working on chap 7 hw #1

HW

@@ -8,3 +8,11 @@ Chap 9: 7, 11, 12, 13, 14
 
 Chap 7: 5, 7, 8, one from last class, 11
     due friday 3-18, the class one is: show L̂𝓏 and p² commute
+
+Chap 7: 10, 13, 18, 23, 30 due april 1
+
+
+
+Exam 2: harmonic oscillator 1d, harmonic oscillator 3d, square well, angular momentum
+
+

concepts

@@ -0,0 +1,22 @@
+the meaning of an operator - how it relates to an action that is a measurement and also an action on the state
+
+the meaning of an eigenstate and an eigenvalue for an operator
+
+the concept of a complete set of orthogonal operators that commute
+    - this says something about having common states
+
+    superposition states
+
+    unitary operators:
+
+    ❙Ψ❭ = ∑ ❙n❭❬n❙Ψ❭ = ∑Cₙ❙n❭
+
+    ❬n❙Ψ❭ = ∫ dx ❬n❙x❭❬x❙Ψ❭
+
+          = ∫ dx n†(x) Ψ(x)
+
+time evolution
+
+schrodinger equation - specific cases: infinite or finite well, harmonic oscillator, hydrogen atom
+
+angular momentum operators

lecture_notes/2-29/overview

@@ -0,0 +1,58 @@
+Problems with known polarization
+    
+    (pic) Uniformly polarized sphere
+
+    Gauss' Law in Dielectric
+
+        ε₀∇⋅𝐄 = ρ = ρfree + ρbound
+                = ρfree - ∇⋅𝐏
+
+        ε₀∇⋅𝐄 + ∇⋅𝐏 = ρfree
+
+        ∇⋅(ε₀𝐄 + 𝐏) = ρfree/ε₀
+
+        𝐃 ≡ ε₀𝐄 + 𝐏
+
+    Bar Electret
+
+        - Analogous to Magnetic bar
+
+        Two limits and an intermediate case.
+
+        ∇×𝐃 = ∇×(ε₀𝐄 + 𝐏) = ∇×𝐏
+
+        𝐄 for bar electret, properties:
+
+            (𝐄₂ - 𝐄₁)⋅𝐧̂₁→₂ = σ/ε₀ with σ = σᵦ + σfree
+
+            Curl eq shows 𝐄"₂ - 𝐄"₁ = 0 (tangential components).
+
+            (𝐃₂ - 𝐃₁)⋅𝐧̂₁→₂ = σfree
+
+            (𝐃"₂ - 𝐃"₁) = 𝐏"₂ - 𝐏"₁
+
+            𝐏 = ε₀χₑ𝐄
+
+            • χₑ is the susceptibility of a linear dielectric material.
+
+            𝐃 = ε₀𝐄 + 𝐏 = ε₀𝐄 + ε₀χₑ𝐄 = ε₀(1+χₑ) 𝐄
+
+            • ε₀(1+χₑ) ≡ permittivity
+            • 1+χₑ = dielectric constant
+
+    Metal Sphere
+
+        (pic) What is V(0)?
+
+        (pic) ε becomes relevant in electric field
+
+
+        𝐄 = ⎧ 𝐃/ε = Q𝐫̂/(4πεr²), R<r<b
+            ⎨ 
+            ⎩ 𝐃/ε = Q𝐫̂/(4πε₀r²), b<r 
+
+                        R       b        ∞
+        V(∞) - V(0) = -∫ 𝐄⋅d𝐥 - ∫ 𝐄⋅d𝐥 - ∫ 𝐄⋅d𝐥
+                        0       R        b
+
+                    =   0 - ...

