## Section 13.5: Particle on a Ring

n1 = | n2 =

Please wait for the animation to completely load.

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 xL . (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) = AeikRθ, 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 ψn(θ) = (2π)-1/2einθ.  We also find that the energy is quantized En = 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 energy eigenfunction. In the animation, ħ = 2m = 1 and the time evolution of the energy eigenfunction, which is also shown in the animation, can be easily shown to obey

ψn(θ,t) = (2π)−1/2 eiEnt/ħ einθ . (13.16)

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 energy eigenfunction becomes

ψn(x) = (2πR)−1/2einx/R = (L)−1/2 einx/L(13.17)

We also find, as before, that

En = n2ħ2/2mR2 = 4[n2π2ħ2/2mL2], (13.18)

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.

OSP Projects: Open Source Physics - EJS Modeling Tracker Physlet Physics Physlet Quantum Physics STP Book