Quantized Energy Levels

A particle cannot have any energy it likes when it is confined. That is the central surprise of the infinite square well, and understanding why is the goal of this lesson.

The infinite square well

The infinite square well (or particle in a box) is the simplest quantum confinement model. The potential energy is

V(x) = \begin{cases} 0, & 0 \le x \le L, \\ \infty, & \text{otherwise.} \end{cases}

Because the walls are impenetrable, the wavefunction must vanish at both ends: $\psi(0) = 0$ and $\psi(L) = 0$ . Inside the well the particle is free, so the time-independent Schrödinger equation reduces to

-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\,\psi,

whose general solution is $\psi(x) = A\sin(kx) + B\cos(kx)$ with $k = \sqrt{2mE}/\hbar$ .

Boundary conditions force quantization

The left boundary $\psi(0) = 0$ forces $B = 0$ , leaving $\psi(x) = A\sin(kx)$ . The right boundary $\psi(L) = 0$ then requires

\sin(kL) = 0 \implies kL = n\pi, \quad n = 1, 2, 3, \ldots

Negative integers give the same states (just a sign flip), and $n = 0$ would make $\psi$ vanish everywhere — no particle. So only the positive integers survive, and the allowed wave numbers are

k_n = \frac{n\pi}{L}.

Substituting back into $k = \sqrt{2mE}/\hbar$ and solving for $E$ gives the quantized energy levels:

E_n = \frac{\hbar^2 k_n^2}{2m} = \frac{n^2\pi^2\hbar^2}{2mL^2}, \qquad n = 1, 2, 3, \ldots

This is the key result. The integer $n$ is the quantum number of the state; it is forced on us by the requirement that the wavefunction be continuous and vanish at impenetrable walls — not by any external assumption.

Structure of the spectrum

Three features of $E_n = n^2 E_1$ are worth memorizing:

There is a ground state energy. The lowest allowed energy is $E_1 \gt 0$ . A classical particle can sit still at the bottom of the well; a quantum particle cannot. The residual motion is a direct consequence of the uncertainty principle: if position is confined to a region of size $L$ , momentum must fluctuate, and kinetic energy cannot be zero.
The levels grow as $n^2$ , not linearly. The spacing between adjacent levels increases with $n$ : $E_{n+1} - E_n = (2n+1)E_1$ . High- $n$ states are more spread out in energy than low- $n$ states.
All lengths scale identically. Shrinking $L$ by a factor of 2 increases every $E_n$ by a factor of 4. This is why electrons confined to atomic-scale distances (tenths of nanometers) have energies in the electron-volt range, while a macroscopic box has negligibly spaced levels that mimic a continuum.

A numerical check

For an electron ( $m = 9.109 \times 10^{-31}$ kg) in a well of width $L = 1$ nm, the ground-state energy is

E_1 = \frac{\pi^2 (1.055 \times 10^{-34})^2}{2 \times (9.109 \times 10^{-31}) \times (10^{-9})^2} = 6.024 \times 10^{-20}\,\text{J} = 0.376\,\text{eV}.

The factor $\pi^2 \approx 9.87$ and the exact numerical result follow directly from the formula derived above. The second level is $E_2 = 4E_1 \approx 1.50$ eV, the third is $E_3 = 9E_1 \approx 3.39$ eV, and so on.

Try it

This is a numerical exercise — return a number. An electron is trapped in an infinite square well of width $L = 1\,\text{nm}$ . Compute $E_2$ — the energy of the $n = 2$ state — in electron-volts.

Run your code to see the quantum state.