renamed text.pdf, starting chap 9

notes/harmonic oscillator

@@ -0,0 +1,4 @@
+Ĥ = T̂ + V̂ = -_͟ħ͟²͟ ∇² + V(𝐫,t). 
+              2m    
+
+V(𝐫,t) = ½kx²              

notes/particle operators

@@ -0,0 +1,13 @@
+    Ĥ = T̂ + V̂
+with
+    V̂ = V = V(𝐫,t) (potential energy operator)
+and
+    T̂ = _͟𝐩̂͟⋅͟𝐩̂͟ = _͟p̂͟²͟ = -_͟ħ͟²͟ ∇² 
+         2m     2m     2m
+since
+    p̂ = -ιħ∇.
+
+
+Ĥ = T̂ + V̂ = -_͟ħ͟²͟ ∇² + V(𝐫,t). 
+              2m    
+

notes/random

@@ -0,0 +1,35 @@
+
+In position space,
+             ⌠ ∞
+    ❬φₙ❙Ψ❭ ≐ ⎮   dx φₙ⃰(x) Ψ(x).
+             ⌡-∞
+
+But for just 3L/4 to L, 
+    ⌠L
+    ⎮   dx φₙ⃰(x) Ψ(x).
+    ⌡3L/4
+
+Since this is a particle in an infinite square well,
+
+    φₙ(x) = √⎛2͟⎞ sin⎛n͟π͟x͟⎞ and Ψ(x) = ∑ cₙ φₙ(x).
+             ⎝L⎠    ⎝ L ⎠            ⁿ
+
+So the probability to calculate is
+
+    ⌠L
+    ⎮   dx √⎛2͟⎞ sin⎛n͟π͟x͟⎞ ∑ cₙ φₙ(x).
+    ⌡3L/4   ⎝L⎠    ⎝ L ⎠ ⁿ
+
+
+
+
+
+
+
+The evaluation of the last piece of this expression is elucidated by the identity
+
+    sin(u)sin(v) = 1/2 (cos(u-v) - cos(u+v)), i.e.
+
+    sin(2nπ) sin(³/₂nπ) = ¹/₂(cos(1/2nπ) - cos(7/2nπ)).
+
+So when n is odd, this product disappears, and when n is even one of the sin² terms disappears. The other sin² will always return 0, since regardless of the integer n, its argument is an integer multiple of π.

solutions/chap5/class_prob

@@ -0,0 +1,19 @@
+The task is to graph the following expression, developed in class. We had a lot of trouble with this, because while it seems straight forward, the parameter values of sa were difficult to pin down for some reason. The produced graph of transmission vs. displacement drops off too quickly according to our standard, but it is within the right ballpark, so I thought it worthwhile to at least include the incorrect result. The following function is plotted:
+
+The transmission
+
+	T = _____͟1͟_____ ,
+	    1 + sinh(sa)
+
+with s = √(2͟m͟ₑ͟E͟) and a the displacement from the origin.
+            ħ
+
+	E = 5 eV
+	mₑ = 5.11*10⁶ eV/c²
+	c = 299792458 m/s
+	ħ = 6.582*10⁽⁻¹⁶⁾ eV s
+
+So, s = 3.62*10¹⁰ m⁻¹ = 3.62 A⁻¹.
+
+The parameter space is defined from a=0 to a=4 angstroms.
+

solutions/chap5/prob2

@@ -45,8 +45,8 @@ So,
      π͟²͟ħ͟².          𝓟₁=¹/₃
      2mL²    
 
-    4͟π͟²͟ħ͟².          𝓟₂=¹/₃
-     2mL²                        
+    2͟π͟²͟ħ͟².          𝓟₂=¹/₃
+     mL²                         
 
     9͟π͟²͟ħ͟².          𝓟₃=¹/₃
      2mL²                         
@@ -87,16 +87,32 @@ Because the hamiltonian is time-independent, the state vector progresses timewis
     ❙Ψ′(t)❭ = exp(-ι͟Ĥ͟t͟ )❙Ψ′(t=0)❭ = _͟1͟ exp(-ι͟Ĥ͟t͟ ) (❙φ₁❭ - ❙φ₂❭ + ι ❙φ₃❭).
                     ħ               √3       ħ                       
 
