Superposition

Question 1

General solution as a linear combination of two stationary states

Answer 1

Ψ(x,t) = A₁ψ₁(x)exp(-iE₁t/ħ) + A₂ψ₂(x)exp(-iE₂t/ħ)

Question 2

Normalisation of superposition coefficients

Answer 2

If Ψ, ψ₁, and ψ₂ are all normalized, then
|A₁|² + |A₂|² = 1

Question 3

Energy expectation in superposition of two states

Answer 3

⟨E⟩ = ∫Ψ(x,t) * H Ψ(x,t) dx
=E₁|A₁|² + E₂|A₂|²

Measuring energy gives either E value, with probabilities |A₁|² and |A₂|²

Question 4

Prove that measuring energy twice gives same answer twice

Answer 4

Apply H² operator, same as applying Ĥ operator twice.
Results should give:
⟨E⟩ = E₁|A₁|² + E₂|A₂|²
⟨E²⟩ = E₁²|A₁|² + E₂²|A₂|²
as opposed to the classical expectation of ⟨E²⟩ = ⟨E⟩².
This proves that waefunction “collapses” into one value once measured.