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

2024
Views:

222

Pages:

5
Author:

Jabin F.

Problem: The wave function of a particle of mass m in 1-D position space is given by ψ(x) = Ae^(ik₀x), where k₀ is a real constant. Find the expectation value of its linear momentum. Continuation from the previous image:

= ∫ ψ*(x) (-iħ d/dx) ψ(x) dx / ∫ ψ*(x) ψ(x) dx

= (-iħ) * ∫ ψ*(x) d/dx ψ(x) dx / ∫ |ψ(x)|² dx

= (-iħ) * ∫ e^(-ik₀x) * (ik₀) * e^(ik₀x) dx / ∫ |Ae^(ik₀x)|² dx

= (-iħ) * (ik₀) * ∫ e^(-ik₀x) * e^(ik₀x) dx / ∫ |A|² * |e^(ik₀x)|² dx

= (-iħ) * (ik₀) * ∫ dx / ∫ |A|² dx

= (-iħ) * (ik₀) * x |₀^∞ / |A|²

= (-iħ) * (ik₀) * (∞ - 0) / |A|²

= ħk₀ Additional Notes: ψ(x) = Ae^(ikx) p̂ = -iħ d/dx p̂ψ(x) = -iħ d/dx (Ae^(ikx)) = -iħ * ikAe^(ikx) = ħkAe^(ikx)

= ħk Expectation Value of Momentum

= ∫ ψ*(x) (-iħ d/dx) ψ(x) dx / ∫ ψ*(x) ψ(x) dx

= (-iħ) * ∫ ψ*(x) d/dx ψ(x) dx / ∫ |ψ(x)|² dx

= (-iħ) * ∫ e^(-ikx) * (ik) * e^(ikx) dx / ∫ |Ae^(ikx)|² dx

= (-iħ) * (ik) * ∫ e^(-ikx) * e^(ikx) dx / ∫ |A|² * |e^(ikx)|² dx

= (-iħ) * (ik) * ∫ dx / ∫ |A|² dx

= (-iħ) * (ik) * x |₀^∞ / |A|²

= (-iħ) * (ik) * (∞ - 0) / |A|²

= ħk Expectation Value of Momentum in Combined State

= ∫ ψ*(x) (-iħ d/dx) ψ(x) dx / ∫ ψ*(x) ψ(x) dx

= (-iħ) * ∫ ψ*(x) d/dx ψ(x) dx / ∫ |ψ(x)|² dx

= (-iħ) * ∫ (ψ₁*(x)ψ₂*(x)) * d/dx (ψ₁(x)ψ₂(x)) dx / ∫ (ψ₁*(x)ψ₂*(x)) * (ψ₁(x)ψ₂(x)) dx

= (-iħ) * ∫ (ψ₁*(x)ψ₂*(x)) * (ψ₁'(x)ψ₂(x) + ψ₁(x)ψ₂'(x)) dx / ∫ (ψ₁*(x)ψ₂*(x)) * (ψ₁(x)ψ₂(x)) dx

= (-iħ) * ∫ (ψ₁*(x)ψ₂*(x)) * ψ₁'(x)ψ₂(x) dx / ∫ (ψ₁*(x)ψ₂*(x)) * (ψ₁(x)ψ₂(x)) dx + (-iħ) * ∫ (ψ₁*(x)ψ₂*(x)) * ψ₁(x)ψ₂'(x) dx / ∫ (ψ₁*(x)ψ₂*(x)) * (ψ₁(x)ψ₂(x)) dx

= + Expectation Value of Momentum The expectation value of linear momentum of a particle in a state (that represents a bound state) is zero. Given: ψ(x) = χe^(-x²) Find: Δp Solution: ψ(x) is real ψ'(x) = χ'e^(-x²) + χ(-2x)e^(-x²) ψ'(x) is real

= 0 (since ψ(x) and ψ'(x) are real) Δp = √( -

²) Δp = √() Given: Wave function: ψ(x) = (1/√(4a²)) * (a² - x²) for -a ≤ x ≤ a, 0 otherwise Find: Δp Calculations:

= 0 (for bound states) = ∫ ψ*(x) (-ħ²/2m) d²/dx² ψ(x) dx / ∫ ψ*(x) ψ(x) dx = (-ħ²/2m) * ∫ (1/(4a²)) * (a² - x²) * d²/dx² ((1/(4a²)) * (a² - x²)) dx / ∫ (1/(4a²)) * (a² - x²)² dx Simplifying: = (-ħ²/2m) * (1/(4a²)) * ∫ (a² - x²) * (-2) dx / (1/(4a²)) * ∫ (a² - x²)² dx = (ħ²/2m) * 2 * ∫ (a²x - x³) dx / ∫ (a² - x²)² dx Calculating Integrals: ∫ (a²x - x³) dx = (a²/2)x² - (1/4)x⁴ + C ∫ (a² - x²)² dx = (a⁴/3)x - (a²/2)x³ + (1/4)x⁴ + C Substituting and Evaluating: = (ħ²/2m) * 2 * ((a²/2)x² - (1/4)x⁴) |₀^a / ((a⁴/3)x - (a²/2)x³ + (1/4)x⁴) |₀^a = (ħ²/2m) * 2 * ((a⁴/2) - (a⁴/4)) / ((a⁵/3) - (a⁵/2) + (a⁵/4)) = (ħ²/2m) * 2 * (a⁴/4) / (a⁵/12) = (ħ²/2m) * 6/a = 3ħ²/ma² Calculating Δp: Δp = √( -

²) Δp = √((3ħ²/ma²) - 0) Δp = √(3ħ²/ma²) Δp = √(3) * ħ / a