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

2024
Views:

351

Pages:

6
Author:

Jabin F.

Classically forbidden region k^2(x) = (2m/ħ^2) * (V(x) - E) ψ(x) = (A/sqrt(P(x))) * e^(∫k(x')dx') + (B/sqrt(P(x))) * e^(-∫k(x')dx') for classical allowed region from eqn (i) α = (1/ħ) * ∫P(x)dx ψ(x) = (1/sqrt(P(x))) * (Ae^(iαx) + Be^(-iαx)) = (1/sqrt(P(x))) * ((Acosαx + iAsinαx) + (Bcosαx - iBsinαx)) = (1/sqrt(P(x))) * ((A+B)cosαx + (i(A-B))sinαx) ψ(x) = (1/sqrt(P(x))) * (Ccosαx + Dsinαx) C = A + B, D = i(A - B) Ψ(x) = √(1/P(x)) * (Acosαx + Bsinαx) Ψ(x) = √(1/P(x)) * sin(x + β) Ψ''(x) = (A/√P(x)) * sin{ (1/ħ)∫P(x)dx + β } * exp(V(x)/ħ²) ↳ for classically allowed region By using sine formula θ₁θ₂ = π/2, m = 1, 3, 5 - Sinθ₁ = ± Sinθ₂ θ₁ + θ₂ = (2n + 1)π, m = 1, 3, 5 - Sinθ₁ ≠ ± Sinθ₂ = 0 θ₁ + θ₂ = (2n)π, m = 2, 4, 6 -> θ₁ + θ₂ = nπ, n = 1, 2, 3... θ₁ + θ₂ = (2n + 1)π, n = 0, 1, 2... (1/ħ) * ∫√(2m(E - V(x)))dx + β₁ + (1/ħ) * ∫√(2m(E - V(x)))dx + β₂ = { nπ, n = 1, 2, 3... { (n + 1)π, n = 0, 1, 2... Energy quantization Cond'n β = phase factor = { 0, if Turning point lies at Rigid Boundary (V=∞) { π/4, if Turning point lies at non rigid boundary (V≠0) Prob. Find the energy quantization Cond'n and energy of m-th state of a Particle mass m moving in 1-D pot'l V(x) = (1/2)mω²x² by using W.K.B approx. method. Sol'n At turning point V(x) = E E = (1/2)mω²a² a² = 2E/mω² x = +a (1/ħ) * ∫[-a,a] √(2m(E - (1/2)mω²x²)) dx + π/4 + (1/ħ) * ∫[a,∞] √(2m(E - (1/2)mω²x²)) dx + π/4 = nπ, n = 1, 2, 3... (1/ħ) * ∫[-a,a] √(2m(E - (1/2)mω²x²)) dx + π/2 = nπ, n = 1, 2, 3... (1/ħ) * ∫[-a,a] √(2m(E - (1/2)mω²x²)) dx = (n - 1/2)π, n = 1, 2, 3... → (1/ħ) * ∫[-a,a] √(2m(E - (1/2)mω²x²)) dx + π/2 = { nπ, n = 1, 2, 3... { (n + 1)π, n = 0, 1, 2... for energy quantization for energy (2/ħ) * ∫[0,a] √(2m(E - (1/2)mω²x²)) dx = { (n - 1/2)π, n = 1, 2... { (n + 1/2)π, n = 0, 1, 2... → (1/ħ) * ∫[0,a] √(2m(E - (1/2)mω²x²)) dx = = (1/ħ) * ∫[0,a] √(2m((1/2)mω²a² - (1/2)mω²x²)) dx = mω/ħ * ∫[0,a] √(a² - x²) dx x = a sin θ dx = a cos θ dθ = mωa²/ħ * ∫[0,π/2] cos²θ dθ = mωa²/ħ * ∫[0,π/2] (1 + cos2θ)/2 dθ = mωa²/ħ * [(θ/2 + (sin2θ)/4)] = mωa²/ħ * [(π/4 + 0)] = (π/4) * mωa²/ħ put. (2/ħ) * (π/4) * mωa² = { (n - 1/2)π, n = 1, 2, 3... { (n + 1/2)π, n = 0, 1, 2... Em = { (n - 1/2)ħω, n = 1, 2, 3... { (n + 1/2)ħω, n = 0, 1, 2... Ans. Prob. Find energy quantization and energy... By using W.K.B. V(x) = { (1/2)mω²x², x >= 0 { ∞, x < 0 Sol'n ... x = 0 x = a (1/ħ) * ∫[0,a] √(2m(E - (1/2)mω²x²)) dx + 0 + π/4 = { (n - 1/2)π, n = 1, 2, 3... { (n + 1/2)π, n = 0, 1, 2... ∫[0,a] √(2m(E - (1/2)mω²x²)) dx = { (n - 1/4) * π * ħ, n = 1, 2, 3... { (n + 3/4) * π * ħ, n = 0, 1, 2... → (π/4) * mωa² = { (n - 1/4) * π * ħ, n = 1, 2, 3... { (n + 3/4) * π * ħ, n = 0, 1, 2... π * mω²a² * (2E / mω²) = { (n - 1/4) * 4 * π * ħ { (n + 3/4) * 4 * π * ħ En = { (2n - 1/2) * ħω, n = 1, 2, 3... { (2n + 3/2) * ħω, n = 0, 1, 2... = (2n ± 1/2) * ħω