lots of lectures. Class is now over.

lecture_notes/4-13/overview

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+                        Exam 2 Problem 2
+━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
+
+(pic) Finished the problem using the boundary conditions
+
+    boundary conditions limit the number of possible spherical harmonics.
+
+
+
+                        Two Similar Particles
+━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
+
+❙a,b❭ = ❙a❭⊗❙b❭
+
+                            Exchange Operator
+──────────────────────────────────────────────────────────────────────────
+
+𝓟₁₂❙a,b❭ = ❙a,b❭
+
+𝓟₁₂(❙a❭⊗❙b❭) = ❙b❭⊗❙a❭
+
+If particles are indistinguishable 
+
+𝓟₁₂❙a,b❭ = exp(iδ) ❙a,b❭ = λ ❙a,b❭
+
+𝓟²₁₂❙a,b❭ = λ² ❙a,b❭ = ❙a,b❭ ⇒ λ=±1
+
+
+                            Symmetry States
+──────────────────────────────────────────────────────────────────────────
+(pic) Two cases: symmetric, antisymmetric
+
+Symmetric States, λ=1
+
+    𝓟₁₂❙a,b❭
+
+(pic) Constructed the exchange operator in matrix form, then found eigen states
+
+(pic) Showed that the exchange operator leads to the Pauli Exclusion Principle
+
+
+
+

lecture_notes/4-15/Overview

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+                          Two Spin-½ Particles
+━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
+uncoupled basis
+───────────────
+❙s₁ s₂ m₁ m₂❭
+❙+ +❭, ❙- -❭, ❙+ -❭, ❙- +❭ 
+
+coupled basis
+─────────────
+❙S Mₛ❭
+❙1 1❭, ❙1 -1❭, ❙1 0❭, ❙0 0❭
+
+Given the state ❙1 0❭ = 1/√2(❙+ -❭ + ❙- +❭) :
+    P₁₂❙1 0❭ = ❙1 0❭ (symmetric)
+
+❙0 0❭ = 1/√2(❙+ -❭ - ❙- +❭) :
+    P₁₂❙0 0❭ = -❙0 0❭ (antisymmetric)
+
+
+                         Spatial Representation
+──────────────────────────────────────────────────────────────────────────
+❙Ψ❭ = ❙Ψₛₚₐₜᵢₐₗ❭❙Ψₛₚᵢₙ❭
+
+Symmetric:
+❙Ψ❭ˢ = ❙Ψₛₚₐₜᵢₐₗ❭ˢ❙Ψₛₚᵢₙ❭ˢ OR ❙Ψₛₚₐₜᵢₐₗ❭ᴬ❙Ψₛₚᵢₙ❭ᴬ
+
+Antisymmetric:
+❙Ψ❭ᴬ = ❙Ψₛₚₐₜᵢₐₗ❭ᴬ❙Ψₛₚᵢₙ❭ˢ OR ❙Ψₛₚₐₜᵢₐₗ❭ˢ❙Ψₛₚᵢₙ❭ᴬ
+
+
+                              Helium Atom
+──────────────────────────────────────────────────────────────────────────
+Fermions, so overall must be antisymmetric
+
+Configuration       Term        Energy
+    1s²              1s            0
+        ❙Ψ❭ = 1/√2 (❙a❭₁ₛ❙b❭₁ₛ + ❙b❭₁ₛ❙a❭₁ₛ)❙0 0❭
+
+    1s²s             3s            1.46
+        ❙Ψ❭ = 1/√2(❙a❭₁ₛ❙b❭₁ₛ - ❙b❭₁ₛ❙a❭₁ₛ)❙1 m❭, m=-1,0,1
+
+    1s2s             1s            1.52
+        ❙Ψ❭ = 1/√2 (❙a❭₁ₛ❙b❭₁ₛ + ❙b❭₁ₛ❙a❭₁ₛ)❙0 0❭
+
+    1s2p             3pᵒ           ~1.60
+        ❙Ψ❭ = 1√2 
+
+    1s²p             1pᵒ           ~1.65
+
+    

lecture_notes/4-18

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+∮𝐁⋅d𝐥 = B(s) sπs = μ₀[∫𝐉⋅n̂da + ε₀ d/dt ∫𝐄⋅n̂da]
+
+∫𝐉⋅n̂da = 0
+
+For capacitor:
+│𝐄│ = σ/ε₀ ⇒ B(s) 2πs = μ₀ ε₀ d/dt σ(t)/ε₀ πs²

solutions/chap2/prob2

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+S𝓍 ≐ ħ͟⎛0 1⎞
+     2⎝1 0⎠
+
+Diagonalize the matrix...
+
+First, find the characteristic equation and solve for the eigenvalues.
+
+S𝓍 ≐ ħ͟⎛0 1⎞
+     2⎝1 0⎠
+
+ħ͟ ⎛-λ  1⎞
+2 ⎝ 1 -λ⎠
+
+    λ² - ħ²/4 = 0 
+    λ² = ħ²/4 
+
+    λ = ±ħ/2 
+
+So the eigenvalues are those expected for the measurement of a spin-1/2 component.
+
+The eigenvalue equations are
+
+S𝓍❙Ψ❭ = λ❙Ψ❭
+
+which is represented by
+
+ħ͟⎛0 1⎞⎛a⎞ = λ⎛a⎞
+2⎝1 0⎠⎝b⎠    ⎝b⎠
+
+

solutions/chap3/prob7

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+After the first Stern-Gerlach magnet, the entire wave function will be in a single state:
+
+❙Ψ₁❭ = ❙S₊❭