Orthogonality of Box Wavefunctions

Introduction

The particle in a box has an infinite ladder of wavefunctions, one for each quantum number n. These functions are not independent of each other in an arbitrary way: any two different ones are orthogonal , meaning their overlap integral is exactly zero. Orthogonality is one of the most important structural facts in quantum chemistry. It underpins how molecular orbitals are built from atomic orbitals, why spectroscopic selection rules exist and how any complicated wavefunction can be expanded as a sum of simpler ones.

Core explanation

Definition. Two wavefunctions ψₘ and ψₙ are orthogonal if their overlap integral vanishes:

∫ ψₘ ψₙ dτ = 0 (m ≠ n)

The integral runs over all space; for the one-dimensional box, over 0 ≤ x ≤ L. If each function is also normalised, the set is orthonormal , which is summarised compactly as

∫ ψₘ ψₙ dτ = δₘₙ

where δₘₙ is the Kronecker delta.

Proof for the box. The box wavefunctions are ψₙ = (2/L)^½ sin(nπx/L). The overlap integral for m ≠ n is

Sₘₙ = (2/L) ∫₀ᴸ sin(mπx/L) sin(nπx/L) dx

Use the product identity sin A sin B = ½[cos(A − B) − cos(A + B)]:

Sₘₙ = (1/L) ∫₀ᴸ [cos((m − n)πx/L) − cos((m + n)πx/L)] dx

Each cosine integrates to a sine: ∫₀ᴸ cos(kπx/L) dx = (L/kπ) sin(kπ), and sin(kπ) = 0 for any non-zero integer k. Since m − n and m + n are both non-zero integers, both terms vanish and Sₘₙ = 0. When m = n, the first cosine becomes cos 0 = 1, which integrates to L, giving Sₙₙ = 1: normalisation is recovered.

A general theorem. Orthogonality is not a lucky feature of sine functions. Eigenfunctions of any Hermitian operator that belong to different eigenvalues are automatically orthogonal. The Hamiltonian is Hermitian, and each box state has a different energy, so orthogonality was guaranteed before we did the integral. When eigenvalues are equal (degenerate), the eigenfunctions need not be orthogonal, but they can always be combined into orthogonal ones.

Symmetry shortcut. Measure positions from the centre of the box. States with odd n are symmetric (even) about the centre; states with even n are antisymmetric (odd). The product of an even and an odd function is odd, and the integral of an odd function over a symmetric interval is zero. So ψ₁ and ψ₂, or ψ₂ and ψ₃, are orthogonal by symmetry alone. States of the same symmetry, such as ψ₁ and ψ₃, are still orthogonal, but that follows from the integral, with positive and negative regions of the product cancelling exactly.

Formulae

∫₀ᴸ ψₘψₙ dx = δₘₙ; sin A sin B = ½[cos(A − B) − cos(A + B)]; ∫₀ᴸ cos(kπx/L) dx = 0 for non-zero integer k.

Step-by-step reasoning

To show that two box wavefunctions are orthogonal:

1. Write the overlap integral with the normalisation constants outside. 2. Convert the product of sines into a difference of cosines. 3. Integrate each cosine between 0 and L. 4. Note that sin(kπ) = 0 for integer k, so each term vanishes. 5. Alternatively, check parity about the centre: an even times an odd function integrates to zero.

Visual explanation

Plot ψ₁ and ψ₂ on the same axes, then plot their product. In the left half the product is positive; in the right half it is an exact mirror image with opposite sign. The shaded positive and negative areas are equal, so the total area, the overlap integral, is zero.

Real-world analogy

Think of the directions north and east on a map. Walking any distance north moves you no distance east: the two directions are independent. Orthogonal wavefunctions are like perpendicular directions in an abstract space with infinitely many axes, one for each quantum number.

Real-world example

The same mathematics appears in Fourier analysis, used in NMR and infrared spectrometers. Because sine and cosine waves of different frequencies are orthogonal, a detector signal can be decomposed uniquely into its separate frequency components, turning a time-domain signal into a spectrum.

Why?

Why should orthogonality matter to chemists? Because it lets any state be written as a unique combination of eigenfunctions, Ψ = Σ cₙψₙ, with each coefficient found simply as cₙ = ∫ψₙ Ψ dx. Without orthogonality, the coefficients would be tangled together and the expansion would be ambiguous.

Common misconception

"Orthogonal wavefunctions do not overlap in space." ψ₁ and ψ₂ occupy the same box and are both non-zero almost everywhere. Orthogonality means their integrated product is zero because positive and negative contributions cancel, not that they avoid each other.

Worked example

Question: Show that ψ₁ and ψ₃ in a box are orthogonal.

Reasoning: S₁₃ = (1/L) ∫₀ᴸ [cos(2πx/L) − cos(4πx/L)] dx. The first term gives (1/L)(L/2π) sin(2π) = 0, and the second gives (1/L)(L/4π) sin(4π) = 0.

Answer: S₁₃ = 0, so the functions are orthogonal even though both are symmetric about the centre.

Quick check

1. Why is the overlap of ψ₂ and ψ₅ zero without doing any integration? Answer: ψ₂ is antisymmetric about the centre and ψ₅ is symmetric, so their product is odd and integrates to zero over the box.

Exam focus

Be ready to prove orthogonality using the product-to-sum identity and to state the orthonormality condition with the Kronecker delta. Quote the general theorem: eigenfunctions of a Hermitian operator with different eigenvalues are orthogonal.

Advanced insight

The box eigenfunctions form a complete orthonormal set: any well-behaved function that vanishes at the walls can be written as a sum of them. This is the foundation of basis-set methods in computational chemistry, where unknown molecular wavefunctions are expanded in a chosen set of functions, and of time-dependent quantum dynamics, where a wavepacket is a superposition of stationary states.

Summary

Two wavefunctions are orthogonal when their overlap integral is zero. The particle-in-a-box functions satisfy ∫ψₘψₙ dx = δₘₙ, which can be proved with the identity sin A sin B = ½[cos(A − B) − cos(A + B)]. Orthogonality follows generally from the Hermitian nature of the Hamiltonian, and parity arguments give a quick check for states of opposite symmetry.

Practice questions

1. State the orthonormality condition for the box wavefunctions. Answer: ∫₀ᴸ ψₘψₙ dx = δₘₙ, which equals 1 when m = n and 0 when m ≠ n. 2. Which of the pairs (ψ₁, ψ₄), (ψ₂, ψ₄) and (ψ₃, ψ₆) are orthogonal by parity alone? Answer: (ψ₁, ψ₄) and (ψ₃, ψ₆), because each pairs an odd-n (symmetric) function with an even-n (antisymmetric) one. (ψ₂, ψ₄) are both antisymmetric, so their orthogonality needs the full integral. 3. A state is Ψ = 0.6ψ₁ + 0.8ψ₂. Use orthogonality to show that Ψ is normalised. Answer: ∫Ψ² dx = 0.36 + 0.64 + 2(0.6)(0.8)∫ψ₁ψ₂ dx = 1.00 + 0 = 1. 4. What general property of the Hamiltonian guarantees orthogonality of non-degenerate eigenfunctions? Answer: It is a Hermitian operator, and eigenfunctions of a Hermitian operator with different eigenvalues are orthogonal.