<math> \frac{h}{2m}\nabla^2 \phi=E\phi. </math>
This must be solved under the circularity condition. Let the ring's radius be r, then in one dimension:
<math> \nabla = \frac{\partial}{\partial x}=\frac{1}{r}\frac{\partial}{\partial \alpha}, </math>
where α is the position angle on the ring. We get
<math> \frac{h}{2mr^2}\frac{\partial^2 \phi}{\partial \alpha^2} = E \phi. </math>
The solution of this is
<math> \phi = \exp(\pm i \sqrt{\frac{2mE}{h}} r\alpha). </math>
The circularity condition is now, that
<math> \sqrt{\frac{2mE}{h}} r \alpha = 2 n \pi \alpha </math>
eli
<math> E = \frac{h}{2m}\frac{4n^2\pi^2}{r^2}=\frac{2h\pi^2}{mr^2}n^2. </math>
Now the wave function becomes
<math> \phi=\exp(\pm i 2 n \pi \alpha). </math>
Quantum states found:
====<math>n=0</math>:====
====<math>n=1</math>:====
====<math>n=2</math> (and higher):====
Conclusion: every quantum state is filled by a total of <math>2n + 1</math> particles.
In organic chemestry, aromatic compounds contain atomic rings, such as benzene rings (the Kekulé structure) consisting of five or six, usually carbon, atoms. So does the surface of "buckyballs" (buckminsterfullerene). These molecules are exceptionally stable.
The above explaines why: the ring behaves like a circular wave guide. The excess (valency) electrons spin around in both directions.
Every energy level is filled by <math>2\times(2n+1)</math> electrons, as electrons have additionally two possible orientations of their spins.
The rule that <math>4n+2</math> excess electrons in the ring produces an exceptionally stable ("aromatic") compound, is know as the Hückel rule.
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