Initialization

Postulates

@@ -0,0 +1,14 @@
+(1) A general state vector is represented by |Ψ> (a ket)
+
+(2) A physical observable is represented by an operator A that acts on |Ψ>
+
+(3) The possible measurements of an observable are the eigenvalues aₙ of the operator A
+
+(4) The probability of measuring eigenvalue aₙ is given by Paₙ=|<aₙ|Ψ>|²
+
+(5) After a measurement A that yields aₙ, the quantum system is in a new state that is the normalized projection of the original system ket onto the ket (or kets) corresponding to the result of the measurement:
+    |Ψ'> = Pn|Ψ> / √(<Ψ|Pn|Ψ>)
+
+(6) The time-evolution of the quantum system is determined by the Hamiltonian operator H(t) through the Schrodinger equation
+    ιħ(d/dt)|Ψ(t)> = H(t)|Ψ(t)>
+

book_notes/chap2/S² operator

@@ -0,0 +1,10 @@
+𝐒² = S𝓍² + S𝓎² + S𝓏²
+
+𝐒²|Ψ〉 = ¾ħ²|Ψ〉
+
+〈𝐒〉 = ¾ħ²
+
+|𝐒| = √〈𝐒²〉 = √(¾)ħ
+
+Vector Model, pg. 58
+

book_notes/chap2/commutators

@@ -0,0 +1,5 @@
+The commutator of two operators A and B is notated as and defined to be
+    [A,B] = AB - BA .
+
+A and B commute when [A,B] = 0, i.e. AB = BA.
+

book_notes/chap2/operators

@@ -0,0 +1,35 @@
+Eigenvalue equations for the S𝓏 operator in a spin-1/2 system:
+    S𝓏|+> = +ħ/2|+>
+    S𝓏|-> = -ħ/2|->
+
+Matrix form of an operator:
+    S𝓏 ≐ ( a b )
+         ( c d )
+
+Eigenvalue equations in matrix form:
+
+ ( a b )( 1 ) = +ħ/2 ( 1 )
+ ( c d )( 0 )        ( 0 )
+
+ ( a b )( 0 ) = -ħ/2 ( 0 )
+ ( c d )( 1 )        ( 1 )
+
+Can show with matrix multiplication operations that
+    S𝓏 ≐ ħ/2 ( 1  0 )
+             ( 0 -1 )
+
+Some Properties:
+     An operator is always diagonal in its own basis.
+     Eigenvectors are unit vectirs in their own basis.
+
+    S𝓏 ≐ ħ/2 ( 1  0 )  |+> = ( 1 )   |-> = ( 0 )
+             ( 0 -1 )        ( 0 )         ( 1 )
+
+
+Diagonalizing an Operator
+     Diagonize a matrix --> find the eigenvalues and eigenvector
+         
+     S𝓍 ≐ ħ/2 ( 0 1 )
+              ( 1 0 )
+
+     (1) 1⁄2   ⁵⁄ₐ

book_notes/chap2/spin-1 system

@@ -0,0 +1,17 @@
+Eigenvalue Equations for Spin-1 System:
+
+    𝐒𝓏|1〉 = ħ|1〉
+    𝐒𝓏|0〉 = 0ħ|0〉
+    𝐒𝓏|-1〉 = ħ|-1〉
+
+      ħ  ( 1 0 0 )
+S𝓏 ≐ --- ( 0 1 0 )
+     √2  ( 0 0 1 )
+
+      ħ  ( 0 1 0 )
+S𝓍 ≐ --- ( 1 0 1 )
+     √2  ( 0 1 0 )
+
+      ħ  ( 0  -ι   0  )
+S𝓎 ≐ --- ( ι   0  -ι  )
+     √2  ( 0   ι   0  )

