Recent content by MrPhoton

ψ1(x)* = 1/√2π∫φ*(k)e-ikx dk ψ2(x) = 1/√2π∫φ(k)eikx dk Therefore applying LHS of Initial equation, we get: ∫ ψ1(x)*ψ2(x) dx = 1/2π∫φ*(k)e-ikx φ(k)eikx dk ∫ ψ1(x)*ψ2(x) dx = 1/2π∫φ*(k) φ(k) dk ∫ ψ1(x)*ψ2(x) dx = 1/2π∫φ*(k) φ(k) dk = 0 (By Orthogonality) ∫ ψ1(x)*ψ2(x) dx = ∫φ*(k) φ(k) dk = 0 =...

Homework Statement Show that ∫ ψ1(x)*ψ2(x) dx = ∫φ1(k)*φ2(k) dk (Where the integrations are going from -∞ to ∞) Homework Equations 1. Plancherel Theorem: ψ(x) = 1/√2π∫φ(k)eikx dk ⇔ φ(k) = 1/√2π∫ψ(x)e-ikx dx The Attempt at a Solution It is clear that Plancherel's theorem must be used to...