finished exam 1 prob 1, started prob 2

solutions/exam1/prob1

@@ -1,37 +1,33 @@
-The characteristic equation for the spin operator S𝓏 is
+If S𝓏 and S𝓍 are spin-1 operators in the z basis, what are the results if S𝓏 (S𝓏 + ħ)(S𝓏 - ħ) and S𝓍 (S𝓍 + ħ)(S𝓍 - ħ) are evaluated? 
 
-    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 ⎠
+The spin eigenstates and eigenvalues are known from experiment for a spin-1 system, and the spin-z and spin-x operators in the z basis, S𝓏 and S𝓍, have the following matrix representations:
 
 
-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.
+S𝓏 ≐
+        ⎛ 1 0  0 ⎞
+      ħ ⎜ 0 0  0 ⎟ and
+        ⎝ 0 0 -1 ⎠
+S𝓍 ≐
+        ⎛ 0 1 0 ⎞
+      ħ ⎜ 1 0 1 ⎟.
+        ⎝ 0 1 0 ⎠
 
-            ⎛ cosθ  0  sinθ ⎞   ⎛ 0  0  1 ⎞   
-        R = ⎜   0   1   0   ⎟ = ⎜ 0  1  0 ⎟  
-            ⎝-sinθ  0  cosθ ⎠   ⎝-1  0  0 ⎠
+Using the matrix representations, the expressions can be evaluated. For the spin-z operator, the expression S𝓏 (S𝓏 + ħ)(S𝓏 - ħ) ≐
 
-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. 
+      ⎛ 1 0  0 ⎞ ⎧   ⎛ 1 0  0 ⎞       ⎛ 1 0 0 ⎞ ⎫ ⎧   ⎛ 1 0  0 ⎞      ⎛ 1 0 0 ⎞ ⎫  
+    ħ ⎜ 0 0  0 ⎟ ⎪ ħ ⎜ 0 0  0 ⎟  +  ħ ⎜ 0 1 0 ⎟ ⎪ ⎪ ħ ⎜ 0 0  0 ⎟  - ħ ⎜ 0 1 0 ⎟ ⎪,
+      ⎝ 0 0 -1 ⎠ ⎩   ⎝ 0 0 -1 ⎠       ⎝ 0 0 1 ⎠ ⎭ ⎩   ⎝ 0 0 -1 ⎠      ⎝ 0 0 1 ⎠ ⎭  
 
+which simplifies to the matrix multiplication operation, where 𝟘 represents the 0 matrix,
 
+      ⎛ 1 0  0 ⎞   ⎛ 2 0 0 ⎞   ⎛ 0  0 0 ⎞ 
+    ħ ⎜ 0 0  0 ⎟ ħ ⎜ 0 1 0 ⎟ ħ ⎜ 0 -1 0 ⎟ = 𝟘. 
+      ⎝ 0 0 -1 ⎠   ⎝ 0 0 0 ⎠   ⎝ 0  0 0 ⎠ 
 
+The multiplication operation apparently returns 𝟘 because the third factor will nullify any terms besides center terms, and the first factor will nullify any center terms.
 
+Similarly, S𝓍 (S𝓍 + ħ)(S𝓍 - ħ) ≐
 
+       ⎛ 0 1 0 ⎞ ⎛ 1 1 0 ⎞ ⎛ -1  1  0 ⎞      ⎛ 1 1 1 ⎞ ⎛ -1  1  0 ⎞      ⎛ 0 1 0 ⎞
+    ħ³ ⎜ 1 0 1 ⎟ ⎜ 1 1 1 ⎟ ⎜  1 -1  1 ⎟ = ħ³ ⎜ 1 2 1 ⎟ ⎜  1 -1  1 ⎟ = ħ³ ⎜ 1 0 1 ⎟.
+       ⎝ 0 1 0 ⎠ ⎝ 0 1 1 ⎠ ⎝  0  1 -1 ⎠      ⎝ 1 1 1 ⎠ ⎝  0  1 -1 ⎠      ⎝ 0 1 0 ⎠

solutions/exam1/prob2

@@ -1,47 +1,7 @@
-͟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
+Show that ͟d͟〈͟p͟〉͟ = -〳͟d͟V͟(͟x͟)͟〵 when a particle is subjected to a potential 〈V(x)〉.
+           dt     〵  dx 〳
 
 
-      ∞
-〈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)