Wavefunctions, probability density and superposition
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: What does a wavefunction tell us about where a particle may be found?
The wavefunction ψ is a probability amplitude; |ψ|² is probability density. Probability within an interval is area under the |ψ|² graph, and a normalised state has total area one. Superposed amplitudes can reinforce or cancel before squaring, which produces interference.
Interpret probability amplitude carefully
The wavefunction ψ is a probability amplitude. It may be positive, negative or complex and is not itself a directly measured probability. The probability density is |ψ|², so probability in a small interval is approximately |ψ|²Δx.
A physical one-dimensional wavefunction is normalised so ∫|ψ|² dx = 1. For example, if ψ = A throughout a region of width L and zero elsewhere, A²L = 1 and A = 1/√L. Nodes where ψ = 0 have zero probability density; opposite signs of ψ can still have equal probability densities but matter when amplitudes interfere.
Check your understanding: A constant wavefunction has value A over a region of width L. What normalises it?
A²L = 1, so A = 1/√L. Probability density is |ψ|² and is never negative.
Use superposition before squaring
If alternatives are indistinguishable, their probability amplitudes add and the total probability follows |ψ₁ + ψ₂|². Cross terms produce interference. Adding |ψ₁|² and |ψ₂|² first would lose this phase information.
Measurement outcomes are probabilistic even when the wavefunction evolves predictably. A plotted |ψ|² distribution tells where repeated detections are likely, not the path followed by one particle.
Check your understanding: Why do two amplitudes sometimes cancel?
Their phases or signs can oppose before the sum is squared, giving a small or zero resultant amplitude.
Key ideas to keep
- ψ itself may be negative or complex; probability density cannot be negative.
- Add amplitudes before finding probability.
- A node has ψ = 0 and therefore zero probability density.
See the reasoning
Worked example
Normalise a sinusoidal wavefunction
Question: Normalise ψ = A sin(πx/L) on 0 < x < L.
Step 1: Write the probability condition
Why: A normalised particle must be found somewhere in the well.
Working: ∫₀ᴸ|ψ|²dx = 1.
Step 2: Use the sine-squared integral
Why: Its mean value over the interval is one half.
Working: A²∫₀ᴸsin²(πx/L)dx = A²L/2 = 1.
Step 3: Solve and interpret symmetry
Why: The normalisation factor is chosen positive here.
Working: A = √(2/L); symmetry gives probability 1/2 in either half.
Answer: Set ∫₀ᴸ|ψ|²dx = A²L/2 = 1, giving A = √(2/L). Symmetry then gives probability 1/2 in either half.
Check: A has units L⁻¹/² so that |ψ|²dx is dimensionless.
Use a hint if needed
Practise with support
Try this
A normalised ψ is multiplied by −1. State the effect on probability density.
Hint: Probability uses modulus squared.
Check your answer
|−ψ|² = |ψ|², so measured position probabilities are unchanged.
Now work without the hint
Practise independently
Your turn
Explain ψ, |ψ|², normalisation and superposition for a square or sinusoidal wavefunction.
Check your answer
ψ represents the quantum state. |ψ|² is probability density and ∫|ψ|²dx = 1 fixes the normalisation factor. Amplitudes superpose before taking modulus squared, producing interference terms and standing-wave solutions.
Avoid these traps
Common mistakes
Common mistake
ψ itself is position probability.
What is wrong with this reasoning?
Show better thinking
|ψ|² is probability density; probability over an interval is its integral.
Common mistake
Probabilities from two paths are added before interference.
What is wrong with this reasoning?
Show better thinking
Superpose amplitudes first, ψ = ψ₁ + ψ₂, then calculate |ψ|².
Write for the examiner
Exam guidance
State whether a graph shows ψ or |ψ|² before interpreting its sign or area.
Exam-style practice [6 marks]
For ψ = A sin(2πx/L) on 0 < x < L, find positive A and state the number of internal nodes.
Plan before you answer
- Normalise using the full interval.
- Identify all zeros of the sine.
- Separate boundary and internal nodes.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
The sine-squared integral is L/2, so A = √(2/L). There is one internal node at x = L/2, in addition to boundary nodes.
Come back in three days
Check what stayed with you
Recall question
What integral gives probability between a and b?
Check the answer
P(a ≤ x ≤ b) = ∫ₐᵇ|ψ(x)|²dx.
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(f) / Topic 19(g) · 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