 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s3/3/2/solution

∣ ψ (x) ∣^{2} d x

is the probability of a position measurement in an interval of width

d x

. For a normalized bound state

\int ∣ ψ ∣^{2} d x = 1

⟨ x ⟩ = \int x ∣ ψ (x) ∣^{2} d x, ⟨ x^{2} ⟩ = \int x^{2} ∣ ψ (x) ∣^{2} d x .

The integrals run over the allowed domain. If the given state is not normalized, divide both by

\int ∣ ψ ∣^{2} d x

. Its variance is

⟨ x^{2} ⟩ - ⟨ x ⟩^{2}

; neither

ψ

itself nor its overall phase is a directly measured position probability.

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