lecture_notes/3-25/overview

@@ -0,0 +1,65 @@
+Bound states of a central potential
+━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
+
+
+For any central potential V(r) = V(│r│) the eigenfunctions of H can be separated as
+
+    ❬r❙E,l,mₗ❭ = Rₑ﹐ₗ(r) Yₗ﹐ₘ(θ,ϕ)
+
+The radial S.E. is 
+
+    ⎡−͟ħ͟² ⎛ ∂͟²͟  +  2͟∂͟  ⎞ + l͟(l͟+͟1͟)ħ͟² + V(│r│) ⎤ Rₑ﹐ₗ(r) = E Rₑ﹐ₗ(r)
+    ⎣2m  ⎝ ∂r²    r∂r ⎠    2 m r²           ⎦
+
+
+    Rₑ﹐ₗ(r) = U͟ₑ͟﹐͟ₗ͟(r)
+            r
+
+            (pic) ...
+
+(pic) Developed radial schrodinger equation  using U(r) replacement
+
+    - Developed normalization condition
+
+If V(r) is not more singular at the origin than 1/r^2 then the SE has power series solutions.
+
+    Thus for small r we take U(r) → rˢ
+
+(pic) substitute U(r) = rˢ into S.E.
+
+    -ħ²/2m [(s(s-1) + l(l+1)] + V r² = E r²
+    
+    For r → 0
+        r² → 0
+        V(r) r² → 0
+    
+    ⇒ s(s-1) + l(l+1) = 0
+        ⇒ s = l+1 or s = -l
+
+    If s = -l, the normalization conditions
+                                        
+         ∞                             │∞ 
+        ∫ r⁻²ˡ dr = 1/(2l-1) 1/(r²ˡ⁻¹) │  → diverges
+         0                             │0
+
+
+    Uₑ﹐ₗ(r) → (r→0) → rˡ⁺¹
+
+    Rₑ﹐ₗ(r) → (r→0) → rˡ
+
+
+The Hydrogen Atom
+━━━━━━━━━━━━━━━━━
+
+    V(r) = -e²/r
+
+    For a hydrogenic ion with nuclear charge Z
+
+    V(r) = -Ze²/r
+
+    Eigenfunctions:
+
+        Ψₑ﹐ₗ﹐ₘ(r,θ,φ) = Rₑ﹐ₗ(r) Yₗ﹐ₘ(θ,φ) = Uₑ﹐ₗ/r Yₗ﹐ₘ(θ,φ)
+
+
+

solutions/chap7/prob5

@@ -47,6 +47,61 @@ The eigenvalue equations for the L̂𝓏 operator are simplified because L̂𝓏
         √2 ⎜ 1 0 1 ⎟
            ⎝ 0 1 0 ⎠.
 
+The general eigenvalue equation is 
+
+    L̂𝓍❙λ,mₗ❭ = λ❙λ,mₗ❭, where the eigenvalues λ are the possible measured values of L̂𝓍. The eigenvalues can be obtained from the secular equation
+
+    det│L̂𝓍 - λ𝕀│ = 0
+
+    ħ͟  ⎛ 0 1 0 ⎞ - ⎛ λ 0 0 ⎞ = ⎛   -λ  ħ/√2   0  ⎞        
+    √2 ⎜ 1 0 1 ⎟   ⎜ 0 λ 0 ⎟   ⎜  ħ/√2  -λ  ħ/√2 ⎟     
+       ⎝ 0 1 0 ⎠   ⎝ 0 0 λ ⎠   ⎝    0  ħ/√2  -λ  ⎠.    
+
+    │⎛   -λ  ħ/√2   0  ⎞│ = (-λ(λ² - ħ²/2) + (ħ²/2) λ) = -λ³ + ħ²λ.
+    │⎜  ħ/√2  -λ  ħ/√2 ⎟│                                   
+    │⎝    0  ħ/√2  -λ  ⎠│
+
+    λ(-λ² + ħ²) = -λ(λ² - ħ²)) = 0.        
+
+One eigenvalue is immediately obvious: λ = 0. The other two are given by
+
+    λ² = ħ², so the eigenvalues are
+    λ = 0,±ħ.
+
+These are exactly the expected measured values for a spin component.
+
+The eigenvalue equations are 
+
+    L̂𝓍❙1  1❭𝓍 =  ħ❙1  1❭𝓍 ,
+    L̂𝓍❙1  0❭𝓍 =  0❙1  1❭𝓍 , and
+    L̂𝓍❙1 -1❭𝓍 = -ħ❙1 -1❭𝓍 .
+
+Matrix analysis can be used to find the eigenvectors for these eigenstates. The first one is
+
+    ħ͟  ⎛ 0 1 0 ⎞ ⎛ a ⎞ = ħ ⎛ a ⎞, which gives the system
+    √2 ⎜ 1 0 1 ⎟ ⎜ b ⎟     ⎜ b ⎟
+       ⎝ 0 1 0 ⎠ ⎝ c ⎠     ⎝ c ⎠
+
+⎧ b = √2 a    
+⎨ (a + c) = √2 b
+⎩ b = √2 c
+     
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
 Applying the operator to the states in Ψ,
 
     L̂𝓍❙1 1❭ ≐ 
@@ -87,14 +142,20 @@ Normalizing the function,
     C⎛⎛-2͟ ⎞² + ⎛ι 3͟ ⎞² + ⎛ι 3͟ ⎞²⎞ = 1.
      ⎝⎝√58⎠    ⎝ √58⎠    ⎝ √58⎠ ⎠
 
+    C⎛4 - 9 - 9⎞ = 58 = C(14).
+     ⎝         ⎠
 
- STOPPED HERE 
+    C = 58/14 = 29/7.
 
