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135 lines
3.9 KiB
Plaintext
135 lines
3.9 KiB
Plaintext
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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Some differential forms may come in handy.
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∂/∂θ:
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⎧ ∂x = r cosϕ cosθ ∂θ
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⎪
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⎨ ∂y = r sinϕ cosθ ∂θ
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⎪
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⎩ ∂z = - r sinθ ∂θ
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∂/∂ϕ:
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⎧ ∂x = - r sinθ sinϕ ∂ϕ
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⎪
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⎨ ∂y = r sinθ cosϕ ∂ϕ
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⎪
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⎩ ∂z = 0 ∂ϕ
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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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⎪ ∂z ∂y
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⎪
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⎨ L̂𝓎 = zp𝓍 - xp𝓏 = -ιħ (z ∂͟_ - x ∂͟_ )
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⎪ ∂x ∂z
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⎪
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⎪ L̂𝓏 = xp𝓎 - yp𝓍 = -ιħ (x ∂͟_ - y ∂͟_ )
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⎩ ∂y ∂x
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Substituting 7.35 into 7.47,
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⎧ L̂𝓍 = -ιħ (r sinθ sinϕ ∂͟_ - r cosθ ∂͟_ )
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⎪ ∂z ∂y
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⎪
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⎨ L̂𝓎 = -ιħ (r cosθ ∂͟_ - r sinθ cosϕ ∂͟_ )
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⎪ ∂x ∂z
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⎪
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⎪ L̂𝓏 = -ιħ (r sinθ cosϕ ∂͟_ - r sinθ sinϕ ∂͟_ )
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⎩ ∂y ∂x
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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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⎪
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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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The spherical representation, i.e. ending place, is the set
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⎧ L̂𝓍 = ιħ (sinϕ ∂͟_ + cosϕ cotθ ∂͟_ )
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⎪ ∂θ ∂ϕ
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⎪
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⎨ L̂𝓎 = ιħ (-cosϕ ∂͟_ + sinϕ cotθ ∂͟_ )
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⎪ ∂θ ∂ϕ
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⎪
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⎪ L̂𝓏 = -ιħ ∂͟_
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⎩ ∂ϕ
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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. |