finished most of chap 7 (a) homework

solutions/chap7/prob5

@@ -85,84 +85,41 @@ Matrix analysis can be used to find the eigenvectors for these eigenstates. The
 ⎧ b = √2 a    
 ⎨ (a + c) = √2 b
 ⎩ b = √2 c
-     
+
+Following this to conclusion just like with spin operators will provide the eigenstates, and then from that the wave function can be expressed using the x basis, and probabilities obtained.
 
 
+I need to stop here, but I will produce at least sthe histogram from part a:
 
+(𝐜)
 
+𝓟(L̂𝓏)     
 
-
+       ╭─────────────────────────╮
-
+       │                         │ 
-
+       │                         │ 
-
+       │                         │ 
-
+       │                         │ 
-
+       │                         │ 
-
+       │                         │ 
-
+       │                         │    
-
+ ¹⁶/₂₉ ├   ▓                     │    
-
+       │   ▓                     │    
-
+       │   ▓                     │   
-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 
+  ⁹/₂₉ ├   ▓        ▓            │   
-                   ⎝ 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 
+  ⁴/₂₉ ├   ▓        ▓        ▓   │   
-                   ⎝ 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 
+           -ħ       0       ħ    
-                   ⎝ 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⎠ ⎠
-
-    C⎛4 - 9 - 9⎞ = 58 = C(14).
-     ⎝         ⎠
-
-    C = 58/14 = 29/7.
-
-The probability of measuring one of the possible angular momenta will be given by 
-
-    │❬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
-
-
-
-
-
-    L̂𝓍❙Ψ❭ =  ⎛  + ι 3͟ L̂𝓍❙1 0❭ - 4͟ L̂𝓍❙1 -1❭ ⎞
-             ⎝     √29        √29         ⎠             
-
-
 

solutions/chap7/prob7

@@ -83,6 +83,3 @@ Histogram of probabilities:
    ╰─────────────────┼────────────────╯
        2ħ²     6ħ²       0       ħ    
             𝐋̂²               L̂𝓏
-
-            h
-

solutions/chap7/prob8

@@ -1,4 +1,4 @@
-7.35 is nothing more than a definition of spherical coordinates.
+The transformation from polar coordinates to cartesian is the set of equations
 
 ⎧    x = r sinθ cosϕ
 ⎪
@@ -6,21 +6,142 @@
 ⎪
 ⎩    z = r cosθ
 
+The transformation from cartesian to polar coordinates is the set
+
+⎧   r = √(x²+y²+z²)
+⎪
+⎪   cos(θ) = z͟ =     _͟z͟
+⎨            r   √(x²+y²+z²)
+⎪
+⎪   tan(ϕ) = y͟
+⎩            x
+
 Some differential forms may come in handy.
 
+⎧   dcos(θ) = -sin(θ) dθ
+⎨
+⎪   dtan(ϕ) =  _͟1͟   dϕ
+⎩            cos²(ϕ)
+
 ∂/∂θ:
-⎧    ∂x = r cosϕ cosθ ∂θ
+
+⎧   ∂͟x = r cosϕ cosθ 
+⎪   ∂θ
 ⎪
-⎨    ∂y = r sinϕ cosθ ∂θ
+⎨   ∂͟y = r sinϕ cosθ 
+⎪   ∂θ
 ⎪
-⎩    ∂z = - r sinθ ∂θ
+⎪   ∂͟z = -r sinθ 
+⎩   ∂θ
 
