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PS3/doc.tex

@@ -318,5 +318,180 @@ S_u = [1, s; s, 1];
                 only need consider the range \(k \in [-\frac{\pi}{a}, \frac{\pi}{a}]\).
             \end{enumerate}
         \end{enumerate}
+        \newpage
+
+    \section*{Problem 3}
+        \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] 
+            \item %3a
+            Given the geometry and interaction rules stated, the interaction matrices \([H]_{nm}\) are zero except for
+            \begin{align*}
+                [H]_{nn} &= \begin{bmatrix}
+                                E_0 & t_i \\
+                                t_i & E_0 \\
+                                \end{bmatrix} \\
+                [H]_{n,m*} &= \begin{bmatrix}
+                                t_A & 0   \\
+                                0   & t_B \\
+                                \end{bmatrix}
+            \end{align*}
+            where \(m*\) corresponds to any of the four nearest neighbor square-shaped unit cells to cell \(n\).
+            For cell \(n\), let cell \(a\) be above, cell \(b\) be to the right, cell \(c\) be below, and cell \(d\) be to the left.
+            Given that each cell is spaced a length \(a\) apart, the associated phase factors for nonzero \(H_{nm}\),
+            \(e^{i\vec{k}\cdot(\vec{d_m}-\vec{d_n})}\), are then:
+            \begin{align*}
+                (n):& \quad 1 \\
+                (a):& \quad e^{ik_y a} \\
+                (b):& \quad e^{ik_x a} \\
+                (c):& \quad e^{-ik_y a} \\
+                (d):& \quad e^{-ik_x a} \\
+            \end{align*}
+            with \(\vec{k} = k_x\hat{x} + k_y\hat{y}\). The Bloch matrix then follows:
+            \begin{align*}
+                [h(\vec{k})]    &=  \sum_m [H]_{nm}  e^{i\vec{k}\cdot(\vec{d_m}-\vec{d_n})} \\
+                                &=  \begin{bmatrix}
+                                        E_0 + t_A\left(e^{ik_x a} + e^{-ik_x a} + e^{ik_y a} + e^{-ik_y a}\right) & t_i \\
+                                        t_i & E_0 + t_B\left(e^{ik_x a} + e^{-ik_x a} + e^{ik_y a} + e^{-ik_y a}\right) \\
+                                    \end{bmatrix} \\
+                                &=  \begin{bmatrix}
+                                        E_0 + 2t_A\left(\cos(k_x a) + \cos(k_y a)\right) & t_i \\
+                                        t_i & E_0 + 2t_B\left(\cos(k_x a) + \cos(k_y a)\right) \\ 
+                                    \end{bmatrix}
+            \end{align*}
+
+            \item %3b
+
+            To find the E-\textbf{k} relationship in terms of cosine functions, we must first
+            find \( h_0 \) in terms of trigonometric functions.
+            We will require the following trigonometric identities to do so:
+
+            \begin{align}
+                sin(\alpha \pm \beta) = sin(\alpha)cos(\beta) \pm sin(\beta)cos(\alpha)
+                \label{eq:sin_sum}
+            \end{align}
+
+            \begin{align}
+                cos(\alpha \pm \beta) = cos(\alpha)cos(\beta) \mp sin(\alpha)sin(\alpha)
+                \label{eq:cos_sum}
+            \end{align}
+
+            \begin{align}
+                1 = sin^2(\theta) cos^2(\theta)
+                \label{eq:sum_squares}
+            \end{align}
+
+            As well as the following general sine/cosine properties:
+            \begin{align}
+                cos(\theta) &= cos(-\theta)
+                sin(\theta) &= -sin(-\theta)
+                \label{eq:neg_sin_cos}
+            \end{align}
+
+            First we can rewrite the exponentials in \( h_0 \) using Euler's identity:
+
+            \begin{align*}
+                h_0 = -t_c \left( 1 + cos(-k_x a - k_y b) + i sin(-k_x a - k_y b) + cos(-k_x a + k_y b) + i sin(-k_x a + k_y b) \right)
+            \end{align*}
+
+            Using identities (\ref{eq:sin_sum}) and (\ref{eq:cos_sum}), we can expand this relationship further.
+
+            
+            % TODO no idea how to format this...
+            \begin{align*}
+                h_0 = -t+c ( 1 &+ cos(-k_x a)cos(-k_y b) - sin(-k_x a)sin(-k_y b) + i sin(-k_x a)cos(-k_y b) + i sin(-k_y b)cos(-k_x a) \\
+                               &+ cos(k_y b)cos(k_x a) + sin(k_y b)sin(k_x a) + i sin(k_y b)cos(k_x a) - i sin(k_x a)cos(k_y b) )
+            \end{align*}
+
+            Here we can use the properties from (\ref{eq:neg_sin_cos}) to reduce this equation. Since \( \left| h_0 \right| ^2 = h_0 h^*_0 \),
+            we also obtain a simple expression for \( h^*_0 \).
