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\begin{document}
\centerline{\textbf{Lösungen zum Übungsblatt 9}  }
\vspace{0,5cm} 
Johannes Dörr, Jan Schumann-Bischoff \hfill Übungsgruppe 2, Jürgen Lampe\\
\mbox{}\hrulefill\\
\vspace {1cm}

\textbf{Problem 1: Particle number fluctuations} \\ 
	We concider our system as a grand canonical ensemble because the particle number $N$ is not given exactly. We want to compute the particle number fluctuations $\mean{(N-\Nq)^2}$ with $\mean{N}=\N$ at fixed temperature $T$.
	
	The following approach was splitten up in several steps. In step (i) we show the Maxwell relation ($V$: Volume, $P$: pressure):
	\begin{equation}
		\pd{\mu}{V}{T,N} = -\pd{P}{N}{T,V}
		\label{eq:maxwell}
	\end{equation}
	With this result we'll show the following relation for homogeneous systems with the isothermal compressibility $\kappa_T$ (step(ii)):
	\begin{equation}
		\pd{\mu}{N}{T,V} = \frac{V}{N^2} \frac{1}{\kappa_T}
		\label{eq:homogen}
	\end{equation}
	
	\emph{Proof to (i):} \\
	Given is the differential for the free energy:
	\begin{gather*}
		\diff{F} = -S \diff{T}-P\diff{V}+\mu N \\
		\Rightarrow \pd{F}{N}{T,V} = \mu \mathrm{ ,    } \pd{F}{V}{T,N}=-P \\
		\Rightarrow \pd{\mu}{V}{T,N} = \frac{\partial^2 F}{\partial V \partial N}
					= \frac{\partial^2 F}{\partial N \partial V} = -\pd{P}{N}{T,V}
	\end{gather*}
	In the last step we avail that the order of differentiations is equal. 
	\begin{flushright}
	q.e.d.
	\end{flushright}
	
	\emph{Proof to (ii):} \\
	Our system is a homogeneous system. It means, that for the pressure (for example) we can write:
	\begin{displaymath}
		P(T,V,N) = P(T, \alpha V,\alpha N)
	\end{displaymath}
	If we make our system bigger, the volume and the particle number are growing proportional to $\alpha$. $T$ and $P$ don't. They are \textit{intensive} variables. After differentiation to $\alpha$ we obtain (at the point $\alpha=1)$:
	\begin{align}
		\frac{\diff{P(T,V,N)}}{\diff{\alpha}}	&=	\frac{\diff{P(T, \alpha V,\alpha N)}}{\diff{\alpha}} \notag \\
		&=	\left( \frac{\partial P}{\partial T} \frac{\diff{T}}{\diff{\alpha}}
			+ \frac{\partial P}{\partial \alpha V} \frac{\diff{\alpha V}}{\diff{\alpha}}
			+ \frac{\partial P}{\partial \alpha N} \frac{\diff{\alpha N}}{\diff{\alpha}} \right)_{\alpha=1} \notag \\
		&=	\pd{P}{V}{T,N}V + \pd{P}{N}{T,V}N=0
		\label{eq:zw1} 
	\end{align}
	Analogue we can show for $\mu(T,V,N)=\mu(T,\alpha V,\alpha N)$ ($\mu$ is also an intensive variable):
	\begin{displaymath}
		\pd{\mu}{V}{T,N}V + \pd{\mu}{N}{T,V}N = 0
	\end{displaymath}
	With the definition of the isothermal compressibility $\kappa_T = -\frac{1}{V} \pd{V}{P}{T,N}$ and \eqref{eq:maxwell} we obtain:
	\begin{align}
		\pd{\mu}{N}{T,V}	&=	-\frac{V}{N} \pd{\mu}{V}{T,N}	\notag \\
		&=	\frac{V}{N} \pd{P}{N}{T,N} \notag \\
		&=	-\frac{V^2}{N^2} \pd{P}{V}{T,N} \notag \\
		&=	\frac{V}{N^2} \frac{1}{\kappa_T} \notag
	\end{align}
	\begin{flushright}
	q.e.d.
	\end{flushright}
	