 ∂/∂ϕ:
-⎧    ∂x = - r sinθ sinϕ ∂ϕ
+
+⎧   ∂͟x = -r sinθ sinϕ 
+⎪   ∂ϕ
 ⎪
-⎨    ∂y = r sinθ cosϕ ∂ϕ
+⎨   ∂͟y = r sinθ cosϕ 
+⎪   ∂ϕ    
+⎪    
+⎪   ∂͟z = 0 
+⎩   ∂ϕ
+
+∂/∂x: 
+
+⎧   ∂͟r = √(x²+y²+z²) = x͟
+⎪   ∂x                 r
 ⎪
-⎩    ∂z = 0 ∂ϕ
+⎪   ∂͟cos(θ) = -z͟x͟ 
+⎨   ∂x         r³ 
+⎪
+⎪   ∂͟tan(ϕ) = -y͟
+⎩   ∂x         x²
+
+∂/∂y: 
+
+⎧   ∂͟r = √(x²+y²+z²) = y͟
+⎪   ∂y                 r
+⎪
+⎪   ∂͟cos(θ) = -z͟y͟ 
+⎨   ∂y         r³ 
+⎪
+⎪   ∂͟tan(ϕ) = 1͟
+⎩   ∂y        x
+
+∂/∂z: 
+
+⎧   ∂͟r = √(x²+y²+z²) = z͟
+⎪   ∂z                 r
+⎪
+⎪   ∂͟cos(θ) = z͟ = 1͟  ⎛r ∂͟z͟ - z ∂͟r͟⎞ = 1͟ - z͟²
+⎨   ∂z        r   r² ⎝  ∂z     ∂z⎠   r   r³
+⎪
+⎪   ∂͟tan(ϕ) = 0
+⎩   ∂z          
+
+
+These differential forms can be used to transform each cartesian differentiation operator. 
+
+    ∂͟ = ∂͟r͟ ∂͟ + ∂͟c͟o͟s͟θ͟ ∂͟     +  ∂͟t͟a͟n͟ϕ͟ ∂͟    ;
+    ∂x  ∂x ∂r  ∂x    ∂cosθ    ∂x    ∂tanϕ
+
+    ∂͟ = x͟ ∂͟ -  z͟x͟  _͟1͟   ∂͟  - y͟ cos²(ϕ) ∂͟ ;  
+    ∂x  r ∂r   r³ -sinθ ∂θ   x²        ∂ϕ
+
+    ∂͟ = sin(θ) cos(ϕ) ∂͟ 
+    ∂x                ∂r
+                + 1͟ sin(θ) cos(ϕ) cos(θ) _͟1͟    ∂͟  
+                  r                     sin(θ) ∂θ     
+                        - r͟  s͟i͟n͟(ϕ͟)  s͟i͟n͟(θ͟)  cos²(ϕ͟) ∂͟  ;  
+                          r² cos²(ϕ) sin²(θ)         ∂ϕ 
+
+    ∂͟ = sin(θ) cos(ϕ) ∂͟  + 1͟ cos(ϕ) cos(θ) ∂͟  - 1͟ s͟i͟n͟(ϕ͟) ∂͟  .
+    ∂x                ∂r   r               ∂θ   r sin(θ) ∂ϕ
+
+    ───────────────────────────────────────────────
+
+    ∂͟ = ∂͟r͟ ∂͟ + ∂͟c͟o͟s͟θ͟ ∂͟     +  ∂͟t͟a͟n͟ϕ͟ ∂͟    ;
+    ∂y  ∂y ∂r  ∂y    ∂cosθ    ∂y    ∂tanϕ
+
+    ∂͟ = y͟ ∂͟  - z͟y͟ ∂͟     + 1͟ ∂͟    ;
+    ∂y  r ∂r   r³ ∂cosθ   x ∂tanϕ
+
+    ∂͟ = sin(θ) sin(ϕ) ∂͟  - 1͟ sin(θ) cos(θ) s͟i͟n͟(ϕ͟) ∂͟  +  c͟o͟s͟(ϕ͟)  ∂͟  ;
+    ∂y                ∂r   r              -sin(θ) ∂θ   r sin(θ) ∂ϕ
+
+    ∂͟ = sin(θ) sin(ϕ) ∂͟  + 1͟ cos(θ) s͟i͟n͟(ϕ͟) ∂͟  +  c͟o͟s͟(ϕ͟)  ∂͟  .
+    ∂y                ∂r   r               ∂θ   r sin(θ) ∂ϕ
+
+    ───────────────────────────────────────────────
+
+    ∂͟ = ∂͟r͟ ∂͟ + ∂͟c͟o͟s͟θ͟ ∂͟     +  ∂͟t͟a͟n͟ϕ͟ ∂͟    ;
+    ∂z  ∂z ∂r  ∂z    ∂cosθ    ∂z    ∂tanϕ
+
+    ∂͟ = z͟ ∂͟ + ⎛1͟ - z͟²⎞∂͟     +  0 ;
+    ∂z  r ∂r  ⎝r   r³⎠∂cosθ    
+
+    ∂͟ = cos(θ) ∂͟ + 1͟⎛1 - cos²(θ)⎞   _͟1͟    ∂͟  ;
+    ∂z         ∂r  r⎝           ⎠ -sin(θ) ∂θ    
+
+    ∂͟ = cos(θ) ∂͟ -  _͟1͟_   ⎛1 - cos²(θ)⎞ ∂͟  ;
+    ∂z         ∂r  rsin(θ)⎝           ⎠ ∂θ    
+
+    ∂͟ = cos(θ) ∂͟ -  _͟1͟_    sin²(θ) ∂͟  ;
+    ∂z         ∂r  rsin(θ)         ∂θ    
+
+    ∂͟ = cos(θ) ∂͟ - s͟i͟n͟(θ͟) ∂͟  .
+    ∂z         ∂r    r    ∂θ    
+
+
+The set of differential operator transformations is
+
+⎧   ∂͟ = sin(θ) cos(ϕ) ∂͟  + 1͟ cos(ϕ) cos(θ) ∂͟  - 1͟ s͟i͟n͟(ϕ͟) ∂͟ 
+⎪   ∂x                ∂r   r               ∂θ   r sin(θ) ∂ϕ
+⎪
+⎨   ∂͟ = sin(θ) sin(ϕ) ∂͟  + 1͟ cos(θ) s͟i͟n͟(ϕ͟) ∂͟  +  c͟o͟s͟(ϕ͟)  ∂͟ 
+⎪   ∂y                ∂r   r               ∂θ   r sin(θ) ∂ϕ
+⎪
+⎪   ∂͟ = cos(θ) ∂͟ - s͟i͟n͟(θ͟) ∂͟ 
+⎩   ∂z         ∂r    r    ∂θ
 
