started on chap 5 hw, sent in exam1 #1 and #4

notes/particles in a box

@@ -0,0 +1,67 @@
+❙Ψ❭ ≐ Ψ(x)
+Ψ(x) = ❬x❙Ψ❭
+𝓟(x) = │Ψ(x)│²
+𝓟(x) = ⎮Ψ(x)⎮²
+
+            ⌠ ∞
+1 = ❬Ψ❙Ψ❭ = ⎮  │Ψ(x)│² dx = 1
+            ⌡-∞
+
+❙Ψ❭ → Ψ(x)
+❬Ψ❙ → Ψ⃰(x)
+
+Â → A(x)
+           ⌠b     
+𝓟(a<x<b) = ⎮ │Ψ(x)│² dx                  
+           ⌡a
+                
+                    │⌠∞              │²     
+𝓟(Eₙ) = │❬Eₙ❙Ψ❭│² = │⎮ Eₙ⃰(x) Ψ(x) dx │
+                    │⌡-∞             │
+
+x̂ = x
+
+p̂ = ι͟ ∂͟
+    ħ ∂x                         
+
+
+⎛- ħ͟²͟ d͟²͟ + V(x)⎞ φₙ(x) = E φₙ(x)
+⎝  2m dx²      ⎠ 
+
+Boundary conditions:
+
+    1) φₙ(x) is continuous.
+    2) d φₙ(x) is continuous unless V = ∞.
+       dx
+
+Infinite square potential energy well:
+
+    Eₙ = n͟²͟π͟²͟ħ͟²,    n = 1, 2, 3, ...
+          2mL²
+        
+    φₙ(x) = √⎛2͟⎞ sin⎛n͟π͟x͟⎞,  n = 1, 2, 3, ...
+             ⎝L⎠    ⎝ L ⎠ 
+
+
+Energy eigenstates obey the following properties:
+
+    Bra-ket Notation       Wavefunction Notation
+
+    Normalization
+
+                           ⌠∞       
+    ❬Eₙ❙Eₙ❭ = 1            ⎮ │φₙ(x)│² dx = 1
+                           ⌡-∞           
+
+    Orthogonality
+
+                           ⌠∞                    
+    ❬Eₙ❙Eₘ❭ = δₙₘ          ⎮ φₙ⃰(x) φₘ(x) dx = δₙₘ
+                           ⌡-∞                    
+
+    Completeness
+                           ⌠∞                    
+    ❬Eₙ❙Eₘ❭ = δₙₘ          ⎮ φₙ⃰(x) φₘ(x) dx = δₙₘ
+                           ⌡-∞                    
+
+

solutions/chap5/prob2

@@ -0,0 +1,36 @@
+A particle in an infinite square well has an initial state vector
+
+    ❙Ψ(t=0)❭ = A(❙φ₁❭ - ❙φ₂❭ + ι❙φ₃❭).
+
+where ❙φₙ❭ are the energy eigenstates. This also means
+
+    ❬Ψ(t=0)❙ = A⃰(❬φ₁❙ - ❬φ₂❙ + ι❬φ₃❙)
+
+
+
+
+    ❙Ψ(t=0)❭ =  _͟A͟  (αβ❙φ₁❭ - βγ❙φ₂❭ + αγι❙φ₃❭)
+               αβγ
+
+In the energy basis,
+
+    ❙φ₁❭ ≐ ⎛1⎞  ❙φ₂❭ ≐ ⎛0⎞  and ❙φ₃❭ ≐ ⎛0⎞
+           ⎜0⎟         ⎜1⎟             ⎜0⎟
+           ⎝0⎠,        ⎝0⎠,            ⎝1⎠.
+
+So, 
+    ❙Ψ(t=0)❭ ≐ ⎛ A ⎞
+               ⎜-A ⎟
+               ⎝ιA ⎠.
+
+(𝐚) Multiplying the state vector by its magnitude normalizes it.
+
+    ❙Ψ′(t=0)❭ ≐  __͟A͟__ ⎛ 1 ⎞ =  _͟1͟ ⎛ 1 ⎞    
+                √(3A²) ⎜-1 ⎟    √3 ⎜-1 ⎟.  
+                       ⎝ ι ⎠       ⎝ ι ⎠    
+
+
+
+
+
+

solutions/exam1/prob1

@@ -35,7 +35,7 @@ Similarly, S𝓍 (S𝓍 + ħ)(S𝓍 - ħ) ≐
 Performing the multiplication operation on the last two matrices returns the expression
 
   ͟ħ͟³͟   ⎛ 0 1 0 ⎞ ⎛ -1  0  1 ⎞
-   2√2 ⎜ 1 0 1 ⎟ ⎜  0  0  0 ⎟ = 𝟘.
+  2√2  ⎜ 1 0 1 ⎟ ⎜  0  0  0 ⎟ = 𝟘.
        ⎝ 0 1 0 ⎠ ⎝  1  0 -1 ⎠
 
-It is quite obvious that this operation returns 𝟘 since there are no components that will not match with a 0 throughout the multiplication of these matrices. Therefore, the second expression is also equivalent to the zero matrix 𝟘.
+Because each pair of row and column in this matrix has alternating 0s and ±1s, every multiplication operation will return 0. The second expression is therefore equivalent to the zero matrix 𝟘.

solutions/exam1/prob4

@@ -7,13 +7,13 @@ Â ≐                   B̂ ≐
 
 B exhibits a degenerate spectrum when it has repeated eigenvalues. The eigenvalues of B are obtained from its characteristic equation.
  
