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

2024
Views:

239

Pages:

6
Author:

Jabin F.

=> Eigen kets of Ln for l = 1 Ln = [0, h/sqrt(2), 0; h/sqrt(2), 0, h/sqrt(2); 0, h/sqrt(2), 0] = (h/sqrt(2)) * [0, 1, 0; 1, 0, 1; 0, 1, 0] |Lx - λI| = 0 λ = +ħ, 0, -ħ Eigenfn: Lx |ψx> = λ |ψx> Lx |ψx(+ħ)> = ħ |ψx(+ħ)> ħ/sqrt(2) * [0 1 0; 1 0 1; 0 1 0] * [a; b; c] = ħ * [a; b; c] b/sqrt(2) + c/sqrt(2) = b -(i) b/sqrt(2) = c -(ii) b/sqrt(2) = a -(iii) from (i), (iii) => a = c put in (ii) 2a/sqrt(2) = b => b = sqrt(2)a |ψx(+ħ)> = [a; sqrt(2)a; a] <ψx(+ħ)|ψx(+ħ)> = 1 => |ψx(+ħ)> = A * [1; sqrt(2); 1] A^* [a^* sqrt(2) a^* a^*] A [a; sqrt(2)a; a] = 1 |A|^2 = 1 / (|a|^2 + 2|a|^2 + |a|^2) |A| = 1 / 2 |a|, a = +/- real => |ψ(+ħ)> = (1/2) [1; sqrt(2); 1] = [1/2; sqrt(2)/2; 1/2] For |ψ(0)> Lx |ψ(0)> = λ |ψ(0)> (ħ/sqrt(2)) [0 1 0; 1 0 1; 0 1 0] [a; b; c] = 0 [a; b; c] => b = 0 a + c = 0 => c = -a => |ψ(0)> = [a; 0; -a] (Not normalized) |ψ(0)> = (1/sqrt(2)) [1; 0; -1] => Lx |ψ(-ħ)> = -ħ |ψ(-ħ)> a = c, b = -sqrt(2)a => |ψ(-ħ)> = (1/sqrt(2)) [1; -sqrt(2); 1] Ly = (ħ/sqrt(2)) [0 -i 0; i 0 -i; 0 i 0] => |ψ(y +ħ)> = |m⟩ → Yₘ(φ) |l, m⟩ → Yₗₘ(θ, φ) |l, m, m⟩ → Yₗₘᵐ(r, θ, φ) => Lz |l, m⟩ = mħ |l, m⟩ L² |l, m⟩ = l(l+1)ħ² |l, m⟩ => Lz Yₗₘ(θ, φ) = Lz Yₗₘ(θ, φ) = mħ Yₗₘ(θ, φ) L² Yₗₘ(θ, φ) = l(l+1)ħ² Yₗₘ(θ, φ) Prob. Check whether the state ∣ψ⟩=∣1,+1⟩+∣1,0⟩−i∣1,−1⟩ Where ∣l,m⟩ is eigen state of L 2 and L z is simultaneous eigen state of L 2 and L z with eigen values l(l+1)ℏ 2 and mℏ respectively. Soln L 2 ∣ψ⟩=L 2 ∣1,+1⟩+L 2 ∣1,0⟩−iL 2 ∣1,−1⟩=αℏ 2 ∣1,+1⟩+2ℏ 2 ∣1,0⟩−i2ℏ 2 ∣1,−1⟩=αℏ 2 (∣1,+1⟩+∣1,0⟩−i∣1,−1⟩)=αℏ 2 ∣ψ⟩ So ∣ψ⟩ is an eigen state of L 2 . L z ∣ψ⟩=L z ∣1,+1⟩+L z ∣1,0⟩−iL z ∣1,−1⟩=mℏ∣1,+1⟩+0∣1,0⟩+ζℏ∣1,−1⟩L z ∣ψ⟩=ℏ(∣1,+1⟩+i⟨∣1,−1⟩)  =ℏ∣ψ⟩ ∣ψ⟩ is not an eigen state of L z = = 1/4i = 1/4i ( - + + ) = 1/4i (0 - 0 - <(L^2 - Lz^2 + ħLz)> + <(L^2 - Lz^2 + ħLz)>) = 1/4i = 1/4i (0 - 0 - (l(l+1) - m^2 + m)ħ^2 + (l(l+1) - m^2 - m)ħ^2) = -1/2i (mħ^2) Similarly = +1/2i (mħ^2) => <(LxLy + LyLx)> = 0