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% header
\begin{center}
 {\LARGE {\bf Materials Physics II}\\}
 \vspace{8pt}
 Prof. B. Stritzker\\
 SS 2008\\
 \vspace{8pt}
 {\Large\bf Tutorial 3 - proposed solutions}
\end{center}

\vspace{8pt}

\section{Specific heat in the classical theory of the harmonic crystal -\\
         The law of Dulong and Petit}

\begin{enumerate}
 \item Energy:
       \begin{eqnarray}
       w&=&-\frac{1}{V}\frac{\partial}{\partial \beta}
       ln \int d\Gamma \exp(-\beta H)
       =-\frac{1}{V}\frac{1}{\int d\Gamma \exp(-\beta H)}
       \frac{\partial}{\partial \beta} \int d\Gamma \exp(-\beta H)\nonumber\\
       &=&-\frac{1}{V}\frac{1}{\int d\Gamma \exp(-\beta H)}
       \int d\Gamma \frac{\partial}{\partial \beta} \exp(-\beta H)\nonumber\\
       &=&-\frac{1}{V}\frac{1}{\int d\Gamma \exp(-\beta H)}
       \int d\Gamma \exp(-\beta H) (-H) \qquad \textrm{ q.e.d.} \nonumber
       \end{eqnarray}
 \item Potential energy:
       \[
       U=\frac{1}{2}\sum_{{\bf RR'}}\Phi({\bf r}({\bf R})-{\bf r}({\bf R'}))
        =\frac{1}{2}\sum_{{\bf RR'}}
         \Phi({\bf R}-{\bf R'}+{\bf u}({\bf R})-{\bf u}({\bf R'}))
       \]
       Using Taylor and
       $U_{\text{eq}}=\frac{1}{2}\sum_{{\bf R R'}} \Phi({\bf R}-{\bf R'})$:
       \[
       U=U_{\text{eq}}+
         \frac{1}{2}\sum_{{\bf RR'}}({\bf u}({\bf R})-{\bf u}({\bf R'}))
	 \nabla\Phi({\bf R}-{\bf R'})+
	 \frac{1}{4}\sum_{{\bf RR'}}
	 [({\bf u}({\bf R})-{\bf u}({\bf R'})) \nabla]^2
	 \Phi({\bf R}-{\bf R'}) + \mathcal{O}(u^3)
       \]
       Linear term:\\
       The coefficient of ${\bf u}({\bf R})$ is
       $\sum_{\bf R'}\nabla\Phi({\bf R}-{\bf R'})$
       which is minus the force excerted on atom ${\bf R}$
       by all other atoms in equlibrium positions.
       There is no net force on any atom in equlibrium.
       The linear term is zero.\\\\
       Harmonic term:\\
       $(a\nabla)^2 \Phi=
        a\nabla a\nabla \Phi=
	a\nabla \sum_u a_u \frac{\partial\Phi}{\partial r_u}=
	\sum_v \frac{\partial \sum_u a_u
	\frac{\partial\Phi}{\partial r_u}}{\partial r_v} a_v=
	\sum_{uv}\frac{\partial}{\partial r_v} a_u
	\frac{\partial \Phi}{\partial r_u} a_v=
	\sum_{uv}a_u \frac{\partial^2\Phi}{\partial r_u \partial r_v} a_v$\\
       \[\Rightarrow
       U_{\text{harm}}=\frac{1}{4}\sum_{\stackrel{{\bf R R'}}{\mu,v=x,y,z}}
       [u_{\mu}({\bf R})-u_{\mu}({\bf R'})]\Phi_{\mu v}({\bf R}-{\bf R'})
       [u_v({\bf R})-u_v({\bf R'})],
       \quad \Phi_{\mu v}({\bf r})=
        \frac{\partial^2 \Phi({\bf r})}{\partial r_{\mu}\partial r_v}.
       \]
 \item Change of variables:
       \[
       {\bf u}({\bf R})=\beta^{-1/2}\bar{{\bf u}}({\bf R}), \qquad
       {\bf P}({\bf R})=\beta^{-1/2}\bar{{\bf P}}({\bf R})
       \]
       \[
       \Rightarrow
       d{\bf u}({\bf R})=\beta^{-3/2}d\bar{{\bf u}}({\bf R}), \qquad
       d{\bf P}({\bf R})=\beta^{-3/2}d\bar{{\bf P}}({\bf R}), \qquad
       \]
       Kinetic energy contribution:
       \[
       H_{\text{kin}}=\frac{{\bf P}({\bf R})^2}{2M}
       \]
       Integral (using change of variables):
       \begin{eqnarray}
       \int d\Gamma \exp(-\beta H)&=&
       \int d\Gamma \exp\left[-\beta\left(\sum \frac{{\bf P}({\bf R})^2}{2M}+
       U_{\text{eq}} + U_{\text{harm}}\right)\right]\nonumber\\
       &=&
       \exp(-\beta U_{\text{eq}})\beta^{-3N}
       \LARGE(\int\prod_{{\bf R}}d\bar{{\bf u}}({\bf R})d\bar{{\bf P}}({\bf R})
       \nonumber\\
       &&\times \exp\left[
       -\sum\frac{1}{2M}\bar{{\bf P}}({\bf R})^2
       -\frac{1}{4}\sum
       [\bar{u}_{\mu}({\bf R})-\bar{u}_{\mu}({\bf R'})]
       \Phi_{\mu v}({\bf R}-{\bf R'})
       [\bar{u}_v({\bf R})-\bar{u}_v({\bf R'})]
       \right]\LARGE)\nonumber
       \end{eqnarray}
       \[
       \Rightarrow w=-\frac{1}{V}\frac{\partial}{\partial \beta}
       ln\left((\exp(-\beta U_{\text{eq}})\beta^{-3N} \times \text{const}
       \right)
       =\frac{U_{\text{eq}}}{V}+3\frac{N}{V}k_{\text{B}}T
       =u_{\text{eq}}+3nk_{\text{B}}T
       \]
       \[
       \Rightarrow
       c_{\text{V}}=\frac{\partial w}{\partial T}=3nk_{\text{B}}
       \]
\end{enumerate}