-|⎛ b-λ  0    0 ⎞|
+│⎛ b-λ  0    0 ⎞│
-|⎜ 0   -λ  -ιb ⎟| = 0, i.e.
+│⎜ 0   -λ  -ιb ⎟│ = 𝟘, i.e.
-|⎝ 0   ιb   -λ ⎠|
+│⎝ 0   ιb   -λ ⎠│
 
 (b - λ)(λ² + ι²b²) = (b - λ)(λ² - b²) = (b - λ)(b - λ)(b + λ) = 0.
 
-(a) The eigenvalues for this operator are therefore λ = b,b,-b. Since b appears twice, the operator exhibits a degenerate spectrum.
+(𝗮) The eigenvalues for this operator are therefore λ = b,b,-b. Since b appears twice, the operator exhibits a degenerate spectrum.
 
 To find if A and B commute, their commutator need be evaluated. They commute if the value is 0. The commutator of two operators is defined as
 
@@ -28,10 +28,10 @@ For the given operators, then, the commutator is
 which reduces to
 
     ⎛ ab    0    0 ⎞   ⎛ ab    0    0 ⎞    
-    ⎜ 0     0  ιab ⎟ - ⎜ 0     0  ιab ⎟ = 0.
+    ⎜ 0     0  ιab ⎟ - ⎜ 0     0  ιab ⎟ = 𝟘.
     ⎝ 0  -ιab    0 ⎠   ⎝ 0  -ιab    0 ⎠
 
-(b) Therefore, these operators commute.
+(𝗯) Therefore, these operators commute.
 
 
 Since the operators commute, they share a set of common eigenstates. The eigenstates of  are apparent from inspection:
@@ -48,14 +48,28 @@ For B̂, the eigenvalues are already known (λ = b,b,-b.), and using the eigenva
     ⎜ 0   0 -ιb ⎟ ⎜ β ⎟ = ⎜ -ι b γ ⎟ =  b  ⎜ β ⎟ 
     ⎝ 0  ιb   0 ⎠ ⎝ γ ⎠   ⎝  ι b β ⎠       ⎝ γ ⎠
 
-reveals -ι γ = β, which, combined with the normalization condition, will allow the determination of two eigenstates of the B̂. One eigenstate is obvious from inspection,
+dicates two possible eigenstates (for the eigenvalue b). One eigenstate is obvious from inspection:
 
 	|b₁〉 ≐ ⎛1⎞
 	       ⎜0⎟
 	       ⎝0⎠.
 
-If β = 1 is chosen, then γ = -ι, and if β = ι, then γ = -1. The two additional eigenstates of B̂ are therefore, after normalizing,
+The eigenvalue equation also reveals -ι γ = β. If β = 1 is chosen, then γ = -ι, revealing a second eigenstate, after normalizing:
 
-	|b₂〉 ≐ 1  ⎛ 0 ⎞     |b₃〉 ≐ 1  ⎛ 0  ⎞
+    |b₂〉 ≐ 1  ⎛ 0 ⎞
-           √2 ⎜ 1 ⎟            √2 ⎜ ι  ⎟
+           √2 ⎜ 1 ⎟
-              ⎝-ι ⎠ and           ⎝ -1 ⎠.
+              ⎝-ι ⎠.
+
+Similarly, when the eigenvalue -b is used, the eigenvalue equation reveals ι γ = β. So, if β = ι, γ = 1. The third eigenstate is therefore, after normalizing,
+
+    |b₃〉 ≐ 1  ⎛0⎞
+           √2 ⎜ι⎟
+              ⎝1⎠
+
+The complete set of eigenstates of the operator B̂ is
+
+    |b₁〉 ≐ ⎛1⎞  |b₂〉 ≐ 1  ⎛ 0 ⎞      |b₃〉 ≐ 1  ⎛0⎞
+           ⎜0⎟         √2 ⎜ 1 ⎟             √2 ⎜ι⎟
+           ⎝0⎠,           ⎝-ι ⎠, and           ⎝1⎠.
+
+(𝗰) These eigenstates can be expressed as linear combinations of the eigenstates of Â, so they are shared eigenstates between these operators, and so this basis is simultaneously a basis of  and B̂.