Normalisation of Wavefunctions

Introduction

The Born interpretation says that ψ ² dτ is the probability of finding a particle in a small volume. Since the particle must be found somewhere, all these probabilities must add up to exactly one. A wavefunction obtained by solving the Schrödinger equation usually contains an arbitrary multiplying constant, and that constant must be chosen to satisfy this requirement. The process is called normalisation , and it is the first thing to do before using a wavefunction to calculate probabilities or average values.

Core explanation

The normalisation condition. A wavefunction is normalised if

∫ ψ ψ dτ = 1

where the integral is taken over all space accessible to the particle. For a particle on a line this means ∫ ψ ² dx from −∞ to +∞; for a particle in a box from 0 to L; for an electron in an atom over all r, θ and φ.

Why a constant is free. The Schrödinger equation is linear: if ψ is a solution, so is Nψ for any constant N. The equation therefore fixes the shape of ψ but not its overall size. Normalisation fixes the size.

Finding N. Suppose an unnormalised function ψ gives ∫ ψ ² dτ = I. Then Nψ satisfies ∫ Nψ ² dτ = N²I = 1 when

N = 1/√I

The normalised wavefunction is ψ/√I. Conventionally N is chosen real and positive, because any phase factor is unobservable.

Relative probabilities are unchanged. Multiplying ψ by N multiplies every probability by N². Ratios such as "twice as likely here as there" are the same before and after normalisation. Normalisation converts relative likelihoods into absolute probabilities.

Useful standard integrals.

- ∫₀ᴸ sin²(nπx/L) dx = L/2 - ∫₀^∞ rⁿ e^(−br) dr = n!/bⁿ⁺¹ - ∫₋∞^∞ e^(−bx²) dx = (π/b)^½ - ∫ over all angles of sin θ dθ dφ = 4π

Three standard results.

1. Particle in a box, ψ = N sin(nπx/L): N²(L/2) = 1, so N = (2/L)^½. 2. Gaussian, ψ = N e^(−ax²): ∫ e^(−2ax²) dx = (π/2a)^½, so N = (2a/π)^¼. 3. Hydrogen 1s, ψ = N e^(−r/a₀): the angular integral gives 4π and the radial integral ∫ r² e^(−2r/a₀) dr = 2!/(2/a₀)³ = a₀³/4. Thus N² × 4π × a₀³/4 = N²πa₀³ = 1, giving N = (1/πa₀³)^½.

Functions that cannot be normalised. If ∫ ψ ² dτ is infinite, no finite N works. A free-particle wave e^(ikx) extends uniformly over all space and is not normalisable in the ordinary sense; physically, real particles are described by wave packets of finite extent. Functions that blow up, such as e^(+r/a₀), are never acceptable. The requirement that ψ be square-integrable is one of the conditions for an acceptable wavefunction.

Formulae

∫ψ ψ dτ = 1. N = 1/(∫ ψ ² dτ)^½. Box: N = (2/L)^½. Gaussian e^(−ax²): N = (2a/π)^¼. Hydrogen 1s: N = (πa₀³)^(−½).

Step-by-step reasoning

To normalise a wavefunction:

1. Write ψ with an unknown constant N. 2. Form ψ ² = N²ψ ψ. 3. Write the correct volume element and the limits for all accessible space. 4. Evaluate the integral I of ψ ψ, using standard integrals. 5. Set N²I = 1 and solve for N = I^(−½). 6. Check units: N must give ψ the correct dimensions (m⁻¹ᐟ² in one dimension).

Visual explanation

Draw an unnormalised curve for ψ ² whose area is 3.7. Normalising simply rescales it vertically until the area under the curve is exactly 1. The shape, the node positions and the ratio of heights all stay the same.

Real-world analogy

A recipe written for "3 parts flour to 1 part sugar" gives proportions but not amounts. Deciding to make exactly 1 kg of mixture fixes the actual quantities — 750 g and 250 g — without changing the ratio. Normalisation fixes the total probability at one in the same way.

Real-world example

Quantum chemistry programs build molecular orbitals from normalised basis functions and check normalisation at every step. A tiny error in normalisation would give molecules the wrong number of electrons when the electron density is integrated, so the code verifies that integrating the density returns exactly the number of electrons in the molecule.

Why?

Why can we not normalise a wavefunction that grows without limit? The integral of ψ ² would be infinite, and no finite constant can scale an infinite total down to one. Such a function cannot describe a real particle.

Common misconception

"Normalising a wavefunction changes where the particle is likely to be." Normalisation multiplies the whole function by one constant, so the shape of the probability distribution is unchanged; only the absolute scale is fixed.

Worked example

Question: A trial function for a particle in a box of length L is ψ = N x(L − x) for 0 ≤ x ≤ L. Find N.

Reasoning: ∫₀ᴸ x²(L − x)² dx = ∫₀ᴸ (L²x² − 2Lx³ + x⁴) dx = L⁵/3 − L⁵/2 + L⁵/5 = L⁵/30. Setting N² × L⁵/30 = 1 gives N = (30/L⁵)^½.

Answer: N = (30/L⁵)^½, with units of m⁻⁵ᐟ² so that ψ has units of m⁻¹ᐟ².

Quick check

1. An unnormalised wavefunction gives ∫ ψ ² dx = 4.0 nm. What normalisation constant is needed? Answer: N = 1/√4.0 = 0.50 nm⁻¹ᐟ², so the normalised function is 0.50 nm⁻¹ᐟ² × ψ.

Exam focus

Show the integral, the limits and the volume element explicitly. In three dimensions, forgetting the r² sin θ factor or the 4π from the angular integration is the most common source of lost marks. Always state N in its final form.

Advanced insight

Physicists sometimes "box normalise" a free particle by confining it to a large region of length L and letting L become very large at the end of a calculation, or they use delta-function normalisation. These devices let non-normalisable plane waves be used as convenient building blocks, while physical states remain normalisable wave packets.

Summary

A wavefunction is normalised when ∫ψ ψ dτ = 1 over all space, reflecting certainty that the particle is somewhere. Because the Schrödinger equation is linear, solutions carry an arbitrary constant, fixed by N = 1/(∫ ψ ² dτ)^½. Normalisation preserves relative probabilities and requires ψ to be square-integrable. Standard results: (2/L)^½ for the box and (1/πa₀³)^½ for hydrogen 1s.

Practice questions

1. Normalise ψ = N sin(2πx/L) for a particle in a box of length L. Answer: ∫₀ᴸ sin²(2πx/L) dx = L/2, so N = (2/L)^½. 2. Find the normalisation constant for ψ = N e^(−x²/2) on the whole x-axis (x dimensionless). Answer: ∫ e^(−x²) dx = √π, so N = π^(−¼) ≈ 0.751. 3. Normalise ψ = N r e^(−r/2a₀) (angular factor ignored, angular integral 4π included). Answer: ∫ r⁴ e^(−r/a₀) dr = 24a₀⁵, so N² × 4π × 24a₀⁵ = 1 and N = (96πa₀⁵)^(−½). 4. Explain why ψ = e^(+x) on the whole x-axis cannot represent a physical state. Answer: It grows without limit as x becomes large, so ∫ ψ ² dx is infinite and the function cannot be normalised.