# Boundary conditions: Particle in a ring

Particle in ring is a well-known example where a solution of the Schrodinger equation exists. My question is: In principle we also want that $\psi'(\theta) = \psi'(\theta + 2\pi)$. The thing is that this condition is never explicitly stated ( probably because it is fulfilled anyway, but in principle we would also need this condition, right?

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Yes, we expect the condition to hold for a ring. – JamalS Apr 29 '14 at 8:42
What is $\psi'$? – John Rennie Apr 29 '14 at 8:43
@JohnRennie the first derivative of the solution to the wavefunction. – Xin Wang Apr 29 '14 at 8:50
Oops, yes, of course. Sorry for the silly question :-) – John Rennie Apr 29 '14 at 8:52
The condition $\psi'(\theta) = \psi'(\theta + 2\pi)$ is a consequence of $\psi(\theta) = \psi(\theta + 2\pi)$ whenever $\psi$ is differentiable. – Qmechanic Apr 29 '14 at 9:50

A particle in a ring corresponds to a configuration space $S^{1}$ which is simply a circle. The solution to the Schrödinger equation is given by (in natural units):

$$\psi_{\pm} = \frac{1}{\sqrt{2\pi}}e^{\pm ir \sqrt{2mE}\theta}$$

Clearly, we must identify $\theta$ with $\theta +2\pi n$. Differentiating the solution yields,

$$\psi_{\pm}' =\pm ir \sqrt{\frac{mE}{\pi}}e^{\pm ir \sqrt{2mE}\theta}$$

The function $\psi'_{\pm}$ differs by $\psi_{\pm}$ only by a constant, hence it is also periodic in $\theta$ with period $2\pi$, i.e.

$$\psi'_{\pm}(\theta)=\psi'_{\pm}(\theta+2\pi )$$

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