+
+            \begin{align*}
+                h_0 &= -t_c ( 1 + 2cos(k_x a)cos(k_y b) -  2 i sin(k_x a)cos(k_y b)) \\
+                h^*_0 &= -t_c ( 1 + 2cos(k_x a)cos(k_y b) + 2 i sin(k_x a)cos(k_y b))
+            \end{align*}
+            
+            We can now find \( \left| h_0 \right| ^2 = h_0 h^*_0 \):
+
+            \begin{align*}
+                h_0 h^*_0 = 1 + 4cos(k_x a)cos(k_y b) + 4cos^2(k_y b)
+            \end{align*}
+
+            From this, we can finally obtain an expression for \( E(k) \):
+
+            \begin{align}
+                E(k) = E_0 \pm t_c \sqrt{1 + 4cos(k_x a)cos(k_y b) 4 cos^2(k_y b)}
+            \end{align}
+
+        \end{enumerate}
+
+    \section*{Problem 4}
+        \begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)] 
+            \item %4a
+
+            We can substitute (from the assignment) equation (24) into equation (23):
+
+            \begin{align}
+                \sum _k i \hbar \frac{\delta}{\delta t} c_k(t) \phi_k(x) e^{-i [E(k) / \hbar] t} &= \sum _k c_k(t) \hat{H_0} \phi(x)e^{-i [E(k) \ \hbar] t} \\
+                 &+ \sum _k c_k(t) U_s(x, t) \phi _k (x) e^ {-i [E(k) / \hbar] t}
+                 \label{eq:phi}
+            \end{align}
+
+            We can partially evaluate the derivative on the left hand side:
+
+            \begin{align*}
+                \frac{\delta}{\delta t} c_k(t) \phi _k(t)e^{-i [E(k) / \hbar]t} = \frac{\delta c_k(t)}{\delta t} \phi _k(t)e^{-i [E(k) / \hbar]t} - \frac{i}{\hbar} c_k(t)E(k)\phi _k (x) e^{-i [E(k) / \hbar]t}
+            \end{align*}
+
+            Expanding the rightmost term out into the summation, we get the expression \( \sum _k c_k(t) E(k) \phi _k(x) e^ {-i [E(k) / \hbar]t} \).
+            Since \( \hat{H_0} \phi _k(x) = E(k) \phi_k(x) \), we can see that this cancels out the term with \( \hat{H_0} \) in our first substition (\ref{eq:phi}),
+            and we get the final differential equation:
+
+            \begin{align*}
+                \sum_k c_k(t) U_s(x, t) \phi _k (x) e^{-i [E(k) / \hbar]t} = \sum_k i \hbar \frac{\delta c_k(t)}{\delta t} \phi_k(x) e^{-i [E(k) / \hbar] t}
+            \end{align*}
+
+            \item %4b
+            \begin{align*}
+                \sum_k c_k(t) \left[\int\phi_{k_f}^*(x)U_s(x,t) \phi_k(x)\:dx\right] e^{-i[E(k)/\hbar]t} &= \sum_k i\hbar \frac{\partial c_k(t)}{\partial t} \left[\int\phi_{k_f}^*(x)\phi_k(x)\:dx\right] e^{-i[E(k)/\hbar]t} \\
+                \sum_k c_k(t) I_{k_f k} e^{-i[E(k)/\hbar]t} &= \sum_k i\hbar \frac{\partial c_k(t)}{\partial t} \delta_{k_f k} e^{-i[E(k)/\hbar]t} \\
+                \sum_k c_k(t) I_{k_f k} e^{-i[E(k)/\hbar]t} &= i\hbar \frac{\partial c_k(t)}{\partial t} e^{-i[E(k)/\hbar]t} \\
+            \end{align*}
+            where \(I_{k_f k}\) is as defined in the assignment and \(\delta_{k_f k}\) is the Kronecker delta.
+            \item %4c
+
+            Starting with the initial equation
+
+            \begin{align*}
+                \sum _k c_k(t) I_{k_f k} e ^ {-i [ E(k) / \hbar ] t } = i \hbar \frac{\delta c_{k_f}(t)}{\delta t} e ^ {-i [ E(k_f) / \hbar ] t}
+            \end{align*}
+
+            Since we approximated that \( c_k = 1 \), only when \( k  = k_i \), and 0 otherwise, we can simplify
+            the sum on the left hand side to a single element, and divide out the exponentials.
+
+            \begin{align*}
+                I_{k_f k_i} e^{ i [- E(k_i) + E(k_f) / \hbar] t} = i \hbar \frac{\delta c_k(t)}{\delta t}
+            \end{align*}
+
+            Defining a new symbol \( \Lambda = [E(k_f) - E(k_i)] / \hbar \), we can simplify the equation to it's final form.
+
+            \begin{align}
+                I_{k_f k_i} e^{i \Lambda t} = i \hbar \frac{\delta c_k(t)}{\delta t}
+            \end{align}
+
+            \item %4d
+            \begin{align*}
+                \int \frac{\partial c_{k_f}(t)}{\partial t} &= \frac{1}{i\hbar} I_{k_fk_i} \int e^{i\Lambda t} \\
+                c_{k_f}(t) &= \frac{1}{i\hbar} I_{k_fk_i} \frac{1}{i\Lambda} e^{i\Lambda t} + C \\
+                c_{k_f}(t) &= \frac{1}{i\hbar} I_{k_fk_i} e^{i\Lambda t/2} \left[\frac{1}{i\Lambda} e^{i\Lambda t/2} + C e^{-i\Lambda t/2} \right]
+            \end{align*}
+            Let \(C=-\frac{1}{i\Lambda}\). We then have:
+            \begin{align*}
+                c_{k_f}(t) &= \frac{1}{i\hbar} I_{k_fk_i} e^{i\Lambda t/2} \frac{1}{i\Lambda} \left[ e^{i\Lambda t/2} - e^{-i\Lambda t/2} \right] \\
+                c_{k_f}(t) &= \frac{1}{i\hbar} I_{k_fk_i} e^{i\Lambda t/2} \frac{\sin(\Lambda t/2)}{\Lambda/2}.
+            \end{align*}
+
+            \item %4e
+
+
+        \end{enumerate}
 
 \end{document}