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

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

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.