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        \end{eqnarray}
 176
 177  \item Evaluate the expression of the energy density.
 178        {\bf Hint:}
 179        The energy levels of a harmonic crystal of N ions
 180        can be regarded as 3N independent oscillators,
 181        whose frequencies are those of the 3N classical normal modes.
 182        The contribution to the total energy of a particular normal mode
 183        with angular frequency $\omega_s({\bf k})$
 184        ($s$: branch, ${\bf k}$: wave vector) is given by
 185        $(n_{{\bf k}s} + \frac{1}{2})\hbar\omega_s({\bf k})$ with the
 186        excitation number $n_{{\bf k}s}$ being restricted to integers greater
 187        or equal zero.
 188        The total energy is given by the sum over the energies of the individual
 189        normal modes.
 190        Use the totals formula of the geometric series to expcitly calculate
 191        the sum of the exponential functions.
 192  \item Separate the above result into a term vanishing as $T$ goes to zero and
 193        a second term giving the energy of the zero-point vibrations of the
 194        normal modes.
 195  \item Write down an expression for the specific heat.
 196        Consider a large crystal and thus replace the sum over the discrete
 197        wave vectors with an integral.
 198  \item Debye replaced all branches of the vibrational spectrum with three
 199        branches, each of them obeying the dispersion relation
 200        $w=ck$.
 201        Additionally the integral is cut-off at a radius $k_{\text{D}}$
 202        to have a total amount of N allowed wave vectors.
 203        Determine $k_{\text{D}}$.
 204        Evaluate the simplified integral and introduce the
 205        Debye frequency $\omega_{\text{D}}=k_{\text{D}}c$
 206        and the Debye temperature $\Theta_{\text{D}}$ which is given by
 207        $k_{\text{B}}\Theta_{\text{D}}=\hbar\omega_{\text{D}}$.
 208        Write down the resulting expression for the specific heat.
 209 \end{enumerate}
 210
 211 \end{document}