diff --git a/PS3/doc.pdf b/PS3/doc.pdf index ea50fe1..a193376 100644 Binary files a/PS3/doc.pdf and b/PS3/doc.pdf differ diff --git a/PS3/doc.tex b/PS3/doc.tex index e4dc6fd..3bae04d 100644 --- a/PS3/doc.tex +++ b/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} \ No newline at end of file