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finished most of chap 7 (a) homework
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@ -85,84 +85,41 @@ Matrix analysis can be used to find the eigenvectors for these eigenstates. The
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⎧ b = √2 a
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⎨ (a + c) = √2 b
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⎩ b = √2 c
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Following this to conclusion just like with spin operators will provide the eigenstates, and then from that the wave function can be expressed using the x basis, and probabilities obtained.
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I need to stop here, but I will produce at least sthe histogram from part a:
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(𝐜)
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𝓟(L̂𝓏)
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Applying the operator to the states in Ψ,
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L̂𝓍❙1 1❭ ≐
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ħ͟ ⎛ 0 1 0 ⎞⎛1⎞ = ħ͟ ⎛0⎞ = ħ͟ ❙1 0❭.
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√2 ⎜ 1 0 1 ⎟⎜0⎟ √2 ⎜1⎟ √2
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⎝ 0 1 0 ⎠⎝0⎠ ⎝0⎠
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L̂𝓍❙1 0❭ ≐
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ħ͟ ⎛ 0 1 0 ⎞⎛0⎞ = ħ͟ ⎛1⎞ = ħ͟ (❙1 1❭ + ❙1 -1❭), and
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√2 ⎜ 1 0 1 ⎟⎜1⎟ √2 ⎜0⎟ √2
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⎝ 0 1 0 ⎠⎝0⎠ ⎝1⎠
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L̂𝓍❙1 -1❭ ≐
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ħ͟ ⎛ 0 1 0 ⎞⎛0⎞ = ħ͟ ⎛0⎞ = ħ͟ ❙1 0❭.
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√2 ⎜ 1 0 1 ⎟⎜0⎟ √2 ⎜1⎟ √2
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⎝ 0 1 0 ⎠⎝1⎠ ⎝0⎠
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L̂𝓍❙Ψ❭ = ⎛ 2͟ L̂𝓍❙1 1❭ + ι 3͟ L̂𝓍❙1 0❭ - 4͟ L̂𝓍❙1 -1❭ ⎞
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⎝ √29 √29 √29 ⎠
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2͟ L̂𝓍❙1 1❭ = 2͟ ħ❙1 0❭,
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√29 √58
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ι 3͟ L̂𝓍❙1 0❭ = ι 3͟ ħ (❙1 1❭ + ❙1 -1❭), and
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√29 √58
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4͟ L̂𝓍❙1 -1❭ = 4͟ ħ❙1 0❭.
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√29 √58
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Then,
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L̂𝓍❙Ψ❭ = ħ ⎛ -2͟ ❙1 0❭ + ι 3͟ (❙1 1❭ + ❙1 -1❭)⎞
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⎝ √58 √58 ⎠
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Normalizing the function,
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C⎛⎛-2͟ ⎞² + ⎛ι 3͟ ⎞² + ⎛ι 3͟ ⎞²⎞ = 1.
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⎝⎝√58⎠ ⎝ √58⎠ ⎝ √58⎠ ⎠
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C⎛4 - 9 - 9⎞ = 58 = C(14).
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⎝ ⎠
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C = 58/14 = 29/7.
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The probability of measuring one of the possible angular momenta will be given by
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│❬l m❙L̂𝓍❙Ψ❭│².
