Questions: Wave Function Normalization and Orthogonality

Question 1

A wave function ψ is found to satisfy ∫|ψ|² dτ = 4 over all space. What must be done, and what changes?

• A. Multiply ψ by 4 so the integral equals 1
• B. Multiply ψ by 1/2 (the normalization constant N = 1/√4 = 1/2) so that ∫|Nψ|² dτ = 1
• C. The wave function cannot be normalized and must be discarded
• D. Divide ψ by 4; since ∫|ψ|² = 4, the maximum value of ψ must be set to 1

Answer: (B)

Normalization requires ∫|Nψ|² dτ = N²∫|ψ|² dτ = 1, so N²×4 = 1 gives N = 1/2. You multiply ψ by this normalization constant, which rescales the overall amplitude without changing the physics — the probability density |Nψ|² retains the same shape and all predictions are unchanged. Option D reveals the common misconception that normalization constrains the maximum value of ψ; it does not. The integral is constrained, not the peak.

Question 2

Two wave functions ψₘ and ψₙ are orthogonal. What can you physically conclude?

• A. Their probability densities do not overlap anywhere in space
• B. If a particle is measured to be in state ψₘ, there is zero probability of measuring it in state ψₙ
• C. The two states describe particles at perpendicular positions in space
• D. Their energy eigenvalues must be equal

Answer: (B)

Orthogonality means ∫ψₘ*ψₙ dτ = 0 — the inner product vanishes. Physically, this means the two states are completely independent: a system definitely in state ψₘ has zero probability of being found in state ψₙ upon measurement. Option A is wrong — orthogonal wave functions can have overlapping probability densities (e.g., the 1s and 2p orbitals of hydrogen overlap spatially but are orthogonal). Option C confuses orthogonality in function space with geometric perpendicularity. Option D is wrong; degenerate states with the same energy can also be orthogonalized.

Question 3

For a complete orthonormal basis, the sum of |cₙ|² over all basis states equals 1, where cₙ are the expansion coefficients of a normalized quantum state.

• T. True
• F. False

Answer: True

If Ψ = Σ cₙψₙ is expanded in an orthonormal basis and Ψ is itself normalized (∫|Ψ|² dτ = 1), then Σ|cₙ|² = 1. Each |cₙ|² gives the probability of measuring the system in state ψₙ, and since all probabilities must sum to 1, the completeness relation ensures this holds. This is the quantum mechanical analog of Parseval's theorem and is the foundation for why probabilities are well-defined in quantum mechanics.

Question 4

A normalized wave function should have a maximum value of exactly 1, since the probability of finding the particle somewhere should equal 1.

• T. True
• F. False

Answer: False

Normalization constrains the *integral* of |ψ|², not the maximum value of ψ itself. For the particle-in-a-box, the normalized wave function is ψₙ(x) = √(2/L)·sin(nπx/L). When L is small (e.g., L = 0.1 nm), √(2/L) ≈ 141 nm⁻¹/², so ψ takes values much larger than 1. When L is large, ψ values are much less than 1. ψ is a probability amplitude, not a probability; its integral is dimensionally compensated by the volume element dτ.

Question 5

Why is it physically significant that eigenstates of a quantum system are orthogonal, rather than just a convenient mathematical property?

Model answer: Orthogonality ensures that distinct quantum states are genuinely independent — measuring the system to be in one eigenstate gives zero probability for any other eigenstate. This means quantum states partition the probability space completely without redundancy. Mathematically, it allows arbitrary quantum states to be uniquely decomposed into basis states with well-defined probabilities (|cₙ|²). Without orthogonality, the expansion coefficients would mix together, and measuring one observable would not yield definite, reproducible probabilities. Orthogonality is the mathematical expression of the fact that quantum measurement produces definite, distinguishable outcomes.

The physical content of orthogonality is the independence of quantum states. If two states were not orthogonal, knowing you measured the system in one state would not exclude the other, and probability calculations would give inconsistent results. The entire structure of quantum measurement theory depends on the orthogonality of eigenstates.