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The Uncertainty Principle

Key idea: H3 Quantum Mechanics: what Δ means in ΔxΔp ≥ ħ/2, why the uncertainty is intrinsic to the state, and how to use it in estimates.

  • Advanced Physics
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Learning objectives

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

The Heisenberg uncertainty principle is not about faulty instruments. It is a statement about what is physically possible for a given quantum state.

Key recall

Matter waves superpose. A wavepacket that is tightly localised in space needs a broader mix of wavelengths (and hence momenta).

1. Statement and meaning

For a particle in 1D, the spreads (standard deviations) in repeated measurements of position and momentum satisfy:

Δ x Δ p ≥ ħ/2

  • Δ x is the standard deviation of position outcomes for many measurements on identically prepared systems.
  • Δ p is the standard deviation of momentum outcomes.
  • The inequality is a property of the state, not the measuring device.
Precise wording

“A state cannot be prepared in which both position and momentum are simultaneously known to arbitrary precision.”

2. General form

For any two observables A and B:

Δ A Δ B ≥ (1/2)|⟨[A,B]⟩|

where [A,B] = AB-BA is the commutator. For position and momentum, [x,p] = iħ, which gives Δ x Δ p ≥ ħ/2.

3. Wavepacket intuition

To localise a particle (small Δ x), its wavefunction must be built from many wavelengths. Many wavelengths means a wide spread of wave numbers k and hence momentum, because:

p = ħ k

The two-panel sketch links the reciprocal spreads directly: tighter localisation in x requires a broader momentum spread.

Wavepacket width and momentum spreadTwo-column schematic comparing narrow and wide spatial wavepackets and their corresponding momentum spreads.Narrow packet in xWide packet in xx|ψ|small Δxpdistributionlarge Δpx|ψ|large Δxpdistributionsmall ΔpΔx↓ implies Δp↑
Scroll diagram horizontally to read all labels.
Narrow spatial localisation (small Δx) needs a broad momentum spectrum (large Δp); wider packets in x correspond to smaller momentum spread.

4. What the principle does and does not mean

  • You can measure position very precisely — but then the state after measurement has a large momentum spread.
  • The uncertainties do not arise from “instrument errors”; they come from the quantum structure of states.
  • The inequality constrains preparation: you cannot prepare a state with both Δ x → 0 and Δ p → 0.
  • Equality (Δ x Δ p = ħ/2) occurs only for special minimum-uncertainty states; in exam estimates, it is often used to get a minimum.

5. Calculation workflow

  1. Identify/estimate Δ x (often “size of confinement”).
  2. Use Δ pₘᵢₙ ≈ ħ/(2Δ x).
  3. If asked for minimum kinetic energy: Kₘᵢₙ ≈ (Δ pₘᵢₙ)²/(2m).
  4. Check units: Δ x in m, Δ p in kg m s⁻¹, energy in J (convert to eV if needed).

6. Worked Examples

Modelled example 1

Worked problem

Core

Problem

An electron is confined to Δ x ≈ 1.0 × 10⁻¹⁰ m. Estimate the minimum momentum uncertainty and corresponding minimum kinetic energy.
Study the worked solution

  1. Estimate momentum spread

    Method

    Δ pₘᵢₙ ≈ 5.3 × 10⁻²⁵ kg m s⁻¹.

    Reason

    Equality gives the minimum-uncertainty estimate.

    Working

    Δ pₘᵢₙ ≈ ħ/(2Δ x) = 5.3 × 10⁻²⁵ kg m s⁻¹

  2. Convert to kinetic energy

    Method

    Kₘᵢₙ ≈ 1.5 × 10⁻¹⁹ J = 0.95 eV.

    Reason

    Use the non-relativistic kinetic energy associated with the momentum scale.

    Working

    Kₘᵢₙ ≈ ((Δ pₘᵢₙ)²)/2mₑ = 1.5 × 10⁻¹⁹ J

7. Common traps

  1. Using h where ħ is required (ħ = h/2π).
  2. Treating Δ x as “measurement error” rather than the spread of outcomes for a state.
  3. Forgetting unit conversions (nm → m, eV ↔ J).
  4. Assuming equality always holds (use it as a minimum when appropriate).

8. Quick check

Mind stretcher 1: Confinement scaling checkExtension

A. If Δ x decreases by a factor of 2, what happens to the minimum possible Δ p?

Answer

It doubles, because Δ pₘᵢₙ ∝ 1/Δ x.

B. A proton is confined to Δ x = 2.0 fm. Estimate Δ pₘᵢₙ.

Answer

Use 2.0 fm = 2.0 × 10⁻¹⁵ m.

Δ pₘᵢₙ ≈ ħ/(2Δ x)

Δ pₘᵢₙ = (1.055 × 10⁻³⁴)/(2(2.0 × 10⁻¹⁵))

Δ pₘᵢₙ ≈ 2.6 × 10⁻²⁰ kg m s⁻¹.

Continue with the next resource in this course.

Course and syllabus information
Course
Advanced Physics
Edition
Advanced Physics