\section{Specific heat in the quantum theory of the harmonic crystal -\\
         The Debye model}

\[
w=\frac{1}{V}\frac{\sum_i E_i \exp(-\beta E_i)}{\sum_i \exp(-\beta E_i)}.
\]
\begin{enumerate}
 \item Energy: $\rightarrow$ 1(a)
       \[
   w=-\frac{1}{V}\frac{\partial}{\partial \beta} ln \sum_i \exp(-\beta E_i).
       \]
 \item \begin{itemize}
        \item Total energy contribution of a particular normal mode:
              $(n_{{\bf k}s} + \frac{1}{2})\hbar\omega_s({\bf k})$
	      with $n_{{\bf k}s}=0,1,2,\ldots$
        \item A state of the crystal is specified by the excitation numbers
              of the 3N normal modes.
        \item The total energy is the sum of the energies of the individual
	      normal modes:\\
	      $E=\sum_{{\bf k}s}(n_{{\bf k}s}+
	       \frac{1}{2})\hbar\omega_s({\bf k})$
       \end{itemize}
       \begin{eqnarray}
       \Rightarrow
       w&=&-\frac{1}{V}\frac{\partial}{\partial \beta} ln\left(
       \prod_{{\bf k}s}(\exp(-\beta\hbar\omega_s({\bf k})/2)+
                        \exp(-3\beta\hbar\omega_s({\bf k})/2)+
                        \exp(-5\beta\hbar\omega_s({\bf k})/2)+
			\ldots)
       \right)\nonumber\\
       &=&-\frac{1}{V}\frac{\partial}{\partial \beta} ln \prod_{{\bf k}s}
       \frac{\exp(-\beta\hbar\omega_s({\bf k})/2)}
            {1-\exp(-\beta\hbar\omega_s({\bf k}))}\nonumber\\
       &=&-\frac{1}{V}\frac{\partial}{\partial \beta} \sum_{{\bf k}s} ln
       \frac{\exp(-\beta\hbar\omega_s({\bf k})/2)}
            {1-\exp(-\beta\hbar\omega_s({\bf k}))}\nonumber\\
       &=&-\frac{1}{V}\sum_{{\bf k}s}
       \frac{1-\exp(-\beta\hbar\omega_s({\bf k}))}
            {\exp(-\beta\hbar\omega_s({\bf k})/2)}\nonumber\\
       &&\times
       \frac{(1-e^{-\beta\hbar\omega_s({\bf k})})
             e^{-\beta\hbar\omega_s({\bf k})/2}(-\hbar\omega_s({\bf k})/2)+
	     e^{-\beta\hbar\omega_s({\bf k})/2}
	     e^{-\beta\hbar\omega_s({\bf k})}(-\hbar\omega_s({\bf k}))}
            {(1-e^{-\beta\hbar\omega_s({\bf k})})^2}\nonumber\\
       &=&-\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})
       \frac{{\color{red}-}e^{-\beta\hbar\omega_s({\bf k})}-
             \frac{1}{2}(1-e^{-\beta\hbar\omega_s({\bf k})})}
	    {1-e^{-\beta\hbar\omega_s({\bf k})}}\nonumber\\
       &=&-\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})
       \frac{{\color{red}-}\frac{1}{2}e^{-\beta\hbar\omega_s({\bf k})}-\frac{1}{2}}
	    {1-e^{-\beta\hbar\omega_s({\bf k})}}
       =\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})\frac{1}{2}
       \frac{e^{-\beta\hbar\omega_s({\bf k})}+1}
            {1-e^{-\beta\hbar\omega_s({\bf k})}}\cdot
       \frac{e^{\beta\hbar\omega_s({\bf k})}}{e^{\beta\hbar\omega_s({\bf k})}}
       \nonumber\\
       &=&\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})\frac{1}{2}
       \frac{1+e^{\beta\hbar\omega_s({\bf k})}}
            {e^{\beta\hbar\omega_s({\bf k})}-1}
       =\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})\frac{1}{2}
       \frac{2+e^{\beta\hbar\omega_s({\bf k})}-1}
            {e^{\beta\hbar\omega_s({\bf k})}-1}\nonumber\\
       &=&\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})
       (\frac{1}{e^{\beta\hbar\omega_s({\bf k})}-1}
        +\frac{e^{\beta\hbar\omega_s({\bf k})}-1}
              {2(e^{\beta\hbar\omega_s({\bf k})}-1)})
       =\frac{1}{V}\sum_{{\bf k}s}\hbar\omega_s({\bf k})
       (\underbrace{\frac{1}{e^{\beta\hbar\omega_s({\bf k})}-1}}_{n_s({\bf k})}
        +\frac{1}{2})\nonumber
       \end{eqnarray}
       $n_s({\bf k})$: Mean excitation number of the normal mode ${\bf k}s$ at
                       temperature $T$.

