Normalised Box Wavefunctions

Introduction

The boundary conditions fix the shape of each box wavefunction as A sin(nπx/L) but leave the amplitude A undetermined. To use the wavefunctions for predictions — the probability of finding the particle in a region, or an expectation value — we must choose A so that the total probability of finding the particle somewhere is exactly one. This process, normalisation , gives the complete set of box wavefunctions ψₙ = (2/L)^½ sin(nπx/L). This page derives that result and shows how to use it.

Core explanation

The normalisation condition. The particle must be somewhere, so the probability density integrated over all space must equal 1. Outside the box ψ = 0, so only the interior contributes:

∫₀ᴸ A ² sin²(nπx/L) dx = 1

Evaluating the integral. Use the identity sin²θ = ½(1 − cos 2θ):

∫₀ᴸ sin²(nπx/L) dx = ½∫₀ᴸ dx − ½∫₀ᴸ cos(2nπx/L) dx

The first term gives L/2. The second term integrates the cosine over exactly n complete periods, which gives zero, because the positive and negative lobes cancel. So the integral equals L/2 for every n.

The constant. Then A ²(L/2) = 1, giving A ² = 2/L. We choose A to be real and positive:

A = (2/L)^½

and the normalised wavefunctions are

ψₙ(x) = (2/L)^½ sin(nπx/L), for 0 ≤ x ≤ L, n = 1, 2, 3, …

Independent of n. The constant is the same for every state. Physically, sin² averages to ½ over any whole number of half-wavelengths, so every state has the same average value of ψ ², namely 1/L — the value for a particle spread uniformly across the box.

Units. In one dimension, ψ ² dx is a probability and is dimensionless, so ψ ² has units of m⁻¹ and ψ has units of m^(−½). The factor (2/L)^½ supplies exactly these units.

Phase freedom. We could equally choose A = −(2/L)^½ or multiply by any complex number of modulus 1, such as i. These choices change no observable, because ψ ² is unaffected. Taking the real positive root is a convention.

Using the wavefunctions. The probability of finding the particle between x = a and x = b is

P(a, b) = (2/L) ∫ₐᵇ sin²(nπx/L) dx = [(x/L) − (1/2nπ) sin(2nπx/L)] evaluated from a to b

This expression lets you answer questions such as "what is the chance of finding the particle in the left quarter of the box?" for any state.

Formulae

ψₙ = (2/L)^½ sin(nπx/L). ∫₀ᴸ ψₙ² dx = 1. P(a to b) = [x/L − (1/2nπ) sin(2nπx/L)] from a to b. Useful identity: sin²θ = ½(1 − cos 2θ).

Step-by-step reasoning

To normalise a wavefunction and use it:

1. Write ψ with an unknown constant A. 2. Set up ∫ ψ ² dx = 1 over the region where ψ is non-zero. 3. Evaluate the integral, using trigonometric identities where needed. 4. Solve for A, taking the real positive root by convention. 5. Integrate ψ ² over any region of interest to find probabilities.

Visual explanation

Plot sin²(πx/L) and sin²(3πx/L) across the box. The first is one broad hump; the second is three narrower humps. Although their shapes differ, the total area under each curve is L/2, so both need the same scaling factor 2/L to make the area exactly one.

Real-world analogy

Normalisation is like converting raw vote counts into percentages. However many ballots were cast, dividing by the total ensures the shares add up to 100%. The constant (2/L)^½ is the conversion factor that makes the probabilities across the box add up to exactly one.

Real-world example

In the free-electron model of a conjugated dye, normalised box wavefunctions are used to estimate how strongly the molecule absorbs light. The intensity depends on an integral of ψ xψ between two states, and meaningful intensities are obtained only if both wavefunctions are properly normalised.

Why?

Why is normalisation necessary at all, when the Schrödinger equation is satisfied for any A? The equation is linear, so any multiple of a solution is also a solution. The Born interpretation, however, links ψ ² to probability, and probabilities must add to one. Normalisation selects the one scaling consistent with that interpretation.

Common misconception

"Higher states need a larger normalisation constant because they have more humps." Each extra hump is narrower, so the total area under sin² stays L/2. The constant (2/L)^½ is the same for every n.

Worked example

Question: For the n = 1 state, find the probability of finding the particle in the left quarter of the box, 0 ≤ x ≤ L/4.

Reasoning: P = [x/L − (1/2π) sin(2πx/L)] from 0 to L/4 = 1/4 − (1/2π) sin(π/2) − 0 = 0.250 − 0.159 = 0.091.

Answer: About 0.091, or 9.1%. This is much less than the 25% expected for a uniform distribution, because the n = 1 state is concentrated near the centre.

Quick check

1. What is the probability of finding the particle in the left half of the box for any state n? Answer: Exactly one half, because ψₙ ² is symmetric about the centre of the box for every value of n.

Exam focus

Show the normalisation integral explicitly, including the use of sin²θ = ½(1 − cos 2θ) and the vanishing cosine term. Remember the limits run only over the box, and give the units of ψ as m^(−½) in one dimension.

Advanced insight

In three dimensions the normalisation constant for a cubic box becomes (2/L)^(3/2), the product of three one-dimensional factors, because the wavefunction factorises as ψ(x)ψ(y)ψ(z). This product structure, and the resulting product of normalisation constants, recurs whenever a Hamiltonian separates into independent parts.

Summary

Box wavefunctions are normalised by requiring ∫₀ᴸ ψ ² dx = 1. Since ∫₀ᴸ sin²(nπx/L) dx = L/2 for all n, the constant is (2/L)^½ and ψₙ = (2/L)^½ sin(nπx/L). The constant is independent of n, has units of m^(−½), and its sign or phase is a convention. Integrating ψₙ ² over a region gives the probability of finding the particle there.

Practice questions

1. Show that ∫₀ᴸ sin²(2πx/L) dx = L/2. Answer: Using sin²θ = ½(1 − cos 2θ), the integral is L/2 − ½∫₀ᴸ cos(4πx/L) dx; the cosine completes two full periods and integrates to zero, leaving L/2. 2. What is the value of ψ₁ at the centre of a box of length 1.0 nm? Answer: ψ₁ = (2/1.0 × 10⁻⁹)^½ sin(π/2) ≈ 4.5 × 10⁴ m^(−½). 3. For the n = 2 state, what is the probability of finding the particle between 0 and L/4? Answer: P = 1/4 − (1/4π) sin π = 0.25, because the sine term is zero at x = L/4. 4. Why can ψₙ = −(2/L)^½ sin(nπx/L) also be used? Answer: Multiplying by −1 does not change ψ ², so all probabilities and observables are the same.