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❬1 1❙L̂𝓍❙Ψ❭ = ❬1 1❙2͟9͟⎛ -2͟ ❙1 0❭ + ι 3͟ (❙1 1❭ + ❙1 -1❭)⎞
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7 ⎝ √58 √58 ⎠
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= 2͟9͟ ι͟3͟ ❬1 1❙1 1❭ = ι 8͟7͟
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7 √58 7√58
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L̂𝓍❙Ψ❭ = ⎛ + ι 3͟ L̂𝓍❙1 0❭ - 4͟ L̂𝓍❙1 -1❭ ⎞
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⎝ √29 √29 ⎠
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╭─────────────────────────╮
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│ │
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│ │
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│ │
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│ │
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│ │
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│ │
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│ │
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¹⁶/₂₉ ├ ▓ │
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│ ▓ │
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│ ▓ │
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│ ▓ │
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│ ▓ │
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│ ▓ │
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│ ▓ │
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⁹/₂₉ ├ ▓ ▓ │
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│ ▓ ▓ │
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│ ▓ ▓ │
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│ ▓ ▓ │
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│ ▓ ▓ │
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⁴/₂₉ ├ ▓ ▓ ▓ │
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│ ▓ ▓ ▓ │
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│ ▓ ▓ ▓ │
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│ ▓ ▓ ▓ │
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╰─────────────────────────╯
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-ħ 0 ħ
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@ -83,6 +83,3 @@ Histogram of probabilities:
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╰─────────────────┼────────────────╯
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2ħ² 6ħ² 0 ħ
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𝐋̂² L̂𝓏
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h
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@ -1,4 +1,4 @@
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7.35 is nothing more than a definition of spherical coordinates.
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The transformation from polar coordinates to cartesian is the set of equations
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⎧ x = r sinθ cosϕ
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⎪
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@ -6,21 +6,142 @@
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⎪
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⎩ z = r cosθ
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The transformation from cartesian to polar coordinates is the set
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⎧ r = √(x²+y²+z²)
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⎪
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⎪ cos(θ) = z͟ = _͟z͟
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⎨ r √(x²+y²+z²)
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⎪
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⎪ tan(ϕ) = y͟
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⎩ x
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Some differential forms may come in handy.
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⎧ dcos(θ) = -sin(θ) dθ
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⎨
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⎪ dtan(ϕ) = _͟1͟ dϕ
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⎩ cos²(ϕ)
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∂/∂θ:
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⎧ ∂x = r cosϕ cosθ ∂θ
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⎧ ∂͟x = r cosϕ cosθ
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⎪ ∂θ
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⎪
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⎨ ∂y = r sinϕ cosθ ∂θ
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⎨ ∂͟y = r sinϕ cosθ
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⎪ ∂θ
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⎪
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⎩ ∂z = - r sinθ ∂θ
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⎪ ∂͟z = -r sinθ
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⎩ ∂θ
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∂/∂ϕ:
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⎧ ∂x = - r sinθ sinϕ ∂ϕ
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⎧ ∂͟x = -r sinθ sinϕ
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⎪ ∂ϕ
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⎪
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⎨ ∂y = r sinθ cosϕ ∂ϕ
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⎨ ∂͟y = r sinθ cosϕ
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⎪ ∂ϕ
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⎪
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⎪ ∂͟z = 0
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⎩ ∂ϕ
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∂/∂x:
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⎧ ∂͟r = √(x²+y²+z²) = x͟
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⎪ ∂x r
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⎪
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⎩ ∂z = 0 ∂ϕ
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⎪ ∂͟cos(θ) = -z͟x͟
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⎨ ∂x r³
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⎪
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⎪ ∂͟tan(ϕ) = -y͟
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⎩ ∂x x²
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∂/∂y:
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⎧ ∂͟r = √(x²+y²+z²) = y͟
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⎪ ∂y r
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⎪
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⎪ ∂͟cos(θ) = -z͟y͟
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⎨ ∂y r³
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⎪
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⎪ ∂͟tan(ϕ) = 1͟
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⎩ ∂y x
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∂/∂z:
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⎧ ∂͟r = √(x²+y²+z²) = z͟
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⎪ ∂z r
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⎪
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⎪ ∂͟cos(θ) = z͟ = 1͟ ⎛r ∂͟z͟ - z ∂͟r͟⎞ = 1͟ - z͟²
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⎨ ∂z r r² ⎝ ∂z ∂z⎠ r r³
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⎪
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⎪ ∂͟tan(ϕ) = 0
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⎩ ∂z
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These differential forms can be used to transform each cartesian differentiation operator.
