# Quantum Ring

Haxx0rm4ster
When solving Schrodinger's eqn for a quantum ring, what would be the boundary conditions?

The solution (polar) should be
Ψ(Φ) = A exp(ikΦ) + B exp(-ikΦ)

And I believe the boundary conditions are
Ψ(0) = Ψ(2pi)
Ψ(0) = A + B
Ψ(2pi) = A exp(ik*2π) + B exp(ik*2π)

and I suppose I can safely say that
Ψ(0) = 0

From these conditions/assumptions,
k = n, n = 1,2,3...
B = -A,

And therefore
Ψ(Φ) = A exp(inΦ) - A exp(-inΦ)

Which can be expressed as
Ψ(Φ) = 2Ai sin(nΦ)

I'm just wondering, are my boundary condition assumptions complete and correct?

Thanks

## Answers and Replies

nnnm4
No the phi(0) = 0 boundary condition is not a good one. This eliminates and cosine behavior in the wavefunction. Your second boundary condition is that the wavefunction be continuously differentiable.

Nano-Passion
Hey, sorry to hijack your thread but whats a quantum ring? I'm curious.

Haxx0rm4ster
By quantum ring, I'm referring to something like a quantum wire looped around in the shape of a ring.