new overview and HW, prob 8

HW

@@ -7,4 +7,4 @@ Chap 5: 2, 6, 8, 12, in-class assignment
 Chap 9: 7, 11, 12, 13, 14
 
 Chap 7: 7, 8, one from last class, 11
-    due friday 3-18
+    due friday 3-18, the class one is: show L̂𝓏 and p² commute

lecture_notes/3-23/overview

@@ -0,0 +1,35 @@
+Two dimensional harmonic oscillator
+───────────────────────────────────
+
+    This is an oscillator with potential V(x,y) = μ/2 ω (x² + y²)
+
+    The hamiltonian here leaves us with a 3-dimensional differential equation
+
+        Ĥ = ιp²/2μ + μ/2 ω² (x² + y²) = 1/2μ (p²𝓍 + p²𝓎 + p²𝓏) + μ/2 ω²(x² + y²)
+
+        (pic) This is then split into x,y, and z parts.
+
+            (pic) Solved Ψ(z)
+
+
+        Put together solutions of Ψ(x,y) and Ψ(z).
+
+        (pic) To find position space representation of Ψ(x,y), recall the Hermitian Polynomials solution
+
+            !!! STUDY THIS !!!
+
+    Developed the harmonic oscillator in polar coordinates
+
+    Ĥ𝓍𝓎 = -ħ²\2μ ∇² + μ/2 ω² r²
+
+    (pic) Can be solved using separation of variables.
+
+        Ψ(r,θ) = R(r) Θ(θ)
+
+        The problem is invariant under rotations about z, i.e. L̂𝓏 commutes with Ĥ, so the solutions must be eigenstates of both Ĥ and L̂𝓏.
+
+        Don't use L̂𝓏 Θ = ±ιmħ Θ
+
+        Use L̂𝓏² Θ = -m²ħ² Θ ⇒ Θ(θ) = exp(±imθ)
+
+        (pic) further developed hamiltonian using this information

solutions/chap7/prob7

@@ -1,6 +1,6 @@
 Angular momentum system is prepared in the state
 
-    ❙Ψ❭ = 1͟  ❙11❭ - 2͟ ❙10❭ + ι͟2͟ ❙22❭ + ι1͟ ❙20❭
+    ❙Ψ❭ = 1͟  ❙11❭ - 2͟ ❙10❭ + ι͟2͟ ❙22❭ + ι͟1͟ ❙20❭
          √10        √10      √10       √10 
 
 
@@ -18,7 +18,7 @@ For lm = 22 & 20, 𝐋̂² = 6ħ².
 
 Since the same 𝐋̂² is measures for two states, the sum of the probabilities of measuring those states is the probability of measuring that squared angular momentum.
 
-𝓟(𝐋̂²=2ħ²) = │❬11❙Ψ❭│² + │❬10❙Ψ❭│².
+    𝓟(𝐋̂²=2ħ²) = │❬11❙Ψ❭│² + │❬10❙Ψ❭│².
 
     ❬11❙Ψ❭ = ❬11❙⎛  1͟  ❙11❭ - 2͟ ❙10❭ + ι͟2͟ ❙22❭ + ι1͟ ❙20❭ ⎞   
                  ⎝ √10        √10      √10       √10     ⎠.
@@ -41,29 +41,29 @@ No need to calculate the other set: since the vector is normalized, the probabil
     𝓟(𝐋̂²=6ħ²) = ½.
 
 
-For 𝐋̂𝓏 the eigenvalue equation is
+For L̂𝓏 the eigenvalue equation is
 
-    𝐋̂𝓏❙lm❭ = mħ❙lm❭,
+    L̂𝓏❙lm❭ = mħ❙lm❭,
 
 so the expected measurements are,
 
-for lm = 11 & 21, 𝐋̂𝓏 = ħ.
+for lm = 11 & 21, L̂𝓏 = ħ.
 
-for lm = 10 & 20, 𝐋̂𝓏 = 0.
+for lm = 10 & 20, L̂𝓏 = 0.
 