-    _͟1͟ ⎛exp(-ι͟E͟₁͟t͟ )❙φ₁❭ - exp(-ι͟E͟₂͟t͟ )❙φ₂❭ + ι exp(-ι͟E͟₃͟t͟ )❙φ₃❭⎞.
-    √3 ⎝       ħ                 ħ                   ħ       ⎠
+    _͟1͟ ⎛exp(-ι͟E͟₁͟t͟ )❙φ₁❭ - exp(-ι͟E͟₂͟t͟ )❙φ₂❭ + ι exp(-ι͟E͟₃͟t͟ )❙φ₃❭⎞
+    √3 ⎝       ħ                    ħ                   ħ    ⎠
+
+(𝐝)
+    = _͟1͟ ⎛exp(-ι π͟²͟ħ͟t͟ )❙φ₁❭ - exp(-ι 2͟π͟²͟ħ͟t͟ )❙φ₂❭ + ι exp(-ι 9͟π͟²͟ħ͟t͟ )❙φ₃❭⎞
+      √3 ⎝       2mL²                 mL²                    2L²       ⎠.
 
 
+What are the possible measurements at time t = ħ/E₁?
 
-The coefficients for the states have the same magnitudes, so the probability of measuring a particular state at time t=0 is 1/3.
+(𝐞) The same value of energy will be measured for each state. Since there has been no change to the coefficients besides a change in phase, and the phase term goes to 1 under the modulus, the probabilities remain the same. 
 
+❙Ψ′(t=ħ/E₁)❭ = _͟1͟ ⎛exp(-ι)❙φ₁❭ - exp(-ι E͟₂͟ )❙φ₂❭ + ι exp(-ι E͟₃͟ )❙φ₃❭⎞ 
+               √3 ⎝                     E₁                  E₁      ⎠
+
+    Energy          Probability     
+
+     π͟²͟ħ͟².          𝓟₁=¹/₃
+     2mL²    
+
+    2͟π͟²͟ħ͟².          𝓟₂=¹/₃
+     mL²                         
+
+    9͟π͟²͟ħ͟².          𝓟₃=¹/₃
+     2mL²  
 
-    ❙Ψ(t=0)❭ =  _͟A͟  (αβ❙φ₁❭ - βγ❙φ₂❭ + αγι❙φ₃❭)
-               αβγ
 
 
 

solutions/chap5/prob6

@@ -0,0 +1,68 @@
+For a particle in an infinite square well, calculate the probability of finding the particle in the range 3L/4 < x < L for each of the first three energy eigenstates.
+
+
+The first three energy eigenstates are
+
+❙φ₁❭, ❙φ₂❭, and ❙φ₃❭.
+
+The probability of finding a particle in a state n is
+
+    │❬φₙ❙φₙ❭│² = │⌠ ∞              │²
+                 │⎮   dx │φₙ(x)│². │ 
+                 │⌡-∞              │.
+
+But for just 3L/4 to L, 
+              ⌠L
+    ❬φₙ❙φₙ❭ = ⎮    dx │φₙ(x)│².
+              ⌡³/₄L
+
+Since this is a particle in an infinite square well, the position space representation of an energy eigenstate n is
+
+    φₙ(x) = √⎛2͟⎞ sin⎛n͟π͟x͟⎞, so │φₙ(x)│² = ⎛2͟⎞ sin²⎛n͟π͟x͟⎞ 
+             ⎝L⎠    ⎝ L ⎠                ⎝L⎠     ⎝ L ⎠.
+
+Therefore, with the power reduction identity,
+
+                  ⌠L                   ⌠L     
+    ❬φₙ❙φₙ❭ = ⎛2͟⎞ ⎮ dx sin²⎛n͟π͟x͟⎞ = ⎛2͟⎞ ⎮  dx ⎛1͟ - 1͟ cos(2͟n͟π͟x͟)⎞
+              ⎝L⎠ ⎮        ⎝ L ⎠   ⎝L⎠ ⎮     ⎝2   2       L  ⎠.
+                  ⌡³/₄L                ⌡³/₄L 
+
+    ❬φₙ❙φₙ❭ = ⎛2͟⎞ ⎛_͟1͟L͟ - ⎛ _͟1͟L͟ sin(2nπ) - _͟1͟L͟ sin(3͟n͟π͟) ⎞ ⎞
+              ⎝L⎠ ⎝ 8    ⎝ 4nπ            4nπ      2   ⎠ ⎠
+
+        =  _͟1͟ L͟  ⎛sin(2nπ) - sin(3͟n͟π͟)⎞
+           16 nπ ⎝                2  ⎠.
+
+In general, then, the probability of finding the particle in a particular energy state n is given by
+
+    │❬φₙ❙Ψ❭│² = │_͟1͟ L͟  ⎛sin(2nπ) - sin(3͟n͟π͟)⎞│²                         
+                │16 nπ ⎝                 2 ⎠│
+
+        = _͟1͟  ⎛_͟L͟²͟ ⎞ │⎛sin(2nπ) - sin(3͟n͟π͟)⎞│²
+          256 ⎝n²π²⎠ │⎝                 2 ⎠│
+
+        = _͟1͟  ⎛_͟L͟²͟ ⎞ ⎛sin(2nπ) - sin(3͟n͟π͟)⎞ ⎛sin(2nπ) - sin(3͟n͟π͟)⎞
+          256 ⎝n²π²⎠ ⎝                2  ⎠ ⎝                2  ⎠
+
+        = _͟1͟  ⎛_͟L͟²͟ ⎞ ⎛sin²(2nπ) + sin²(3͟n͟π͟) + 2 sin(2nπ) sin(3͟n͟π͟)⎞ 
+          256 ⎝n²π²⎠ ⎝                  2                     2  ⎠ .
+
+The third term in the parentheses will always return 0, since one function or the other returns 0 depending on whether n is even or odd. The first term will also always be zero, since regardless of the value of n, the argument is an integer multiple of π. The final term will disappear when n is even, so the probability of finding a particle in the specified region for an energy state n is therefore
+
+     ⎧   
+     ⎪ 0, n even
+P(n) ⎨
+     ⎪ _͟1͟  ⎛_͟L͟²͟ ⎞ ⎛sin²(3͟n͟π͟)⎞  
+     ⎩ 256 ⎝n²π²⎠ ⎝      2  ⎠ , n odd 
+
+This is wrong, because for n=2, there should be some probability of finding the particle within this region, since the only node is at the center, and the wave function is non-zero across the region and does not cross the axis. This function will return a probability of 0 for n=2, which does not make sense with my statement. For n=1,3, at least a probability can be calculated:
+
+
+
+P(n=1) = _͟1͟  ⎛_͟L͟²͟⎞ sin²(3͟π) = _͟1͟  ⎛_͟L͟²͟⎞ = 3.96*10⁻⁴ L² 
+         256 ⎝ π²⎠      2     256 ⎝ π²⎠  
+
+P(n=3) = _͟1͟  ⎛_͟L͟²͟ ⎞ ⎛sin²(9͟π͟)⎞ = _͟1͟   ⎛_͟L͟²͟⎞ = 4.40*10⁻⁵ L²
+         256 ⎝9 π²⎠ ⎝      2 ⎠   2304 ⎝ π²⎠
+

