Particle in a Box: Problem-Solving Workshop

Introduction

The particle in a box is the first quantum model you can solve completely, and it rewards practice. Almost every problem reduces to three tools: the energy formula, the normalised wavefunction and the probability integral. This workshop gathers those tools, shows a reliable order in which to use them, and works through the traps that catch students most often, from unit conversions to forgetting that n starts at 1.

Core explanation

Tool 1: energies. For a particle of mass m in a one-dimensional box of length L with infinitely high walls,

E n = n²h²/(8mL²), n = 1, 2, 3, …

It is efficient to calculate the ground-state energy E₁ = h²/(8mL²) once and then scale: E₂ = 4E₁, E₃ = 9E₁, and so on. For an electron in a box of length 1.00 nm, E₁ = (6.626 × 10⁻³⁴)² / (8 × 9.109 × 10⁻³¹ × (1.00 × 10⁻⁹)²) = 6.03 × 10⁻²⁰ J, or about 0.376 eV.

Tool 2: transitions. A photon absorbed or emitted between levels carries energy ΔE = (n f² − n i²)E₁, and its wavelength is λ = hc/ΔE. Note that the gap grows with n: the 1→2 gap is 3E₁, but the 2→3 gap is 5E₁. In general, adjacent levels are separated by (2n + 1)E₁.

Tool 3: probabilities. The normalised wavefunction is ψ n = (2/L)^½ sin(nπx/L). The probability of finding the particle between x = a and x = b is the integral of ψ n² over that range. Using sin²θ = (1 − cos 2θ)/2, the result is

P(a to b) = (b − a)/L − (1/2nπ)[sin(2nπb/L) − sin(2nπa/L)]

The first term is the classical answer (uniform probability); the second is the quantum correction, which becomes negligible for large n.

Scaling rules. Many questions need no numbers at all. Because E ∝ n²/(mL²), doubling L divides every level by 4; replacing an electron by a proton (about 1836 times heavier) divides the energies by 1836; and multiplying n by 3 multiplies E by 9. Checking your numerical answer against these rules is the quickest way to catch an error.

Symmetry shortcuts. Every box wavefunction is either symmetric or antisymmetric about the centre, so ψ² is always symmetric. The probability of being in the left half is therefore exactly 1/2 for every n, and the expectation value of position is always L/2.

Formulae

E n = n²h²/(8mL²); ΔE = (n f² − n i²)h²/(8mL²); λ = hc/ΔE; ψ n = (2/L)^½ sin(nπx/L); number of nodes inside the box = n − 1.

Step-by-step reasoning

A dependable routine for any box problem:

1. Convert every length to metres and every mass to kilograms before substituting. 2. Calculate E₁ = h²/(8mL²) once. 3. Multiply by n² for individual levels or by (n f² − n i²) for a transition. 4. For a wavelength, use λ = hc/ΔE and convert to nanometres at the end. 5. For a probability, write down the limits a and b, use the integrated formula and check the answer lies between 0 and 1.

Visual explanation

Sketch the box with its energy ladder: rungs at 1, 4, 9 and 16 units, spreading apart as they rise. On each rung draw the wavefunction, a half-wave for n = 1, a full wave for n = 2 and so on. Shade the region between L/3 and 2L/3 to see at once why n = 1 has a large probability there and n = 2 a small one.

Real-world analogy

Solving box problems is like converting a recipe for a different number of guests. Once you know the amount for one serving (E₁), you do not start again from scratch; you multiply by a simple factor. The n² factor is the scaling rule of the quantum recipe.

Real-world example

Linear dye molecules called cyanines are often analysed with exactly these tools. Measuring the length of the conjugated chain, counting the π electrons, filling the box levels two electrons at a time and calculating the HOMO–LUMO gap gives absorption wavelengths within about 10–20% of the observed values, which is remarkable for such a simple model.

Why?

Why does the classical probability term (b − a)/L appear in the quantum result? At high n the wavefunction oscillates so rapidly that ψ² averages to 1/L over any region of reasonable width. The oscillating correction term shrinks as 1/n, so the quantum answer smoothly approaches the classical one, as the correspondence principle requires.

Common misconception

"The lowest level is n = 0, with zero energy." Putting n = 0 into ψ gives zero everywhere, which describes no particle at all. The lowest allowed level is n = 1, and it has a non-zero zero-point energy.

Worked example

Question: An electron is in a box of length 1.00 nm. (a) What wavelength of light excites it from n = 1 to n = 2? (b) In the ground state, what is the probability of finding it in the middle third of the box?

Reasoning: (a) ΔE = 3E₁ = 3 × 6.03 × 10⁻²⁰ = 1.81 × 10⁻¹⁹ J. λ = hc/ΔE = (6.626 × 10⁻³⁴ × 2.998 × 10⁸)/(1.81 × 10⁻¹⁹) = 1.10 × 10⁻⁶ m. (b) With n = 1, a = L/3, b = 2L/3: P = 1/3 − (1/2π)[sin(4π/3) − sin(2π/3)] = 1/3 − (1/2π)(−1.732) = 0.333 + 0.276 = 0.609.

Answer: (a) About 1100 nm, in the near infrared. (b) About 0.61, well above the classical value of 0.33.

Quick check

1. If the length of a box is tripled, by what factor does the wavelength of the 1→2 transition change? Answer: It increases by a factor of 9, because the transition energy falls as 1/L² and wavelength is inversely proportional to energy.

Exam focus

Show the calculation of E₁ explicitly with units, then scale. Common mark losses come from leaving L in nanometres, using the mass in grams, or confusing the level energy with the transition energy. For probability questions, state the integral you are evaluating before quoting the result.

Advanced insight

For n = 2 the same middle-third probability is only about 0.20, because a node sits exactly at the centre. Comparing states this way shows that quantum probabilities depend strongly on the shape of the wavefunction, not merely on the size of the region. For a three-dimensional box the probability integral factorises into a product of three one-dimensional integrals, so the same formula does most of the work.

Summary

Box problems use three tools: E n = n²h²/(8mL²), the normalised sine wavefunctions and the probability integral. Calculate E₁ once and scale by n² or by (n f² − n i²) for transitions; energies fall as 1/L² and 1/m. Probabilities are integrals of ψ² and approach the classical value (b − a)/L at high n. Careful units and quick scaling checks prevent most errors.

Practice questions

1. What is the energy of the n = 3 level of an electron in a 1.00 nm box, in joules? Answer: E₃ = 9 × 6.03 × 10⁻²⁰ J = 5.42 × 10⁻¹⁹ J. 2. How does the energy of the n = 1 level change if the electron is replaced by a proton in the same box? Answer: It falls by a factor of about 1836, the ratio of the proton mass to the electron mass. 3. What is the probability of finding a particle in the left half of the box when n = 5? Answer: Exactly 0.5, because ψ² is symmetric about the centre of the box for every value of n. 4. Show that the gap between adjacent levels n and n + 1 equals (2n + 1)E₁. Answer: (n + 1)²E₁ − n²E₁ = (n² + 2n + 1 − n²)E₁ = (2n + 1)E₁.