The Energy Eigenfunctions

From boundary conditions to eigenfunctions

In the previous lesson you found that inside an infinite square well of width $L$ (walls at $x = 0$ and $x = L$ ), the time-independent Schrödinger equation reduces to

$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\,\psi, \quad 0 \lt x \lt L,$

with the boundary conditions $\psi(0) = 0$ and $\psi(L) = 0$ imposed by the infinitely high walls. Writing $k^2 = 2mE/\hbar^2$ , the general solution inside the well is

$\psi(x) = A\sin(kx) + B\cos(kx).$

The condition $\psi(0) = 0$ immediately forces $B = 0$ , leaving only the sine term. Applying $\psi(L) = 0$ then gives

$A\sin(kL) = 0.$

Because $A = 0$ would make the wave function trivially zero everywhere (no particle), we must have $\sin(kL) = 0$ , which means $kL = n\pi$ for a positive integer $n$ . That is exactly how the quantum numbers appear: not by assumption, but as a logical consequence of the boundary conditions.

The family of eigenfunctions

Substituting $k_n = n\pi/L$ gives one eigenfunction for every positive integer $n$ :

$\psi_n(x) = A\sin\!\left(\frac{n\pi x}{L}\right), \quad n = 1, 2, 3, \ldots$

The constant $A$ is fixed by normalization (the next lesson covers that); for now it is enough to know the shape of each $\psi_n$ .

A few features worth noting:

$\psi_1$ has no nodes (zero crossings) inside the well — it is a single arch.
$\psi_2$ has one node at $x = L/2$ — the function passes through zero there.
In general, $\psi_n$ has exactly $n - 1$ interior nodes.

Each $\psi_n$ is an energy eigenfunction: a state where measuring the energy always returns the corresponding eigenvalue $E_n = n^2\pi^2\hbar^2/(2mL^2)$ with certainty.

A concrete example

Take the ground state $n = 1$ with $L = 1\,\text{m}$ (convenient units). The eigenfunction is $\psi_1(x) = A\sin(\pi x)$ . At the midpoint $x = L/2 = 0.5\,\text{m}$

$\psi_1\!\left(\tfrac{L}{2}\right) = A\sin\!\left(\pi \cdot \tfrac{1}{2}\right) = A\sin\!\left(\tfrac{\pi}{2}\right) = A \cdot 1 = A.$

The ground-state wave function reaches its maximum at the center of the well, consistent with having no interior nodes. For $n = 2$ the midpoint is a node: $\sin(2\pi \cdot \tfrac{1}{2}) = \sin(\pi) = 0$ .

Try it

This is a numerical exercise — your code should return a number. Evaluate the unnormalized ground-state eigenfunction $\psi_1(x) = \sin(\pi x/L)$ at $x = L/2$ with $L = 1$ .

Run your code to see the quantum state.