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90 lines
2.4 KiB
Plaintext
90 lines
2.4 KiB
Plaintext
Rotational Invariance
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𝐩² and 𝐫 are invariant under rotation.
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Consider
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R̂(dϕ k̂) = 1 - ι/ħ L̂𝓏 dϕ
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❙x - ydϕ,y + xdϕ,z❭ = ❙x,y,z❭ + ∂/∂x ❙ ❭ dϕ + ∂/∂y ❙ ❭ dϕ
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= [1 - ι/ħ p𝓍 (-ydϕ)][1 - ι/ħ p𝓎 (xdϕ)] ❙x,y,z❭
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= [1 - ι/ħ (x p𝓎 - y p𝓍)dϕ] ❙x,y,z❭
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(x p𝓎 - y p𝓍) ≝ L̂𝓏 (angular momentum in z axis)
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L̂𝓏 is the z component of L̂ = 𝐫×𝐩̂
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Commutation:
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[L̂𝓏,p̂𝓏] = [x p𝓎 - y p𝓍,p𝓏]
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= (xp𝓎p𝓍 - yp𝓍p𝓍) - (p𝓍xp𝓎 - p𝓍yp𝓍)
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= [x,p𝓍]p𝓎 - [y,p𝓍]p𝓍 = [x,p𝓍]p𝓎 = ιħp𝓎
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↓
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0
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[L̂𝓏,p̂𝓍] = ιħp̂𝓎
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[L̂𝓏,p̂𝓎] = -ιħp̂𝓎
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[L̂𝓏,p̂𝓏] = 0
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[L̂𝓏,p̂²] = 0
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i.e. the kinetic energy commutes with the L̂𝓏, so one can measure angular momentum and kinetic energy without disturbing the other.
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Proof:
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[L̂𝓏,p̂²] = [L̂𝓏,p̂𝓍² + p̂𝓎² + p̂𝓏²]
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= [L̂𝓏,p̂𝓍²] + [L̂𝓏,p̂𝓎²] + [L̂𝓏,p̂𝓏²]
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= p̂𝓍[L̂𝓏,p̂𝓍] + [L̂𝓏,p̂𝓍]p̂𝓍
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+ p̂𝓎[L̂𝓏,p̂𝓎] + [L̂𝓏,p̂𝓎]p̂𝓎
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+ p̂𝓏[L̂𝓏,p̂𝓏] + [L̂𝓏,p̂𝓏]p̂𝓏 note¹
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= ιħp̂𝓍p̂𝓎 + ιħp̂𝓎p̂𝓍 - ιħp̂𝓎p̂𝓍 - ιħp̂𝓍p̂𝓎 = 0
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[L̂𝓏,r̂²] = 0 (homework) → [L̂𝓏,1/r²] = 0
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→ [L̂𝓏,V(│r│)] = 0
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So,
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[L̂𝓏,Ĥ] = 0.
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L̂𝓏 is therefore a constant of motion (time-independent)
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- Must do work to change?
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[L̂²,Ĥ] = 0, so L̂² is also a constant of motion.
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note¹
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[L̂𝓏,p̂𝓍²] = [L̂𝓏,p̂𝓍 p̂𝓍]
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= L̂𝓏 p̂𝓍 p̂𝓍 - p̂𝓍 p̂𝓍 L̂𝓏
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= ?
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= p̂𝓍[L̂𝓏,p̂𝓍] + [L̂𝓏,p̂𝓍]p̂𝓍
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= (p̂𝓍 L̂𝓏 p̂𝓍 - p̂𝓍 p̂𝓍 L̂𝓏) + (L̂𝓏 p̂𝓍 p̂𝓍 - p̂𝓍 L̂𝓏 p̂𝓍)
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= L̂𝓏 p̂𝓍 p̂𝓍 - p̂𝓍 p̂𝓍 L̂𝓏
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✓ = [L̂𝓏,p̂𝓍²] = [L̂𝓏,p̂𝓍 p̂𝓍]
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