For a particle on a sphere, is zero energy possible?

[blaisem]

In my introduction to quantum mechanics, I learned about the particle in a box, followed by the quantum harmonic oscillator. In both instances, zero energy was not possible; the ground states had non-zero energy.

However, in deriving the solutions to the Schrödinger equation for a particle on a field-free sphere, the energy was found proportional to:

E ∝ (l² + l )

Furthermore, the ground-state spherical harmonic corresponds to l = 0, which does indeed yield zero energy as the eigenvalue of the hamiltonian. Since the particle is confined to the surface of the sphere, dρ = 0, and the radial component also contributes zero energy.

Questions:

1. Doesn't zero energy violate the uncertainty principle?
   
2. For a particle in a box, n = 0 was not allowed because it violated the uncertainty principle—why then is l = 0 permitted for a particle on a sphere?

Thank you!

 

[Nugatory]

I haven't checked your solution to confirm that it does yield zero energy, but...

Doesn't zero energy violate the uncertainty principle?

No, just as any other eigenstate of the Hamiltonian doesn't violate the uncertainty principle. Prepare the system in such a state and it won't (in general) be in an eigenstate of any operator that doesn't commute with the Hamiltonian - and that's all that the uncertainty principle requires.

For a particle in a box, n = 0 was not allowed because it violated the uncertainty principle—why then is l = 0 permitted for a particle on a sphere?

The problem with $n=0$ for a particle in a box isn't that it violates the uncertainty principle, it is that that wave function is zero everywhere when $n=0$.

 

[PeroK]

In my introduction to quantum mechanics, I learned about the particle in a box, followed by the quantum harmonic oscillator. In both instances, zero energy was not possible; the ground states had non-zero energy.

However, in deriving the solutions to the Schrödinger equation for a particle on a field-free sphere, the energy was found proportional to:

E ∝ (l² + l )

Furthermore, the ground-state spherical harmonic corresponds to l = 0, which does indeed yield zero energy as the eigenvalue of the hamiltonian. Since the particle is confined to the surface of the sphere, dρ = 0, and the radial component also contributes zero energy.

Questions:

1. Doesn't zero energy violate the uncertainty principle?
   
2. For a particle in a box, n = 0 was not allowed because it violated the uncertainty principle—why then is l = 0 permitted for a particle on a sphere?

Thank you!

The value of energy depends on the arbitrary energy of your potential. You can always add a constant. There's no physical significance to zero energy.

For example, taking electric potential energy to be zero at infinity, the ground state of hydrogen has an energy of $-13.6eV$. If you take the potential to be $13.6eV$ at infinity, then the ground state would have zero energy.

 

[bhobba]

Well being on a sphere then physical chemistry comes into it and you have stuff like van der Waals forces etc. I would not even guess at a solution without more detail.

Thanks
Bill

 

[hilbert2]

Even a particle constrained on a circular path on 2d plane, with Hamiltonian

$H=-\frac{\hbar^2}{2mR^2}\frac{\partial^2}{\partial \theta^2}$

has an energy eigenfunction $\psi(\theta)$ which is a $\theta$-independent constant and is therefore most conveniently described as having zero energy.

The difference to a particle in a box is that here the boundary condition doesn't require the wave function to be zero at any points, it only requires $\psi(\theta)$ to be continuous.

The free-particle eigenstates have been studied for many different topologies, like particle on a torus or even on a Möbius strip.

 

[vanhees71]

Indeed, and it's always important to remember that for a particle on a sphere the translations are the rotations, i.e., you have angular momentum. Also note that position observables are at least problematic on the sphere. So there's no analogue of the position-momentum uncertainty relation for a particle on a sphere, and thus zero energy for a free particle is no contradiction to any well-defined uncertainty relation (e.g., between the components of angular momentum).