# Predicting the value of $n$ for the energy of a wavefunction in the infinite square well by inspection

$$E=\frac{n^2 \pi ^2 \hbar^2}{2ma^2}$$

Is it always possible to tell the value of $n$ by inspecting the shape of the wavefunction in the infinite square well no matter what the value of $a$ is?

Right now, I'm numerically generating a wavefunction. I look at the shape and it has zero nodes (not counting $\psi (0)$ or $\psi(a)$. Can I conclude that this must be the ground state energy? Let's say that I change the value of $a$ to something else. As long as I generate a wavefunction with zero nodes, that wavefunction must have the ground state energy, right?

Take a look at this image: Notice that we can tell which one $n=1$ is simply by counting the nodes. To restate my question - no matter what I change my value of $a$ to, as long as I generate a wavefunction with zero nodes, then that must be the ground state energy, correct?

• this is a nice question; some easy properties of the solutions are often overlooked. Mar 6, 2017 at 3:02

Yes. One can prove the so-called "oscillatory" theorem, which shows that the number of nodes in the eigenfunctions describing bound states must increase with energy. Thus, if you have an eigenfunction with $0$ nodes, it must be the ground state wavefunction.
I do not know of examples where the number of nodes of eigenfunction number $n+1$ is not exactly one more than eigenfunction number $n$, but I suppose there could be strange examples where the number of nodes of consecutive solutions goes up by $2$.