---
title: "The Infinite Square Well and Energy Quantisation"
description: "In a one-dimensional infinite square well, the wall conditions select stationary waves with whole numbers of half wavelengths. Their energies rise as the square of the state number. Solving the bounda"
canonical: https://lightmysky.com/learn/science/the-infinite-square-well-and-energy-quantisation-mt_VV9ItoW7ew
source: https://lightmysky.com/learn/science/the-infinite-square-well-and-energy-quantisation-mt_VV9ItoW7ew.md
retrieved: 2026-11-12
---

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# The Infinite Square Well and Energy Quantisation

In a one-dimensional infinite square well, the wall conditions select stationary waves with whole numbers of half wavelengths. Their energies rise as the square of the state number. Solving the boundary-value problem gives both the wavefunctions and their allowed energies.

Subject: Science · Area: Quantum & Modern Physics · Ages 19 to 20
Page: https://lightmysky.com/learn/science/the-infinite-square-well-and-energy-quantisation-mt_VV9ItoW7ew

## Ready when they can

- Explains why the wall conditions select discrete energies and calculates an electron's energy using a supplied infinite-well formula
- Identifies interior nodes and determines the probability density at a wavefunction zero
- Predicts how the energy levels shift when the well is made narrower

## Lesson: A trapped wave can only sing some notes

For a particle of mass m, set U=0 between x=0 and x=L, with infinite walls outside. The stationary equation is -(ℏ²/2m)ψ''=Eψ, where ℏ=h/(2π). For E>0, write ψ=A cos(kx)+B sin(kx), with E=ℏ²k²/(2m). The wall at 0 gives A=0. The wall at L requires sin(kL)=0, so k=nπ/L for n=1,2,3,... . The n=0 wave is zero everywhere and cannot be normalised.

Substitute k=nπ/L to obtain E_n=n²π²ℏ²/(2mL²)=n²E1. Here m is particle mass and L is well width. The ground state has n=1; n=2 is the first excited state. For example, if E1=0.8 eV, then E2=4×0.8=3.2 eV. The first three energies are E1, 4E1 and 9E1.

*(drawing: These bars compare energies in units of E1. Their heights are not wave amplitudes.)*

A normalised real spatial wave is ψ_n(x)=√(2/L)sin(nπx/L). It has n lobes with alternating signs and n-1 interior nodes. Squaring gives probability density |ψ_n|², with nonnegative lobes and the same zeros. For n=3, the interior nodes are L/3 and 2L/3. For n=2, the centre is a node. The ground-state density peaks at the centre. Integrating density over an interval gives the probability for that interval.

At fixed mass and state number, E_n varies as 1/L². Halving the width makes every level four times higher; doubling it makes every level one quarter as high. Gaps between a fixed pair of levels have the same width factor. At fixed width and state number, energy varies as 1/m.

**Example.** A transition needs an interaction that transfers energy. If one emitted photon carries all the energy lost from n=2 to n=1, its energy is E2-E1=3E1. With E1=0.8 eV, this gives 3.2-0.8=2.4 eV. Energy conservation fixes the difference, but does not require every downward transition to emit a photon.

A superposition can combine different stationary states. If it contains only n=1 and n=2, with both present, an energy measurement can return E1 or E2. It cannot return an intermediate energy. The state sets the chances of the two results. Its probability density can change with time.

**Recap.** The walls select k=nπ/L and E_n=n²E1. State n has n-1 interior nodes. At fixed mass, halving the width multiplies every energy by four.

*Lesson and questions written for this dataset. OpenStax University Physics Volume 3, sections 7.3 and 7.4, were consulted for physics facts and methods.*

## Practice

8 questions on this page, each with its working shown.

## Needs first

- [The Time-Independent Schrödinger Equation](https://lightmysky.com/learn/science/the-time-independent-schrodinger-equation-mt_zeBVPery5T)

## Opens up

- [The Finite Well and Wavefunction Penetration](https://lightmysky.com/learn/science/the-finite-well-and-wavefunction-penetration-mt_Vs4pWnXW7l)
