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

2024
Views:

376

Pages:

5
Author:

Jabin F.

Perturbation Theory Question 19 Page-87 GATE-2020 H_0 = [E 0 0; 0 E 0; 0 0 E], λ = E, E, E E_n^(0) = E_1, 3-fold deg., α = 1, 2, 3 H' = [0 K 0; K 0 K; 0 K 0] H^ = H_0 + H' |ϕ_n1> = (i 0 0)^T, |ϕ_n2> = (0 i 0)^T, |ϕ_n3> = (0 0 i)^T ⇒ H_p11 = <ϕ_n1|H'|ϕ_n1> = (1 0 0) [0 K 0; K 0 K; 0 K 0] (i 0 0)^T = 0 ⇒ H_p22 = <ϕ_n2|H'|ϕ_n2> = (0 1 0) [0 K 0; K 0 K; 0 K 0] (0 i 0)^T = 0 ⇒ H_p33 = <ϕ_n3|H'|ϕ_n3> = (0 0 1) [0 K 0; K 0 K; 0 K 0] (0 0 i)^T = 0 H_p12 = <ϕ_n1|H'|ϕ_n2> = (1 0 0) [0 K 0; K 0 K; 0 K 0] (0 1 0)^T = K H_p21 = (H_p12)^* = K H_p23 = <ϕ_n2|H'|ϕ_n3> = (0 1 0) [0 K 0; K 0 K; 0 K 0] (0 0 1)^T = 0 H_p32 = (H_p23)^* = 0 H_p31 = <ϕ_n3|H'|ϕ_n1> = (0 0 1) [0 K 0; K 0 K; 0 K 0] (1 0 0)^T = K H_p13 = (H_p31)^* = K Determinant: |E - E_n(1) K 0; K E - E_n(1) K; 0 K E - E_n(1)| = 0 => E_n(1) = 0, ±√2K The energy corrected to first order E_n = E_n(0) + E_n(1) = E ± √2K second excited first excited ground state max energy E + √2K = 3 2 + √2k = 3 => k = 1/√2 = 0.707 Page-25 n=2, l=1, m= +1, 0, -1 |1m⟩ = |210⟩, |211⟩, |21-1⟩ H' = (α/ħ) Lx <210|H'|210> - E(2) <210|1|210> <210|H'|211> <210|H'|21-1> <211|H'|210> <211|H'|211> - E(2) <211|1|211> <211|H'|21-1> = 0 <21-1|H'|210> <21-1|H'|211> <21-1|H'|21-1> - E(2) <21-1|1|21-1> <210|H'|210> = (α/2ħ) <210|(L+ + L-)|210> = 0 <211|H'|210> = 0 = <21-1|H'|21-1> <210|H'|211> = (α/2ħ) <210|(L+ + L-)|211> = (α/2ħ) <210|(L+)|211> = α/√2 => <211|H'|210> = <210|H'|211>* = α/√2 => <210|H'|21-1> = (α/ħ) <210|Lx|21-1> = α/√2 => <21-1|H'|210> = <210|H'|21-1>* = α/√2 <211|H'|211> = 0 <21-1|H'|211> ≠ 0 => <21-1|H'|21-1> = α |1 - E(2) α/√2 α/√2; α/√2 -E(2) 0; α/√2 0 -E(2)| = 0 E(2) = 0, ±α (option-a) Question 20 H_0 = (p_x^2 + p_y^2 + p_z^2)/(2m) + (1/2)k(x^2 + y^2 + z^2) E_n(0) = (n + 3/2)ħω, n = 0, 1, 2, 3, ... ∞ B = B_0 z H_p = -μ.B H_p = (-lħ/2m)μB for second excited E = (7/2)ħω E_n(0) = (n + 3/2)ħω, n = 9/2, ... ∞ = (2N + 1 + 3/2)ħω, N, J = 0, 1, 2, ... ∞ for second excited state 2N + 1 = 2 N, J = (0, 2), (1, 0) l = 0, 2 l = 0, m = 0, E_2^(0) = 0 l = 2, m = 0, ±1, ±2, E_2^(0) = 0, ±(1/2ħ^2) ±(1/2ħ^2) 5 different levels. (option-a)