book_notes/chap3/Bohr Frequency

@@ -0,0 +1,14 @@
+Ψ(t) is the time-revolving state function,
+    Ψ(t) = c₁ exp(-ι E₁/ħ t) + c₂ exp(-ι E₂/ħ t) + ...
+
+The eigenstate of A corresponding to the eigenvalue a₁  
+    |a₁〉 = α₁|E₁〉 + α₂|E₂〉
+
+The probability of measuring the eigenvalue a₁ is
+    P(a₁) = |〈a₁|Ψ(t)〉|²
+          ↓ Derivation on page 71
+    P(a₁) = |α₁|²|c₁|² + |α₂|²|c₂|² + Re(α₁c₁⃰α₂⃰c₂ exp(-ι (E₂-E₁)/ħ t))
+
+The Bohy Frequency (angular frequency) of two states E₁ and E₂ is thus revealed:
+    ω₂₁ = (E₂ - E₁)/ħ
+

book_notes/chap3/Time-Independent Hamiltonian Recipe

@@ -0,0 +1,7 @@
+(1) Diagonalize H (find the eigenvalues of Eₙ and eigenvectors |Eₙ〉).
+
+(2) Write |Ψ(0)〉 in terms of the energy eigenstates |Eₙ〉.
+
+(3) Multiple each eigenstate coefficient by exp(-ι Eₙ/ħ t) to get |Ψ(t)〉.
+
+(4) Calculate the probability P(aⱼ) = |〈aⱼ|Ψ(t)〉|²

lecture_notes/1-11/Heisenberg

@@ -0,0 +1,29 @@
+Talked about Movie.
+
+Ket vectors/ Basis using hero/villain (pic)
+
+Orthogonal Basis using Bra Ket notation
+
+Orthogonal basis
+
+	< villain | villain > = < hero | hero > = 1
+
+Determining coefficients using projections in braket notation (pic)
+
+Probability from complex numbers
+
+	C_v* C_v = |C_v|^2
+
+Playing around with braket notation
+
+----------------------------------------------------
+
+Observables and Operators
+
+	Eigen-kets, or eigen-states (pic)
+
+	Operators change the state when applied to a state
+
+	Operators refer to measurements (pic)
+
+	

lecture_notes/1-13/Basis

@@ -0,0 +1,14 @@
+Quantum State Basis
+-------------------
+
+	|Ψ> = C_+ |+> + C_- |-> 	with |C_+|^2 + |C_-|^2 = 1
+
+	Example (pic)
+
+	Spin in x, y, and z (pic)
+
+	Coefficients in general
+		|+x> = C_+ |+z> + C_- |+z>
+		C_+ = e^(i δ_+) / √2 ,  C_- = e^(i δ_-) / √2
+
+		<+x| S_z |+x> = <S_z> = expectation value

lecture_notes/1-13/Stern-Gerlach

@@ -0,0 +1,10 @@
+Explanation of STern-Gerlach Experiment (pic)
+=============================================
+
+	- Electrons have intrinsic magnetic moment
+
+	Showed how magnetic moment is related to angular momentum (pic)
+
+	mu-bar_s = g q /2 /m /c S-bar	g is measured for any particle, S-bar is spin angular momentum
+
+	S_z = ± h-bar / 2

lecture_notes/1-15/Applications

@@ -0,0 +1,19 @@
+Revisited Stern-Gerlach setup, now using three machines in series.
+
+Introduced incompatible operators.
+
+Reintroduced normalized and phased notation, using δ. (pic)
+
+Combined y and x to view result of <+y|+x>
+
+
+Talked about projections on other axes (pic x2)
+
+Practice problem (pic) performed partially in class
+
+	what is the probability that the S_y will be measured to be +ħ∕2
+		Should be .93
+		To find, calculate <+y|Ψ> to find probability.
+
+	What is <S_y>? I.e. the expecation value for S_y?
+		<Ψ|S_y|Ψ> gives expectation value, or use <+y|Ψ> and <-y|Ψ> to find coefficients. (pic)

