How is the energy of a particle trapped in a one-dimensional rigid box derived using Schrödinger's equation?

In one line

Solving Schrödinger's equation for a particle confined between two infinitely high walls a distance \(L\) apart gives quantised energies \(E_n = n^2h^2/(8mL^2)\) with \(n = 1, 2, 3, \dots\).

Explained simply

Think of a guitar string fixed at both ends: it can only vibrate in patterns that fit a whole number of half-waves between the ends. A particle trapped in a box behaves the same way. Only certain wavelengths fit, and since wavelength fixes momentum and energy, only certain energies are allowed.

How it works

1. Potential: \(V = 0\) for \(0 < x < L\), \(V = \infty\) outside. The particle cannot be outside, so \(\psi = 0\) there.
2. Inside: \(\dfrac{d^2\psi}{dx^2} + k^2\psi = 0\), with \(k^2 = \dfrac{2mE}{\hbar^2} = \dfrac{8\pi^2mE}{h^2}\).
3. General solution: \(\psi = A\sin kx + B\cos kx\).
4. Boundary \(\psi(0) = 0\) gives \(B = 0\).
5. Boundary \(\psi(L) = 0\) gives \(A\sin kL = 0\); since \(A \ne 0\), \(kL = n\pi\), so \(k = n\pi/L\).
6. Energy: \(E_n = \dfrac{\hbar^2k^2}{2m} = \dfrac{n^2h^2}{8mL^2}\).
7. Normalisation gives \(A = \sqrt{2/L}\), so \(\psi_n = \sqrt{\dfrac{2}{L}}\sin\dfrac{n\pi x}{L}\).
8. \(n = 0\) is not allowed because \(\psi\) would be zero everywhere (no particle).

Results: energy is quantised; the lowest energy \(E_1 = h^2/8mL^2\) is not zero (zero-point energy); levels go as 1 : 4 : 9 : ...; spacing increases with \(n\); the \(n\)th state has \(n-1\) nodes inside the box.

Key formula

\(E_n = \dfrac{n^2h^2}{8mL^2}\). \(n\): quantum number (1, 2, 3, ...), \(h = 6.63\times10^{-34}\) J s, \(m\): particle mass (kg), \(L\): box width (m), \(E_n\) in joules (divide by \(1.6\times10^{-19}\) for eV).

Quick example

An electron in a box of width 1 Å: \(E_1 = (6.63\times10^{-34})^2/(8\times9.11\times10^{-31}\times10^{-20}) \approx 6.0\times10^{-18}\) J \(\approx 37.6\) eV. Then \(E_2 = 4E_1 \approx 150\) eV and \(E_3 = 9E_1 \approx 338\) eV.

Common mistakes

• Allowing \(n = 0\), which gives no particle at all.
• Writing \(E_n \propto n\) instead of \(n^2\).
• Using \(\hbar\) and \(h\) inconsistently (\(n^2h^2/8mL^2 = n^2\pi^2\hbar^2/2mL^2\)).

How MU asks it

A very common 5-mark derivation: "Obtain the energy eigenvalues and eigenfunctions of a particle in a one-dimensional rigid box". Draw the potential well, apply both boundary conditions, derive \(E_n\), normalise \(\psi_n\), and sketch \(\psi\) and \(|\psi|^2\) for \(n = 1, 2, 3\).