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Particle In A Box (Infinite Square Well)

Key idea: Use standing-wave wavefunctions and En = n^2 h^2 / (8mL^2) for a particle in a 1D infinite square well (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Interpret wavefunctions, probability density and superposition.
  • Apply uncertainty and infinite-square-well energy quantisation.

1. Definitions (Must Know)

  • Infinite square well (1D) of width L:
    • V(x) = 0 for 0 < x < L, V(x) = ∞ otherwise
  • Boundary conditions (wavefunction must be zero where V = ∞):
    • ψ(0) = ψ(L) = 0
  • Allowed standing-wave wavefunctions:
    • ψₙ(x) = square root of (2/L) sin((nπ x)/L) (0 < x < L), n = 1,2,3,…
  • Allowed energies:
    • Eₙ = n²h²/8mL²

2. Key Ideas (What Earns Marks)

  • “Particle in a box” is a standing wave condition:
    • the wave must “fit” in the box with nodes at both ends
    • L = nλ/2 ⇒ λₙ = 2L/n
  • Quantisation comes from boundary conditions:
    • n can only be 1,2,3,… (no n = 0)
  • Useful derived results:
    • momentum magnitude: pₙ = h/λₙ = nh/2L
    • energy: Eₙ = pₙ²/2m = n²h²/8mL²
  • Scaling:
    • Eₙ ∝ n² (higher states much higher energy)
    • Eₙ ∝ 1/L² (bigger box → much smaller energy spacing)

Energy levels in a 1D infinite well (scaled)

A plot showing En proportional to n^2, so higher levels are increasingly spaced.

Scroll across the graph to read all labels.

A plot showing En proportional to n^2, so higher levels are increasingly spaced.A plot showing En proportional to n^2, so higher levels are increasingly spaced.
Because En grows as n², the gaps between successive energy levels get larger as n increases.
Open full-size graph
View figure data
Values for Energy levels in a 1D infinite well (scaled)
Quantum number, n (arbitrary units)En/E1 = n²
11
24
39
416
525

3. Detailed Explanations

A. Why the solutions are standing waves (superposition)

In the box, the particle is described by a wavefunction. Reflections at the infinitely high walls mean the wave travelling right and the wave travelling left superpose, producing a standing wave.

The walls enforce nodes at the ends: ψ(0) = ψ(L) = 0

B. Allowed wavelengths

For a standing wave with nodes at both ends: L = nλ/2 ⇒ λₙ = 2L/n

So only discrete wavelengths are allowed.

C. From wavelength to energy levels

Use the de Broglie relation: p = h/λ

So: pₙ = h/λₙ = h/(2L/n) = nh/2L

For a non-relativistic particle:

Eₙ = pₙ²/2m = (1/2m)(nh/2L)²; = n²h²/8mL²

D. What the wavefunctions look like

ψₙ(x) = square root of (2/L) sin((nπ x)/L)

  • n = 1: one “hump” (no internal nodes)
  • n = 2: two humps (one internal node)
  • in general: n-1 internal nodes

4. Common Mistakes

  • Using n = 0 (not allowed in an infinite square well).
  • Forgetting that energy depends on 1/L² (doubling L quarters the energies).
  • Using the wrong mass (electron vs nucleon) when calculating Eₙ.
  • Mixing h and ħ formulas without converting (ħ = h/2π).

5. Exam Tips

  • Start with the standing-wave condition: L = nλ/2.
  • Quote the energy formula clearly before substitution:
    • Eₙ = n²h²/8mL²
  • Always state the unit (J or eV) and show the conversion if using eV.

6. Worked Examples

Modelled example 1

Ratio of energy levels

Core

Problem

For a particle in the same one-dimensional infinite well, find E₃/E₁.
Study the worked solution

  1. Cancel common box factors

    Method

    For fixed m and L, energy is proportional to n².

    Reason

    The factors h²/(8mL²) are identical for both states and cancel in a ratio.

    Working

    E₃/E₁ = (3²h²/(8mL²))/(1²h²/(8mL²))

  2. Evaluate

    Method

    E₃/E₁ = 9.