lecture_notes/1-22/Matrix Forms

@@ -0,0 +1,15 @@
+States can be represented as vectors
+
+|Ψ> = ½|+> + (i√3)/2|-> =  ½ ( 1 ) + (i√3)/2 ( 0 ) = (   1/2    )
+      	     		     ( 0 )   	     ( 1 )   ( (i√3)/2  )
+
+Operators are matrices
+
+Showed how Â|Ψ>=|φ> translates to a matrix in the z basis (pic x2)
+
+Example with S𝓏 operator (pic)
+Example with S𝓍 operator (pic)
+
+Often we know operator but not basis. How to find basis? (pic)
+      i.e. find the eigen-basis S|α>=λ|α>
+      Only non-trivial if determinant is zero. Solve for basis. (pic)

lecture_notes/1-27/Expectation Value Time Dependence

@@ -0,0 +1,13 @@
+Computed Derivative (pic) d/dt <Â> (pic)
+
+Discovered/Introduced Commutator
+Cpmpatible Observables occur when commutator is equal to 0.
+
+[Ĥ,Â] = ĤÂ - ÂĤ
+
+Constant of Motion
+------------------
+    if d/dt  = 0 and [Ĥ,Â] = 0,
+    then  is a constant of motion
+
+    Thm: Two operators such that [Â,B̂] always have common Eigenstates

lecture_notes/1-27/Spin-Half Precession

@@ -0,0 +1,20 @@
+Precession of a Spin-1/2 Particle in a Constant Magnetic Field
+==============================================================
+    Find Hamiltoninan (pic)
+    Find eigenstates of the Hamiltonian (pic)
+    Find time-progression expression
+
+    Û(t) = exp(-ι Ĥ t/ħ) = exp(-ι ωₒt S𝓏/ħ) = exp(-ι θ S𝓏/ħ)
+    Û(t) = R̂(σk̂) 
+           ⸜   ⸝
+        Rotational Operator
+
+    So for arbitrary t (pic)
+
+
+    Projection along 𝓏-axis? <+𝓏|Ψ(t)>
+
+--->Why for (-x)-axis does this turn out to be sin²(ω₀ t/2)
+
+Hw->Find projection of 𝓎-axis
+

lecture_notes/1-27/Time-Independent SE

@@ -0,0 +1,14 @@
+Time-Independent Schrodinger Equation
+
+Û(t) = lim [1 - ι/ħ H t/n] ^n (pic)
+        n→∞
+
+Û(t) = exp(-ι Ĥ t/n) 
+
+|Ψ(t)> = exp(-ι Ĥ t/n) 
+
+Ĥ|E> = E|E>
+
+exp(-ι Ĥ t/n) |E> = exp(-ι E t/n) |E> (pic)
+
+|E> = ?

lecture_notes/1-28/Work a Hamiltonian Problem

@@ -0,0 +1,18 @@
+Magnetic Resonance Problem (pic 1)
+
+    This problem has time dependence (pic 2)
+
+Time-dependent Ψ? (pic 3)
+
+Plug in to Schrodinger Equation (pic 3)
+
+Solve S.E. with unknown functions. (pic 4,5)
+   Plug back into S.E. from matrices    
+     ↓
+   Nice concise equation system (pic 6)
+     ↓ Time derivative of both sides (pic 7)
+    Final Solution? Yes, and apply initial conditions.
+
+    Minus sign is switched somewhere in these pictures
+    
+    

