update

lecture_notes/1-13/Basis

@@ -11,4 +11,4 @@ Quantum State Basis
 		|+x> = C_+ |+z> + C_- |+z>
 		C_+ = e^(i δ_+) / √2 ,  C_- = e^(i δ_-) / √2
 
-		<+x| S_z |+x> = <S_z> = expectation value
+		<+x| S_z |+x> = <S_z> = expectation value

solutions/chap1/prob14/draft

@@ -0,0 +1,11 @@
+c₊²=.36
+
+
+       ___         ___
+|Ψ〉 = √.90 |+z〉 + √.10 |-z〉
+       ___         ___
+|Ψ〉 = √.20 |+y〉 + √.80 |-y〉 = √.20 ( )
+                ___         ___      ___         ___
+〈Ψ|Ψ〉 = ( 〈+y| √.20 + 〈-y| √.80 ) ( √.90 |+z〉 + √.10 |-z〉 )
+
+〈+x|Ψ〉 = d

solutions/chap4/draft

@@ -28,7 +28,7 @@ The Hamiltonian is measureable, and therefore is an operator. The magnetic field
 Therefore, the Hamiltonian
 
     Ĥ ≐ -k B𝓏 ħ/2  ( 1  0 )
-	          ( 0 -1 ) .
+                  ( 0 -1 ) .
 
 The commutator can now be computed. This computation is included on an attached sheet. The computation indicates that Ŝ𝓏Ĥ = ĤŜ𝓏, and therefore the commutator is zero. The Hamiltonian and the spin operator in the z direction commute. A similar computation for the x and y directions should indicate a lack of commutability.
 

solutions/exam1/prob1

@@ -0,0 +1,37 @@
+The characteristic equation for the spin operator S𝓏 is
+
+    S𝓏 (S𝓏 + ħ)(S𝓏 - ħ) = 0.
+
+The eigenvalues, which are the roots of this equation, are
+
+    λ = 0, ±ħ.
+
+The S𝓏 operator is already diagonalized in its own basis, so the matrix has the immediately constructable form of
+
+    S𝓏 ≐
+            ⎛ 1 0 0  ⎞
+          ħ ⎜ 0 0 0  ⎟
+            ⎝ 0 0 -1 ⎠
+
+
+To produce the operator S𝓏, one can apply the rotation matrix for a rotation about the y axis., where the angle of rotation is π/2.
+
+            ⎛ cosθ  0  sinθ ⎞   ⎛ 0  0  1 ⎞   
+        R = ⎜   0   1   0   ⎟ = ⎜ 0  1  0 ⎟  
+            ⎝-sinθ  0  cosθ ⎠   ⎝-1  0  0 ⎠
+
+S𝓍 = S𝓏 R ≐ 
+           ⎛ 1 0 0  ⎞ ⎛ 0  0  1 ⎞   ⎛ 0  0  1 ⎞
+         ħ ⎜ 0 0 0  ⎟ ⎜ 0  1  0 ⎟ = ⎜ 0  1  0 ⎟
+           ⎝ 0 0 -1 ⎠ ⎝-1  0  0 ⎠   ⎝-1  0  0 ⎠
+
+
+
+
+
+The same is true for the S𝓍 operator in its basis. To express this operator in the z basis, however, it must be diagonalized. 
+
+
+
+
+

solutions/exam1/prob2

@@ -0,0 +1,47 @@
+͟d͟ ͟\<͟p͟\>͟ = - 〈͟d͟ ͟V͟(͟X͟)͟〉
+  dt           dx
+
+  If a particle is subject to potential 〈V(x)〉
+
+The potential is known, and in the absense of any non-conservative influences, the Hamiltonian is equal to the potential.
+
+H(x) = V(x) 
+
+H|E〉 = E|E〉
+
+Ĥ = ͟p̂͟²͟ + V(x̂)
+     2m
+
+x̂ ≐ x
+
+p̂ ≐ -ι ħ  ͟d͟
+          dx
+
+p̂² ≐ -ħ²  ͟d͟²
+          dx²
+
+Ĥ = ͟p̂͟²͟ + V(x̂)
+     2m
+
+
+      ∞
+〈p̂〉 = ∫ dx p(x) = 
+     -∞
+
+d/dt ∫ dx p(x) = 
+      
+
+
+      
+Probably start here:
+
+〈V(x)〉 = ∫ dx V(x) 
+
+〈d/dx V(x)〉 = ∫ d/dx V(x) dx = V(x)
+
+d/dt 〈p〉 = d/dt ∫ dx p(x) 
+
+the definition of momentum in function space is 
+
+    d/dt p(x) = -d/dx V(x)
+

solutions/exam1/prob3

@@ -0,0 +1,6 @@
+|a〉 and |b〉 are eigenstates of a Hermitian operator A with eigenvalues a and b, a ≠ b. The Hamiltonian operator is
+
+    Ĥ = |a〉 δ 〈b| + |b〉 δ 〈a|, with δ a real number.
+
+a) 
+