solid_state_physics/tutorial/2_03s.tex

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  30 \begin{document}
  31
  32 % header
  33 \begin{center}
  34  {\LARGE {\bf Materials Physics II}\\}
  35  \vspace{8pt}
  36  Prof. B. Stritzker\\
  37  SS 2008\\
  38  \vspace{8pt}
  39  {\Large\bf Tutorial 3 - proposed solutions}
  40 \end{center}
  41
  42 \vspace{8pt}
  43
  44 \section{Specific heat in the classical theory of the harmonic crystal -\\
  45          The law of Dulong and Petit}
  46
  47 \begin{enumerate}
  48  \item Energy:
  49        \begin{eqnarray}
  50        w&=&-\frac{1}{V}\frac{\partial}{\partial \beta}
  51        ln \int d\Gamma \exp(-\beta H)
  52        =-\frac{1}{V}\frac{1}{\int d\Gamma \exp(-\beta H)}
  53        \frac{\partial}{\partial \beta} \int d\Gamma \exp(-\beta H)\nonumber\\
  54        &=&-\frac{1}{V}\frac{1}{\int d\Gamma \exp(-\beta H)}
  55        \int d\Gamma \frac{\partial}{\partial \beta} \exp(-\beta H)\nonumber\\
  56        &=&-\frac{1}{V}\frac{1}{\int d\Gamma \exp(-\beta H)}
  57        \int d\Gamma \exp(-\beta H) (-H) \qquad \textrm{ q.e.d.} \nonumber
  58        \end{eqnarray}
  59  \item Potential energy:
  60        \[
  61        U=\frac{1}{2}\sum_{{\bf RR'}}\Phi({\bf r}({\bf R})-{\bf r}({\bf R'}))
  62         =\frac{1}{2}\sum_{{\bf RR'}}
  63          \Phi({\bf R}-{\bf R'}+{\bf u}({\bf R})-{\bf u}({\bf R'}))
  64        \]
  65        Using Taylor and
  66        $U_{\text{eq}}=\frac{1}{2}\sum_{{\bf R R'}} \Phi({\bf R}-{\bf R'})$:
  67        \[
  68        U=U_{\text{eq}}+
  69          \frac{1}{2}\sum_{{\bf RR'}}({\bf u}({\bf R})-{\bf u}({\bf R'}))
  70          \nabla\Phi({\bf R}-{\bf R'})+
  71          \frac{1}{4}\sum_{{\bf RR'}}
  72          [({\bf u}({\bf R})-{\bf u}({\bf R'})) \nabla]^2
  73          \Phi({\bf R}-{\bf R'}) + \mathcal{O}(u^3)
  74        \]
  75        Linear term:\\
  76        The coefficient of ${\bf u}({\bf R})$ is
  77        $\sum_{\bf R'}\nabla\Phi({\bf R}-{\bf R'})$
  78        which is minus the force excerted on atom ${\bf R}$
  79        by all other atoms in equlibrium positions.
  80        There is no net force on any atom in equlibrium.
  81        The linear term is zero.\\\\
  82        Harmonic term:\\
  83        $(a\nabla)^2 \Phi=
  84         a\nabla a\nabla \Phi=
  85         a\nabla \sum_u a_u \frac{\partial\Phi}{\partial r_u}=
  86         \sum_v \frac{\partial \sum_u a_u
  87         \frac{\partial\Phi}{\partial r_u}}{\partial r_v} a_v=
  88         \sum_{uv}\frac{\partial}{\partial r_v} a_u
  89         \frac{\partial \Phi}{\partial r_u} a_v=
  90         \sum_{uv}a_u \frac{\partial^2\Phi}{\partial r_u \partial r_v} a_v$\\
  91        \[\Rightarrow
  92        U_{\text{harm}}=\frac{1}{4}\sum_{\stackrel{{\bf R R'}}{\mu,v=x,y,z}}
  93        [u_{\mu}({\bf R})-u_{\mu}({\bf R'})]\Phi_{\mu v}({\bf R}-{\bf R'})
  94        [u_v({\bf R})-u_v({\bf R'})],
  95        \quad \Phi_{\mu v}({\bf r})=
  96         \frac{\partial^2 \Phi({\bf r})}{\partial r_{\mu}\partial r_v}.