	
	Now we concider our grand canonical enselmble with the partition sum $Z_G$:
	\begin{displaymath}
		Z_G = \sum\limits_{N=1}^{\infty} \int [\mathrm{d}\Gamma ]e^{-\beta (H - \mu N)}
	\end{displaymath}
	We know for the mean of $N$:
	\begin{displaymath}
		\N = \frac{1}{\beta Z_G} \frac{\partial Z_G}{\partial \mu}
	\end{displaymath}
	We compute $\frac{\partial Z_G}{\partial \mu}$ and $\frac{\partial^2 Z_G}{\partial \mu^2}$because 
	we'll need it for the next calculation:
	\begin{align}
		\frac{\partial Z_G}{\partial \mu}
		&=	\beta \sum\limits_{N=1}^{\infty} \int [\mathrm{d}\Gamma ]N e^{-\beta (H - \mu N)}
			= \beta \N Z_G  \notag \\
		\frac{\partial^2 Z_G}{\partial \mu^2}
		&=	\beta^2 \sum\limits_{N=1}^{\infty} \int [\mathrm{d}\Gamma ]N^2 e^{-\beta (H - \mu N)}
			= \beta^2 \mean{N^2} Z_G \notag \\
	\end{align}
	Differentation of $\N$ to $\mu$:
	\begin{align}
		\pd{\N}{\mu}{T,V}
		&=	\beta^{-1} \left[Z_G^{-1} \frac{\partial^2 Z_G}{\partial^2 \mu}  
			+ \frac{\partial Z_G^{-1}}{\partial \mu} \frac{\partial Z_G}{\partial \mu} \right] \notag \\
		&=	\beta^{-1} \left[Z_G^{-1} \frac{\partial^2 Z_G}{\partial^2 \mu}  
			- Z_G^{-2} \frac{\partial Z_G}{\partial \mu} \frac{\partial Z_G}{\partial \mu} \right] \notag \\
		&=	\beta \left( \mean{N^2} - \N^2 \right) \notag \\
		&=	\beta \mean{(N-\mean{N})^2}
		\label{eq:zw2}
	\end{align}
	After comparing \eqref{eq:homogen} and \eqref{eq:zw2} we obtain ($\beta = k_BT$):
	\begin{displaymath}
		\kappa_T = \frac{k_BTV}{\N^2}\mean{(N-\mean{N})^2}
	\end{displaymath}
	With this equation you can compute the variance of $N$ by known $V$ and $\N$. \\


\textbf{Problem 2: Thermodynamical potentials} \\
	In the considered system with Volume $V$, particle number $N$ and temperature $T$,
	the free energy $F$ is given by:
	\begin{align}
		F(T, V)	= - N \kb T \ln(C_0 V) - N \kb T \ln(C_1[\kb T]^\alpha)
	\label{eq:FreeEnergy}
	\end{align}
	with $C_0, C_1 = \tx{const} >1$. We regard $N$ as constant in the following.
	
	\begin{itemize}
		\item[(a)] The entropy $S(T,V)$ is given by:
 							 \begin{align}
								 S(T, V) &= - \left( \frac{\partial F}{\partial T} \right)_{V} \notag \\
								 				 &= N \kb \cdot \left(\ln(C_0 V) + \ln(C_1 [\kb T]^\alpha) + \alpha \right) \label{eq:ResA} \ \ .
							 \end{align} 
							 
		\item[(b)] With the definition of the free energy
 							 \begin{align*}
								 F(T, V) &= E(T, V) - T \cdot S(T,V) \ \ ,
							 \end{align*} 		
							 we obtain using \eqref{eq:FreeEnergy} and \eqref{eq:ResA}:
 							 \begin{align}
								 E(T, V) &= F(T, V) + T \cdot S(T,V) \notag \\
								 				 &= \alpha N \kb T \label{eq:ResB} \ \ .
							 \end{align}
		\item[(c)] Now we have to replace $T$ by $S$ in \eqref{eq:ResB}, what is done using \eqref{eq:ResA}:
 							 \begin{align*}
								 T &= \frac{1}{\kb} \cdot \exp \left(\frac{S}{\alpha N \kb} - 1\right) \cdot \left(\frac{1}{C_0 C_1 V}\right) \\
								 \Rightarrow \ \ E(S,V) &= \alpha N \kb \frac{1}{\kb} \cdot \exp \left(\frac{S}{\alpha N \kb} - 1\right) 
								 													\cdot \left(\frac{1}{C_0 C_1 V}\right) \ \ .
							 \end{align*} 									 		
		\item[(d)] The pressure is computed using \eqref{eq:FreeEnergy} as follows:
 							 \begin{align}
								 p(T, V) &= -\left(\frac{\partial F}{\partial V}\right)_T \notag \\
								 				 &= \frac{N \kb T}{V} \label{eq:pressure} \ \ .
							 \end{align}		
		\item[(e)] With \eqref{eq:pressure}, we obtain:
 							 \begin{align}
								 V(T,p) &= \frac{N \kb T}{p} \ \ .
							 \label{eq:volume}
							 \end{align}					
							 The free enthalpy is given by:
 							 \begin{align*}
								 G(T, p) &= F + pV \\
								 				 &= -N \kb T \cdot \left(\ln(C_0 V) + \ln(C_1[\kb T]^\alpha) - 1\right) \\
								 				 &= N \kb T \cdot \left(1 - \ln(C_0 C_1 V[\kb T]^\alpha)\right) \\
								 				 &= N \kb T \cdot \left(1 - \ln \left(C_0 C_1 \frac{N}{p}[\kb T]^{\alpha+1}\right)\right) \ \ .
							 \end{align*}	
		\item[(f)] To get the specific heat, we proceed as follows:
 							 \begin{align*}
								 c_V &= \frac{T}{m} \left(\frac{\partial S}{\partial T}\right)_V \\
								 		 &= \frac{\alpha N \kb}{m} \ \ ,
							 \end{align*}
							 where $m$ is the mass of all $N$ particles.
		\item[(g)] With its definition, we get for the isothermal compressibility using \eqref{eq:volume}:
 							 \begin{align*}
								 \kappa_T &= - \frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_T \\
								 					&= \frac{1}{p} \ \ .
							 \end{align*}												 
	\end{itemize}
	