 
 7.47 is the set of algebraic conditions expressed by the vector definition 𝐋 = 𝐫 × 𝐩.
@@ -45,91 +166,87 @@ Substituting 7.35 into 7.47,
 ⎪    L̂𝓏 = -ιħ (r sinθ cosϕ ∂͟_ - r sinθ sinϕ ∂͟_ )
 ⎩                          ∂y               ∂x
 
+Substituting the transformed differentiation operators,
 
-
+⎧   L̂𝓍 = -ιħ⎛r sinθ sinϕ ⎛cos(θ) ∂͟ - s͟i͟n͟(θ͟) ∂͟ ⎞ 
-Geometry is shown on the attached notes page.
+⎪           ⎝            ⎝       ∂r    r    ∂θ⎠ 
-    
+⎪               
-For L̂𝓍:
+⎪         - r cosθ ⎛sin(θ) sin(ϕ) ∂͟  + 1͟ cos(θ) s͟i͟n͟(ϕ͟) ∂͟  +  c͟o͟s͟(ϕ͟)  ∂͟ ⎞⎞
-
+⎪                  ⎝              ∂r   r               ∂θ   r sin(θ) ∂ϕ⎠⎠
-    ∂͟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 _͟1͟  ∂͟_ 
-⎪            cosϕ ∂θ
 ⎪
-⎪    L̂𝓏 = -2ιħ ∂͟_ 
+⎪   L̂𝓎 = -ιħ⎛r cosθ⎛sin(θ) cos(ϕ) ∂͟  + 1͟ cos(ϕ) cos(θ) ∂͟  - 1͟ s͟i͟n͟(ϕ͟) ∂͟ ⎞
-⎩              ∂ϕ
+⎪          ⎝      ⎝              ∂r   r               ∂θ   r sin(θ) ∂ϕ⎠
-   
+⎨           
+⎪               - r sinθ cosϕ ⎛cos(θ) ∂͟ - s͟i͟n͟(θ͟) ∂͟ ⎞⎞           
+⎪                             ⎝       ∂r    r    ∂θ⎠⎠
+⎪
+⎪
+⎪   L̂𝓏 = -ιħ⎛r sinθ cosϕ⎛sin(θ) sin(ϕ) ∂͟  + 1͟ cos(θ) s͟i͟n͟(ϕ͟) ∂͟  +  c͟o͟s͟(ϕ͟)  ∂͟ ⎞ 
+⎪          ⎝           ⎝              ∂r   r               ∂θ   r sin(θ) ∂ϕ⎠ 
+⎪
+⎪               - r sinθ sinϕ ⎛sin(θ) cos(ϕ) ∂͟  + 1͟ cos(ϕ) cos(θ) ∂͟  - 1͟ s͟i͟n͟(ϕ͟) ∂͟ ⎞⎞
+⎩                             ⎝              ∂r   r               ∂θ   r sin(θ) ∂ϕ⎠⎠
 