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∂͟ = ∂͟r͟ ∂͟ + ∂͟c͟o͟s͟θ͟ ∂͟ + ∂͟t͟a͟n͟ϕ͟ ∂͟ ;
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∂x ∂x ∂r ∂x ∂cosθ ∂x ∂tanϕ
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∂͟ = x͟ ∂͟ - z͟x͟ _͟1͟ ∂͟ - y͟ cos²(ϕ) ∂͟ ;
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∂x r ∂r r³ -sinθ ∂θ x² ∂ϕ
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∂͟ = sin(θ) cos(ϕ) ∂͟
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∂x ∂r
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+ 1͟ sin(θ) cos(ϕ) cos(θ) _͟1͟ ∂͟
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r sin(θ) ∂θ
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- r͟ s͟i͟n͟(ϕ͟) s͟i͟n͟(θ͟) cos²(ϕ͟) ∂͟ ;
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r² cos²(ϕ) sin²(θ) ∂ϕ
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∂͟ = sin(θ) cos(ϕ) ∂͟ + 1͟ cos(ϕ) cos(θ) ∂͟ - 1͟ s͟i͟n͟(ϕ͟) ∂͟ .
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∂x ∂r r ∂θ r sin(θ) ∂ϕ
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───────────────────────────────────────────────
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∂͟ = ∂͟r͟ ∂͟ + ∂͟c͟o͟s͟θ͟ ∂͟ + ∂͟t͟a͟n͟ϕ͟ ∂͟ ;
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∂y ∂y ∂r ∂y ∂cosθ ∂y ∂tanϕ
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∂͟ = y͟ ∂͟ - z͟y͟ ∂͟ + 1͟ ∂͟ ;
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∂y r ∂r r³ ∂cosθ x ∂tanϕ
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∂͟ = sin(θ) sin(ϕ) ∂͟ - 1͟ sin(θ) cos(θ) s͟i͟n͟(ϕ͟) ∂͟ + c͟o͟s͟(ϕ͟) ∂͟ ;
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∂y ∂r r -sin(θ) ∂θ r sin(θ) ∂ϕ
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∂͟ = sin(θ) sin(ϕ) ∂͟ + 1͟ cos(θ) s͟i͟n͟(ϕ͟) ∂͟ + c͟o͟s͟(ϕ͟) ∂͟ .
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∂y ∂r r ∂θ r sin(θ) ∂ϕ
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───────────────────────────────────────────────
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∂͟ = ∂͟r͟ ∂͟ + ∂͟c͟o͟s͟θ͟ ∂͟ + ∂͟t͟a͟n͟ϕ͟ ∂͟ ;
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∂z ∂z ∂r ∂z ∂cosθ ∂z ∂tanϕ
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∂͟ = z͟ ∂͟ + ⎛1͟ - z͟²⎞∂͟ + 0 ;
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∂z r ∂r ⎝r r³⎠∂cosθ
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∂͟ = cos(θ) ∂͟ + 1͟⎛1 - cos²(θ)⎞ _͟1͟ ∂͟ ;
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∂z ∂r r⎝ ⎠ -sin(θ) ∂θ
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∂͟ = cos(θ) ∂͟ - _͟1͟_ ⎛1 - cos²(θ)⎞ ∂͟ ;
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∂z ∂r rsin(θ)⎝ ⎠ ∂θ
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∂͟ = cos(θ) ∂͟ - _͟1͟_ sin²(θ) ∂͟ ;
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∂z ∂r rsin(θ) ∂θ
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∂͟ = cos(θ) ∂͟ - s͟i͟n͟(θ͟) ∂͟ .
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∂z ∂r r ∂θ
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The set of differential operator transformations is
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⎧ ∂͟ = sin(θ) cos(ϕ) ∂͟ + 1͟ cos(ϕ) cos(θ) ∂͟ - 1͟ s͟i͟n͟(ϕ͟) ∂͟
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⎪ ∂x ∂r r ∂θ r sin(θ) ∂ϕ
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⎪
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⎨ ∂͟ = sin(θ) sin(ϕ) ∂͟ + 1͟ cos(θ) s͟i͟n͟(ϕ͟) ∂͟ + c͟o͟s͟(ϕ͟) ∂͟
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⎪ ∂y ∂r r ∂θ r sin(θ) ∂ϕ
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⎪
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⎪ ∂͟ = cos(θ) ∂͟ - s͟i͟n͟(θ͟) ∂͟
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⎩ ∂z ∂r r ∂θ
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7.47 is the set of algebraic conditions expressed by the vector definition 𝐋 = 𝐫 × 𝐩.