notes/eigenstates

@@ -0,0 +1,22 @@
+A(x) is a real-space vector
+Ψ(x) is a wave function
+A acts on Ψ
+
+Some Ψ, when acted upon by A, result in a multiple of the original Ψ. These Ψ are called eigenstates and may be denoted Ψₐ. In algebraic terms,
+
+    A Ψₐ(x) = a Ψₐ(x), where a is complex.
+
+Ψₐ is called an eigenstate of Α corresponding to the eigenvalue a.
+
+If A is a Hermitian operator corresponding to some physical dynamical variable:
+      ∞                 ∞
+〈A〉 = ∫ Ψₐ⃰ A Ψₐ dx = a ∫ Ψₐ˟ Ψₐ dx = a
+     -∞                -∞
+ₐ
+       ∞                  ∞                  ∞
+〈A²〉 = ∫ Ψₐ⃰ A² Ψₐ dx = a ∫ Ψ⃰ₐ A Ψₐ dx = a² ∫ Ψₐ˟ Ψₐ dx = a² 
+      -∞                 -∞                 -∞
+
+σ² = 〈A²〉 - 〈A〉² = a² - a² = 0
+ ᴬ
+

solutions/chap3/prob4/draft

@@ -0,0 +1,65 @@
+A spin-1/2 particle has a magnetic moment 𝛍 and is placed in a uniform magnetic field 𝐁, which is aligned with the z axis, so 𝐁 = B𝓏 ẑ. It is known that the Hamiltonian operator for this sytem commutes with the spin component operator in the z direction but not with spin component operators in the x and y directions. The student is tasked with demonstrating this prediction.
+
+The commutator is defined as [û,v̂] = ûv̂-v̂û, where û and v̂ are operators.
+
+If the product of û and v̂ is independent of order, the commutator will be zero, and the operators can be said to commute. To test whether the Hamiltonian and the spin operators commute, one need express them in a common basis and compute the commutator. In this case, the question is whether the Hamiltonian and spin operators commute in the z direction, so they shall be expressed in the z basis.
+
+The spin operator in the z basis Ŝ𝓏 is well-known and diagonalized in the z basis because the magnetic field is oriented in the z direction. 
+
+    Ŝ𝓏 ≐ ħ/2  ( 1   0 )
+             ( 0  -1 ) .
+
+The Hamiltonian H is the total energy of the system. The only contributing term for this system is the potential energy V = -𝛍⋅𝐁, where 𝛍 and 𝐁 are the physical terms described in the introduction. The magnetic moment of a magnetic dipole, such as that seen with a spin-1/2 particle, is
+    
+    𝛍 = g q/2mₑ 𝐒,
+
+with 𝐒 the intrinsic spin vector for the case of no orbital angular momentum, and the remaining factors constant, intrinsic values of the particle. Then,
+
+    𝛍 = k 𝐒,
+
+The Hamiltonian H is therefore, because  𝐁 = B𝓏 ẑ,
+
+    H = V = -𝛍⋅𝐁 = -k S𝓏 B𝓏.
+
+The Hamiltonian is measureable, and therefore is an operator. The magnetic field strength is constant in this system, so the Hamiltonian operator is related to the spin operator in the z basis by
+
+    Ĥ = -k B𝓏 Ŝ𝓏.
+
+Therefore, the Hamiltonian
+
+    Ĥ ≐ -k B𝓏 ħ/2  ( 1  0 )
+	          ( 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.
+
+The spin operators in the x direction and z direction in the z basis, with ι the imaginary unit,
+
+    Ŝ𝓍 ≐ ħ/2 ( 0  1 )
+            ( 1  0 )
+
+    Ŝ𝓎 ≐ ħ/2 ( 0 -ι )
+            ( ι  0 )
+
+As with the spin component in the z direction, when multiplying the Hamiltonian and spin in x or spin in y operators, the multiplicative factors associate outside the matrices, and only the matrix multiplication determines the commutability of the operators, so, the operators are commutable if their matrix components are commutable. 
+
+For Ĥ and Ŝ𝓍:
+
+    ( 1  0 ) ( 0  1 )  =  ( 0   1 )
+    ( 0 -1 ) ( 1  0 )     ( -1  0 )
+
+    ( 0  1 ) ( 1  0 )  =  ( 0  -1 )
+    ( 1  0 ) ( 0 -1 )     ( 1   0 )
+
+Since these matrix products are not equal, their difference is not zero. Therefore, these operators do not commute.
+
+For Ĥ and Ŝ𝓎:
+
+    ( 1  0 ) ( 0 -ι )  =  ( 0  -ι )
+    ( 0 -1 ) ( ι  0 )     ( -ι  0 )
+
+    ( 0 -ι ) ( 1  0 )  =  ( 0  ι )
+    ( ι  0 ) ( 0 -1 )     ( ι  0 )
+
+Again, these matrix products are not equal, so these operators do not commute.
+
+