 In this case, the probabilities will be
 
-    𝓟(𝐋̂𝓏=ħ) = │❬11❙Ψ❭│² + │❬21❙Ψ❭│² and
+    𝓟(L̂𝓏=ħ) = │❬11❙Ψ❭│² + │❬21❙Ψ❭│² and
 
-    𝓟(𝐋̂𝓏=0) = │❬10❙Ψ❭│² + │❬20❙Ψ❭│².
+    𝓟(L̂𝓏=0) = │❬10❙Ψ❭│² + │❬20❙Ψ❭│².
 
 The method has already been demonstrated, so taking the probability components and summing,
 
 (𝐛)
 
-    𝓟(𝐋̂𝓏=ħ) = ²/₁₀.
+    𝓟(L̂𝓏=ħ) = ²/₁₀.
 
-    𝓟(𝐋̂𝓏=0) = ⁸/₁₀.
+    𝓟(L̂𝓏=0) = ⁸/₁₀.
 
 Histogram of probabilities:
 
@@ -72,17 +72,17 @@ Histogram of probabilities:
    ╭─────────────────┬────────────────╮
 1  │                 │                │ 1 
    │                 │                │   
-.8 │                 │           ▧    │ .8
+.8 │                 │           ▓    │ .8
-   │                 │           ▧    │   
+   │                 │           ▓    │   
-   │                 │           ▧    │   
+   │                 │           ▓    │   
-.5 │   ▧        ▧    │           ▧    │ .5
+.5 │   ▓        ▓    │           ▓    │ .5
-   │   ▧        ▧    │           ▧    │    
+   │   ▓        ▓    │           ▓    │    
-   │   ▧        ▧    │           ▧    │    
+   │   ▓        ▓    │           ▓    │    
-   │   ▧        ▧    │   ▧       ▧    │    
+   │   ▓        ▓    │   ▓       ▓    │    
-.1 │   ▧        ▧    │   ▧       ▧    │ .1
+.1 │   ▓        ▓    │   ▓       ▓    │ .1
    ╰─────────────────┼────────────────╯
-       2ħ²     6ħ²   │   0       ħ    
+       2ħ²     6ħ²       0       ħ    
-            𝐋̂²       │       𝐋̂𝓏
+            𝐋̂²               L̂𝓏
 
             h
 

solutions/chap7/prob8

@@ -0,0 +1,40 @@
+7.35 is nothing more than a definition of spherical coordinates.
+
+⎧    x = r sinθ cosϕ
+⎪
+⎨    y = r sinθ sinϕ
+⎪
+⎩    z = r cosθ
+
+
+7.47 is the set of algebraic conditions expressed by the vector definition 𝐋 = 𝐫 × 𝐩.
+
+⎧    L̂𝓍 = yp𝓏 - zp𝓎 = -ιħ (y ∂͟_  - z ∂͟_ )
+⎪                           ∂y      ∂y
+⎪
+⎨    L̂𝓎 = zp𝓍 - xp𝓏 = -ιħ (z ∂͟_  - x ∂͟_ )
+⎪                           ∂y      ∂y
+⎪
+⎪    L̂𝓏 = xp𝓎 - yp𝓍 = -ιħ (x ∂͟_  - y ∂͟_ )
+⎩                           ∂y      ∂y
+
+Substituing 7.35 into 7.47,
+
+⎧    L̂𝓍 = -ιħ (r sinθ sinϕ ∂͟_  - r cosθ ∂͟_ )
+⎪                          ∂y           ∂y
+⎪
+⎨    L̂𝓎 = -ιħ (r cosθ ∂͟_  - r sinθ cosϕ ∂͟_ )
+⎪                     ∂y                ∂y
+⎪
+⎪    L̂𝓏 = -ιħ (r sinθ cosϕ ∂͟_ - r sinθ sinϕ ∂͟_ )
+⎩                          ∂y               ∂y
+
+
+⎧    L̂𝓍 = -ιħ (r sinθ sinϕ ∂͟_  - r cosθ ∂͟_ )
+⎪                          ∂y           ∂y
+⎪
+⎨    L̂𝓎 = ιħ (r cosθ ∂͟_ + r sinθ cosϕ ∂͟_ )
+⎪                    ∂y               ∂y
+⎪
+⎪    L̂𝓏 = -ιħ (r sinθ cosϕ ∂͟_ - r sinθ sinϕ ∂͟_ )
+⎩                          ∂y               ∂y