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

2024
Views:

75

Pages:

5
Author:

Jabin F.

For a system of N - identical particles -> There are N! different linearly independent wave functions with same energy. i.e. energy is N! fold degenerate. (Exchange degeneracy) Consider a system consisting of N particles: Non - identical particles:- H = (P1^2)/2m1 + (P2^2)/2m2 + ... + (PN^2)/2mN + V1(r1) + V2(r2) + ... + Vn(rn) (ii) non - identical & non - interacting --> H = (P1^2)/2m1 + (P2^2)/2m2 + ... + (PN^2)/2mN + V1(|r1|) + V2(|r2|) + ... + Vn(|rn|) H = Σ(i=1 to N) [(Pi^2)/2mi + Vi(ri)] = Σ(i=1 to N) Hi additive Ψ(r1, r2, ..., rn) = Ψ1(r1)Ψ2(r2)...ΨN(rn) E = Σ(i=1 to N) Ei ↓ additive 3. Classical Identical & non-interacting particles. Intrinsic quantity: mass H = (p1^2)/2m + (p2^2)/2m + ... + (pN^2)/2m + V1(r1) + V2(r2) + ... + VN(rn) H = Σ(i=1 to N) Hi E = Σ(i=1 to N) Ei Ψ(r1, r2, ..., rn) = Π(i=1 to N) Ψi(ri) 4. Quantum Identical & non-interacting Particles: H = (P1^2)/2m + (P2^2)/2m + (P3^2)/2m + ... + (PN^2)/2m + V1(r1) + ... + VN(rn) E = Σ(i=1 to N) Ei Ψ(r1, r2, ..., rn) = Π(i=1 to N) Ψi(ri) Definite symmetric -Symmetric Anti-symmetric Total wave fn of the system consisting of quantum identical particles have definite symmetry Identical Fermions Fill according to Pauli's Exclusion Principle. → The symmetry character of wave function does not change with the exchange of particles in a pair. The wave function that remains unchanged under the exchange of particles in a pair is called symmetric wave function. The wave function that changes sign under the exchange of particles in a pair is called anti-symmetric wave function. Page - 73 Ψ = √(4/3π) cos²Φ = √(4/3π)(1 + cos2Φ / 2) Ψ = √(2/3)[Φ₀(Φ) + 1/2(Φ₊₂(Φ) + Φ₋₂(Φ))] Φ = √(2/3)(1/√2π + 1/2(1/√2π)e^(iλΦ) + 1/√2π)e^(-iiiq)) Φ = 1/√2πe^(imΦ) = i/2 √(4/3π)(1 + e^(2iΦ) + e^(-2i'Φ) / 2 ...) if Lz is measured, the possible outcomes are 0, ±25 P(0) = 1 / (1 + (1/2)^2 + (1/2)^2) = 2/3 P(2ℏ) = (1/2)^2 / (1 + (1/2)^2 + (1/2)^2) = 1/6 P(-2ℏ) = (1/2)^2 / (1 + (1/2)^2 + (1/2)^2) = 1/6 Conclusion: Lz can equal 0, ±2ℏ. P(0) = 2/3 P(±2ℏ) = 1/6 Therefore, the answer is (d).