-    C = 58⎛⎛1͟⎞ - ⎛ι 3 ⎞⁻² + ⎛ι 3 ⎞⁻²⎞
+The probability of measuring one of the possible angular momenta will be given by 
-          ⎝⎝4⎠   ⎝    ⎠     ⎝    ⎠  ⎠
-So,
 
-    \|❬1 1❙L̂𝓍❙Ψ❭\|^2 = 
+    │❬l m❙L̂𝓍❙Ψ❭│².
+
+    ❬1 1❙L̂𝓍❙Ψ❭ =  ❬1 1❙2͟9͟⎛ -2͟  ❙1 0❭ + ι 3͟ (❙1 1❭ + ❙1 -1❭)⎞
+                       7 ⎝ √58         √58                 ⎠
+
+        = 2͟9͟ ι͟3͟ ❬1 1❙1 1❭ = ι 8͟7͟ 
+           7 √58             7√58
 
 
 

solutions/chap7/prob8

@@ -6,35 +6,130 @@
 ⎪
 ⎩    z = r cosθ
 
+Some differential forms may come in handy.
+
+∂/∂θ:
+⎧    ∂x = r cosϕ cosθ ∂θ
+⎪
+⎨    ∂y = r sinϕ cosθ ∂θ
+⎪
+⎩    ∂z = - r sinθ ∂θ
+
+∂/∂ϕ:
+⎧    ∂x = - r sinθ sinϕ ∂ϕ
+⎪
+⎨    ∂y = r sinθ cosϕ ∂ϕ
+⎪
+⎩    ∂z = 0 ∂ϕ
+
 
 7.47 is the set of algebraic conditions expressed by the vector definition 𝐋 = 𝐫 × 𝐩.
 
 ⎧    L̂𝓍 = yp𝓏 - zp𝓎 = -ιħ (y ∂͟_  - z ∂͟_ )
-⎪                           ∂y      ∂y
+⎪                           ∂z      ∂y
 ⎪
 ⎨    L̂𝓎 = zp𝓍 - xp𝓏 = -ιħ (z ∂͟_  - x ∂͟_ )
-⎪                           ∂y      ∂y
+⎪                           ∂x      ∂z
 ⎪
 ⎪    L̂𝓏 = xp𝓎 - yp𝓍 = -ιħ (x ∂͟_  - y ∂͟_ )
-⎩                           ∂y      ∂y
+⎩                           ∂y      ∂x
 
-Substituing 7.35 into 7.47,
+Substituting 7.35 into 7.47,
 
 ⎧    L̂𝓍 = -ιħ (r sinθ sinϕ ∂͟_  - r cosθ ∂͟_ )
-⎪                          ∂y           ∂y
+⎪                          ∂z           ∂y
 ⎪
 ⎨    L̂𝓎 = -ιħ (r cosθ ∂͟_  - r sinθ cosϕ ∂͟_ )
-⎪                     ∂y                ∂y
+⎪                     ∂x                ∂z
 ⎪
 ⎪    L̂𝓏 = -ιħ (r sinθ cosϕ ∂͟_ - r sinθ sinϕ ∂͟_ )
-⎩                          ∂y               ∂y
+⎩                          ∂y               ∂x
 
