314 lines
11 KiB
TeX
314 lines
11 KiB
TeX
\documentclass{article}
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\lstset{language=Matlab,%
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%emph=[2]{word1,word2}, emphstyle=[2]{style},
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% }
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\newcommand{\angstrom}{\textup{\AA}}
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\title{ECE 456 - Problem Set 3 (Part 1)}
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\date{2021-03-31}
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\author{David Lenfesty \\ lenfesty@ualberta.ca
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\and Phillip Kirwin \\ pkirwin@ualberta.ca}
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\pagestyle{fancy}
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\fancyhead[L]{\textbf{ECE 456} - Problem Set 3 (Part 1)}
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\fancyhead[R]{David Lenfesty and Phillip Kirwin}
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\fancyfoot[C]{Page \thepage}
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\renewcommand{\headrulewidth}{1pt}
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\renewcommand{\footrulewidth}{1pt}
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\begin{document}
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\doublespacing
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\maketitle
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\thispagestyle{empty}
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\singlespacing
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\newpage
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\section*{Problem 1}
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\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
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\item %1a
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %1bi
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The matrix equation is
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\begin{equation*}
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[\hat{H}]\left\{\phi\right\} = E \left\{\phi\right\},
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\end{equation*}
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where \([\hat{H}]\) is an \(N\)-by-\(N\) matrix with \([\hat{H}]_{nm} = 0\) except for the following elements:
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\begin{align*}
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&[\hat{H}]_{nn} = 2t_0 + U_n \\
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&[\hat{H}]_{n,n \pm 1} = -t_0 \\
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&[\hat{H}]_{0,N} = [\hat{H}]_{N,0} = -t_0,
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\end{align*}
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with \(t_0 = \hbar^2/(2ma^2)\) and \(U_n = U(na)\). The \(N\)-vector \(\left\{\phi\right\}\) has elements \(\phi_n\) which each represent
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the value of the eigenvector at the point \(na = x_n\).
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\item %1bii
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The expression of the wave function \( \phi (x) \) as a sum of basis functions is as below:
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\begin{equation*}
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\phi (x) = \sum _{n = 1} ^{N} \phi _n u_n(x)
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\end{equation*}
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The derived matrix equation:
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\begin{equation*}
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[\hat{H}] _u \{ \phi \} = [S]_u \{ \phi \}
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\end{equation*}
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Where \( [\hat{H}]_u \) is a matrix with the elements:
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\begin{equation*}
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H_{nm} = \int u _n ^* (x) \hat{H} u_m(x) dx
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\end{equation*}
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and \( [S]_u \) is a matrix with the elements:
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\begin{equation*}
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S_{nm} = \int u _n ^* (x) u_m(x) dx
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\end{equation*}
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\([\hat{H}]_u\) and \([S]_u\) are both of size \(N\)-by-\(N\). The elements
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of \(\left\{\phi\right\}\), \(\phi_n\), are the expansion coefficients of \(\phi(x)\).
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\end{enumerate}
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\item %1b
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %1bi
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code:
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\begin{lstlisting}
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%constants
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E1 = -13.6;
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R = 0.074;
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a0 = 0.0529;
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%matrix elements
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R0 = R/a0;
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a = 2*E1*(1-(1+R0)*exp(-2*R0))/R0;
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b = 2*E1*(1+R0)*exp(-R0);
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s = exp(-R0)*(1+R0+(R0^2/3));
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%matrices
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H_u = [E1 + a, E1*s+b; E1*s+b, E1 + a];
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S_u = [1, s; s, 1];
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%find eigenvalues and eigenvectors
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[vectors,energies] = eig(inv(S_u)*H_u);
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\end{lstlisting}
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The bonding and antibonding eigenenergies are \(\boxed{\SI{-32.2567}{eV}}\) and \(\boxed{\SI{-15.5978}{eV}}\) respectively.
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\item %1bii
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Neglecting normalization, we have the following expressions for \(\phi_B(z)\) and \(\phi_A(z)\):
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\begin{align*}
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\phi_B(z) = u_L(z) + u_R(z) \\
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\phi_A(z) = u_L(z) - u_R(z).
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\end{align*}
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We obtain the following plot:
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\begin{minipage}[t]{\linewidth}
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\centering
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\adjustbox{valign=t}{
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\includegraphics[width=0.5\textwidth]{q1bii.png}
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}
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\captionof{figure}{non-normalized probability densities for bonding and antibonding solutions.}
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\end{minipage}
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\end{enumerate}
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\end{enumerate}
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\newpage
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\section*{Problem 2}
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\begin{enumerate}[align=left,leftmargin=*,labelsep=1em,label=\bfseries(\alph*)]
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\item %2a
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %2ai
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\begin{minipage}[t]{\linewidth}
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\centering
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\adjustbox{valign=t}{
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\includegraphics[width=0.5\textwidth]{q2ai.png}
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}
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% TODO still don't like this caption
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\captionof{figure}{Energy vs. wave vector relationship (note: not discretized to account for N)}
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\end{minipage}
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\item %2aii
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Energy values from \( -2 \) eV to \(2\) eV are allowed.