-The spherical representation, i.e. ending place, is the set
+Simplifying...
+
+⎧   L̂𝓍 = -ιħ⎛r sinθ sinϕ cosθ ∂͟ - sinϕ sin²θ ∂͟  
+⎪          ⎝                 ∂r             ∂θ 
+⎪               
+⎪         -  r cosθ sinθ sinϕ ∂͟  - cos²θ sinϕ ∂͟  - c͟o͟s͟θ͟ cosϕ ∂͟ ⎞
+⎪                             ∂r              ∂θ   sinθ      ∂ϕ⎠
+⎪
+⎪
+⎪   L̂𝓎 = -ιħ⎛r cosθ sinθ cosϕ ∂͟  + cosθ cosϕ cosθ ∂͟  - c͟o͟s͟θ͟ sinϕ ∂͟ 
+⎪          ⎝                 ∂r                  ∂θ   sinθ      ∂ϕ
+⎨           
+⎪               - r sinθ cosϕ cosθ ∂͟ + sinθ cosϕ sinθ ∂͟ ⎞           
+⎪                                  ∂r                 ∂θ⎠
+⎪
+⎪
+⎪    L̂𝓏 = -ιħ⎛r sin²θ cosϕ sinϕ ∂͟  + sinθ cosθ sinϕ cosϕ ∂͟  + cosϕ c͟o͟s͟ϕ͟  ∂͟  
+⎪           ⎝                  ∂r                       ∂θ              ∂ϕ 
+⎪
+⎪               - r sin²θ sinϕ cosϕ ∂͟  - sinθ cosθ sinϕ cosϕ  ∂͟  + sinθ sinϕ s͟i͟n͟ϕ͟ ∂͟ ⎞
+⎩                                   ∂r                        ∂θ             sinθ ∂ϕ⎠
+
+
+⎧   L̂𝓍 = -ιħ⎛- sinϕ sin²θ ∂͟  - cos²θ sinϕ ∂͟  - c͟o͟s͟θ͟ cosϕ ∂͟ ⎞
+⎪          ⎝             ∂θ              ∂θ   sinθ      ∂ϕ⎠
+⎪
+⎨   L̂𝓎 = -ιħ⎛cosθ cosϕ cosθ ∂͟  - c͟o͟s͟θ͟ sinϕ ∂͟  + sinθ cosϕ sinθ ∂͟ ⎞
+⎪          ⎝               ∂θ   sinθ      ∂ϕ                  ∂θ⎠
+⎪           
+⎪    L̂𝓏 = -ιħ⎛cosϕ c͟o͟s͟ϕ͟  ∂͟  + sinθ sinϕ s͟i͟n͟ϕ͟ ∂͟ ⎞
+⎩           ⎝           ∂ϕ             sinθ ∂ϕ⎠
+
+
+⎧   L̂𝓍 = ιħ⎛ sinϕ sin²θ ∂͟  + cos²θ sinϕ ∂͟  + c͟o͟s͟θ͟ cosϕ ∂͟ ⎞
+⎪         ⎝            ∂θ              ∂θ   sinθ      ∂ϕ⎠
+⎪
+⎨   L̂𝓎 = ιħ⎛-cos²θ cosϕ  ∂͟ - sin²θ cosϕ  ∂͟ + c͟o͟s͟θ͟ sinϕ ∂͟  ⎞
+⎪         ⎝             ∂θ              ∂θ  sinθ      ∂ϕ ⎠
+⎪           
+⎪    L̂𝓏 = -ιħ⎛cos²ϕ  ∂͟  + sin²ϕ ∂͟ ⎞
+⎩           ⎝       ∂ϕ         ∂ϕ⎠
+
+
+⎧   L̂𝓍 = ιħ⎛ sinϕ  ∂͟ + cotθ cosϕ ∂͟ ⎞
+⎪         ⎝       ∂θ            ∂ϕ⎠
+⎪
+⎨   L̂𝓎 = ιħ⎛-cosϕ ∂͟ + cotθ sinϕ ∂͟  ⎞
+⎪         ⎝      ∂θ            ∂ϕ ⎠
+⎪           
+⎪   L̂𝓏 = -ιħ∂͟ 
+⎩          ∂ϕ 
+
+The spherical representation is the following set, which perfectly matches the obtained result.
 
 ⎧    L̂𝓍 = ιħ (sinϕ ∂͟_  + cosϕ cotθ ∂͟_ )
-⎪                  ∂θ              ∂ϕ
+⎪                 ∂θ              ∂ϕ
 ⎪
 ⎨    L̂𝓎 = ιħ (-cosϕ ∂͟_ + sinϕ cotθ ∂͟_ )
-⎪                   ∂θ             ∂ϕ
+⎪                  ∂θ             ∂ϕ
 ⎪
 ⎪    L̂𝓏 = -ιħ ∂͟_ 
-⎩             ∂ϕ
+⎩            ∂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.