3.3 · Quantum fundamentals

The particle in a box

The simplest quantum system: quantized energies from standing waves.

By the end you should be able to:

Use the particle-in-a-box model: energies, transitions, nodes and scaling with box length

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Particle in a box

Change n and the box length to see wavefunctions, nodes and quantized energy levels.

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Key idea

The model

A particle of mass $m$ moves freely along a line of length $L$ between two infinitely high walls. The wavefunction must be zero at both walls, so only standing waves that fit a whole number of half-wavelengths are allowed:

$L = n\,\frac{\lambda}{2}, \qquad n = 1, 2, 3, \ldots$

Combining this with $\lambda = h/p$ gives a set of allowed (quantized) energies. Confinement causes quantization: a free particle can have any energy, but a trapped one cannot.

Formula

Energies and wavefunctions

$E_n = \frac{n^2h^2}{8mL^2}, \qquad n = 1, 2, 3, \ldots$

$\psi_n(x) = \sqrt{\frac{2}{L}}\,\sin\!\left(\frac{n\pi x}{L}\right)$

$E \propto n^2$ : $E_2 = 4E_1$ and $E_3 = 9E_1$ .
$E \propto 1/L^2$ : doubling the box length quarters every energy.
$E \propto 1/m$ : heavier particles have more closely spaced levels.
$\psi_n$ has $n - 1$ nodes (the walls are not counted), and $\psi_n^2$ is the probability density.

Key idea

Zero-point energy and nodes

$n = 0$ is not allowed: $\psi$ would be zero everywhere, meaning no particle. The lowest energy is therefore $E_1 = \frac{h^2}{8mL^2} > 0$ , the zero-point energy. A confined particle is never at rest, as the uncertainty principle requires.

Each level adds one node: $\psi_1$ has 0, $\psi_2$ has 1 (at the centre, so in $n = 2$ the particle is never found exactly at $L/2$ ), $\psi_3$ has 2. More nodes means a shorter wavelength and a higher energy.

The levels spread apart as $n$ grows: $E_{n+1} - E_n = (2n + 1)\,\frac{h^2}{8mL^2}$ .

Formula

Transitions between levels

$\Delta E = E_{n_f} - E_{n_i} = \frac{(n_f^2 - n_i^2)\,h^2}{8mL^2}, \qquad \lambda = \frac{hc}{|\Delta E|}$

A photon of exactly this energy is absorbed (or emitted) when the particle moves between the two levels.

In module 8 the π electrons of conjugated molecules are modelled as particles in a box whose length is the number of conjugated carbon atoms multiplied by 0.140 nm. The HOMO → LUMO gap sets the absorbed wavelength, and longer conjugated chains absorb at longer wavelengths.

Method

Solving a particle-in-a-box problem

Convert $L$ to metres (1 nm = $10^{-9}$ m) and use $m_e = 9.109\times10^{-31}$ kg for an electron.
Compute $E_1 = \dfrac{h^2}{8mL^2}$ once.
Any level: $E_n = n^2E_1$ . Any gap: $\Delta E = (n_f^2 - n_i^2)\,E_1$ .
Photon wavelength: $\lambda = hc/\Delta E$ , then convert to nm.
Sanity check: a smaller box or a lighter particle gives larger gaps and shorter wavelengths.

Common mistake

Particle-in-a-box traps

Wrong: starting at $n = 0$ or saying the ground-state energy is zero. Right: $n$ starts at 1 and $E_1 > 0$ .

Wrong: squaring $L$ in nm and forgetting the $10^{-18}$ . Right: $L = 1.00$ nm gives $L^2 = 1.00\times10^{-18}$ m².

Wrong: counting the walls as nodes (saying $\psi_2$ has 3 nodes). Right: $\psi_n$ has $n - 1$ interior nodes.

Wrong: assuming evenly spaced levels. Right: the gaps grow: $E_2 - E_1 = 3E_1$ , $E_3 - E_2 = 5E_1$ .

Worked example

Worked example: an electron in a 1.00 nm box

$E_1 = \frac{(6.626\times10^{-34})^2}{8(9.109\times10^{-31})(1.00\times10^{-9})^2} = 6.02\times10^{-20}\ \mathrm{J}$

For the $n = 1 \to 2$ transition, $\Delta E = (2^2 - 1^2)E_1 = 3E_1 = 1.81\times10^{-19}$ J, so

$\lambda = \frac{(6.626\times10^{-34})(2.998\times10^{8})}{1.81\times10^{-19}} = 1.10\times10^{-6}\ \mathrm{m}$

Answer: $\lambda = 1.10\times10^{-6}$ m, about 1100 nm (infrared). Halving the box to 0.500 nm makes every energy 4 times larger, and the same transition then absorbs at 275 nm (ultraviolet).

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