# 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?

-
You might want to read [particles in a ring][1], it seems to be a legitimate solution that occurs in organic chemistry..., [1]: en.wikipedia.org/wiki/Particle_in_a_ring – innisfree May 3 '13 at 23:39
Doesn't this state get in trouble with the Uncertainty Principle, though? It appears to have precisely zero momentum... – Rococo May 4 '13 at 2:10
@BenCrowell But all the basis states $|x + 2n \pi\rangle$ with $n$ integer are physically equivalent. I would argue that the correct variance of the position is finite ($2\pi / \sqrt{3}$ if my algebra's right). Presumably the apparent conflict with the canonical position-momentum uncertainty relation is resolved by properly deriving the analogous expression on a compact space with periodic BCs. I'm going to bed but will try to formalise this tomorrow. Or leave it as a challenge... – Mark Mitchison May 4 '13 at 3:33
@Mark Mitchison: The issue of how to construe the uncertainty relation in a space that wraps around is interesting, but I think it's a completely separate question. – Ben Crowell May 5 '13 at 3:42
@BenCrowell Fair point. In case you're interested, though, this problem was apparently considered way back in '63. The upshot is that one can modify the uncertainty relation in this case to allow for a state $\psi(x) = \sqrt{1/2\pi}$ of definite angular momentum. See On the uncertainty relation for Lz and φ by D. Judge, Physics Letters 5 p. 189 (1963). – Mark Mitchison May 5 '13 at 18:22

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.

-

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.

-
I suppose this is true, but doesn't changing the potential change the form of the solutions as well as the energies? Unlike the gravitational potential say, the quantum potential determines the form of the allowed solutions. Doesn't changing it lead to a different problem? – Tom Oldfield May 4 '13 at 8:57
@TomOldfield: Yes, adding a constant to the potential changes the form of the solutions. However, that isn't really all that surprising. A Galilean boost changes the form of the solutions to the Schrodinger equation in a similar way. – Ben Crowell May 5 '13 at 3:43