  97        \]
  98  \item Change of variables:
  99        \[
 100        {\bf u}({\bf R})=\beta^{-1/2}\bar{{\bf u}}({\bf R}), \qquad
 101        {\bf P}({\bf R})=\beta^{-1/2}\bar{{\bf P}}({\bf R})
 102        \]
 103        \[
 104        \Rightarrow
 105        d{\bf u}({\bf R})=\beta^{-3/2}d\bar{{\bf u}}({\bf R}), \qquad
 106        d{\bf P}({\bf R})=\beta^{-3/2}d\bar{{\bf P}}({\bf R}), \qquad
 107        \]
 108        Kinetic energy contribution:
 109        \[
 110        H_{\text{kin}}=\frac{{\bf P}({\bf R})^2}{2M}
 111        \]
 112        Integral (using change of variables):
 113        \begin{eqnarray}
 114        \int d\Gamma \exp(-\beta H)&=&
 115        \int d\Gamma \exp\left[-\beta\left(\sum \frac{{\bf P}({\bf R})^2}{2M}+
 116        U_{\text{eq}} + U_{\text{harm}}\right)\right]\nonumber\\
 117        &=&
 118        \exp(-\beta U_{\text{eq}})\beta^{-3N}
 119        \LARGE(\int\prod_{{\bf R}}d\bar{{\bf u}}({\bf R})d\bar{{\bf P}}({\bf R})
 120        \nonumber\\
 121        &&\times \exp\left[
 122        -\sum\frac{1}{2M}{\bf P}({\bf R})^2
 123        -\frac{1}{4}\sum
 124        [\bar{u}_{\mu}({\bf R})-\bar{u}_{\mu}({\bf R'})]
 125        \Phi_{\mu v}({\bf R}-{\bf R'})
 126        [\bar{u}_v({\bf R})-\bar{u}_v({\bf R'})]
 127        \right]\LARGE)\nonumber
 128        \end{eqnarray}
 129        \[
 130        \Rightarrow w=-\frac{1}{V}\frac{\partial}{\partial \beta}
 131        ln\left((\exp(-\beta U_{\text{eq}})\beta^{-3N} \times \text{const}
 132        \right)
 133        =\frac{U_{\text{eq}}}{V}+3\frac{N}{V}k_{\text{B}}T
 134        =u_{\text{eq}}+3nk_{\text{B}}T
 135        \]
 136        \[
 137        \Rightarrow
 138        c_{\text{V}}=\frac{\partial w}{\partial T}=3nk_{\text{B}}
 139        \]
 140 \end{enumerate}
 141
 142 \section{Specific heat in the quantum theory of the harmonic crystal -\\
 143          The Debye model}
 144
 145 \[
 146 w=\frac{1}{V}\frac{\sum_i E_i \exp(-\beta E_i)}{\sum_i \exp(-\beta E_i)}.
 147 \]
 148 \begin{enumerate}
 149  \item Energy: $\rightarrow$ 1(a)
 150        \[
 151    w=-\frac{1}{V}\frac{\partial}{\partial \beta} ln \sum_i \exp(-\beta E_i).
 152        \]
 153  \item \begin{itemize}
 154         \item Total energy contribution of a particular normal mode:
 155               $(n_{{\bf k}s} + \frac{1}{2})\hbar\omega_s({\bf k})$
 156               with $n_{{\bf k}s}=0,1,2,\ldots$
 157         \item A state of the crystal is specified by the excitation numbers
 158               of the 3N normal modes.
 159         \item The total energy is the sum of the energies of the individual
 160               normal modes:\\
 161               $E=\sum_{{\bf k}s}(n_{{\bf k}s}+
 162                \frac{1}{2})\hbar\omega_s({\bf k})$
 163        \end{itemize}
 164        \begin{eqnarray}
 165        \Rightarrow
 166        w&=&-\frac{1}{V}\frac{\partial}{\partial \beta} ln\left(
 167        \prod_{{\bf k}s}(\exp(-\beta\hbar\omega_s({\bf k})/2)+
 168                         \exp(-3\beta\hbar\omega_s({\bf k})/2)+
 169                         \exp(-5\beta\hbar\omega_s({\bf k})/2)+
 170                         \ldots)
 171        \right)\nonumber\\
 172        &=&-\frac{1}{V}\frac{\partial}{\partial \beta} ln \prod_{{\bf k}s}
 173        \frac{\exp(-\beta\hbar\omega_s({\bf k})/2)}
 174             {1-\exp(-\beta\hbar\omega_s({\bf k}))}\nonumber\\
 175        &=&-\frac{1}{V}\frac{\partial}{\partial \beta} \sum_{{\bf k}s} ln
 176        \frac{\exp(-\beta\hbar\omega_s({\bf k})/2)}
 177             {1-\exp(-\beta\hbar\omega_s({\bf k}))}\nonumber\\
 178        &=&-\frac{1}{V}\sum_{{\bf k}s}
 179        \frac{1-\exp(-\beta\hbar\omega_s({\bf k}))}
 180             {\exp(-\beta\hbar\omega_s({\bf k})/2)}\nonumber\\
 181        &&\times
 182        \frac{(1-e^{-\beta\hbar\omega_s({\bf k})})
 183              e^{-\beta\hbar\omega_s({\bf k})/2}(-\hbar\omega_s({\bf k})/2)+
 184              e^{-\beta\hbar\omega_s({\bf k})/2}
 185              e^{-\beta\hbar\omega_s({\bf k})}(-\hbar\omega_s({\bf k}))}