    Reason

    The third state has nine times the ground-state energy, not three times.

    Working

    E₃/E₁ = 3²/1² = 9

Guided practice 2

Electron in a 1.0 nm box

About 7 min

Problem

An electron is in a one-dimensional infinite square well of width L = 1.0 nm. Find E₁ and E₂ in eV. Use h = 6.63 × 10⁻³⁴ J s, mₑ = 9.11 × 10⁻³¹ kg and 1 eV = 1.60 × 10⁻¹⁹ J.

Try this before viewing the solution

Hints

Hint 1: calculate the ground state first
Convert L to metres and use E₁ = h²/(8mₑL²).
Hint 2: reuse the scaling
After converting E₁ to eV, use E₂/E₁ = 2².
View solution step by step

  1. Calculate ground-state energy

    Method

    E₁ = 6.0 × 10⁻²⁰ J.

    Reason

    The ground state has n = 1 and the box width must be in metres.

    Working

    E₁ = ((6.63 × 10⁻³⁴)²)/(8(9.11 × 10⁻³¹)(1.0 × 10⁻⁹)²) = 6.0 × 10⁻²⁰ J

  2. Convert to electronvolts

    Method

    E₁ ≈ 0.38 eV.

    Reason

    Divide the energy in joules by the joules per electronvolt.

    Working

    E₁ = (6.0 × 10⁻²⁰)/(1.60 × 10⁻¹⁹) eV ≈ 0.38 eV

  3. Scale to the second state

    Method

    E₂ ≈ 1.5 eV.

    Reason

    At fixed mass and width, Eₙ ∝ n².

    Working

    E₂ = 2²E₁ = 4(0.38) eV ≈ 1.5 eV

Common misconception 3

Why n = 0 is not the ground state

Find and correct the mistake

Learner claim

A learner substitutes n = 0 into Eₙ = n²h²/(8mL²) and concludes that the ground state has zero energy. Diagnose the claim using the wavefunction and normalisation.

Try this before viewing the solution

Lowest physical quantum number

View solution step by step

  1. Substitute into the standing wave

    Method

    For n = 0, ψ₀(x) = 0 everywhere.

    Reason

    sin(0π x/L) = 0 at every point, not merely at the walls.

    Working

    ψ₀(x) = square root of (2/L) sin(0) = 0

  2. Test normalisation

    Method

    The zero function cannot represent a particle in the box.

    Reason

    Its total probability is zero rather than one.

    Working

    ∫₀^L|ψ₀|²dx = 0 ≠ 1

  3. Identify the ground state

    Method

    The lowest physical state is n = 1, with non-zero energy.

    Reason

    It is the first standing wave satisfying both boundary conditions and normalisation.

    Working

    E₁ = h²/8mL² > 0

Challenge 4

Effect of changing the box width

Minimal support

Independent transfer

The width of an infinite square well is doubled from L to 2L, while the particle and quantum number remain unchanged. Determine the factor by which E₁ changes.

Try this before viewing the solution

New ground-state energy

Hints

Hint 1: isolate the width dependence
From Eₙ = n²h²/(8mL²), identify the power of L before substituting the factor of two.
View solution step by step

  1. State the scaling

    Method

    E₁ ∝ 1/L².

    Reason

    All other quantities are held fixed.

    Working

    E_(1,new)/E_(1,old) = L²/(2L)²

  2. Evaluate

    Method

    E_(1,new) = E_(1,old)/4.

    Reason

    Doubling a denominator that is squared increases it by a factor of four.

    Working

    E_(1,new) = (1/2²)E_(1,old) = (1/4)E_(1,old)

7. Mind Stretchers

Mind stretcher 1: Example: Photon emitted in a transitionExtension

A particle in a 1D box drops from n = 3 to n = 2 and emits a photon. Show that the photon frequency is: f = 5h/8mL²

Show Answer

Energy difference: Δ E = E₃-E₂ = 9h²/8mL²-4h²/8mL² = 5h²/8mL²

Photon energy: Δ E = hf: hf = 5h²/8mL² ⇒ f = 5h/8mL²

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027