Probability Distributions in the Box

Introduction

Solving the Schrödinger equation for the particle in a box gives wavefunctions and energies, but a chemist really wants to know something more physical: where is the particle? Quantum mechanics cannot give a single answer. Instead, the Born interpretation tells us that the square of the wavefunction is a probability density. On this page we turn the box wavefunctions into probability maps, calculate the chance of finding the particle in any region, and see how strongly these maps differ from the picture of a classical ball rattling back and forth.

Core explanation

The probability density. For a particle of mass m in a one-dimensional box of length L with infinitely high walls at x = 0 and x = L, the normalised wavefunctions are

ψₙ(x) = (2/L)^½ sin(nπx/L), n = 1, 2, 3, …

Because ψₙ is real, the probability density is simply its square:

Pₙ(x) = ψₙ² = (2/L) sin²(nπx/L)

Pₙ(x) dx is the probability of finding the particle between x and x + dx. The density has units of inverse length (m⁻¹), which is why only its integral over a finite interval is a true, dimensionless probability.

Shape of the distributions.

- n = 1 (ground state): a single hump with its maximum at the centre, x = L/2. The particle is most likely to be found in the middle and least likely near the walls, where ψ² falls to zero. - n = 2: two humps with maxima at x = L/4 and 3L/4 and a node at x = L/2. The probability of finding the particle exactly at the centre is zero, even though it is found with equal likelihood in the left and right halves. - General n: n humps of equal height 2/L separated by n − 1 interior nodes. Every state has zero density at both walls.

Probability in a region. The probability of finding the particle between x = a and x = b is

P(a ≤ x ≤ b) = ∫ₐᵇ (2/L) sin²(nπx/L) dx

Using sin²θ = ½(1 − cos 2θ), the integral from 0 to some point x is

P(0 → x) = x/L − (1/2nπ) sin(2nπx/L)

This single expression answers almost every "how likely is the particle to be here?" question.

Comparison with a classical particle. A classical particle moving at constant speed between perfectly elastic walls spends equal time in every equal length of the box, so its probability density is uniform, 1/L. The quantum ground state is far from uniform: it piles the probability into the centre. The two descriptions agree only in the limit of large n, which is the subject of the correspondence principle.

Formulae

Pₙ(x) = (2/L) sin²(nπx/L); P(0 → x) = x/L − (1/2nπ) sin(2nπx/L); positions of maxima: x = (2k − 1)L/2n for k = 1, 2, …, n; classical density = 1/L.

Step-by-step reasoning

To find the probability of locating the particle in a given region:

1. Write the normalised wavefunction for the stated quantum number n. 2. Square it to obtain the probability density (2/L) sin²(nπx/L). 3. Set the limits a and b of the region of interest. 4. Evaluate P(0 → b) − P(0 → a) using the integrated formula. 5. Check that the answer lies between 0 and 1 and makes sense from a sketch of ψ².

Visual explanation

Sketch the box as a horizontal line from 0 to L. Above it, draw ψ₁² as a smooth dome peaking at the centre, ψ₂² as two equal domes touching zero in the middle, and ψ₃² as three domes. Shade the area under a curve between two positions: that shaded area is the probability of finding the particle there, and the total area under every curve is exactly 1.

Real-world analogy

Imagine a guitar string plucked to vibrate in its fundamental mode. The middle of the string swings widely while the fixed ends do not move at all. The probability density of the ground-state particle follows the same pattern: large in the middle, zero at the clamped ends.

Real-world example

In a conjugated molecule such as a linear polyene, the π electrons behave roughly like particles in a box. The electron density of the highest occupied orbital, obtained from ψ², shows where the molecule is most reactive towards electrophiles, and regions near nodes of that orbital are correspondingly less reactive.

Why?

Why is the ground-state particle most likely to be found in the middle? The wavefunction must vanish at the walls to join smoothly with the zero wavefunction outside. The lowest-energy way to do this is half a sine wave, which is largest at the centre, so its square concentrates probability there.

Common misconception

"A node in the middle means the particle cannot get from one side to the other." A node is a point of zero probability density, not a wall. The particle is not a tiny ball travelling through the box; it is described by a standing wave spread over both halves at once.

Worked example

Question: For a particle in the ground state (n = 1), what is the probability of finding it in the middle third of the box, between L/3 and 2L/3?

Reasoning: P(0 → x) = x/L − (1/2π) sin(2πx/L). At x = 2L/3: 2/3 − (1/2π) sin(4π/3) = 0.6667 + 0.1378 = 0.8045. At x = L/3: 1/3 − (1/2π) sin(2π/3) = 0.3333 − 0.1378 = 0.1955. Difference: 0.8045 − 0.1955 = 0.609.

Answer: about 0.61, compared with 0.33 for a classical particle, confirming the concentration of probability at the centre.

Quick check

1. Where are the most probable positions of a particle in the n = 2 state of a box of length L? Answer: At x = L/4 and x = 3L/4, where sin²(2πx/L) reaches its maximum value of 1.

Exam focus

Examiners often ask for the probability in a region such as 0 to L/4, or for the positions of maxima and nodes. Show the squared wavefunction, the integral with limits, and the use of sin²θ = ½(1 − cos 2θ). Remember that ψ² at a point is a density, not a probability.

Advanced insight

For every state the probability of finding the particle in the left half of the box is exactly one half, because ψₙ² is symmetric about the centre. Yet the fine structure within each half changes with n. As n grows, the humps become so narrow that any measuring device of finite resolution averages them out, and the observed distribution becomes the uniform classical value 1/L.

Summary

The probability density for the particle in a box is (2/L) sin²(nπx/L). The state n has n equal maxima and n − 1 interior nodes, with zero density at the walls. Probabilities for regions come from integrating the density, giving x/L − (1/2nπ) sin(2nπx/L) from the left wall. The ground state concentrates the particle in the centre, unlike the uniform classical distribution.

Practice questions

1. Calculate the probability of finding a particle in the n = 1 state between x = 0 and x = L/4. Answer: P = 1/4 − (1/2π) sin(π/2) = 0.250 − 0.159 = 0.091. 2. Repeat the calculation for the n = 2 state between 0 and L/4, and explain the difference. Answer: P = 1/4 − (1/4π) sin(π) = 0.25. In n = 2 the first hump lies entirely in the region 0 to L/2, so a quarter of the box holds a quarter of the total probability, far more than in the ground state. 3. How many interior nodes does ψ₄² have, and where are they? Answer: Three, at x = L/4, L/2 and 3L/4. 4. What are the units of the probability density in a one-dimensional box, and why? Answer: m⁻¹ (inverse length), because the density must be multiplied by a length dx to give a dimensionless probability.