Particle in a Three-Dimensional Box

Introduction

Molecules in a gas move in three dimensions, bouncing around a container. Treating each molecule as a particle in a three-dimensional box gives the quantised translational energy levels that underlie the kinetic theory of gases and statistical thermodynamics. The mathematics is a direct extension of the two-dimensional case: one more coordinate, one more separated equation and one more quantum number. The key chemical insight is just how extraordinarily closely spaced these levels are for real gases.

Core explanation

Separation in three dimensions. For a cuboid with sides a, b and c, the Hamiltonian inside the box is a sum of three kinetic-energy terms, one for each coordinate. The wavefunction is therefore a product

ψ(x, y, z) = X(x)Y(y)Z(z)

and each factor satisfies its own one-dimensional box equation. Applying the boundary conditions on all six faces gives

ψ = (8/abc)^½ sin(nₓπx/a) sin(n yπy/b) sin(n zπz/c)

E = (h²/8m)(nₓ²/a² + n y²/b² + n z²/c²)

with nₓ, n y and n z each equal to 1, 2, 3, … The normalisation constant (8/abc)^½ is the product of the three one-dimensional constants (2/a)^½(2/b)^½(2/c)^½.

Cubic box. When a = b = c = L,

E = (h²/8mL²)(nₓ² + n y² + n z²)

The ground state (1, 1, 1) has a zero-point energy of 3h²/8mL², three times the one-dimensional value, because the particle is confined in three independent directions.

Nodal planes. The factor sin(nₓπx/a) contributes nₓ − 1 nodal planes perpendicular to x, and similarly for y and z. Higher states have more nodal planes and more kinetic energy.

Translational states of gases. Consider a nitrogen molecule (m = 4.65 × 10⁻²⁶ kg) in a cubic vessel of side 0.10 m. The energy unit is

h²/8mL² = (6.626 × 10⁻³⁴)² / (8 × 4.65 × 10⁻²⁶ × 0.010) ≈ 1.2 × 10⁻⁴⁰ J

At 298 K, kT = 4.1 × 10⁻²¹ J, around 10¹⁹ to 10²⁰ times larger. A typical molecule has a translational energy of (3/2)kT, which requires nₓ² + n y² + n z² ≈ 5 × 10¹⁹, so each quantum number is of order 10⁹ to 10¹⁰. The levels are so densely packed that translational motion can be treated as continuous: this is the correspondence principle at work.

Counting states. Although individual levels are invisible, the number of accessible states matters greatly in thermodynamics. The translational partition function is proportional to the volume V and to T^(3/2), which is why the entropy of a gas increases when it expands or is heated.

Formulae

E = (h²/8m)(nₓ²/a² + n y²/b² + n z²/c²); ψ = (8/abc)^½ sin(nₓπx/a) sin(n yπy/b) sin(n zπz/c); cubic box: E = (h²/8mL²)(nₓ² + n y² + n z²); zero-point energy (cube) = 3h²/8mL².

Step-by-step reasoning

To estimate whether translational quantisation matters:

1. Find the molecular mass in kilograms (molar mass ÷ Avogadro constant). 2. Calculate h²/8mL² for the container size. 3. Compute kT at the temperature of interest. 4. Compare the two: a ratio kT/(h²/8mL²) far above 1 means a quasi-continuous ladder. 5. Estimate typical quantum numbers from (3kT/2)/(h²/8mL²) = nₓ² + n y² + n z².

Visual explanation

Imagine a lattice of points in three dimensions, one for each set of positive integers (nₓ, n y, n z). The energy of each point is proportional to its squared distance from the origin in a cubic box. States of similar energy lie on the surface of one-eighth of a sphere, and at room temperature the sphere's radius is billions of lattice spacings.

Real-world analogy

A cinema seating plan labels each seat by row, column and level in a three-tier auditorium. Three numbers are needed to specify a seat uniquely, just as three quantum numbers are needed for a state in a three-dimensional box.

Real-world example

Statistical thermodynamics uses the three-dimensional box to derive the Sackur–Tetrode equation for the molar entropy of a monatomic gas. For argon at 298 K and 1 bar it predicts about 155 J K⁻¹ mol⁻¹, in excellent agreement with values measured by calorimetry.

Why?

Why are the translational levels of a gas so closely spaced compared with electronic levels? The energy unit h²/8mL² is inversely proportional to mass and to the square of the box length. A molecule is thousands of times heavier than an electron, and a flask is about a hundred million times larger than a molecule, making the spacing smaller by an enormous factor.

Common misconception

"Because gas molecules follow classical kinetic theory, their energies are not quantised." Translational energies are quantised; the gaps are simply far smaller than kT. The quantum description is still needed to count states correctly, for example in the absolute entropy of a gas.

Worked example

Question: An electron is in a cubic box of side 0.50 nm. Calculate the energy of the (1, 1, 1) state.

Reasoning: h²/8mL² = (6.626 × 10⁻³⁴)² / (8 × 9.109 × 10⁻³¹ × (0.50 × 10⁻⁹)²) = 2.41 × 10⁻¹⁹ J. For (1, 1, 1), the factor is 1 + 1 + 1 = 3.

Answer: E = 3 × 2.41 × 10⁻¹⁹ = 7.2 × 10⁻¹⁹ J, about 4.5 eV.

Quick check

1. What is the energy of the (2, 1, 1) state in a cubic box, in units of h²/8mL²? Answer: 4 + 1 + 1 = 6 units, twice the ground-state energy of 3 units.

Exam focus

Write the three-dimensional energy expression with all three lengths and quantum numbers. Be able to reduce it to the cubic case, calculate a zero-point energy, and compare h²/8mL² with kT for a gas molecule to justify treating translation classically.

Advanced insight

The number of translational states with energy below E grows as E^(3/2) and in proportion to the volume. Differentiating gives a density of states proportional to V E^½. The same density-of-states idea, applied to electrons, explains the heat capacity and conductivity of metals in the free-electron model of solids.

Summary

A particle in a cuboid has a product wavefunction with three sine factors and energies (h²/8m)(nₓ²/a² + n y²/b² + n z²/c²), each quantum number a positive integer. In a cube the zero-point energy is 3h²/8mL². For gas molecules in ordinary containers the level spacing is some 10¹⁹ times smaller than kT, so translation appears continuous, though quantum counting of states remains essential for entropy.

Practice questions

1. Write the normalisation constant for the wavefunction of a particle in a box with sides a, b and c. Answer: (8/abc)^½, the product of (2/a)^½, (2/b)^½ and (2/c)^½. 2. How many nodal planes does the (3, 2, 1) state have? Answer: (3 − 1) + (2 − 1) + (1 − 1) = 3 nodal planes. 3. By what factor does h²/8mL² change if a nitrogen molecule is replaced by a hydrogen molecule in the same box? Answer: It increases by the mass ratio 28/2 = 14, since the energy unit is inversely proportional to mass. 4. Explain why the entropy of a gas increases on expansion, using the box model. Answer: A larger box lowers all translational energies and packs more states below any given energy, so more states are thermally accessible and the entropy rises.