Questions: Probability Amplitude and Born Interpretation

Question 1

Two electrons pass through a double slit one at a time. An interference pattern builds up over many electrons. This result is best explained by:

• A. Each electron passes through one definite slit, but we don't know which — the pattern reflects our ignorance
• B. Electrons repel each other, creating alternating zones of high and low electron density on the screen
• C. Each electron's wavefunction passes through both slits, and the two probability amplitudes interfere — constructively at bright bands, destructively at dark bands
• D. The Born rule assigns higher probability to positions near the center, producing the central bright fringe

Answer: (C)

Interference requires superposition of amplitudes, not particles. Each single electron passes through both slits as a quantum superposition ψ = ψ₁ + ψ₂, and the probability density is |ψ₁ + ψ₂|² = |ψ₁|² + |ψ₂|² + 2Re(ψ₁*ψ₂). The last interference term can be positive or negative depending on the relative phase of the two amplitudes, creating bright and dark bands. Option A is the classical interpretation — it predicts no interference pattern, because if each electron went through one slit, there would be no amplitude from the other slit to interfere with. The actual interference pattern rules out option A.

Question 2

A student writes ψ(x) = C·e^(−x²) as a proposed wavefunction. Before computing any probabilities from it, they must first:

• A. Verify that ψ is real-valued, since complex wavefunctions cannot represent physically realizable particles
• B. Confirm that ψ satisfies the time-independent Schrödinger equation everywhere
• C. Choose C so that ∫|ψ(x)|² dx = 1, ensuring the total probability of finding the particle somewhere equals 1
• D. Check that ψ(0) = 1 at the most probable position

Answer: (C)

The Born rule gives probabilities from |ψ|², but for these to be genuine probabilities they must sum (integrate) to 1. This normalization condition ∫|ψ|² dx = 1 constrains which functions can represent physical quantum states. For ψ = Ce^(−x²), integrating |ψ|² = C²e^(−2x²) gives C²√(π/2), so C = (2/π)^(1/4). Without normalization, |ψ|² gives a probability density that doesn't integrate to 1 — the output of any probability calculation would be meaningless.

Question 3

The complex phase of the wavefunction is physically meaningful in quantum mechanics even though it cannot be directly observed, because it determines interference patterns when amplitudes are superposed.

• T. True
• F. False

Answer: True

For a single wavefunction in isolation, the global phase is unobservable — |e^(iθ)ψ|² = |ψ|². But relative phase between two amplitudes matters crucially. When ψ = ψ₁ + ψ₂, |ψ|² = |ψ₁|² + |ψ₂|² + 2Re(ψ₁*ψ₂). The interference term 2Re(ψ₁*ψ₂) depends on the relative phase of ψ₁ and ψ₂: if they are in phase (relative phase 0), it adds; if out of phase (relative phase π), it cancels. Change the phase of ψ₁ relative to ψ₂ and the entire interference pattern shifts. Phase is physically real even though a single amplitude's absolute phase is not.

Question 4

In quantum mechanics, the Born rule's probability distribution describes our ignorance about a particle's definite but unknown position — the particle is somewhere specific before measurement, and measurement merely reveals it.

• T. True
• F. False

Answer: False

This is the classical interpretation of probability, which quantum mechanics rejects. In quantum mechanics, the wavefunction is the complete description of the particle's state — there is no 'hidden' definite position that measurement reveals. Before measurement, the particle does not have a definite position. The Born rule gives the probability distribution for what position will be found upon measurement, but that outcome is not pre-determined — it is genuinely random in a way that has no classical analogue. This irreducible probabilism (not ignorance about a definite state) is what distinguishes quantum probability from classical statistical mechanics.

Question 5

Why does the complex phase of the wavefunction matter physically, even though |ψ|² — not ψ — gives the measurable probability density?

Model answer: The phase matters because quantum states can be superposed: when two amplitudes are added, the probability density of the combined state is |ψ₁ + ψ₂|², which contains the interference term 2Re(ψ₁*ψ₂). This term depends on the relative phase between ψ₁ and ψ₂ and can be constructive (adding) or destructive (canceling). A change in relative phase changes the entire interference pattern, even though |ψ₁|² and |ψ₂|² individually are unchanged. The phase encodes which-way information and coherence — it is the reason quantum mechanics can produce phenomena (like the double-slit pattern) that have no classical explanation in terms of particles with definite trajectories.

If quantum states were just probability distributions with no phase, superposition would just be probabilistic mixture, and there would be no interference. The complex structure of the wavefunction is precisely what distinguishes quantum superposition from classical uncertainty.