working on chap 7 hw

HW

@@ -6,5 +6,5 @@ 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
+Chap 7: 5, 7, 8, one from last class, 11
     due friday 3-18, the class one is: show L̂𝓏 and p² commute

lecture_notes/3-14/rotational invariance

@@ -42,7 +42,7 @@ Rotational Invariance
 
             = p̂𝓍[L̂𝓏,p̂𝓍] + [L̂𝓏,p̂𝓍]p̂𝓍 
                 + p̂𝓎[L̂𝓏,p̂𝓎] + [L̂𝓏,p̂𝓎]p̂𝓎 
-                    + p̂𝓏[L̂𝓏,p̂𝓏] + [L̂𝓏,p̂𝓏]p̂𝓏  (f.n. 1)
+                    + p̂𝓏[L̂𝓏,p̂𝓏] + [L̂𝓏,p̂𝓏]p̂𝓏  note¹
 
             = ιħp̂𝓍p̂𝓎 + ιħp̂𝓎p̂𝓍 - ιħp̂𝓎p̂𝓍 - ιħp̂𝓍p̂𝓎 = 0
 
@@ -69,7 +69,7 @@ Rotational Invariance
 
 
 
-(f.n. 1)
+note¹
 
 
             [L̂𝓏,p̂𝓍²] = [L̂𝓏,p̂𝓍 p̂𝓍] 

lecture_notes/3-23/overview

@@ -1,7 +1,7 @@
 Two dimensional harmonic oscillator
 ───────────────────────────────────
 
-    This is an oscillator with potential V(x,y) = μ/2 ω (x² + y²)
+    This is an oscillator with potential V(x,y) = μ/2 ω² (x² + y²)
 
     The hamiltonian here leaves us with a 3-dimensional differential equation
 
@@ -14,13 +14,14 @@ Two dimensional harmonic oscillator
 
         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
+        (pic) To find position space representation of Ψ(x,y), recall the Hermitian-- HARMONIC? DUBIOUS Polynomials solution
 
             !!! STUDY THIS !!!
 
     Developed the harmonic oscillator in polar coordinates
 
-    Ĥ𝓍𝓎 = -ħ²\2μ ∇² + μ/2 ω² r²
+    Ĥ𝓍𝓎 = -ħ²/2μ ∇² + μ/2 ω² r²
+        = -ħ²/2μ (∂²/∂r² + 1/r ∂/∂r + 1/r² ∂²/∂θ²) + μ/2 ω² r².
 
     (pic) Can be solved using separation of variables.
 
@@ -33,3 +34,9 @@ Two dimensional harmonic oscillator
         Use L̂𝓏² Θ = -m²ħ² Θ ⇒ Θ(θ) = exp(±imθ)
 
         (pic) further developed hamiltonian using this information
+
+    Ĥ𝓍𝓎 = -ħ²/2μ (∂²/∂r² + 1/r ∂/∂r + 1/r² ∂²/∂θ²) + μ/2 ω² r².
+        = -ħ²/2μ (∂²/∂r² + 1/r ∂/∂r) + -ħ²/(2μr²) L̂𝓏²/ħ² + μ ω² r²/2
+
+    L̂𝓏² ≐ ħ²∂²/∂θ²
+    L̂𝓏 ≐ -ιħ∂/∂θ

solutions/chap7/lectureprob

@@ -0,0 +1,21 @@
+This problem associated with chapter 7 was assigned during lecture.
+
+Does L̂𝓏 commute with 𝐫̂²?
+
+	[L̂𝓏,𝐫̂²] = L̂𝓏 𝐫̂² - 𝐫̂² L̂𝓏.
+
+	L̂𝓏 𝐫̂² - 𝐫̂² L̂𝓏.
+
+Using the position representations, in spherical coordinates,
+
+	L̂𝓏 ≐ -ιħ∂/∂θ and 𝐫̂² ≐ 𝐫²,
+
+	L̂𝓏 𝐫̂² - 𝐫̂² L̂𝓏 = 𝐫² ιħ∂/∂θ - ιħ∂/∂θ 𝐫².
+
+𝐫² has no θ dependence, so it can be separated from any quantity differentiated with respect to theta, I.E.,
+
+    ∂/∂θ 𝐫² = 𝐫² ∂/∂θ.
+
+    𝐫² ιħ∂/∂θ - ιħ∂/∂θ 𝐫² = 𝐫² ιħ∂/∂θ -  𝐫² ιħ∂/∂θ = 0 = [L̂𝓏,𝐫̂²] = 0.
+
+    [L̂𝓏,𝐫̂²] = 0, so these quantities commute.

