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University Quantum Mechanics I

University Quantum Mechanics I · The Infinite Square Well · 6.02

Energy quantisation as the condition for a nontrivial solution

Inside, psi'' = −k² ψ with E = ℏ² k²/(2m). E ≤ 0 leaves only ψ = 0, ψ(0) = 0 kills the cosine, and ψ(L) = 0 forces sin(kL) = 0, so kₙ = n π/L and Eₙ = n² π² ℏ²/(2mL²). n = 0 is not a state, and negative n only flips an overall sign, so n = 1, 2, 3, ... exhausts them.

01

Build the model

Connect the measurement to the mechanism.

Inside, psi'' = −k² ψ with E = ℏ² k²/(2m). E ≤ 0 leaves only ψ = 0, ψ(0) = 0 kills the cosine, and ψ(L) = 0 forces sin(kL) = 0, so kₙ = n π/L and Eₙ = n² π² ℏ²/(2mL²). n = 0 is not a state, and negative n only flips an overall sign, so n = 1, 2, 3, ... exhausts them.

Treat this course-map statement as a claim to test rather than an invitation to import a familiar equation. In The Infinite Square Well, begin from fourier sine expansion of a state, then state the system, observable, assumptions, and evidence before calculating.

Simple definition
Inside, psi'' = −k² ψ with E = ℏ² k²/(2m).
Example
A strong response uses ψₙ and |ψₙ|² stacked against x and states where the model stops being reliable.
Infinite-well energiesEₙ = n²π²ℏ²/(2mL²)

Boundary conditions select half-wavelengths and a discrete spectrum.

n=1,2,3,…; n=0 gives only the zero function.

01

The subsection's claim

Inside, psi'' = −k² ψ with E = ℏ² k²/(2m). E ≤ 0 leaves only ψ = 0, ψ(0) = 0 kills the cosine, and ψ(L) = 0 forces sin(kL) = 0, so kₙ = n π/L and Eₙ = n² π² ℏ²/(2mL²). n = 0 is not a state, and negative n only flips an overall sign, so n = 1, 2, 3, ... exhausts them.

02

How to work with it

Start from fourier sine expansion of a state. Then stating the boundary conditions before integrating, and keeping exp(−i Eₙ t/ℏ) attached to each term rather than pulled out front. Select an equation only after its variables and assumptions match the stated system.

03

What evidence would decide

Does a two-term superposition oscillate at angular frequency (E₂ - E₁)/ℏ, and does a many-term packet return at the revival time 4mL²/(π ℏ)? Useful evidence includes snapshots of |Ψ(x, t)|², ⟨x⟩(t) with its fitted period set against 2 pi ℏ/(E₂ - E₁), |cₙ|² and ⟨H⟩ shown numerically constant, and the revival traced to level gaps that are integer multiples of E₁.

04

Keep the boundary visible

This GioPhysics course map is an adaptable learning sequence, not academic credit, accreditation, or a universal university syllabus. It is a first dedicated quantum-mechanics course, distinct from University Physics V, which covers quantum mechanics alongside atomic, nuclear and particle physics in twenty units: this course is narrower, slower, and teaches the linear algebra it needs rather than assuming it. Hydrogen appears here as an introduction, with the full radial derivation belonging to a second course. A midterm examination is assumed around week 8. Follow your institution's published scope, notation, laboratory programme, and assessment rules. A result should be checked against units, signs, limiting cases, and the conditions under which its model was derived.

02

Change one variable at a time

Make the relationship visible.

Interactive model
2
1.0 nm

Change n to count half-wavelengths and nodes. Change L to test the 1/L² energy dependence.

Interactive physics modelInfinite-well eigenfunction shape and electron energy scaling in one dimension.0 < x < L

ELECTRON ENERGY1.504 eV

INTERNAL NODES1

Live interpretationELECTRON ENERGY: 1.504 eV. INTERNAL NODES: 1

03

Catch the common trap

Explain before calculating.

Why is n=0 excluded for the infinite square well?

Choose an answer to test the model.

04

Practice & worked examples

Reason from the model, then test the result.

Worked calculationAn electron is in a 1.00 nm infinite well. Find E₁ and E₃ using E₁=0.376 eV for this width.
  1. The levels scale as n².
  2. E₁=0.376 eV.
  3. E₃=3²E₁=9(0.376)=3.38 eV.

AnswerE₁=0.376 eV and E₃=3.38 eV.

TransferDesign one observation that separates Energy quantisation as the condition for a nontrivial solution from Normalisation, the ground state, and zero-point energy.
  1. Name the observable central to Energy quantisation as the condition for a nontrivial solution.
  2. Name the contrasting observable or condition in Normalisation, the ground state, and zero-point energy.
  3. Choose a graph feature, sign, scale, or limiting case that would distinguish them.

AnswerThe comparison is useful only if the proposed observation could rule out at least one of the two accounts.