Quantum Ring

  • #1
Haxx0rm4ster
37
0
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

  • #2
nnnm4
113
0
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.
 
  • #3
Nano-Passion
1,291
0
Hey, sorry to hijack your thread but whats a quantum ring? I'm curious.
 
  • #4
Haxx0rm4ster
37
0
By quantum ring, I'm referring to something like a quantum wire looped around in the shape of a ring.
 
  • #5
Avodyne
Science Advisor
1,396
90
The correct boundary condition is periodicity: Ψ(Φ) = Ψ(Φ+2π).
 
  • #6
nnnm4
113
0
Which he already used, so the periodicity of the derivative is the other boundary condition.
 

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