adding new lectures: 3d eigenstates, rotation, angular momentum

lecture_notes/3-14/3d eigenstates

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+❙r❭=❙x,y,z❭
+
+with eigenvalue equations
+
+x̂❙r❭ = x❙r❭
+ŷ❙r❭ = y❙r❭
+ẑ❙r❭ = z❙r❭
+
+An arbitrary state
+
+    ❙Ψ❭ = ∫∫∫ dx dy dz ❙x,y,z❭❬x,y,z❙Ψ❭
+        = ∫ d³r ❙r❭ ❬r❙Ψ❭
+
+
+Understanding a System:
+
+    measure Ĥ, L̂², L̂𝓏 → Constitutes a complete set of commuting observables (except spin)
+
+    I.E., There is a set of eigenstates that are eigenstates of all three operators.
+
+    Ĥ❙E,l,mₗ❭ = E❙E,l,mₗ❭
+
+    L̂²❙E,l,mₗ❭ = l(l+1)ħ²❙E,l,mₗ❭
+
+    L̂𝓏❙E,l,mₗ❭ = mₗħ❙E,l,mₗ❭
+
+    Ĥ must now include angular momentum
+
+    L̂²= (r̂×p̂)(r̂×p̂) = (geometric identity) = r̂²p̂ - (r̂⋅p̂) + ιħr̂⋅p̂

lecture_notes/3-14/rotational invariance

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+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`
+
+            = 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.
+

lecture_notes/3-14/translation operators

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+Translation Operators
+
+    Matrix mechanical expression:
+
+    T̂(a𝓍 î) ❙x,y,z❭ = ❙x+a𝓍,y,z❭
+    T̂(a𝓎 î) ❙x,y,z❭ = ❙x,y+a𝓎,z❭
+    T̂(a𝓏 î) ❙x,y,z❭ = ❙x,y,z+a𝓏❭
+
+    Position space form:
+
+    T(a𝓍 î) = exp(-ι p𝓍 a𝓍/ħ )
+
+    Translation operators commute since momentum operators commute.
+
+    But, [x̂ᵢ,p̂ⱼ]=ιħδᵢⱼ
+
+    T̂(𝐚) = exp(-ι p𝓍 a𝓍/ħ ) exp(-ι p𝓎 a𝓎/ħ ) exp(-ι p𝓏 a𝓏/ħ )
+         = exp(-ι 𝐩 𝐚/ħ )
+
+    ❬r❙p̂❙Ψ❭ = ħ/ι ∇ ❬r❙Ψ❭

lecture_notes/3-16/3d eigenstates

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+../3-14/3d eigenstates

lecture_notes/3-16/Angular Momentum Commutator

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+[L̂𝓏,p̂𝓏] = ιħp̂𝓎
+[L̂𝓏,p̂𝓍] = -ιħp̂𝓎
+[L̂𝓏,p̂𝓏] = 0
+[L̂𝓏,p̂²] = 

lecture_notes/3-16/rotational invariance

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+../3-14/rotational invariance