University:
Santa Fe College
Course:
PHY 2048 | General Physics 1 with Calculus
Academic year:

2024
Views:

134

Pages:

5
Author:

Jabin F.

Problem: Find the expectation value of x² and p for a particle represented by the wave function ψ(x) = Ae^(-x²/4σ²) * e^(2ik₀x), where A, σ, and k₀ are constants. Solution: |ψ(x)|² = |A|² * e^(-x²/2σ²) = 0 (compare with e^(-x²/2σ²)) (Δx)² = σ² / 2 Δx = σ / √2 = ² + (Δx)² = σ² / 2 ΔpΔx = ħ / 2 (Δp)² = ħ² / (4σ²) Δp = ħ / (2σ) Expectation Values =

² +

² = (4ħ²/π₀²) + (-2ħk₀)² = 4ħ²/π₀² + 4ħ²k₀² Problem: Find the expectation value of x² and p² for a particle represented by the wave function ψ(x) = Ae^(-(x-x₀)² / (2σ²)) * e^(ik₀x) Solution: = x₀ (Δx)² = σ² / 2 = ² + (Δx)² = x₀² + σ²/2 ΔpΔx = ħ / 2 (Δp)² = ħ² / (4σ²) Δp = ħ / (2σ) Expectation Value of Momentum Squared = (Δp)² +

² = ħ²/2α Problem: The wave function of a particle in 1-D position space is defined as ψ(x) = Ae^(-α|x|), where A is a constant and α is a suitable unit. (i) Find the value of A so that ψ(x) is normalized to unity. (ii) Find uncertainty in position. Solution (i): ∫ |ψ(x)|² dx = 1 |A|² * ∫ e^(-2α|x|) dx = 1 |A|² * 2 * ∫₀^∞ e^(-2αx) dx = 1 Calculation: ∫₀^∞ e^(-2αx) dx = 1 / (2α) Substituting: 2|A|² * (1 / (2α)) = 1 |A|² = α |A| = √α A = ±√α Given: ψ(x) = √α * e^(-α|x|) Solution (ii): = ∫ ψ*(x) x² ψ(x) dx / ∫ ψ*(x) ψ(x) dx = ∫ e^(-2α|x|) x² dx / ∫ e^(-2α|x|) dx Integration: ∫ e^(-2α|x|) x² dx = (1/4α³) * (2αx + 1) * e^(-2α|x|) Substituting: = (1/4α³) * (2αx + 1) * e^(-2α|x|) |₀^∞ = (1/4α³) * (0 - 1) = 1 / (4α²) Uncertainty in Position (Δx): Δx = √( - ²) Δx = √(1 / (4α²) - 0) Δx = 1 / (2α) Probability Density (f(x)): f(x) = |ψ(x)|² f(x) = (1/α) * e^(-2α|x|) Expectation Value of Position () = ∫ ψ*(x) x ψ(x) dx / ∫ |ψ(x)|² dx = ∫₀^∞ e^(-x²/2λ²) x e^(-x²/2λ²) dx / ∫₀^∞ e^(-x²/2λ²) e^(-x²/2λ²) dx = ∫₀^∞ x e^(-x²/λ²) dx / ∫₀^∞ e^(-x²/λ²) dx Integration: ∫₀^∞ x e^(-x²/λ²) dx = (λ²/2) * ∫₀^∞ e^(-u) du = (λ²/2) * 1 = λ²/2 ∫₀^∞ e^(-x²/λ²) dx = √(πλ²) Substituting: = (λ²/2) / √(πλ²) = λ / (2√π)