Rotational Invariance

    𝐩² and 𝐫 are invariant under rotation.

    Consider

        R̂(dϕ k̂) = 1 - ι/ħ L̂𝓏 dϕ

        ❙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)

        L̂𝓏 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

            = ιħp̂𝓍p̂𝓎 + ιħp̂𝓎p̂𝓍 - ιħp̂𝓎p̂𝓍 - ιħp̂𝓍p̂𝓎 = 0

            [L̂𝓏,r̂²] = 0 (homework) → [L̂𝓏,1/r²] = 0
                                   → [L̂𝓏,V(│r│)] = 0

            So,

            [L̂𝓏,Ĥ] = 0.

            L̂𝓏 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,p̂𝓍 p̂𝓍] 

                = L̂𝓏 p̂𝓍 p̂𝓍 - p̂𝓍 p̂𝓍 L̂𝓏

                    = ?

            = p̂𝓍[L̂𝓏,p̂𝓍] + [L̂𝓏,p̂𝓍]p̂𝓍 

            = (p̂𝓍 L̂𝓏 p̂𝓍 - p̂𝓍 p̂𝓍 L̂𝓏) + (L̂𝓏 p̂𝓍 p̂𝓍 - p̂𝓍 L̂𝓏 p̂𝓍) 

            = L̂𝓏 p̂𝓍 p̂𝓍  - p̂𝓍 p̂𝓍 L̂𝓏 

        ✓   = [L̂𝓏,p̂𝓍²] = [L̂𝓏,p̂𝓍 p̂𝓍]