Do we always ignore zero energy solutions to the (one dimensional) Schrödinger equation?

Question

When we solve the Schrödinger equation on an infinite domain with a given potential $U$, much of the time the lowest possible energy for a solution corresponds to a non-zero energy. For example, for the simple harmonic oscillator with frequency $\omega$, the possible energies are $\hbar\omega(n+\frac12)$ for $n =0,1,\dots$ . Some of the time, solutions with zero energy are possibly mathematically, but boundary conditions would mean that such solutions would be everywhere zero, and hence the probability of finding a particle anywhere would be zero. (For example, an infinite potential well).

However, when solving the Schrödinger equation for a particle moving freely on a circle of lenfth $2\pi$ with periodic boundary conditions $\Psi(0,t)=\Psi(2\pi,t)$ and $\frac{\partial\Psi}{\partial x}(0,t)=\frac{\partial\Psi}{\partial x}(2\pi,t)$, I have found a (normalised) solution $\Psi(x,t)=\frac1{2\pi}$ with corresponding energy $0$. I can't find a way to discount this mathematically, and it seems to make sense physically as well. Is this a valid solution, and so is it sometimes allowable to have solutions with $0$ energy? Or is there something I'm missing?

Answer 1

Your solution is valid. It has zero kinetic energy. It doesn't necessarily have zero energy. It can have any potential energy you'd like. Just because your particle is "freely moving," that doesn't mean the potential is zero. You could have $V(x)=k$ for any constant $k$. The value of $k$ is not observable and has no physical significance.

In general there is no special significance to having zero energy in a solution to the Schrodinger equation. Any solution can be defined to have zero energy simply by changing the potential appropriately like $V\rightarrow V+c$, where $c$ is some constant.

A realistic example involving zero kinetic energy and a constant wavefunction would be some particle-rotor models of nuclei, in which the deformed (prolate) nucleus (rotor) has some orientation in space, specified by one or two angular coordinates. If the odd particle has some component $K$ of its angular momentum along the symmetry axis, you get a rotational band with energies proportional to $J(J+1)$, starting with a ground state at $J=K$. In the ground state for the $K=0$ case, the rotor has zero kinetic energy, and its wavefunction is a constant as a function of the angular coordinates.

Answer 2

In my view, the important question to answer here is a special case of the more general question

Given a space $M$, what are the physically allowable wavefunctions for a particle moving on $M$?

Aside from issues of smoothness wavefunctions (which can be tricky; consider the Dirac delta potential well on the real line for example), as far as I can tell there are precisely two other conditions that one needs to consider:

1. Does the wavefunction in question satisfy the desired boundary conditions?

2. Is the wavefunction in question square integrable?

If a wavefunction satisfies these properties, then I would be inclined to assert that it is physically allowable.

In your case where $M$ is the circle $S^1$, the constant solution is smooth, satisfies the appropriate conditions to be a function on the circle (periodicity), and is square integrable, so it is a physically allowed state. It also happens to be an eigenvector of the Hamiltonian operator with zero eigenvalue; there's nothing wrong with a state having zero energy.