 \item \[
       w=w_{\text{eq}}+
         \frac{1}{V}\sum_{{\bf k}s}\frac{1}{2}\hbar\omega_s({\bf k})+
         \frac{1}{V}\sum_{{\bf k}s}
	 \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
       \]
 \item \[
       c_{\text{V}}=\frac{1}{V}\sum_{{\bf k}s}\frac{\partial}{\partial T}
       \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
       \]
       Large crystal:
       \[
       \lim_{v\rightarrow\infty}c_{\text{V}}=\frac{1}{V}\sum_{{\bf k}s}
       \frac{\partial}{\partial T}
       \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
       =\frac{\partial}{\partial T}
       \sum_s\int\frac{d{\bf k}}{(2\pi)^3}
       \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1}
       \]
 \item \begin{itemize}
        \item {\color{red}3} branches with Debye dispersion relation
	      $w={\color{green}ck}$
	\item Volume of $k$-space per wave vector: $(2\pi)^3/V$\\
	      $\Rightarrow (2\pi)^3N/V=4\pi k_{\text{D}}^3/3
	       \Rightarrow n=\frac{N}{V}=\frac{k_{\text{D}}^3}{6\pi^2}$,
	       $k_{\text{D}}^3=6\pi^2 n$
	\item $dn={\color{blue}\frac{k^2}{2\pi^2}dk}$
	\item Debye frequency: $\omega_{\text{D}}=k_{\text{D}}c$
	\item Debye temperature:
	      $k_{\text{B}}\Theta_{\text{D}}=\hbar\omega_{\text{D}}$,
	      $\Theta_{\text{D}}=\hbar ck_{\text{D}}/k_{\text{B}}$,
	      $\Theta_{\text{D}}^3=\frac{\hbar^3c^3k_{\text{D}}^3}
	      {k_{\text{B}}^3}=
	      \frac{\hbar^3c^3}{k_{\text{B}}^3}6\pi^2n$
       \end{itemize}
       Integral:
       \[
       c_{\text{V}}=\frac{\partial}{\partial T}\, {\color{red}3}\int_0^{k_D}
       {\color{blue}\frac{k^2}{2\pi^2}dk} \frac{\hbar {\color{green}ck}}
       {e^{\beta\hbar {\color{green}ck}}-1}=
       \frac{\partial}{\partial T}\frac{3\hbar c}{2\pi^2}\int_0^{k_D}
       \frac{k^3}{e^{\beta\hbar ck}-1}dk=
       \frac{3\hbar c}{2\pi^2}\int_0^{k_D}
       \frac{k^3e^{\beta\hbar ck}\beta\hbar ck\frac{1}{T}}
       {(e^{\beta\hbar ck}-1)^2}dk
       \]
       Change of variables: $\beta\hbar ck=x$
       \[
       \Rightarrow
       k=\frac{x}{\beta\hbar c} \quad \textrm{, } \quad
       dk=\frac{1}{\beta\hbar c} dx
       \]
       \[
       c_{\text{V}}=\frac{3\hbar c}{2\pi^2}\int_0^{\Theta_D/T}
       \frac{x^3e^xx}{T(\beta\hbar c)^3(e^x-1)^2}\frac{dx}{\beta\hbar c}=
       \frac{3}{2\pi^2T\beta^4\hbar^3 c^3}\int_0^{\Theta_D/T}
       \frac{x^4e^x}{(e^x-1)^2}dx
       \]
       \[
       \frac{3}{2\pi^2T\beta^4\hbar^3 c^3}=
       \frac{3k_{\text{B}}}{2\pi^2\beta^3\hbar^3 c^3}=
       \frac{3k_{\text{B}}T^33n}{\Theta_{\text{D}}^3}
       \]
       \[
       \Rightarrow
       c_{\text{V}}=9nk_{\text{B}}\left(\frac{T}{\Theta_{\text{D}}}
       \right)^3\int_0^{\Theta_D/T}\frac{x^4e^x}{(e^x-1)^2}dx
       \]
\end{enumerate}

\end{document}
