Particle in a Two-Dimensional Box

Introduction

Real molecules are not one-dimensional. Electrons in aromatic rings, in graphene sheets and on metal surfaces move in two dimensions. The next step beyond the one-dimensional box is a particle confined to a rectangle. This problem introduces a technique that runs through the whole of quantum chemistry: separation of variables . It shows how a problem in several dimensions can be broken into independent one-dimensional pieces, and why each extra dimension brings its own quantum number.

Core explanation

The problem. A particle of mass m moves freely inside a rectangle with sides Lₓ and L y, with zero potential energy inside and infinite potential outside. Inside, the Schrödinger equation is

−(ħ²/2m)(∂²ψ/∂x² + ∂²ψ/∂y²) = Eψ

and ψ must vanish on all four edges.

Separation. Because the Hamiltonian is a sum of an x-part and a y-part, try a product wavefunction ψ(x, y) = X(x)Y(y). Substituting and dividing by XY gives

−(ħ²/2m)(1/X)(d²X/dx²) − (ħ²/2m)(1/Y)(d²Y/dy²) = E

The first term depends only on x and the second only on y. For their sum to be constant for all x and y, each term must itself be a constant. Call these Eₓ and E y, with Eₓ + E y = E. Each equation is just the one-dimensional box problem already solved.

Solutions. Applying the boundary conditions separately in each direction gives

X(x) = (2/Lₓ)^½ sin(nₓπx/Lₓ), Y(y) = (2/L y)^½ sin(n yπy/L y)

so the full normalised wavefunction is

ψ(x, y) = (4/LₓL y)^½ sin(nₓπx/Lₓ) sin(n yπy/L y)

with energy

E = (h²/8m)(nₓ²/Lₓ² + n y²/L y²), nₓ, n y = 1, 2, 3, …

Two dimensions require two independent quantum numbers. Neither may be zero, because a zero would make the wavefunction vanish everywhere.

Ground state and zero-point energy. The lowest state is (1, 1), with E = (h²/8m)(1/Lₓ² + 1/L y²). Confinement in each direction contributes its own zero-point energy.

Nodal lines. The wavefunction is zero wherever either sine factor is zero. There are nₓ − 1 interior nodal lines parallel to the y axis and n y − 1 parallel to the x axis. The state (2, 1) has one vertical nodal line down the middle; (2, 2) has a cross dividing the box into four lobes of alternating sign; (3, 1) has two vertical lines. More nodes mean more curvature and higher energy, as in one dimension.

Ordering of levels. For a rectangle with Lₓ ≠ L y, the order of states depends on the aspect ratio. Motion along the longer side costs less energy, so excitations along that direction appear first.

Formulae

ψ = (4/LₓL y)^½ sin(nₓπx/Lₓ) sin(n yπy/L y); E = (h²/8m)(nₓ²/Lₓ² + n y²/L y²); number of interior nodal lines = (nₓ − 1) + (n y − 1).

Step-by-step reasoning

To find the energy levels of a rectangular box:

1. Check that the Hamiltonian is a sum of independent x and y parts. 2. Write ψ = X(x)Y(y) and separate into two one-dimensional equations. 3. Apply the boundary conditions to each factor to obtain nₓ and n y. 4. Add the two one-dimensional energies to obtain E. 5. List the combinations (nₓ, n y) in order of increasing energy.

Visual explanation

Picture a rectangular drumskin. The (1, 1) state is a single dome, highest at the centre. The (2, 1) state has two domes of opposite sign, one rising and one dipping, separated by a straight nodal line. The (2, 2) state looks like a four-square chessboard of alternating up and down lobes.

Real-world analogy

The patterns resemble the vibrations of a rectangular metal plate sprinkled with sand. When the plate vibrates in a particular mode, sand collects along the lines that do not move, tracing out the nodal lines directly.

Real-world example

Electrons trapped in thin semiconductor layers and on certain metal surfaces behave approximately as particles in two-dimensional boxes. Scanning tunnelling microscopes have imaged such standing-wave patterns of surface electrons inside artificial enclosures built from individual atoms.

Why?

Why does separation of variables work here? Because the potential inside the box does not couple x and y: motion in one direction has no effect on the energy of motion in the other. Whenever a Hamiltonian splits into a sum of independent parts, the wavefunction is a product and the energy is a sum.

Common misconception

"A state such as (1, 0) is the lowest excited state." A quantum number of zero makes the sine factor zero everywhere, giving no particle at all. Both quantum numbers must be at least 1, so the lowest state is (1, 1).

Worked example

Question: An electron is in a rectangular box with Lₓ = 1.0 nm and L y = 2.0 nm. Find the energies of the three lowest states in units of h²/8mLₓ².

Reasoning: E = (h²/8mLₓ²)(nₓ² + n y²/4). Candidates: (1, 1) = 1.25; (1, 2) = 2.00; (1, 3) = 3.25; (2, 1) = 4.25.

Answer: (1, 1) at 1.25, (1, 2) at 2.00 and (1, 3) at 3.25 units. Excitations along the longer y direction come first.

Quick check

1. How many interior nodal lines does the (3, 2) state of a two-dimensional box have? Answer: Three: two parallel to the y axis and one parallel to the x axis.

Exam focus

Show the separation step clearly, stating why each part must equal a constant. Write the energy with both quantum numbers and the lengths, and remember that neither quantum number can be zero. Sketches of nodal patterns are commonly requested.

Advanced insight

Separation of variables in Cartesian coordinates works because the box is rectangular. For a circular enclosure, the natural choice is polar coordinates, where the wavefunctions become Bessel functions multiplied by angular factors. The same idea, choosing coordinates that match the symmetry, is what makes the hydrogen atom solvable in spherical coordinates.

Summary

For a particle in a rectangle, the Hamiltonian is a sum of x and y parts, so ψ = X(x)Y(y) and E = Eₓ + E y. The energies are (h²/8m)(nₓ²/Lₓ² + n y²/L y²), with both quantum numbers positive integers. Wavefunctions have nₓ − 1 and n y − 1 nodal lines, and the ordering of levels depends on the relative lengths of the sides.

Practice questions

1. Write the normalised wavefunction for the (2, 1) state of a box with sides Lₓ and L y. Answer: ψ = (4/LₓL y)^½ sin(2πx/Lₓ) sin(πy/L y). 2. Why must the separated terms each equal a constant? Answer: One depends only on x and the other only on y; their sum can stay fixed for all x and y only if each is individually constant. 3. For a box with Lₓ = L and L y = 3L, which is lower in energy, (1, 2) or (2, 1)? Answer: (1, 2), with energy (1 + 4/9) = 1.44 units against (4 + 1/9) = 4.11 units of h²/8mL². 4. What is the zero-point energy of an electron in a 1.0 nm × 1.0 nm box, given h²/8mL² = 6.0 × 10⁻²⁰ J for L = 1.0 nm? Answer: E(1, 1) = 2 × 6.0 × 10⁻²⁰ J = 1.2 × 10⁻¹⁹ J.