Questions: Wavefunction and the Born Rule

Question 1

A particle's wavefunction at position x₀ is ψ = 0.5 + 0.5i. What is the probability density |ψ|² at that point?

• A. 0.25
• B. 0.50
• C. 1.0
• D. 0.5 + 0.5i

Answer: (B)

|ψ|² = ψ*ψ = (0.5)² + (0.5)² = 0.25 + 0.25 = 0.50. The complex value of ψ itself (option 3) cannot be a probability; only the squared modulus yields a real, non-negative probability density. Option 0 would result from squaring only the real part and ignoring the imaginary part.

Question 2

If a valid wavefunction ψ(x) is doubled everywhere to give 2ψ(x), the physical predictions of the theory remain unchanged.

• T. True
• F. False

Answer: False

Doubling ψ multiplies |ψ|² by 4, so every probability density quadruples. This also breaks normalization: ∫|2ψ|² dx = 4 ≠ 1. Only wavefunctions differing by a global phase factor e^(iθ) (which has |e^(iθ)|² = 1) represent the same physical state — rescaling by a real factor changes the physics.

Question 3

Why is the probability density defined as |ψ|² rather than ψ itself?

Model answer: Because ψ is complex-valued and can be negative or imaginary, so it cannot directly represent a probability. The squared modulus |ψ|² is always real and non-negative, and can be normalized to integrate to 1 over all space.

Probability must be a real number between 0 and 1. The wavefunction ψ takes complex values — it could equal something like 3i, which has no meaning as a probability. The Born rule selects |ψ|² = ψ*ψ as the probability density precisely because it is guaranteed to be real and non-negative. The original insight of Born was that ψ encodes amplitude information and |ψ|² extracts the observable probability from it.