The simplest quantum system: a particle trapped between two hard walls a distance $L$ apart. Inside, $U = 0$; the walls are impossible to cross, so $\psi = 0$ at both. The Schrödinger equation plus those two conditions allow only certain energies. It is a fair model for an electron in a long molecule, a nanocrystal (a “quantum dot”) or, roughly, a proton in a nucleus.
🔑 Key idea
A particle of mass $m$ in a box of width $L$ can have only $$E_n = \frac{n^2h^2}{8mL^2},\qquad \psi_n = \sqrt{\frac{2}{L}}\,\sin\frac{n\pi x}{L},\qquad n = 1, 2, 3, \ldots$$ For an electron, $E_n = n^2\,\frac{0.376\ \text{eV·nm}^2}{L^2}$. The lowest energy, $E_1$, is not zero.
💡 Why it works
Inside, the solutions are $\psi = A\sin kx + B\cos kx$. $\psi(0) = 0$ forces $B = 0$. $\psi(L) = 0$ then needs $\sin kL = 0$, so $kL = n\pi$: a whole number of half-wavelengths fit in the box, $\lambda_n = \frac{2L}{n}$, like a guitar string's standing waves. The momentum is $p = \frac{h}{\lambda} = \frac{nh}{2L}$, and $E = \frac{p^2}{2m} = \frac{n^2h^2}{8mL^2}$. Why not $n = 0$? Then $\psi = 0$ everywhere: no particle at all. So the particle always has some kinetic energy, the zero-point energy, as the uncertainty principle demands of anything confined to a width $L$.
The first few states of an electron in a 1 nm box. Energies grow like $n^2$; each state has one more half-wavelength than the one below. Shrink the box and every level rises like $1/L^2$.
Worked examples
✏️ Worked example: An electron in a small box
An electron is confined to a box 0.50 nm wide. Find $E_1$ and $E_2$.
Answer: About 2 MeV, which is why nuclear energies are millions of times atomic ones
Quantum dots put this to work. They are semiconductor crystals only a few nanometers across, and the electrons inside behave much like particles in a box. Shrinking the dot spreads the levels apart, so smaller dots emit bluer light and larger dots redder light, all from the same material. Some television screens use them to make pure, bright colors.
⚠️ Common mistake
Thinking the lowest energy of a particle in a box is zero, as for a ball resting on the floor of a box. $n = 0$ would make $\psi$ zero everywhere. The lowest allowed state, $n = 1$, has a half-wavelength across the box and kinetic energy $\frac{h^2}{8mL^2}$. Check with the uncertainty principle: confined to $\Delta x \approx L$, the particle must have $\Delta p \gtrsim \frac{\hbar}{2L}$, so it cannot be at rest.
⚠️ Common mistake
Forgetting that the energies grow as $n^2$, not $n$: the levels spread apart going up ($E_2 - E_1 = 3E_1$, $E_3 - E_2 = 5E_1$).
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.