Uncertainty and the one-dimensional infinite square well
Key idea: H2 Physics lessons on photons, matter waves, wavefunctions, uncertainty and atomic spectra.
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The core idea
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Big question: Why does confinement force a particle to have quantised energy?
The syllabus uncertainty relation ΔxΔp ≳ h means tighter position confinement requires a wider momentum spread. In an infinite square well, boundary conditions allow only standing-wave states with nodes at the walls, giving discrete energies proportional to n². The lowest state has non-zero energy; a zero-energy wavefunction would vanish everywhere.
Treat uncertainty as a property of the state
The syllabus relation ΔxΔp ≳ h links the spreads of position and momentum. It is not merely instrument error: a state localised more tightly in x necessarily contains a wider range of momenta.
Confinement therefore carries kinetic-energy consequences. Making a region narrower increases the momentum spread and prevents a confined particle from having both exact position and zero momentum.
Check your understanding: If position uncertainty is reduced by a factor of four, what happens to the corresponding momentum-spread estimate?
It increases by a factor of four when the order-of-magnitude product ΔxΔp is kept comparable with h.
Fit standing matter waves into a well
For an infinite well of width L, the wavefunction is zero at the walls and standing-wave conditions allow λ_n = 2L/n. With p = h/λ, the energies are E_n = n²h²/(8mL²), n = 1, 2, 3, ….
There is no n = 0 state: the lowest energy is non-zero. Levels spread farther apart as n increases, and narrowing the well raises every energy as 1/L². The infinite walls are an ideal model, not a literal atomic potential.
Check your understanding: How does the ground-state energy change if well width halves?
It becomes four times larger.
Key ideas to keep
- Uncertainty is intrinsic to the state, not merely poor apparatus.
- Only wavelengths fitting the boundary conditions are allowed.
- Energy-level spacing increases with quantum number in an infinite well.
See the reasoning
Worked example
Connect localisation to the allowed box energies
Question: Explain why localising a particle creates momentum spread and connect this to a box.
Step 1: Interpret localisation
Why: A narrow position distribution requires many spatial wavelengths.
Working: A localised packet has a spread of wave numbers and hence a spread of p = h/λ.
Step 2: Apply the boundary conditions
Why: Infinite walls require the wavefunction to vanish at x = 0 and L.
Working: Only λₙ = 2L/n fits, for n = 1, 2, 3, ….
Step 3: Convert to discrete energies
Why: Use pₙ = h/λₙ in p²/(2m).
Working: Eₙ = n²h²/(8mL²); n = 0 would give ψ = 0 everywhere.
Answer: A localised wave packet needs a superposition of wavelengths, hence a spread of momenta p = h/λ and ΔxΔp ≳ h. Box boundary nodes select discrete standing waves and discrete energies proportional to n²/L².
Check: The non-zero ground energy is consistent with a confined particle not having exact zero momentum.
Use a hint if needed
Practise with support
Try this
Well width doubles at fixed n and m. State the energy factor.
Hint: Do not change n.
Check your answer
Eₙ ∝ 1/L², so energy becomes one quarter.
Now work without the hint
Practise independently
Your turn
Connect localisation, momentum spread, well boundary conditions and the allowed energy equation.
Check your answer
ΔxΔp ≳ h is intrinsic because localisation needs a momentum-component spread. Infinite walls require ψ(0) = ψ(L) = 0, so λₙ = 2L/n, pₙ = nh/(2L), and Eₙ = h²n²/(8mL²) for n = 1, 2, ….
Avoid these traps
Common mistakes
Common mistake
Uncertainty is only instrument imprecision.
What is wrong with this reasoning?
Show better thinking
Localisation intrinsically requires a spread of momentum components.
Common mistake
n = 0 is the ground state of an infinite well.
What is wrong with this reasoning?
Show better thinking
n = 0 makes ψ zero everywhere; the first physical standing wave has n = 1.
Write for the examiner
Exam guidance
Draw the boundary nodes and count half-wavelengths before deriving momentum or energy.
Exam-style practice [6 marks]
An electron is localised to 2.0 × 10⁻¹⁰ m. Estimate minimum Δp using the syllabus relation and find E₃/E₁ in a fixed well.
Plan before you answer
- Use the stated syllabus uncertainty estimate.
- Keep powers of ten explicit.
- Use Eₙ ∝ n² for the ratio.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
Δp ≳ h/Δx = 3.3 × 10⁻²⁴ kg m s⁻¹. Since Eₙ ∝ n², E₃/E₁ = 9.
Come back in three days
Check what stayed with you
Recall question
State the lowest allowed n and why n = 0 fails.
Check the answer
n = 1. n = 0 would give ψ = 0 everywhere, which cannot be normalised to represent a particle.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Use ΔxΔp ≳ h in the form given in the syllabus. Infinite-square-well results apply to a one-dimensional well with ψ zero at both infinite walls and n = 1, 2, …. Photon and matter-wave evidence supports complementary quantum descriptions. X-ray production, solving the Schrödinger equation, finite barriers, tunnelling and scanning tunnelling microscopy are not required here.
- GCE A-Level H2 PhysicsTopic 19(h) / Topic 19(i) / Topic 19(j) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027