solutions/chap7/prob5

@@ -0,0 +1,107 @@
+There is an angular momentum system with the state function
+
+    ❙Ψ❭ = 2͟ ❙1 1❭ + ι 3͟ ❙1 0❭ - 4͟ ❙1 -1❭ 
+         √29         √29       √29       
+
+In general the eigenvalue equation for the L̂𝓏 operator is
+
+    L̂𝓏❙l m❭ = m ħ❙l m❭, where m ħ are the possible measurements.
+
+The possible measurements of this system, then, are, for m = {-1, 0, 1}:
+
+    -ħ, 0, ħ.
+
+The probability for is given by
+
+    │❬1 m′❙Ψ❭│², with m′ = {-1, 0, 1}.
+
+The eigenstates form an orthogonal set such that 
+
+    ❬l′ m′❙l m❭ = δₗₗ′ δₘₘ′.
+
+Then,
+
+    ❬1 1❙Ψ❭ = ❬1 1❙⎛ 2͟ ❙1 1❭ + ι 3͟ ❙1 0❭ - 4͟ ❙1 -1❭ ⎞
+                   ⎝√29         √29       √29       ⎠
+
+            = ❬1 1❙ 2͟ ❙1 1❭ = 2͟ .
+                   √29       √29
+
+(𝐚)
+    │❬1 1❙Ψ❭│² = 4͟ = ⁴/₂₉.
+                 29
+
+Similarly,
+
+    │❬1 0❙Ψ❭│² = 9͟ = ⁹/₂₉ and
+                 29
+
+    │❬1 -1❙Ψ❭│² = 1͟6͟ = ¹⁶/₂₉.
+                  29
+
+
+The eigenvalue equations for the L̂𝓏 operator are simplified because L̂𝓏 is diagonal in the z basis. The L̂𝓍 operator produces the same measurements, but the matrix representation of the L̂𝓍 operator must be applied. It is
+
+    L̂𝓍 ≐
+        ħ͟  ⎛ 0 1 0 ⎞
+        √2 ⎜ 1 0 1 ⎟
+           ⎝ 0 1 0 ⎠.
+
+Applying the operator to the states in Ψ,
+
+    L̂𝓍❙1 1❭ ≐ 
+                ħ͟  ⎛ 0 1 0 ⎞⎛1⎞ = ħ͟ ⎛0⎞ =  ħ͟ ❙1 0❭.
+               √2  ⎜ 1 0 1 ⎟⎜0⎟  √2 ⎜1⎟   √2 
+                   ⎝ 0 1 0 ⎠⎝0⎠     ⎝0⎠ 
+
+    L̂𝓍❙1 0❭ ≐
+                ħ͟  ⎛ 0 1 0 ⎞⎛0⎞ = ħ͟ ⎛1⎞ =  ħ͟ (❙1 1❭ + ❙1 -1❭), and
+               √2  ⎜ 1 0 1 ⎟⎜1⎟  √2 ⎜0⎟   √2 
+                   ⎝ 0 1 0 ⎠⎝0⎠     ⎝1⎠
+
+    L̂𝓍❙1 -1❭ ≐
+                ħ͟  ⎛ 0 1 0 ⎞⎛0⎞ = ħ͟ ⎛0⎞ =  ħ͟ ❙1 0❭.
+               √2  ⎜ 1 0 1 ⎟⎜0⎟  √2 ⎜1⎟   √2 
+                   ⎝ 0 1 0 ⎠⎝1⎠     ⎝0⎠
+
+
+    L̂𝓍❙Ψ❭ =  ⎛  2͟ L̂𝓍❙1 1❭ + ι 3͟ L̂𝓍❙1 0❭ - 4͟ L̂𝓍❙1 -1❭ ⎞
+             ⎝ √29           √29        √29         ⎠
+
+     2͟ L̂𝓍❙1 1❭ =  2͟ ħ❙1 0❭, 
+    √29          √58    
+
+    ι 3͟ L̂𝓍❙1 0❭ = ι 3͟ ħ (❙1 1❭ + ❙1 -1❭), and
+     √29           √58
+
+     4͟ L̂𝓍❙1 -1❭ =  4͟ ħ❙1 0❭.
+    √29           √58
+
+Then,
+
+    L̂𝓍❙Ψ❭ =  ħ⎛ -2͟  ❙1 0❭ + ι 3͟ (❙1 1❭ + ❙1 -1❭)⎞
+              ⎝ √58         √58                 ⎠
+
+Normalizing the function,
+
+    C⎛⎛-2͟ ⎞² + ⎛ι 3͟ ⎞² + ⎛ι 3͟ ⎞²⎞ = 1.
+     ⎝⎝√58⎠    ⎝ √58⎠    ⎝ √58⎠ ⎠
+
+
+ STOPPED HERE 
+ 
+    C = 58⎛⎛1͟⎞ - ⎛ι 3 ⎞⁻² + ⎛ι 3 ⎞⁻²⎞
+          ⎝⎝4⎠   ⎝    ⎠     ⎝    ⎠  ⎠
+So,
+
+    \|❬1 1❙L̂𝓍❙Ψ❭\|^2 = 
+
+
+
+
+
+    L̂𝓍❙Ψ❭ =  ⎛  + ι 3͟ L̂𝓍❙1 0❭ - 4͟ L̂𝓍❙1 -1❭ ⎞
+             ⎝     √29        √29         ⎠             
+
+
+