 186             {(1-e^{-\beta\hbar\omega_s({\bf k})})^2}\nonumber\\
 187        &=&-\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})
 188        \frac{e^{-\beta\hbar\omega_s({\bf k})}-
 189              \frac{1}{2}(1-e^{-\beta\hbar\omega_s({\bf k})})}
 190             {1-e^{-\beta\hbar\omega_s({\bf k})}}\nonumber\\
 191        &=&-\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})
 192        \frac{\frac{1}{2}e^{-\beta\hbar\omega_s({\bf k})}-\frac{1}{2}}
 193             {1-e^{-\beta\hbar\omega_s({\bf k})}}
 194        =\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})\frac{1}{2}
 195        \frac{1+e^{\beta\hbar\omega_s({\bf k})}}
 196             {e^{\beta\hbar\omega_s({\bf k})}-1}\nonumber\\
 197        &=&\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})\frac{1}{2}
 198        \frac{1+e^{\beta\hbar\omega_s({\bf k})}}
 199             {e^{\beta\hbar\omega_s({\bf k})}-1}
 200        =\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})\frac{1}{2}
 201        \frac{2+e^{\beta\hbar\omega_s({\bf k})}-1}
 202             {e^{\beta\hbar\omega_s({\bf k})}-1}\nonumber\\
 203        &=&\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})
 204        (\frac{1}{e^{\beta\hbar\omega_s({\bf k})}-1}
 205         +\frac{e^{\beta\hbar\omega_s({\bf k})}-1}
 206               {2(e^{\beta\hbar\omega_s({\bf k})}-1)})
 207        =\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})
 208        (\underbrace{\frac{1}{e^{\beta\hbar\omega_s({\bf k})}-1}}_{n_s({\bf k})}
 209         +\frac{1}{2})\nonumber
 210        \end{eqnarray}
 211        $n_s({\bf k})$: Mean excitation number of the normal mode ${\bf k}s$ at
 212                        temperature $T$.
 213
 214  \item \[
 215        w=w_{\text{eq}}+
 216          \frac{1}{V}\sum_{{\bf k}s}\frac{1}{2}\hbar\omega_s({\bf k})+
 217          \frac{1}{V}\sum_{{\bf k}s}
 218          \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
 219        \]
 220  \item \[
 221        c_{\text{V}}=\frac{1}{V}\sum_{{\bf k}s}\frac{\partial}{\partial T}
 222        \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
 223        \]
 224        Large crystal:
 225        \[
 226        \lim_{v\rightarrow\infty}c_{\text{V}}=\frac{1}{V}\sum_{{\bf k}s}
 227        \frac{\partial}{\partial T}
 228        \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
 229        =\frac{\partial}{\partial T}
 230        \sum_s\int\frac{d{\bf k}}{(2\pi)^3}
 231        \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
 232        \]
 233  \item \begin{itemize}
 234         \item {\color{red}3} branches with Debye dispersion relation
 235               $w={\color{green}ck}$
 236         \item Volume of $k$-space per wave vector: $(2\pi)^3/V$\\
 237               $\Rightarrow (2\pi)^3N/V=4\pi k_{\text{D}}^3/3
 238                \Rightarrow n=\frac{N}{V}=\frac{k_{\text{D}}^3}{6\pi^2}$
 239               and $dn={\color{blue}\frac{k^2}{2\pi^2}dk}$
 240         \item Debye frequency: $\omega_{\text{D}}=k_{\text{D}}c$
 241         \item Debye temperature:
 242               $k_{\text{B}}\Theta_{\text{D}}=\hbar\omega_{\text{D}}$,
 243               $\Theta_{\text{D}}=\hbar ck_{\text{D}}/k_{\text{B}}$
 244        \end{itemize}
 245        Integral:
 246        \[
 247        c_{\text{V}}=\frac{\partial}{\partial T}\, {\color{red}3}\int_0^{k_D}
 248        {\color{blue}\frac{k^2}{2\pi^2}dk} \frac{\hbar {\color{green}ck}}
 249        {e^{\beta\hbar {\color{green}ck}}-1}=
 250        \frac{\partial}{\partial T}\frac{3\hbar c}{2\pi^2}\int_0^{k_D}
 251        \frac{k^3}{e^{\beta\hbar ck}-1}dk=
 252        \frac{3\hbar c}{2\pi^2}\int_0^{k_D}
 253        \frac{k^3e^{\beta\hbar ck}\beta\hbar ck\frac{1}{T}}
 254        {(e^{\beta\hbar ck}-1)^2}dk
 255        \]
 256        Change of variables: $\beta\hbar ck=x$
 257        \[
 258        \Rightarrow
 259        k=\frac{x}{\beta\hbar c} \quad \textrm{, } \quad
 260        dk=\frac{1}{\beta\hbar c} dx
 261        \]
 262        \[
 263        c_{\text{V}}=
 264        \]
 265 \end{enumerate}
 266
 267 \end{document}