Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
Yes, we expect the condition to hold for a ring. –  JamalS Apr 29 at 8:42
    
What is $\psi'$? –  John Rennie Apr 29 at 8:43
    
@JohnRennie the first derivative of the solution to the wavefunction. –  user180097 Apr 29 at 8:50
    
Oops, yes, of course. Sorry for the silly question :-) –  John Rennie Apr 29 at 8:52
2  
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 at 9:50

1 Answer 1

up vote 3 down vote accepted

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.

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.