Quantum physics: evidence, states and spectra
Key idea: Keep evidence tied to the claim it supports, add probability amplitudes before probabilities, and derive quantised energies from boundary conditions rather than treating them as arbitrary rules.
Before you start: Waves and Superposition objective chainEnergy and Fields objective chain
By the end, you can
- Connect particle and wave evidence to photons, energy, momentum and matter waves.
- Interpret and normalise wavefunctions, use probability density and apply superposition.
- Apply position–momentum uncertainty and infinite-square-well standing-wave quantisation.
- Explain discrete atomic levels, distinguish line spectra and solve photon-transition problems.
Starting-point self-check
1. Check your starting point
Attempt all five groups without notes and mark the first evidence, amplitude, boundary, uncertainty or transition step you cannot justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Uncertainty and the one-dimensional infinite square well 19(h)–(j)
Question 1
A particle is confined in an infinite well of width L. State its ground-state wavelength and energy.
Check the model response
Nodes at both walls give λ₁ = 2L. Thus p₁ = h/(2L) and E₁ = h²/(8mL²); n = 0 is not allowed.
repair
2. Repair the common breaks
Use only the correction matching an error, then retry the corresponding diagnostic.
Uncertainty and the one-dimensional infinite square well 19(h)–(j)
Check this idea
Misconception: Uncertainty is only instrument imprecision.
Repair: Localisation intrinsically requires a spread of momentum components.
Check this idea
Misconception: n = 0 is the ground state of an infinite well.
Repair: n = 0 makes ψ zero everywhere; the first physical standing wave has n = 1.
worked example
3. Follow five worked models
Follow how each solution ties evidence to a claim, normalises amplitudes, applies boundary conditions or selects an allowed transition.
Uncertainty and the one-dimensional infinite square well 19(h)–(j)
Model 1
Explain why localising a particle creates momentum spread and connect this to a box.
Check the model response
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.
guided practice
4. Guided practice
Use each hint only to select the correct proportionality, amplitude rule, boundary condition or level difference.
Uncertainty and the one-dimensional infinite square well 19(h)–(j)
Question 1
Well width doubles at fixed n and m. State the energy factor.
Hint: Do not change n.
Check the model response
Eₙ ∝ 1/L², so energy becomes one quarter.
independent practice
5. Independent practice
Solve without repair notes and state the evidence, normalisation, quantum-number and transition assumptions.
Uncertainty and the one-dimensional infinite square well 19(h)–(j)
Question 1
Connect localisation, momentum spread, well boundary conditions and the allowed energy equation.
Check the model response
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, ….
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Uncertainty and the one-dimensional infinite square well 19(h)–(j)
Question 1
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.
Check the model response
Δp ≳ h/Δx = 3.3 × 10⁻²⁴ kg m s⁻¹. Since Eₙ ∝ n², E₃/E₁ = 9.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Uncertainty and the one-dimensional infinite square well 19(h)–(j)
Question 1
State the lowest allowed n and why n = 0 fails.
Check the model response
n = 1. n = 0 would give ψ = 0 everywhere, which cannot be normalised to represent a particle.