added angular momentum operators and hydrogen atom basics

HW

@@ -0,0 +1,10 @@
+Quantum Mechanics Assignments
+
+Chap 2: 2, 3, 14, 15, 17, 22, 23
+
+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

lecture_notes/3-14/3d eigenstates

@@ -55,7 +55,4 @@ Understanding a System:
 
         With the eigenvalue equation
 
-            ❬r❙E,l,mₗ❭ = E❬r❙E,l,mₗ❭
-
-            
-
+            ❬r❙E,l,mₗ❭ = E❬r❙E,l,mₗ❭

lecture_notes/3-16/Angular Momentum Commutator

@@ -1,4 +0,0 @@
-[L̂𝓏,p̂𝓏] = ιħp̂𝓎
-[L̂𝓏,p̂𝓍] = -ιħp̂𝓎
-[L̂𝓏,p̂𝓏] = 0
-[L̂𝓏,p̂²] = 

lecture_notes/3-16/hydrogen atom

@@ -0,0 +1,20 @@
+V = E͟_͟Z͟ (?)
+     r
+❬r❙E,l,mₗ❭ ≝ R(𝐫) ≝ U͟(𝐫͟)
+                     r
+
+∂²/∂r² (u/r) + 2/r ∂/∂r U/r 
+    = ∂/∂r (1/r (∂/∂r u) - u/r²) + 2/r ((1/r ∂/∂r u) - u/r²)
+
+    = 1/r (∂²/∂r² u) - 2/r^2 ∂/∂r u + 2 u/r³ + 2/r² ∂/∂r u - 2u/r³
+
+    = 1/r (∂²/∂r² u)
+
+
+So,
+
+    (-ħ/2m ∂²/∂r² + Veff(│r│)) = V(│r│) + l͟(l͟+͟1͟)ħ͟²͟
+                                              2mr²
+                                          |   ↓   |
+                                         Centrifugal
+                                           barrier

lecture_notes/3-18/overview

@@ -0,0 +1,43 @@
+Vibrations and rotations of a diatomic molecule
+
+    atoms vibrate about an equilibrium position r₀
+
+    ⊙~~~~~~~~~⊙
+    |←   r₀  →|
+
+    (pic MISSEd) potential, with taylor approximation
+
+    (pic MISSEd) A taylor series solution is appropriate to solve this diffEQ.
+
+    Experiments indicate vibration (E~visible light) has more energy than rotation (E~infrared), so vibrations happen much faster.
+
+The Rotational eigen-value spectrum
+
+    L²,L̂𝓍,L̂𝓎,L̂𝓏 (Hermition operators)
+
+    Say
+        L̂𝓏❙Ψ❭ = mₗħ❙Ψ❭
+
+        (pic) mₗ is bound by λ: m has value from -λ to λ
+
+
+Introduced raising and lower operators
+
+    L̂± = L̂𝓍 ± ιL̂𝓎 = (pic) proof = ±ħL±
+
+    L̂𝓏(L̂ ± ❙λ,mₗ❭) => (pic) proof => L̂± ❙λ,mₗ❭ = ❙λ,mₗ±1❭
+
+    Very important point:
+
+        Let m=l be the maximum value of m
+        L̂₊❙λ,l❭ = 0 (required because wave function goes to 0 in a forbiddin region)
+        This is also true for L̂₋❙λ,l′❭ with l′ the minimum value of m
+
+    (pic) L̂₋L̂₊❙λ,l❭ = ...
+            gives λ = l(l+1)
+
+    (pic) L̂₊L̂₋❙λ,l′❭ = ... 
+            gives λ = l′(l′+1)
+
+    These show that m = -l, -l+1, 0, l
+

solutions/chap7/prob7

@@ -0,0 +1,88 @@
+Angular momentum system is prepared in the state
+
+    ❙Ψ❭ = 1͟  ❙11❭ - 2͟ ❙10❭ + ι͟2͟ ❙22❭ + ι1͟ ❙20❭
+         √10        √10      √10       √10 
+
+
+Possible results of 𝐋² measurement?
+
+The 𝐋̂² operator was developed in lecture and in the text. The related eigenvalue equation is
+
+    𝐋̂²❙lm❭ = l(l+1) ħ²❙lm❭
+
+The possible measurements of 𝐋̂² for this system are those associated with the initial state vector, i.e. the values lm = 11, 10, 22, 20.
+
+For lm = 11 & 10, 𝐋̂² = 2ħ²
+
+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.
+
+    ❬11❙Ψ❭ = ❬11❙⎛  1͟  ❙11❭ - 2͟ ❙10❭ + ι͟2͟ ❙22❭ + ι1͟ ❙20❭ ⎞   
+                 ⎝ √10        √10      √10       √10     ⎠.
+
+The eigenstates for this system are LaPlace's spherical harmonic functions, which comprise an orthogonal set, I.E.:
+
+    ❬lm❙l′m′❭ = δₗₗ′ δₘₘ′.
+
+    │❬11❙Ψ❭│² = │❬11❙  1͟  ❙11❭│² = ¹/₁₀.
+                │     √10     │    
+
+    │❬10❙Ψ❭│² = │❬10❙  2͟  ❙10❭│² = ⁴/₁₀.
+                │     √10     │    
+
+(𝐚)
+    𝓟(𝐋̂²=2ħ²) = ½.
+
+No need to calculate the other set: since the vector is normalized, the probability of measuring 𝐋̂²=6ħ² is also
+
+    𝓟(𝐋̂²=6ħ²) = ½.
+
+
+For 𝐋̂𝓏 the eigenvalue equation is
+
+    𝐋̂𝓏❙lm❭ = mħ❙lm❭,
+
+so the expected measurements are,
+
+for lm = 11 & 21, 𝐋̂𝓏 = ħ.
+
+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=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=0) = ⁸/₁₀.
+
+Histogram of probabilities:
+
+
+𝓟                                      𝓟
+   ╭─────────────────┬────────────────╮
+1  │                 │                │ 1 
+   │                 │                │   
+.8 │                 │           ▧    │ .8
+   │                 │           ▧    │   
+   │                 │           ▧    │   
+.5 │   ▧        ▧    │           ▧    │ .5
+   │   ▧        ▧    │           ▧    │    
+   │   ▧        ▧    │           ▧    │    
+   │   ▧        ▧    │   ▧       ▧    │    
+.1 │   ▧        ▧    │   ▧       ▧    │ .1
+   ╰─────────────────┼────────────────╯
+       2ħ²     6ħ²   │   0       ħ    
+            𝐋̂²       │       𝐋̂𝓏
+
+            h
+