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

2024
Views:

363

Pages:

6
Author:

Jabin F.

presence of Scattering poth assume R(r) = (A * Sin(kr - π/2 + δu)) / kr δu = phase shift total wave fn (presence of Scattering poth) Ψ(r,θ) = Σγ (Aγ * Sin(kγ - γπ/2 + δj)) * Pγ(cosθ) / kγ ψ(γ,θ) = ψ_in + ψ_sc * e^(ikγ) ψ(γ,θ) = ψ_in + f(θ) * e^(ikγ) f(θ) = (1/k) * Σ(l=0 to ∞) [(2l+1) * e^(iδl) * sin(δl) * Pl(cosθ)] f*(θ) = (1/k) * Σ(l'=0 to ∞) [(l'+1) * e^(-iδl') * sin(δl') * Pl'(cosθ)] dσ/dΩ = (1/k^2) * Σ(l'=0 to ∞) Σ(l=0 to ∞) [(l'+1)(2l+1) * e^(-iδl') * sin(δl') * e^(-iδl) * sin(δl) * ∫(0 to π) Pl'(cosθ) * Pl(cosθ) * sinθ dθ] (2/(l+1)) * δl'l σ_T = ∫(0 to π) ∫(0 to 2π) (dσ/dΩ) * sinθ dθ dφ σ_T = (2π/k^2) * Σγ(l'=0 to ∞) Σ(l=0 to ∞) [(l'+1)(2l+1) * e^(-iδl') * sin(δl') * e^(-iδl) * sin(δl) * ∫(0 to π) Pl'(cosθ) * Pl(cosθ) * sinθ dθ] σ_T = (2π/k^2) * Σγ(l=0 to ∞) [(l+1) * sin^2(δl)] σ_T = (4π/k^2) * Σγ(l=0 to ∞) [(l+1) * sin^2(δl)] f(θ=0) = (1/k) * Σγ(l=0 to ∞) [(2l+1) * e^(iδl) * sin(δl)] = (1/k) * Σγ(l=0 to ∞) [(2l+1) * sin(δl) * (cos(δl) + i * sin(δl))] Im f(θ=0) = (1/k) * Σγ(l=0 to ∞) [(2l+1) * sin^2(δl)] σ_T = (4π/k) * Im f(θ=0) -> optical theorem (conservation of particle of flux) f(θ) = f1(θ) + if2(θ) f(θ=0) = f1(θ=0) + if2(θ=0) Two spin 1/2 particles S = |S1 - S2| - |S1 + S2| = 0, 1 |ψ> = Cs|s=0> + Ct|s=1> Â|ψ> = (1)|s=0>, Â -> spin singlet projection operator Â|ψ> = (1)|s=1>, Â -> spin triplet projection operator Option-a Â = 1/4 + S1.S2 Â|ψ> = (1/4 + S1.S2) (Cs|s=0> + Ct|s=1>) ... S1.S2 = {-3/4 -> Singlet, 1/4 -> Triplet} = Cs(1/4 + (-3/4))|s=0> + Ct(1/4 + 1/4)|s=1> = (-2/4)Cs|s=0> + (2/4)Ct|s=1> Â is not projection operator. Option-b Â = 3/4 - S1.S2 Â|ψ> = (3/4 - S1.S2) (Cs|s=0> + Ct|s=1>) = Cs(3/4 - (-3/4))|s=0> + Ct(3/4 - 1/4)|s=1> Â is not proj. operator option-c: Â = 3/4 + S1.S2 ✓ Â|ψ> = Cs(3/4 + (-3/4))|s=0> + Ct(3/4 + 1/4)|s=1> ✓ Â|ψ> = Ct|s=1> -> triplet state Â is operator corresponding to spin triplet state. (option-c) Question-5 page-80 H = A S1.S2 + B(S1z + S2z) S1 = 1, S2 = 1/2 S = |S1 - S2| - |S1 + S2| = 1/2, 3/2 S1.S2 = -ħ²/2 S = 3/2, S1 = 1, S2 = 1/2 S1.S2 = ħ²/2 S = 1/2, S1 = 1, S2 = 1/2 S1.S2 = ħ²/2 max. S1m S2n = ħ/2 ± 3ħ/2 max energy eigen value. = Aħ²/2 + B3ħ²/2 = ± ħ/2 (A + 3B) (Option-a)