solutions/exam1/prob4

@@ -1,4 +1,4 @@
-Two operators' matrix representations are known in the |1〉, |2〉, |3〉 basis, where a and b are real numbers, and ι is the imaginary unit:
+Two operators' matrix representations are known in the |1❭, |2❭, |3❭ basis, where a and b are real numbers, and ι is the imaginary unit:
 
 Â ≐                   B̂ ≐
     ⎛ a  0  0 ⎞           ⎛ b   0   0 ⎞
@@ -36,7 +36,7 @@ which reduces to
 
 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⎞     
+|a₁❭ ≐ ⎛1⎞  |a₂❭ ≐ ⎛0⎞  and |a₃❭ ≐ ⎛0⎞     
        ⎜0⎟         ⎜1⎟             ⎜0⎟
        ⎝0⎠;        ⎝0⎠;            ⎝1⎠.
 
@@ -50,25 +50,25 @@ For B̂, the eigenvalues are already known (λ = b,b,-b.), and using the eigenva
 
 dicates two possible eigenstates (for the eigenvalue b). One eigenstate is obvious from inspection:
 
-	|b₁〉 ≐ ⎛1⎞
-	       ⎜0⎟
-	       ⎝0⎠.
+    |b₁❭ ≐ ⎛1⎞
+           ⎜0⎟
+           ⎝0⎠.
 
 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 ⎞
+    |b₂❭ ≐ 1  ⎛ 0 ⎞
            √2 ⎜ 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⎞
+    |b₃❭ ≐ 1  ⎛0⎞
            √2 ⎜ι⎟
               ⎝1⎠
 
 The complete set of eigenstates of the operator B̂ is
 
-    |b₁〉 ≐ ⎛1⎞  |b₂〉 ≐ 1  ⎛ 0 ⎞      |b₃〉 ≐ 1  ⎛0⎞
+    |b₁❭ ≐ ⎛1⎞  |b₂❭ ≐ 1  ⎛ 0 ⎞      |b₃❭ ≐ 1  ⎛0⎞
            ⎜0⎟         √2 ⎜ 1 ⎟             √2 ⎜ι⎟
            ⎝0⎠,           ⎝-ι ⎠, and           ⎝1⎠.