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

2024
Views:

306

Pages:

6
Author:

Jabin F.

Degenerate Time Independent Perturbation Theory <φ_n|φ_m> = δ_nm Ĥ = Ĥ₀ + Ĥₚ, Ĥₚ << Ĥ₀ or Ĥ Ĥ₀|φ_nγ> = E_n^(0)|φ_nγ>, n = 1, 2, 3, ..., f E_n^(0) = f-fold degenerate energy eigenvalue of Ĥ₀ |φ_nγ> = Eigenf corresponding to f-fold degenerate eigenvalue E_n^(0) |ψ> = c1|φ1> + c2|φ2> + c3|φ3> + ... let |ψ> = Σ(α=1 to f) cα|φ_α> be the eigen fn of H with eigen value E_m H|ψ> = E_m|ψ> H = H_0 + H_p (H_0 + H_p) Σ(α=1 to f) cα|φ_α> = E_m Σ(α=1 to f) cα|φ_α> <φ_n|(H_0 + H_p) Σ(α=1 to f) cα|φ_α> = <φ_n|E_m Σ(α=1 to f) cα|φ_α> Σ(α=1 to f) cα(<φ_n|H_0|φ_α> + <φ_n|H_p|φ_α>) = Σ(α=1 to f) cαE_m<φ_n|φ_α> <φ_n|φ_m> = δ_nm = Σ(α=1 to f) cα(E_n^0<φ_n|φ_α> + <φ_n|H_p|φ_α>) Σ(α=1 to f) cαE_n^0<φ_n|φ_α> = 0 <φ_n|H_p|φ_α> = H_pαβ E_n = E_n^(0) + E_n^(1) Σ(α=1 to f) (H_pαβ - (E_n^(1) - E_m^(1)) δ_αβ) = 0 Σ(α=1 to f) (H_pαβ - E_n^(1) δ_αβ) = 0 x - 2y + z = 0 2x + y - z = 0 3x - y + 5z = 0 => | 1 -2 1 | = 0 | 2 1 -1 | | 3 -1 5 | Secular Determinant | Hp11 - En(1) Hp12 Hp13 ... Hp1f | | Hp21 Hp22 - En(1) Hp23 ... Hp2f | | Hp31 Hp32 Hp33 - En(1) ... Hp3f | | ... ... ... | | Hpf1 Hpf2 Hpf3 ... Hpfp - En(1) | = 0 These det. fn of En(1) with max. f F(En(1)) = 0 fth degree eqn in En(1) => f Roots → (i) If all f roots are same, then there is no removable degeneracy. (ii) If all f roots are different, then there is Complete removable degeneracy. (iii) If out of f roots, Some roots are diff. and some roots are same then there is partial removal of degeneracy. Extracted Text: Prob. Consider a particle of mass 'm' that is moving in 1-D poth V(x) = { 0, 0 = αx * (2V₀ / a) * ∫(0 to a) [sin²(nπx / a) * δ(x - a/2)] dx E_n^(1) = αV₀ sin²(nπ/2) E_n^(1) = { αV₀/2, n odd ; 0, n even } E_n = (n²π²ħ²) / (2ma²) + αV₀ sin²(nπ/2) Prob. Quantum Mechanics page-89 Consider a particle of mass m that is moving in 1-D Harmonic oscillator poth V(x) = (1/2) mω²x², ω is ang. frequency. Find the energy of n-th state corrected to first non-zero correction under the perturbation Hp = λx, λ is constant. Soln E_n^(1) = <φ_n|Hp|φ_n> = λ<φ_n|x|φ_n> for H.O. Position operator <φ_n|x|φ_n> = 0 E_n^(2) = Σ(m≠n) |<φ_m|Hp|φ_n>|^2 / (E_m^(0) - E_n^(0))² = λ² Σ(m≠n) |<φ_m|x|φ_n>|^2 / (E_m^(0) - E_n^(0))² <φ_m|x|φ_n> = (ħ / 2mw) * () = (ħ / 2mw) * ( + ) = (ħ / 2mw) * (√n * δ_m,n-1 + √(n+1) * δ_m,n+1) E_n^(2) = λ² * Σ(m≠n) |(ħ / 2mw) * (√n * δ_m,n-1 + √(n+1) * δ_m,n+1)|² / (E_m^(0) - E_n^(0))² = λ² * (ħ² / (4m²w²)) * Σ(m≠n) [(n * δ_m,n-1² + (n+1) * δ_m,n+1²) / ((m + 1/2)ħw - (n + 1/2)ħw)²] = λ² * (ħ / 2mw²) * (n / (ħw)² + (n+1) / (ħw)²) = λ² / (2mw²) * (n + (n+1)) = λ² / (2mw²) The energy corrected to first non-zero correction: E_n = (n + 1/2)ħw - λ² / (2mw²)