solutions/chap3/prob4/prob4

@@ -0,0 +1,3 @@
+A spin-1/2 particle has a magnetic moment 𝛍 and is placed in a uniform magnetic field 𝐁, which is aligned with the z-axis, so 𝐁 = B𝓏 𝑧̂. It is known that the Hamiltonian operator for this sytem commutes with the spin operator in the z 𝑧̂
+
+

solutions/chap3/prob4/tex/draft.aux

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solutions/chap3/prob4/tex/draft.tex

@@ -0,0 +1,19 @@
+\documentclass{article}
+\usepackage{unicode-math}
+%\usepackage{mathrsfs}
+%\usepackage{amsmath}
+%\usepackage{euscript}
+%\usepackage[utf8x]{inputenc}
+%\usepackage{mathdesign}
+%\setmathfont{DejaVuSansMono.ttf}
+\setmathfont{xits-math.otf}
+%\usepackage{unixode}
+\begin{document}
+
+
+A spin-1/2 particle has a magnetic moment 𝛍 and is placed in a uniform magnetic field 𝐁, which is aligned with the z-axis, so 𝐁 = B𝓏 𝑧̂ = B𝓏 ẑ. It is known that the Hamiltonian operator for this sytem commutes with the spin component operator in the z direction (z basis?) but not with spin component operators in the x and y directions. The following argument should prove that hypothesis.
+
+The Hamiltonian Ĥ = 
+$\mathscr{z}$
+$z$
+\end{document}

solutions/chap3/prob5/draft

@@ -0,0 +1,29 @@
+A spin-1/2 particle with a magnetic moment is known to be in the state |Ψ(t=0)〉 = |+〉.
+
+a) If the observable S𝓍 is measured at t=0, the possible results are ħ/2 and -ħ/2 with an equal probability of measuring either.
+
+b) The system evolves in a uniform magnetic field 𝐁 = B₀ŷ. What is the state of the system at t=T?
+
+Since the magnetic field is oriented along the y axis, the energy eigenstates will be associated with that direction. The eigenstates are |±𝓎〉.
+
+The Hamiltonian for this system is 
+    H = - ω₀ S𝓎, with ω₀ = g q/2mₑ B₀ ≈ e/mₑ B₀. 
+
+The eigenvalue equations in the energy basis are therefore
+
+    Ĥ|E±〉 = ω₀ ±ħ/2 |E±〉 = ω₀ ±ħ/2 |±〉 = E± |±〉
+
+The initial state is prepared to |+〉, which means, in the y basis,
+
+    |Ψ(t=0)〉 = 1/√2 |+𝓎〉 - 1/√2 |-𝓎〉
+
+This Hamiltonian is time independent, so the time evolution is given by multiplying each eigenstate with the time-pdependent phase factor, with ι the imaginary unit.
+
+    |Ψ(t)〉 = exp(-ι E₊ t/ħ)/√2 |+𝓎〉 - exp(-ι E₋ t/ħ)/√2 |-𝓎〉
+
+
+
+If the time-evolution of the general state is known in the energy basis (aligned with the spin y basis), the time-evolved state at some time t in the z basis can be predicted by projecting the z basis onto the state to determine the coefficients in the z basis. In mathematical terms,
+
+    〈+|Ψ(t)〉|+〉 = 
+    〈-|Ψ(t)〉|-〉 = (