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

2024
Views:

474

Pages:

6
Author:

Jabin F.

Presence of E.F. → The energy of first excited level n=2 Corrected to first order E₂ = E₂⁽⁰⁾ + E₂⁽¹⁾ = { E₂⁽⁰⁾ - 3ea₀E₀ { E₂⁽⁰⁾ + 3ea₀E₀ n=2 -13.6/4 + 3ea₀E₀ -13.6/4 E /= 0 -13.6/4 -3ea₀E₀ E₂⁽¹⁾ = 0, 0, +3ea₀Eo, -3ea₀Eo In the presence of Uniform electric field the behaviour of H-atom is equivalent to an electric dipole having dipole moment P which can be oriented In 3 diff. ways→ In one way || to electric field with -3ea₀Eo IP|| = ea₀ induced in presen of E.F. not pertum In one way antiparallel to electric field with energy +3ea₀Eo In two ways ⊥ to electric field with energy zero. |Ψ⟩ = Σcᵢ |nᵢ⟩ |Ψ⟩ = c₁|200⟩ + c₂|210⟩ + c₃|211⟩ + c₄|21-1⟩ A|Ψ⟩ = αc₁|200⟩ + αc₂|210⟩ + αc₃|211⟩ + αc₄|21-1⟩ [(-E(1)2) 3ea₀E₀ 0 0][c₁] [3ea₀E₀ (-E(1)2) 0 0][c₂] = 0 [0 0 E(1)2 0][c₃] [0 0 0 E(1)2][c₄] => [0 3ea₀E₀ 0 0][c₁] [3ea₀E₀ 0 0 0][c₂] =E(1)2[c₁] [0 0 E(1)2 0][c₃] [0 0 0 E(1)2][c₄] for E(1)2 = 3ea₀E₀ [3ea₀E₀ 3ea₀E₀ 0 0][c₁] [3ea₀E₀ 3ea₀E₀ 0 0][c₂] = 0 [0 0 3ea₀E₀ 3ea₀E₀][c₃] [0 0 3ea₀E₀ 3ea₀E₀][c₄] => -3ea₀E₀ c4 = 0 => c4 = 0 => -3ea₀E₀ c₃ = 0 => c₃ = 0 => -3ea₀E₀ c₁ + 3ea₀E₀ c₂ = 0 => c₁ = c₂ => Σ|c_α|² = 1, α=1 => |c₁|² + |c₂|² + |c₃|² + |c₄|² = 1 |c₁|² + |c₁|² = 1 => |c₁|² = 1/2 c₁ = ±1/√2, ±i/√2 |ψ⟩ = (1/√2)|200⟩ + (1/√2)|210⟩ 2dsinθ = xλ λ ~ d = h / sqrt(2mE_k) Ek = ħ^2 / (2md^2) @page-19 H = α * ((1 0 0), (0 2 0), (0 0 3)) + β(t) * ((0 0 1), (0 0 0), (1 0 -2)) t < 0, β(t) = 0 t > 0, β(t) = α H = α * ((1 0 0), (0 2 0), (0 0 3)) H = α * ((1 0 0), (0 2 0), (0 0 3)) λ' = α, 2α, 3α λ = 0, 2α, 2α |ψ(t<0)> = 0 ((0 2 0), (1 0 1)) * ((a), (b), (c)) = 0 a + c = 0 => c = -a |ψ(t<0)> = ce^(-ikt/ħ)|ϕ₁> + c₂e^(-ikt/ħ)|ϕ₂> + c₃e^(iαt/ħ)|ϕ₃> 2b = 0 => b = 0 <ψ(t<0)> = 1/sqrt(2) * ((1), (0), (-i)) |> 1 0 1 0 2 0 1 0 1 at t=0 |Ψ(t<0)> = |Ψ(t>0)> 1/√2 [1 0 -1] = c₁ [1 0 0] + c₂ [0 1 0] + c₃ [0 0 1] [1/√2 , 0, -1/√2 ]=[c_1, c_2, c_3] => c₁ = 1/√2, c₂ = 0, c₃ = -1/√2 |Ψ(t>0)> = 1/√2 e^(-iαt/ħ) (1 0 0) - 1/√2 (e^-i3αt/ħ) (0 0 1) = 1/2 [e^(iαt/ħ) + (e^-i3αt/ħ)] <Ψ(t>0)|Ψ(t>0)> = 1/√2[1 0 -1] 1/√2 1/2 [e^(iαt/ħ) + (e^-i3αt/ħ)] = 1/2 (e^(iαt/ħ) + (e^-i3αt/ħ)) |<Ψ(t<0)|Ψ(t>0)>|^2 = |1/2 (e^-iαt/ħt + e^-i3αt/ħ)| =1/2 (1 + cos(2αt/ħ)) option-a)