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

2024
Views:

79

Pages:

5
Author:

Jabin F.

Probability density [n=2] [n=1] -a [x=0] for n-th state 0 = Snm [integral from 0 to 1 of |psi_n|^2 dx = 1/2] Prob. Same as page-165 But ψ(n) -> ψ(n) = ψ₁(n) + ψ₂(n) |ψ⟩ = |ψ₁⟩ + |ψ₂⟩ Sol'n <π⟩ = (1/2) * (<ψ₁|τ|ψ₁> + <ψ₂|τ|ψ₂> + <ψ₁|ψ₁> + <ψ₁|ψ|ψ₁>) = (1/2) * (a/2 + a/2 - 169/(9π²) - 169/(9π²)) <π⟩ = (9/2 - 164/(9π²)) Prob: A particle of mass m is confined in 1-D Pot'l V(x) = 0, otherwise The particle has the wave fn at t=0 ψ(x,0) = (A/sqrt(a)) * sin(πx/a) + (sqrt(3/5a)) * sin(3πx/a) + (1/sqrt(54)) * sin(5πx/a) Where A is real and positive (i) Find the value of A so that ψ(x,0) is normalized to unity (ii) Find the expectation value of energy of the particle Sol'n ψₙ(x) = { sqrt(2/a) * sin(nπx/a), 0 <= x <= a { 0, otherwise (i) ψ(x,0) = (A/sqrt(a)) * sin(πx/a) + sqrt(3/(5a)) * sin(3πx/a) + sqrt(1/54) * sin(5πx/a) = (A/sqrt(2)) * sqrt(2/a) * sin(πx/a) + sqrt(3/10) * sqrt(2/a) * sin(3πx/a) + sqrt(1/10) * sqrt(2/a) * sin(5πx/a) = (A/sqrt(2)) * ψ₁ + sqrt(3/10) * ψ₃ + sqrt(1/10) * ψ₅ ∑ |Cn|^2 = 1 |A|^2 / 2 + 3/10 + 1/10 = 1 |A|^2 / 2 = 6/10 |A| = sqrt(6/5) A = sqrt(6/5) only +ve (ii) = Σ PnEn = P₁E₁ + P₃E₃ + P₅E₅ ψ(x,0) = sqrt(6/10) * ψ₁ + sqrt(3/10) * ψ₃ + sqrt(1/10) * ψ₅ = (6/10) * ((7^2 * hbar^2) / (2ma^2)) + (3/10) * ((9π^2 * hbar^2) / (2ma^2)) + (1/10) * ((25π^2 * hbar^2) / (2ma^2)) = (58π^2 * hbar^2) / (20ma^2) = (29π^2 * hbar^2) / (10ma^2) Ans. Ques.10 Page-37 V(x) = { 0, 0 <= x <= L { ∞, otherwise ψₙ(x) = sqrt(2/L) * sin(nπx/L), 0 <= x <= L En = (n^2 * π^2 * hbar^2) / (2mL^2) ψ(x) = sqrt(2/L) * [(3/5) * sin((8πx/L)) + (4/5) * sin((4πx/L))] ψ(x) = (3/5) * ψ₂ + (4/5) * ψ₄ 0 <= x <= L The possible outcomes of energy - are eigenvalues - E2 = (4π²ħ²) / (2mL²) = (4ħ²) / (8mL²) = (ħ²) / (2mL²) √ E4 = (16π²ħ²) / (2mL²) = (8ħ²) / (mL²) = (4ħ²) / (2mL²) Ans. = Σ PnEn = (9/25) * (ħ²) / (2mL²) + (16/25) * (4ħ²) / (2mL²) = ((9+64)ħ²) / (50mL²) = (73ħ²) / (50mL²) Ans. (option-a)