How do I arrive at the wave function of a particle in a ring? Why would the answer be ψ = Ae^+-inθ and not one with sine or cosine?

Question

How do I get to the particle wave function? . Consider a particle in a ring. For this example let the circumference line to extend from -L/2 to L2 with the end points glued together to form a complete ring of length L. normalize ψ

Answer 1

To arrive at the wave function of a particle in a ring, we need to solve the Schrödinger equation for the system. The Schrödinger equation for a particle in a ring can be written as:

-(ħ^2/2m) d^2ψ/dθ^2 = Eψ

where ħ is the reduced Planck constant, m is the mass of the particle, E is the energy of the particle, and θ is the angle around the ring.

We can assume that the wave function has the form:

ψ = Ae^±inθ

where A is a normalization constant and n is an integer. The plus sign corresponds to the wave function traveling clockwise around the ring, and the minus sign corresponds to the wave function traveling counterclockwise around the ring.

To verify that this form of the wave function satisfies the Schrödinger equation, we can substitute it into the equation and simplify:

-(ħ^2/2m) d^2/dθ^2 (Ae^±inθ) = E(Ae^±inθ)

= -(ħ^2/2m) (±in)^2 Ae^±inθ = E Ae^±inθ

= (ħ^2n^2/2m) A e^±inθ = E A e^±inθ

So we see that this form of the wave function satisfies the Schrödinger equation, provided that the energy E is given by:

E = (ħ^2n^2/2m)

To normalize the wave function, we need to ensure that the integral of the square of the wave function over the entire ring is equal to 1. This gives us:

∫_0^L |ψ|^2 dθ = ∫_0^L |A|^2 dθ = 1

which implies that:

|A|^2 L = 1

So we can choose A to be:

A = 1/√L

Therefore, the normalized wave function for a particle in a ring is:

ψ = (1/√L) e^±inθ

So, we see that the wave function for a particle in a ring has the form of a complex exponential, rather than a sine or cosine function, because the boundary conditions of the problem require that the wave function be periodic around the ring. The complex exponential provides a simple way to represent a periodic function, which is why it is the natural choice for this problem.