phy-4600/lecture_notes/3-14/rotational invariance
2016-03-24 01:04:06 -04:00

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