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Is there a simple intuitive explanation for why a particle on a ring has no zero point energy? That is, if we write the energy as:

$$ E_n = \frac{n^2\hbar^2}{2mr^2} $$

then the integer $n$ is allowed to take the value zero. If we consider the apparently similar system of a particle in a 1D infinite potential well, where the energy is given by:

$$ E_n = \frac{n^2\pi^2\hbar^2}{2mL^2} $$

then the integer $n$ is not allowed to be zero. Why the difference?

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  • $\begingroup$ It might be worth mentioning here that the particle on a ring is surprisingly rich mathematically, particularly in considerations of its propagator. This is explained beautifully by LS Schulman in Techniques and Applications of Path Integration (Dover, pp. 191-195) as well as in Phys. Rev. 176 1558 (1968). $\endgroup$ – Emilio Pisanty Nov 13 '15 at 19:40
  • $\begingroup$ Good question! As answered below, this is a manifestation of the difference between open and periodic boundary conditions. Another such manifestation is the edge states of a topological insulator at its boundaries. $\endgroup$ – pathintegral Nov 14 '15 at 7:03
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Consider first the particle in a 1D infinite potential well:

1D box

The probability of finding the particle must be zero where the potential is infinite, so the wavefunction $\Psi$ must be zero at the edges of the box. $\Psi$ is non-zero somewhere inside the box, so it must have a form something like the red line. I've just drawn a random squiggle for the red line to illustrate that $\Psi$ must rise from zero at one edge and fall back to zero at the other edge.

The wavefunction is described by the Schrödinger equation:

$$ \frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\psi = E\Psi $$

and inside the box the potential $V$ is zero so this simplifies to:

$$ \frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = E\Psi $$

The only way the energy $E$ can be zero is if the left hand side is zero, which means:

$$ \frac{\partial^2\Psi}{\partial x^2} = 0 $$

But it should be obvious that whatever the shape of the red line its second derivative cannot be zero everywhere and therefore the energy cannot be zero. If $\Psi$ had the form of the green line then $\partial^2\Psi/\partial x^2$ is zero and this would have zero energy. However the wavefunction can't be shaped like the green line because it violates the boundary conditions.

Now consider the particle on a ring.

In this case the particle is free, so the potential $V = 0$, but since the wavefunction can't have two different values at the same point we require that if we go right round the ring the wavefunction must go back to the same value. That is, if we express the wavefunction as a function of the angle round the ring $\theta$ we must have:

$$ \Psi(\theta) = \Psi(\theta + 2\pi) $$

In this case a wave function that is constant is a perfectly good solution.

Particle on a ring

The wavefunction obeys the requirement that $\Psi(\theta) = \Psi(\theta + 2\pi)$ because it's the same everywhere, and it has $\partial^2\Psi/\partial \theta^2 = 0$ so it is a solution to the Schrödinger equation with $E = 0$. So in this case we can have a solution with zero energy.

The big difference between the two is that the 1D box has a boundary condition requiring $\Psi$ to be zero while the ring doesn't. And this is the key to understanding when we get a zero point energy. Any time we have a boundary condition requiring $\Psi = 0$ on some boundary it means $\Psi$ must increase from zero away from the boundary and decrease back to zero at the boundary, and this necessarily means the energy must be non-zero. For example the wavefunction for an electron in a hydrogen atom has to be zero at infinity, so we know immediately that it must have a zero point energy (which it does of course).

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  • $\begingroup$ I'm wondering why the zero energy solution for particle in a ring not violate Heisenberg's Uncertainty Principle? $\endgroup$ – user148792 Jan 2 '18 at 19:49
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There's a meaningful difference between the boundary conditions. For the infinite square well, the probability at the boundaries is required to be 0. This limits us to sines as opposed to cosines (assuming the standard convention of one boundary is at position 0, and there is no nonzero phase in our solutions). Since $\sin 0 = 0$, $\sin \left ( \frac{n \pi x}{L} \right)$ would be $0$ for all $x$ if $n=0$, which is forbidden by the constraint that the total probability must be 1 (i.e., the particle must exist somewhere). For the ring, we have only a periodic boundary condition - the wavefunction must only be the same at $\theta = 0$ and $\theta = 2\pi$ (where $\theta$ measures the angular position on the ring), not necessarily 0. This enables us to have cosine solutions as well, and in particular, it enables the $\cos \left ( n \theta \right )$ solution for $n = 0$ - a constant amplitude all the way around the ring, which can be normalized to total probability 1 - with zero energy.

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