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@ -45,91 +166,87 @@ Substituting 7.35 into 7.47,
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⎪ L̂𝓏 = -ιħ (r sinθ cosϕ ∂͟_ - r sinθ sinϕ ∂͟_ )
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⎩ ∂y ∂x
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Substituting the transformed differentiation operators,
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Geometry is shown on the attached notes page.
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For L̂𝓍:
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∂͟z͟ = -r sinθ
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∂θ
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∂͟y͟ = r sinθ cosϕ
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∂ϕ
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L̂𝓍 = -ιħ ( r sinθ sinϕ ∂͟θ͟ ∂͟_ - r cosθ ∂͟ϕ͟ ∂͟_ )
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∂θ ∂z ∂ϕ ∂y
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L̂𝓍 = ιħ ( -r sinθ sinϕ ∂͟θ͟ ∂͟_ + r cosθ ∂͟ϕ͟ ∂͟_ )
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∂z ∂θ ∂y ∂ϕ
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L̂𝓍 = ιħ ( −͟r͟ s͟i͟n͟θ͟ sinϕ ∂͟_ + c͟o͟s͟θ͟ ∂͟_ )
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-r sinθ ∂θ sinθ cosϕ ∂ϕ
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L̂𝓍 = ιħ ( sinϕ ∂͟_ + c͟o͟t͟θ͟ ∂͟_ )
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∂θ cosϕ ∂ϕ
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For L̂𝓎:
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∂x = r cosϕ cosθ ∂θ
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∂z = 0 ∂ϕ
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L̂𝓎 = -ιħ (r cosθ ∂͟_ - r sinθ cosϕ ∂͟_ )
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∂x ∂z
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L̂𝓎 = ιħ (-r cosθ ∂͟θ͟ ∂͟_ + r sinθ cosϕ ∂͟ϕ͟ ∂͟_ )
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∂θ ∂x ∂ϕ ∂z
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L̂𝓎 = ιħ (-r cosθ ∂͟θ͟ ∂͟_ + r sinθ cosϕ ∂͟ϕ͟ ∂͟_ )
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∂x ∂θ ∂z ∂ϕ
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L̂𝓎 = ιħ (- _͟1͟ cosθ ∂͟_ + r sinθ cosϕ 0 ∂͟_ )
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cosϕ cosθ ∂θ ∂ϕ
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L̂𝓎 = ιħ (- _͟1͟ ∂͟_ )
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cosϕ ∂θ
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L̂𝓎 = -ιħ _͟1͟ ∂͟_
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cosϕ ∂θ
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For L̂𝓏:
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∂x = -r sinθ sinϕ ∂ϕ
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∂y = r sinθ cosϕ ∂ϕ
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L̂𝓏 = -ιħ (r sinθ cosϕ ∂͟͟ϕ͟ ∂͟_ - r sinθ sinϕ ∂͟͟ϕ͟ ∂͟_ )
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∂ϕ ∂y ∂ϕ ∂x
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L̂𝓏 = -ιħ (r͟ s͟i͟n͟θ͟ c͟o͟s͟ϕ͟ ∂͟_ + r͟ s͟i͟n͟θ͟ s͟i͟n͟ϕ͟ ∂͟_ )
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r sinθ cosϕ ∂ϕ r sinθ sinϕ ∂ϕ
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L̂𝓏 = -ιħ ∂͟_ ( 1 + 1 )
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∂ϕ
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L̂𝓏 = -2ιħ ∂͟_
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∂ϕ
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So, according to my calculus, the final solutions should be the set
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⎧ L̂𝓍 = ιħ ( sinϕ ∂͟_ + c͟o͟t͟θ͟ ∂͟_ )
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⎪ ∂θ cosϕ ∂ϕ
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⎧ L̂𝓍 = -ιħ⎛r sinθ sinϕ ⎛cos(θ) ∂͟ - s͟i͟n͟(θ͟) ∂͟ ⎞
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⎪ ⎝ ⎝ ∂r r ∂θ⎠
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⎪
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⎪ - r cosθ ⎛sin(θ) sin(ϕ) ∂͟ + 1͟ cos(θ) s͟i͟n͟(ϕ͟) ∂͟ + c͟o͟s͟(ϕ͟) ∂͟ ⎞⎞
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⎪ ⎝ ∂r r ∂θ r sin(θ) ∂ϕ⎠⎠
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⎪
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⎨ L̂𝓎 = -ιħ _͟1͟ ∂͟_
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⎪ cosϕ ∂θ
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⎪
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⎪ L̂𝓏 = -2ιħ ∂͟_
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⎩ ∂ϕ
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⎪ L̂𝓎 = -ιħ⎛r cosθ⎛sin(θ) cos(ϕ) ∂͟ + 1͟ cos(ϕ) cos(θ) ∂͟ - 1͟ s͟i͟n͟(ϕ͟) ∂͟ ⎞
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⎪ ⎝ ⎝ ∂r r ∂θ r sin(θ) ∂ϕ⎠
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⎨
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⎪ - r sinθ cosϕ ⎛cos(θ) ∂͟ - s͟i͟n͟(θ͟) ∂͟ ⎞⎞
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⎪ ⎝ ∂r r ∂θ⎠⎠
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⎪
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⎪
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⎪ L̂𝓏 = -ιħ⎛r sinθ cosϕ⎛sin(θ) sin(ϕ) ∂͟ + 1͟ cos(θ) s͟i͟n͟(ϕ͟) ∂͟ + c͟o͟s͟(ϕ͟) ∂͟ ⎞
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⎪ ⎝ ⎝ ∂r r ∂θ r sin(θ) ∂ϕ⎠