 
-⎧    L̂𝓍 = -ιħ (r sinθ sinϕ ∂͟_  - r cosθ ∂͟_ )
+
-⎪                          ∂y           ∂y
+Geometry is shown on the attached notes page.
+    
+For L̂𝓍:
+
+    ∂͟z͟ = -r sinθ
+    ∂θ
+    ∂͟y͟ = r sinθ cosϕ 
+    ∂ϕ
+
+    L̂𝓍 = -ιħ ( r sinθ sinϕ ∂͟θ͟ ∂͟_  - r cosθ ∂͟ϕ͟ ∂͟_ )
+                           ∂θ ∂z           ∂ϕ ∂y    
+
+    L̂𝓍 = ιħ ( -r sinθ sinϕ ∂͟θ͟ ∂͟_  + r cosθ ∂͟ϕ͟ ∂͟_ )
+                           ∂z ∂θ           ∂y ∂ϕ
+
+    L̂𝓍 = ιħ ( −͟r͟ s͟i͟n͟θ͟ sinϕ ∂͟_ +   c͟o͟s͟θ͟    ∂͟_ )
+              -r sinθ      ∂θ  sinθ cosϕ  ∂ϕ
+
+    L̂𝓍 = ιħ ( sinϕ ∂͟_ + c͟o͟t͟θ͟ ∂͟_ )
+                   ∂θ   cosϕ ∂ϕ                   
+
+For L̂𝓎:
+
+    ∂x = r cosϕ cosθ ∂θ
+    ∂z = 0 ∂ϕ
+
+    L̂𝓎 = -ιħ (r cosθ ∂͟_  - r sinθ cosϕ ∂͟_ )
+                     ∂x                ∂z
+
+    L̂𝓎 = ιħ (-r cosθ ∂͟θ͟ ∂͟_  + r sinθ cosϕ ∂͟ϕ͟ ∂͟_ )
+                     ∂θ ∂x                ∂ϕ ∂z
+
+    L̂𝓎 = ιħ (-r cosθ ∂͟θ͟ ∂͟_  + r sinθ cosϕ ∂͟ϕ͟ ∂͟_ )
+                     ∂x ∂θ                ∂z ∂ϕ
+
+    L̂𝓎 = ιħ (- _͟1͟  cosθ ∂͟_  + r sinθ cosϕ 0 ∂͟_ )
+              cosϕ cosθ ∂θ                  ∂ϕ
+
+    L̂𝓎 = ιħ (- _͟1͟  ∂͟_ )
+              cosϕ ∂θ
+
+    L̂𝓎 = -ιħ _͟1͟  ∂͟_ 
+            cosϕ ∂θ
+
+For L̂𝓏:
+
+    ∂x = -r sinθ sinϕ ∂ϕ
+    ∂y = r sinθ cosϕ ∂ϕ
+
+    L̂𝓏 = -ιħ (r sinθ cosϕ ∂͟͟ϕ͟ ∂͟_ - r sinθ sinϕ ∂͟͟ϕ͟ ∂͟_ )
+                          ∂ϕ ∂y               ∂ϕ ∂x
+
+    L̂𝓏 = -ιħ (r͟ s͟i͟n͟θ͟ c͟o͟s͟ϕ͟ ∂͟_ + r͟ s͟i͟n͟θ͟ s͟i͟n͟ϕ͟ ∂͟_ )
+              r sinθ cosϕ ∂ϕ   r sinθ sinϕ ∂ϕ 
+
+    L̂𝓏 = -ιħ ∂͟_ ( 1 + 1 )
+             ∂ϕ       
+
+    L̂𝓏 = -2ιħ ∂͟_ 
+              ∂ϕ
+
+So, according to my calculus, the final solutions should be the set
+
+⎧    L̂𝓍 = ιħ ( sinϕ ∂͟_ + c͟o͟t͟θ͟ ∂͟_ )
+⎪                   ∂θ   cosϕ ∂ϕ          
 ⎪
-⎨    L̂𝓎 = ιħ (r cosθ ∂͟_ + r sinθ cosϕ ∂͟_ )
+⎨    L̂𝓎 = -ιħ _͟1͟  ∂͟_ 
-⎪                    ∂y               ∂y
+⎪            cosϕ ∂θ
 ⎪
-⎪    L̂𝓏 = -ιħ (r sinθ cosϕ ∂͟_ - r sinθ sinϕ ∂͟_ )
+⎪    L̂𝓏 = -2ιħ ∂͟_ 
-⎩                          ∂y               ∂y
+⎩              ∂ϕ
+   
+
+The spherical representation, i.e. ending place, is the set
+
+⎧    L̂𝓍 = ιħ (sinϕ ∂͟_  + cosϕ cotθ ∂͟_ )
+⎪                  ∂θ              ∂ϕ
+⎪
+⎨    L̂𝓎 = ιħ (-cosϕ ∂͟_ + sinϕ cotθ ∂͟_ )
+⎪                   ∂θ             ∂ϕ
+⎪
+⎪    L̂𝓏 = -ιħ ∂͟_ 
+⎩             ∂ϕ
+
+
+My set DOES NOT match this. I must be going about this the wrong way. I have to give it more thought. Perhaps a purely geometric approach will improve my answers: I'll try that over the weekend.