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new overview and HW, prob 8
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HW
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HW
@ -7,4 +7,4 @@ Chap 5: 2, 6, 8, 12, in-class assignment
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Chap 9: 7, 11, 12, 13, 14
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Chap 7: 7, 8, one from last class, 11
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due friday 3-18
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due friday 3-18, the class one is: show L̂𝓏 and p² commute
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35
lecture_notes/3-23/overview
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35
lecture_notes/3-23/overview
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@ -0,0 +1,35 @@
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Two dimensional harmonic oscillator
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───────────────────────────────────
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This is an oscillator with potential V(x,y) = μ/2 ω (x² + y²)
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The hamiltonian here leaves us with a 3-dimensional differential equation
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Ĥ = ιp²/2μ + μ/2 ω² (x² + y²) = 1/2μ (p²𝓍 + p²𝓎 + p²𝓏) + μ/2 ω²(x² + y²)
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(pic) This is then split into x,y, and z parts.
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(pic) Solved Ψ(z)
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Put together solutions of Ψ(x,y) and Ψ(z).
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(pic) To find position space representation of Ψ(x,y), recall the Hermitian Polynomials solution
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!!! STUDY THIS !!!
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Developed the harmonic oscillator in polar coordinates
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Ĥ𝓍𝓎 = -ħ²\2μ ∇² + μ/2 ω² r²
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(pic) Can be solved using separation of variables.
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Ψ(r,θ) = R(r) Θ(θ)
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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.
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Don't use L̂𝓏 Θ = ±ιmħ Θ
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Use L̂𝓏² Θ = -m²ħ² Θ ⇒ Θ(θ) = exp(±imθ)
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(pic) further developed hamiltonian using this information
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@ -1,6 +1,6 @@
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Angular momentum system is prepared in the state
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❙Ψ❭ = 1͟ ❙11❭ - 2͟ ❙10❭ + ι͟2͟ ❙22❭ + ι1͟ ❙20❭
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❙Ψ❭ = 1͟ ❙11❭ - 2͟ ❙10❭ + ι͟2͟ ❙22❭ + ι͟1͟ ❙20❭
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√10 √10 √10 √10
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@ -41,29 +41,29 @@ No need to calculate the other set: since the vector is normalized, the probabil
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𝓟(𝐋̂²=6ħ²) = ½.
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For 𝐋̂𝓏 the eigenvalue equation is
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For L̂𝓏 the eigenvalue equation is
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𝐋̂𝓏❙lm❭ = mħ❙lm❭,
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L̂𝓏❙lm❭ = mħ❙lm❭,
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so the expected measurements are,
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for lm = 11 & 21, 𝐋̂𝓏 = ħ.
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for lm = 11 & 21, L̂𝓏 = ħ.
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for lm = 10 & 20, 𝐋̂𝓏 = 0.
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for lm = 10 & 20, L̂𝓏 = 0.
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In this case, the probabilities will be
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𝓟(𝐋̂𝓏=ħ) = │❬11❙Ψ❭│² + │❬21❙Ψ❭│² and
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𝓟(L̂𝓏=ħ) = │❬11❙Ψ❭│² + │❬21❙Ψ❭│² and
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𝓟(𝐋̂𝓏=0) = │❬10❙Ψ❭│² + │❬20❙Ψ❭│².
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𝓟(L̂𝓏=0) = │❬10❙Ψ❭│² + │❬20❙Ψ❭│².
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The method has already been demonstrated, so taking the probability components and summing,
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(𝐛)
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𝓟(𝐋̂𝓏=ħ) = ²/₁₀.
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𝓟(L̂𝓏=ħ) = ²/₁₀.
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𝓟(𝐋̂𝓏=0) = ⁸/₁₀.
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𝓟(L̂𝓏=0) = ⁸/₁₀.
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Histogram of probabilities:
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@ -72,17 +72,17 @@ Histogram of probabilities:
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╭─────────────────┬────────────────╮
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1 │ │ │ 1
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│ │ │
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.8 │ │ ▧ │ .8
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│ │ ▧ │
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│ │ ▧ │
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.5 │ ▧ ▧ │ ▧ │ .5
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│ ▧ ▧ │ ▧ │
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│ ▧ ▧ │ ▧ │
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│ ▧ ▧ │ ▧ ▧ │
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.1 │ ▧ ▧ │ ▧ ▧ │ .1
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.8 │ │ ▓ │ .8
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│ │ ▓ │
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│ │ ▓ │
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.5 │ ▓ ▓ │ ▓ │ .5
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│ ▓ ▓ │ ▓ │
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│ ▓ ▓ │ ▓ │
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│ ▓ ▓ │ ▓ ▓ │
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.1 │ ▓ ▓ │ ▓ ▓ │ .1
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╰─────────────────┼────────────────╯
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2ħ² 6ħ² │ 0 ħ
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𝐋̂² │ 𝐋̂𝓏
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2ħ² 6ħ² 0 ħ
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𝐋̂² L̂𝓏
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h
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40
solutions/chap7/prob8
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40
solutions/chap7/prob8
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7.35 is nothing more than a definition of spherical coordinates.
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⎧ x = r sinθ cosϕ
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⎪
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⎨ y = r sinθ sinϕ
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⎪
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⎩ z = r cosθ
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7.47 is the set of algebraic conditions expressed by the vector definition 𝐋 = 𝐫 × 𝐩.
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⎧ L̂𝓍 = yp𝓏 - zp𝓎 = -ιħ (y ∂͟_ - z ∂͟_ )
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⎪ ∂y ∂y
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⎪
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⎨ L̂𝓎 = zp𝓍 - xp𝓏 = -ιħ (z ∂͟_ - x ∂͟_ )
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⎪ ∂y ∂y
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⎪
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⎪ L̂𝓏 = xp𝓎 - yp𝓍 = -ιħ (x ∂͟_ - y ∂͟_ )
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⎩ ∂y ∂y
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Substituing 7.35 into 7.47,
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⎧ L̂𝓍 = -ιħ (r sinθ sinϕ ∂͟_ - r cosθ ∂͟_ )
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⎪ ∂y ∂y
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⎪
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⎨ L̂𝓎 = -ιħ (r cosθ ∂͟_ - r sinθ cosϕ ∂͟_ )
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⎪ ∂y ∂y
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⎪
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⎪ L̂𝓏 = -ιħ (r sinθ cosϕ ∂͟_ - r sinθ sinϕ ∂͟_ )
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⎩ ∂y ∂y
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⎧ L̂𝓍 = -ιħ (r sinθ sinϕ ∂͟_ - r cosθ ∂͟_ )
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⎪ ∂y ∂y
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⎪
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⎨ L̂𝓎 = ιħ (r cosθ ∂͟_ + r sinθ cosϕ ∂͟_ )
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⎪ ∂y ∂y
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⎪
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⎪ L̂𝓏 = -ιħ (r sinθ cosϕ ∂͟_ - r sinθ sinϕ ∂͟_ )
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⎩ ∂y ∂y
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2452
solutions/chap7/prob8.ps
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2452
solutions/chap7/prob8.ps
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