[Twickel]

Hi
I can understand the particle in a box fine. However, I have no idea why the particle on a ring can have a lowest energy level of zero. ( j=0)
Can someone please explain it in words.


Thank you,

[Jorriss]

What restriction would there be stopping zero from being an energy eigenvalue?

[Twickel]

Am I correct?

 Psi Cos(j) for j = 0 is equal to 1.
psi Sin(j) for j=0 is equal to zero. ( not valid, the lowest energy for the sin wave on a particle on a ring still has to be j=1?)

Both work on a ring, but the sine wave function has to start at one. Sine wave works because its periodic and the cosine wave works but can start at j=0 since cos0 = 1. the Cosine wave is also periodic.

[Twickel]

i tried to edit my post but I could not.

For a particle in a box psi= sin(n) for a particle on a ring psi=cos or sin ( 0 and 1) respectively.


Why can sin j= 0 not work on a particle on a ring if all you need is to fit an integer number of waves and a particle on a ring can be motionless. Shouldnt sin 0 also work for the particle on a ring

If n where to equal zero in a particle in a box, the particle would have no kinetic energy and hence be stationary,. what consequences does that have? does it mean the prob of finding the particle is zero or the same all along the x axis?

[Enthalpy]

I disagree with the zero kinetic energy. As soon as the particle is confined, its wave function contains components with non-zero energy.

The bad joke as the ring is described here is that not all dimensions are told! You may have zero wavelengths, impulse and kinetic energy along the ring but certainly not along other dimensions, like the radial and axial distances. The particle needs non-zero wave vectors there to be confined on a ring.

Sometimes (often) a wave function spanning several atoms (benzene ring, crystal...) is written as a combination of individual wave functions centred on an atom each. The the combination seems to have no wave vector, but this is only because the individual solutions around each atom bring implicitly their wave vector, impulse, kinetic energy.

I do agree that no angular and magnetic momentum is required for some solutions on a ring, but this isn't so special. Spherical orbitals do the same at atoms. And in a box, the least momentum solution depends on how you choose the boundary conditions.