|
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⎪
|
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⎪ - r sinθ sinϕ ⎛sin(θ) cos(ϕ) ∂͟ + 1͟ cos(ϕ) cos(θ) ∂͟ - 1͟ s͟i͟n͟(ϕ͟) ∂͟ ⎞⎞
|
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⎩ ⎝ ∂r r ∂θ r sin(θ) ∂ϕ⎠⎠
|
||||
|
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The spherical representation, i.e. ending place, is the set
|
||||
Simplifying...
|
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|
||||
⎧ L̂𝓍 = -ιħ⎛r sinθ sinϕ cosθ ∂͟ - sinϕ sin²θ ∂͟
|
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⎪ ⎝ ∂r ∂θ
|
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⎪
|
||||
⎪ - r cosθ sinθ sinϕ ∂͟ - cos²θ sinϕ ∂͟ - c͟o͟s͟θ͟ cosϕ ∂͟ ⎞
|
||||
⎪ ∂r ∂θ sinθ ∂ϕ⎠
|
||||
⎪
|
||||
⎪
|
||||
⎪ L̂𝓎 = -ιħ⎛r cosθ sinθ cosϕ ∂͟ + cosθ cosϕ cosθ ∂͟ - c͟o͟s͟θ͟ sinϕ ∂͟
|
||||
⎪ ⎝ ∂r ∂θ sinθ ∂ϕ
|
||||
⎨
|
||||
⎪ - r sinθ cosϕ cosθ ∂͟ + sinθ cosϕ sinθ ∂͟ ⎞
|
||||
⎪ ∂r ∂θ⎠
|
||||
⎪
|
||||
⎪
|
||||
⎪ L̂𝓏 = -ιħ⎛r sin²θ cosϕ sinϕ ∂͟ + sinθ cosθ sinϕ cosϕ ∂͟ + cosϕ c͟o͟s͟ϕ͟ ∂͟
|
||||
⎪ ⎝ ∂r ∂θ ∂ϕ
|
||||
⎪
|
||||
⎪ - r sin²θ sinϕ cosϕ ∂͟ - sinθ cosθ sinϕ cosϕ ∂͟ + sinθ sinϕ s͟i͟n͟ϕ͟ ∂͟ ⎞
|
||||
⎩ ∂r ∂θ sinθ ∂ϕ⎠
|
||||
|
||||
|
||||
⎧ L̂𝓍 = -ιħ⎛- sinϕ sin²θ ∂͟ - cos²θ sinϕ ∂͟ - c͟o͟s͟θ͟ cosϕ ∂͟ ⎞
|
||||
⎪ ⎝ ∂θ ∂θ sinθ ∂ϕ⎠
|
||||
⎪
|
||||
⎨ L̂𝓎 = -ιħ⎛cosθ cosϕ cosθ ∂͟ - c͟o͟s͟θ͟ sinϕ ∂͟ + sinθ cosϕ sinθ ∂͟ ⎞
|
||||
⎪ ⎝ ∂θ sinθ ∂ϕ ∂θ⎠
|
||||
⎪
|
||||
⎪ L̂𝓏 = -ιħ⎛cosϕ c͟o͟s͟ϕ͟ ∂͟ + sinθ sinϕ s͟i͟n͟ϕ͟ ∂͟ ⎞
|
||||
⎩ ⎝ ∂ϕ sinθ ∂ϕ⎠
|
||||
|
||||
|
||||
⎧ L̂𝓍 = ιħ⎛ sinϕ sin²θ ∂͟ + cos²θ sinϕ ∂͟ + c͟o͟s͟θ͟ cosϕ ∂͟ ⎞
|
||||
⎪ ⎝ ∂θ ∂θ sinθ ∂ϕ⎠
|
||||
⎪
|
||||
⎨ L̂𝓎 = ιħ⎛-cos²θ cosϕ ∂͟ - sin²θ cosϕ ∂͟ + c͟o͟s͟θ͟ sinϕ ∂͟ ⎞
|
||||
⎪ ⎝ ∂θ ∂θ sinθ ∂ϕ ⎠
|
||||
⎪
|
||||
⎪ L̂𝓏 = -ιħ⎛cos²ϕ ∂͟ + sin²ϕ ∂͟ ⎞
|
||||
⎩ ⎝ ∂ϕ ∂ϕ⎠
|
||||
|
||||
|
||||
⎧ L̂𝓍 = ιħ⎛ sinϕ ∂͟ + cotθ cosϕ ∂͟ ⎞
|
||||
⎪ ⎝ ∂θ ∂ϕ⎠
|
||||
⎪
|
||||
⎨ L̂𝓎 = ιħ⎛-cosϕ ∂͟ + cotθ sinϕ ∂͟ ⎞
|
||||
⎪ ⎝ ∂θ ∂ϕ ⎠
|
||||
⎪
|
||||
⎪ L̂𝓏 = -ιħ∂͟
|
||||
⎩ ∂ϕ
|
||||
|
||||
The spherical representation is the following set, which perfectly matches the obtained result.
|
||||
|
||||
⎧ L̂𝓍 = ιħ (sinϕ ∂͟_ + cosϕ cotθ ∂͟_ )
|
||||
⎪ ∂θ ∂ϕ
|
||||
⎪ ∂θ ∂ϕ
|
||||
⎪
|
||||
⎨ L̂𝓎 = ιħ (-cosϕ ∂͟_ + sinϕ cotθ ∂͟_ )
|
||||
⎪ ∂θ ∂ϕ
|
||||
⎪ ∂θ ∂ϕ
|
||||
⎪
|
||||
⎪ L̂𝓏 = -ιħ ∂͟_
|
||||
⎩ ∂ϕ
|
||||
|
||||
|
||||
My set DOES NOT match this. I must be going about this the wrong way. I have to give it more thought. Perhaps a purely geometric approach will improve my answers: I'll try that over the weekend.
|
||||
⎩ ∂ϕ
|
||||
|
||||
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Reference in New Issue
Block a user