# Why doesn't the particle in a ring model have a potential energy term?

I was thinking of the "particle on a ring" problem as a subdivision of a bigger problem, i.e. the hydrogen atom. While in the hydrogen atom problem, the Hamiltonian contains a potential energy term, the Hamiltonian in the "particle on a ring" problem does not.

Why is it so? What are its implications in the mathematical solution to the problems?

The short answer is that the particle in a ring is defined to have no potential energy term (at least, not beyond what Karsten Theis has pointed out). Once you introduce something else into the system, then it is no longer really the particle on a ring problem any more.

However, if you want to relate it to the hydrogen atom, then note that in the hydrogen atom, the Coulomb potential is a function of only $$r$$, i.e.

$$V(r) = -\frac{1}{r}$$

in atomic units. In the particle on a ring model system, the particle is already constrained to be at a particular value of $$r$$, so even if we were to hypothetically introduce a Coulomb-type potential $$\propto 1/r$$, this would just be a constant. The effect of having a constant potential energy is just to shift every eigenstate up in energy by the same amount; it has no real physical effect.

What are the differences in the mathematical solutions? Well, you already know it, surely; the eigenstates of the hydrogen atom are atomic orbitals, whereas the eigenstates of the particle on a ring are simply $$\exp(\mathrm im\phi)$$ with $$m \in \mathbb{Z}$$ to satisfy the boundary condition.

Of course, this $$\exp(\mathrm im\phi)$$ term is indeed part of the mathematical form of the atomic orbitals. This is because when you solve the Schrödinger equation for the hydrogen atom, you can successively separate out different degrees of freedom. Firstly you separate the bit which depends on $$r$$, which goes on to become the radial wavefunction $$R(r)$$. Then you need to solve for the angular wavefunction $$Y(\theta,\phi)$$, and the way of doing this is to again use separation of variables, i.e. assume $$Y(\theta,\phi) = f(\theta)g(\phi)$$. That gets you to a differential equation which looks like $$(-1/g)(\mathrm d^2g/\mathrm d \phi^2) = m^2$$, which has $$g = \exp(\mathrm im\phi)$$ as its solutions. I'm glossing over some details here, but that's the general idea; consult e.g. Griffiths' Introduction to Quantum Mechanics for the nitty-gritty bits.

There is a potential energy of some sorts: The potential energy is constant within the ring, and infinitely high outside of it. This is an idealized situation similar to the particle in a box (one-dimensional, two-dimensional, three-dimensional).

Even though these models are artificial and ideal, they can be used to estimate properties of certain sets of molecules. For example, you can discuss the spectra of molecules with linear conjugated double bonds using the model of the one-dimensional particle in the box.

When setting up the Hamiltonian, you implicitly introduced a infinitely high potential by boundary conditions and choice of coordinate systems. The free electron would move in three dimensions. For the one-dimensional particle in the box, the choice of just one coordinate constrains the electron to one dimension, and the boundary conditions of an electron density of zero at the edges of the box constrains the electron to a certain length of box. For the particle in a ring, you can choose a cylindrical coordinate system, keep the radius constant, the height at - say - zero and keep the angle as the only coordinate of the Hamiltonian and the wave function.