψ(θ + 2π).  Therefore, it must be the case that k = n/R where n = 0, ±1, ±2,…..  The normalized wave function is ψ_n(θ) = (2π)^-1/2e^inθ.  We also find that the energy is quantized E_n = ħ²k²/2m = n²ħ²/2mR², where n = 0, ±1, ±2,…..  Note that because of the periodic boundary condition, there are both positive and negative integer solutions for n, and there is a zero-energy solution (n = 0).

The solutions to the ring problem are shown in "Animation 1(θ)".  The periodic boundary condition is already imposed on the wave functions and you can vary the quantum number n to see the effect on the wave function.  In the animation, ħ = 2m = 1 and the time evolution of the wave function, which is also shown in the animation, can be easily shown to obey

ψ_n(θ,t) = (2π)^−1/2 e^−iEn^t/ħ e^inθ .                    (13.19)

Now consider "Animation 1(x)" in which the variable is now x (as opposed to θ). In this case, x repeats every 2πR = L and the conversion between the two variables is θ = x/2πR = x/L.  Given this coordinate transformation, the wave function becomes ψ_n(x) = (2πR)^−1/2e^inx/R = (L)^−1/2 e^2πinx/L.  We also find, as before, that E_n = n²ħ²/2mR² = 4[n²π²ħ²/2mL²], where n = 0, ±1, ±2,…..  Note that because of the periodic boundary condition, the energy is a factor of 4 greater than that of the usual infinite square well of the same length, there are both positive and negative integer solutions for n, and  there is a zero-energy solution (n = 0).

In Animation 2(θ) and Animation 2(x) you can see the time evolution of an equal-mix two-state superposition and also change the two states. Notice how some combinations (like {n₁ = 1,n₂ = 2}) yield time-dependent wave functions, while others (like {n₁ = 1,n₂ = -1}) yield time-independent standing waves.

In Animation 3(θ) and  Animation 3(x) and Animation 4(θ) and  Animation 4(x) you can see how an initial Gaussian wave packet (one without an initial momentum and one with an initial momentum) on a ring behaves. Notice the difference as compared to the wave packet in the infinite square well, Section 10.7. While the wave packet on the ring still revives like the infinite square well case, the differences in the motion are due to the lack of hard walls in the case of the ring.

Optional Visualization
You may also view Animation 3 and Animation 4 in three dimensions if you have Java 3D installed on your computer.  If you do not have Java 3D installed, go to the Sun Java 3D Web site for download: http://java.sun.com/products/java-media/3D/download.html.

© 2006 by Prentice-Hall, Inc. A Pearson Company