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Allowed Energy States of a Trapped Particle

Key idea: Derive infinite-well energy levels and distinguish fixed energy from the two possible momentum outcomes of a standing state.

  • Advanced Physics
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Learning objectives

  • Use state functions, boundary conditions, stationary states, and measurement to analyse introductory quantum systems.

In the infinite square well (“particle in a box”), boundary conditions force a standing wave. This selects discrete wavelengths and energies, together with discrete possible momentum magnitudes.

Prerequisites + Path

1. Allowed momentum outcomes

From the standing-wave condition:

L = nλₙ/2 (n = 1,2,3,…)

so

λₙ = 2L/n

Using de Broglie’s relation |p| = h/λ gives the magnitude

|pₙ| = h/λₙ = nh/2L = nπħ/L.

The energy eigenfunction is a standing wave, equivalent to equal-amplitude travelling waves with momenta + pₙ and -pₙ. Therefore ⟨p⟩ = 0, but a momentum measurement returns either sign and ⟨p²⟩ = pₙ².

2. Quantised energy

Inside the box, V(x) = 0, so total energy equals kinetic energy:

E = K = p²/2m

Substitute p = pₙ:

Eₙ = pₙ²/2m = n²h²/8mL² = n²π²ħ²/2mL²

Key consequence: Eₙ ∝ n², so the spacing between levels increases with n.

Quantised energy levels in a 1D box (relative scale)In the infinite square well, En is proportional to n^2, so the spacing between adjacent energy levels increases with n.Quantised energy levels in a 1D box (relative scale)Energy (E₁)
Since En ∝ n², the jump from n=4 to n=5 is larger than from n=1 to n=2.

3. Ground-state energy

  • Quantised energy: the particle can only have discrete energies labelled by the quantum number n.
  • Ground state: n = 1, so E₁ = h²/8mL² = π²ħ²/2mL² A confined particle cannot be “at rest” with E = 0.

Why n = 0 is not allowed

Boundary-condition viewpoint (most direct)

The walls require ψ(0) = ψ(L) = 0. The allowed solutions inside the box are standing waves, which start at n = 1. The “n = 0 solution” corresponds to ψ = 0 everywhere (no particle).

Uncertainty-principle intuition (good for explanations)

If a particle is confined to a region of size Δ x ∼ L, then the uncertainty principle implies a non-zero momentum spread: Δ p ≳ ħ/2L So the kinetic energy cannot be exactly zero. This argument does not give the exact numerical factor for E₁, but it correctly predicts the key idea: confinement forces a non-zero minimum energy that scales like 1/L².

Limiting case: L → ∞

As L increases, E₁ ∝ 1/L² decreases and the level spacing shrinks. In the limit L → ∞, the particle becomes effectively free and the energy becomes continuous (including E = 0).

4. Common traps

  1. Writing n = 0 as the ground state (ground state is n = 1).
  2. Mixing up “half-wavelengths in the box” (L = nλ/2) with “full wavelengths” (L = nλ).
  3. Confusing ψ with |ψ|² (probability density).
  4. Forgetting to state the reason for quantisation: boundary conditions at the walls.
  5. Saying the standing state has one definite signed momentum pₙ.

6. Worked Examples

Modelled example 1

Electron in a 1.0 nm box

Core

Problem

For an electron in a one-dimensional box of length L = 1.0 nm, calculate E₁ and E₃.
Study the worked solution

  1. Calculate the ground state

    Method

    E₁ = 6.02 × 10⁻²⁰ J = 0.376 eV.

    Reason

    The lowest allowed state is n = 1, not n = 0.

    Working

    E₁ = h²/8mₑL² = 6.02 × 10⁻²⁰ J = 0.376 eV

  2. Scale to the third level

    Method

    E₃ = 3.38 eV.

    Reason

    Eₙ = n²E₁, so the third level is nine times the ground-state energy.

    Working

    E₃ = 9E₁ = 3.38 eV

7. Quick check

Mind stretcher 1: Energy-level scaling checkExtension

A. What is E₂/E₁?

Answer

Since Eₙ ∝ n², E₂/E₁ = 4.

B. If the box length doubles, what happens to E₁?

Answer

E₁ ∝ 1/L², so doubling L makes E₁ four times smaller.

Continue with the next resource in this course.

Course and syllabus information
Course
Advanced Physics
Edition
Advanced Physics