new overview and HW, prob 8

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othocaes 2016-03-23 19:43:48 -04:00
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2
HW
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@ -7,4 +7,4 @@ Chap 5: 2, 6, 8, 12, in-class assignment
Chap 9: 7, 11, 12, 13, 14
Chap 7: 7, 8, one from last class, 11
due friday 3-18
due friday 3-18, the class one is: show L̂𝓏 and p² commute

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Two dimensional harmonic oscillator
───────────────────────────────────
This is an oscillator with potential V(x,y) = μ/2 ω (x² + y²)
The hamiltonian here leaves us with a 3-dimensional differential equation
Ĥ = ιp²/2μ + μ/2 ω² (x² + y²) = 1/2μ (p²𝓍 + p²𝓎 + p²𝓏) + μ/2 ω²(x² + y²)
(pic) This is then split into x,y, and z parts.
(pic) Solved Ψ(z)
Put together solutions of Ψ(x,y) and Ψ(z).
(pic) To find position space representation of Ψ(x,y), recall the Hermitian Polynomials solution
!!! STUDY THIS !!!
Developed the harmonic oscillator in polar coordinates
𝓍𝓎 = -ħ²\2μ ∇² + μ/2 ω² r²
(pic) Can be solved using separation of variables.
Ψ(r,θ) = R(r) Θ(θ)
The problem is invariant under rotations about z, i.e. L̂𝓏 commutes with Ĥ, so the solutions must be eigenstates of both Ĥ and L̂𝓏.
Don't use L̂𝓏 Θ = ±ιmħ Θ
Use L̂𝓏² Θ = -m²ħ² Θ ⇒ Θ(θ) = exp(±imθ)
(pic) further developed hamiltonian using this information

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@ -1,6 +1,6 @@
Angular momentum system is prepared in the state
❙Ψ❭ = 1͟ ❙11❭ - 2͟ ❙10❭ + ι͟2͟ ❙22❭ + ι1͟ ❙20❭
❙Ψ❭ = 1͟ ❙11❭ - 2͟ ❙10❭ + ι͟2͟ ❙22❭ + ι͟1͟ ❙20❭
√10 √10 √10 √10
@ -41,29 +41,29 @@ No need to calculate the other set: since the vector is normalized, the probabil
𝓟(𝐋̂²=6ħ²) = ½.
For 𝐋̂𝓏 the eigenvalue equation is
For L̂𝓏 the eigenvalue equation is
𝐋̂𝓏❙lm❭ = mħ❙lm❭,
L̂𝓏❙lm❭ = mħ❙lm❭,
so the expected measurements are,
for lm = 11 & 21, 𝐋̂𝓏 = ħ.
for lm = 11 & 21, L̂𝓏 = ħ.
for lm = 10 & 20, 𝐋̂𝓏 = 0.
for lm = 10 & 20, L̂𝓏 = 0.
In this case, the probabilities will be
𝓟(𝐋̂𝓏=ħ) = │❬11❙Ψ❭│² + │❬21❙Ψ❭│² and
𝓟(L̂𝓏=ħ) = │❬11❙Ψ❭│² + │❬21❙Ψ❭│² and
𝓟(𝐋̂𝓏=0) = │❬10❙Ψ❭│² + │❬20❙Ψ❭│².
𝓟(L̂𝓏=0) = │❬10❙Ψ❭│² + │❬20❙Ψ❭│².
The method has already been demonstrated, so taking the probability components and summing,
(𝐛)
𝓟(𝐋̂𝓏=ħ) = ²/₁₀.
𝓟(L̂𝓏=ħ) = ²/₁₀.
𝓟(𝐋̂𝓏=0) = ⁸/₁₀.
𝓟(L̂𝓏=0) = ⁸/₁₀.
Histogram of probabilities:
@ -72,17 +72,17 @@ Histogram of probabilities:
╭─────────────────┬────────────────╮
1 │ │ │ 1
│ │ │
.8 │ │ │ .8
│ │
│ │
.5 │ ▧        ▧    │           ▧    │ .5
   │   ▧        ▧    │           ▧    │    
   │   ▧        ▧    │           ▧    │    
   │   ▧        ▧    │   ▧       ▧    │    
.1 │   ▧        ▧    │   ▧       ▧    │ .1
.8 │ │ │ .8
│ │
│ │
.5 │ ▓        ▓    │           ▓    │ .5
   │   ▓        ▓    │           ▓    │    
   │   ▓        ▓    │           ▓    │    
   │   ▓        ▓    │   ▓       ▓    │    
.1 │   ▓        ▓    │   ▓       ▓    │ .1
   ╰─────────────────┼────────────────╯
2ħ² 6ħ² 0 ħ
𝐋̂² 𝐋̂𝓏
2ħ² 6ħ² 0 ħ
𝐋̂² L̂𝓏
h

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7.35 is nothing more than a definition of spherical coordinates.
⎧ x = r sinθ cosϕ
⎨ y = r sinθ sinϕ
⎩ z = r cosθ
7.47 is the set of algebraic conditions expressed by the vector definition 𝐋 = 𝐫 × 𝐩.
⎧ L̂𝓍 = yp𝓏 - zp𝓎 = -ιħ (y ∂͟_ - z ∂͟_ )
⎪ ∂y ∂y
⎨ L̂𝓎 = zp𝓍 - xp𝓏 = -ιħ (z ∂͟_ - x ∂͟_ )
⎪ ∂y ∂y
⎪ L̂𝓏 = xp𝓎 - yp𝓍 = -ιħ (x ∂͟_ - y ∂͟_ )
⎩ ∂y ∂y
Substituing 7.35 into 7.47,
⎧ L̂𝓍 = -ιħ (r sinθ sinϕ ∂͟_ - r cosθ ∂͟_ )
⎪ ∂y ∂y
⎨ L̂𝓎 = -ιħ (r cosθ ∂͟_ - r sinθ cosϕ ∂͟_ )
⎪ ∂y ∂y
⎪ L̂𝓏 = -ιħ (r sinθ cosϕ ∂͟_ - r sinθ sinϕ ∂͟_ )
⎩ ∂y ∂y
⎧ L̂𝓍 = -ιħ (r sinθ sinϕ ∂͟_ - r cosθ ∂͟_ )
⎪ ∂y ∂y
⎨ L̂𝓎 = ιħ (r cosθ ∂͟_ + r sinθ cosϕ ∂͟_ )
⎪ ∂y ∂y
⎪ L̂𝓏 = -ιħ (r sinθ cosϕ ∂͟_ - r sinθ sinϕ ∂͟_ )
⎩ ∂y ∂y

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