still exam 1

solutions/exam1/prob3

@@ -24,7 +24,7 @@ and because H is a hermitian operator, 〈a|b〉 = 〈b|a〉, so
 
     〈a|Ĥ|b〉 = δ (〈a|b〉 + 〈a|b〉) = δ 2〈a|b〉;
 
-because of the hermition property, 
+because of the Hermitian property, 
     〈b|Ĥ|a〉 = 〈a|Ĥ|b〉 = δ 2〈a|b〉; 
 
 finally,
@@ -60,13 +60,27 @@ This gives the equation α₁ + α₁|〈a|b〉|² + 2β₁〈a|b〉 = E₁α₁
     β₁ = ͟α͟₁͟(͟E͟₁͟ ͟-͟ ͟1͟ ͟-͟ ͟|͟〈͟a͟|͟b͟〉͟|͟²͟)͟
              2〈a|b〉
 
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 Using the normalization condition, the values of each constant can be obtained. Plugging the value for α₁ into the equation reveals a quadratic equation.
 
 
 
     |α₁|² + |β₁|² = 1, so
 
-    |α₁|² + | ͟α͟₁͟(͟E͟₁͟ ͟-͟ ͟1͟ ͟-͟ ͟|͟〈͟a͟|͟b͟〉͟|͟²͟)͟ |² = 1.
+    |α₁|² + | ͟α͟₁͟(͟E͟₁͟ ͟-͟ ͟1͟ ͟-͟ ͟|͟〈a͟|͟b͟〉͟|͟²͟)͟ |² = 1.
             |     2〈a|b〉            |
 
 
@@ -77,7 +91,7 @@ Using the normalization condition, the values of each constant can be obtained.
            _________
     α₁ = ±√1 - |β₁|², so
       _________                         _________
-    ±√̅1 - |β₁|² (1 - E₁) + 2β₁〈a|b〉 + ±√1 - |β₁|² |〈a|b〉|² = 0.
+    ±√1 - |β₁|² (1 - E₁) + 2β₁〈a|b〉 + ±√1 - |β₁|² |〈a|b〉|² = 0.
 
     The quadratic formula therefore says that 
 
@@ -86,4 +100,4 @@ Using the normalization condition, the values of each constant can be obtained.
 
 
                                              
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+                                               

solutions/exam1/prob4

@@ -1,9 +1,9 @@
 Two operators' matrix representations are known in the |1〉, |2〉, |3〉 basis, where and b are real numbers:
 
-A ≐                   B ≐
-	⎛ a  0  0 ⎞           ⎛ b   0   0 ⎞
-	⎜ 0 -a  0 ⎟           ⎜ 0   0 -ιb ⎟
-	⎝ 0  0 -a ⎠  and      ⎝ 0  ιb   0 ⎠.
+Â ≐                   B̂ ≐
+    ⎛ a  0  0 ⎞           ⎛ b   0   0 ⎞
+    ⎜ 0 -a  0 ⎟           ⎜ 0   0 -ιb ⎟
+    ⎝ 0  0 -a ⎠  and      ⎝ 0  ιb   0 ⎠.
 
 B exhibits a degenerate spectrum when it has repeated eigenvalues. The eigenvalues of B are obtained from its characteristic equation.
  
@@ -27,21 +27,25 @@ 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  -ιab    0 ⎠   ⎝ 0  -ιab    0 ⎠
+    ⎛ ab    0    0 ⎞   ⎛ ab    0    0 ⎞    
+    ⎜ 0     0  ιab ⎟ - ⎜ 0     0  ιab ⎟ = 0.
+    ⎝ 0  -ιab    0 ⎠   ⎝ 0  -ιab    0 ⎠
 
 (b) Therefore, these operators commute.
 
-Since the operators commute, they share common eigenstates. Therefore, if the eigenstates for one operator can be determined, they are determined for both operators. Since the problem states that a new set of orthonormal kets need be determined, and the given set are the eigenstates related to the operator A, then the eigenstates of the operator B should be determined.
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+Since the operators commute, they share a set of common eigenstates. The eigenstates of  are apparent from inspection:
+
+|a₁〉 ≐ ⎛1⎞  |a₂〉 ≐ ⎛0⎞  and |a₃〉 ≐ ⎛0⎞     
+       ⎜0⎟         ⎜1⎟             ⎜0⎟
+       ⎝0⎠,        ⎝0⎠,            ⎝1⎠.
 
 The eigenvalues are already known (λ = b,b,-b.), and using the eigenvalue equations, the eigenstates can be determined. The eigenvalue equation
 
-	⎛ b   0   0 ⎞ ⎛ α₁ ⎞   ⎛    b α₁ ⎞       ⎛ α₁ ⎞
-	⎜ 0   0 -ιb ⎟ ⎜ β₁ ⎟ = ⎜ -ι b γ₁ ⎟ =  b  ⎜ β₁ ⎟
-	⎝ 0  ιb   0 ⎠ ⎝ γ₁ ⎠   ⎝  ι b β₁ ⎠       ⎝ γ₁ ⎠ reveals
+    ⎛ b   0   0 ⎞ ⎛ α₁ ⎞   ⎛    b α₁ ⎞       ⎛ α₁ ⎞
+    ⎜ 0   0 -ιb ⎟ ⎜ β₁ ⎟ = ⎜ -ι b γ₁ ⎟ =  b  ⎜ β₁ ⎟
+    ⎝ 0  ιb   0 ⎠ ⎝ γ₁ ⎠   ⎝  ι b β₁ ⎠       ⎝ γ₁ ⎠
 
-	-ι γ₁ = β₁, and
-	ι b β₁ = γ₁
+reveals -ι γ₁ = β₁, which, combined with the normalization condition, will allow the determination of two eigenstates of the B̂.