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

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:

enter image description here

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?

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

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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 $2$.

This increase in the number of nodes is a general property of solutions in a given potential (or effective potential if the motion is in 2d or 3d). You can also see the increase in the number of nodes in the solutions to the 1d quantum harmonic oscillator, for instance.

You can read about the oscillatory theorem in many texts, but I know the older textbook by Albert Messiah discusses this and related properties of solutions at length.

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