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\item %2aiii
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The vector \( \{ \phi \} \), which is of length \( N \), and has elements \( n \) is:
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\begin{equation*}
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\{ \phi \} = C e ^{ i k \cdot n a }
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\end{equation*}
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The corresponding wave function is:
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\begin{equation*}
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\phi (x) = \sum _ {n = 1} ^ N C e ^ { i k \cdot n a } u_n (x)
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\end{equation*}
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There is one wave function and thus one energy level for each value of \( k \).
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This means that there is one electronic state per \( k \).
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\end{enumerate}
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\item %2b
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %2bi
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\begin{align*}
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\left\{\phi\right\} = \begin{bmatrix}
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C_Ae^{ika} \\
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C_Be^{ika} \\
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C_Ae^{ik2a} \\
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C_Be^{ik2a} \\
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\vdots \\
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C_Ae^{ikNa} \\
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C_Ae^{ikNa}
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\end{bmatrix}
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\end{align*}
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\item %2bii
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\begin{equation*}
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\phi(x) = \sum_{n=1}^{N} C_Ae^{ikna} u_{nA}(x) + C_Be^{ikna} u_{nB}(x)
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\end{equation*}
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\item %2biii
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\([h(k)]\) is of size 2-by-2. Thus there will be two values of \(E(k)\) for a fixed \(k\).
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This also means there are two \(\phi(x)\) for each \(k\).
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\end{enumerate}
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\item %2c
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\begin{enumerate}[align=left,leftmargin=*,labelsep=0em,label=(\roman*)]
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\item %2ci
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\begin{align*}
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\phi _2 + 2 \phi _3 + \phi _4 &= E \phi_1 \\
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\phi _1 + \phi _3 + 2 \phi _4 &= E \phi_2 \\
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2 \phi _1 + \phi _2 + \phi _4 &= E \phi_3 \\
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\phi _1 + 2 \phi _2 + \phi _3 &= E \phi_4 \\
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\end{align*}
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\item %2cii
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\begin{align*}
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\phi _2 + 2 \phi _3 + \phi _0 &= E \phi_1 \\
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\phi _1 + \phi _3 + 2 \phi _4 &= E \phi_2 \\
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2 \phi _5 + \phi _2 + \phi _4 &= E \phi_3 \\
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\phi _5 + 2 \phi _6 + \phi _3 &= E \phi_4 \\
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\end{align*}
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\item %2ciii
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A generalized form of the $n$th equation is:
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\begin{equation*}
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E \phi_n = \phi_{n-1} + \phi_{n+1} + 2 \phi_{n+2}
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\end{equation*}
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\item %2civ
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\begin{align}
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E C e ^{ikna} &= Ce ^ {ik (n + 1)a} + Ce^{ik (n - 1)a} + 2 Ce^{ik (n + 2)a} \nonumber \\
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E &= e^{ika} + e^{-ika} + 2 e^{2ika} \nonumber \\
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E(k) &= 2 e^{2ika} + 2\cos(ka) \label{eq:e_k}
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\end{align}
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\item %2cv
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Imposing the repeating boundary conditions \( \phi _{n + 4} = \phi _{n} \), We obtain the following relationship:
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\begin{align*}
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Ce^{ikna} &= Ce^{ik(n+4)a} \\
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1 &= e^{i4ka}
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\end{align*}
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For this to hold, \(4ka\) must be some multiple of \(2 \pi \), and this mean \( k = \frac{\pi}{2a} \cdot integer \).
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\item %2cvi
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Using the E-k relationship from equation \ref{eq:e_k},
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we know that \( k \) must always be real, so the \( 2\cos(ka) \) portion of the E-k relationship must be
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real. As well, if we substitute in the equation for \( k \) we obtained in part \nolinebreak (v), we get the following (partial) expression:
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\begin{equation*}
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2e^{i2ka} = 2 e^{i\pi \cdot integer}
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\end{equation*}
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Which we know will always be real (with a value of \( \pm 2 \)).
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\item %2cvii
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Since \(e^{2\pi n} = 1\), we have:
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\begin{align*}
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\phi_n(k+\frac{2\pi}{a}) &= Ce^{i(k+\frac{2\pi}{a} \cdot nA} = Ce^{i(k \cdot nA} \\
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\phi_n(k+\frac{2\pi}{a}) &= \phi_n(k).
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\end{align*}
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Therefore wavefunctions for which \(k\) is separated by \(\frac{2\pi}{a}\) are equivalent, and we
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only need consider the range \(k \in [-\frac{\pi}{a}, \frac{\pi}{a}]\).
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\end{enumerate}
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\end{enumerate}
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\end{document} |