- #1

Haxx0rm4ster

- 37

- 0

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