\textbf{Problem 3: A simple model of thermodynamics of lattice vibrations} \\
	\newcommand{\HOp}[0]{ 
		\hat{H}
	}	
	
	\textbf{Introduction} \\
		We look at a simple model describing the thermodynamical behaviour of solids. The so-called
		{\it Einstein model} uses the following assumptions: each atom is considered as
		harmonnic oscillator without any interaction to other atoms, and all atoms have
		the same frequency of oscillation. We will discuss the validity of this system. \\
		
	\textbf{Methods} \\
		At first, we replace the operator $\hat{a}_i^+ \hat{a}_i=\hat{N}$ in the Hamiltonian with 
		its eigenvalue, since the system is in an eigenstate. Then we calculate the canonical 
		partition sum, that is required for getting energy, average energy, free energy and 
		specific heat. After that, we look at the behaviour of the specific heat for 
		$T \rightarrow 0$ and $T \rightarrow \infty$. \\
		
	\textbf{Results} \\
	  \begin{itemize} 
	    \item[a)] We get the following results:
			  \begin{align*}
				  Z(\beta, N) &= \left(\frac{e^{-\frac{\beta \hbar \omega}{2}}}{1-e^{-\beta \hbar \omega}}\right)^N \\
				  F(\beta, N) &= \frac{\hbar \omega N}{2} + \frac{N}{\beta} \ln \left(1 - e^{-\beta \hbar \omega} \right) \\
					\langle E \rangle &= \frac{1}{2} N \hbar \omega + \frac{N \hbar \omega}{e^{\beta \hbar \omega} -1} \\
				  c_V(T) &= \frac{N \hbar^2 \omega^2 \cdot \exp\left(\frac{\hbar \omega} 
										   			 				 {\kb T}\right)}{\kb T^2 \cdot \exp \left(\frac{\hbar \omega}{\kb T} - 1\right)^2} \\
					E(T,N) &= \frac{1}{2} N \hbar \omega + \frac{N \hbar \omega}{1 - \exp \left(-\frac{\hbar \omega}{\kb T}\right)} \ \ .
			  \end{align*} 
				In Figure \ref{fig:graph_E} and \ref{fig:graph_c_V}, the last two are plotted.

				\begin{figure}[htbp]
					\centering
						\includegraphics[width=0.70\textwidth]{graph_E.png}
					\caption{The energy $E(T,N)/N$ for $\hbar \omega = 1$ and $\kb=1$}
					\label{fig:graph_E}
				\end{figure}

				\begin{figure}[htbp]
					\centering
						\includegraphics[width=0.70\textwidth]{graph_cv.png}
					\caption{The specific heat $c_V(T) / N$ for $\hbar \omega = 1$ and $\kb=1$}
					\label{fig:graph_c_V}
				\end{figure}
	
	    \item[b)] For the cases $T \rightarrow 0$ and $T \rightarrow \infty$, we obtain:
			  \begin{align*}
					\lim_{T \rightarrow 0} c_V(T) &= 0 \\
				  \lim_{T \rightarrow \infty} c_V(T) &= N \kb \ \ .
			  \end{align*} 
	    
	  \end{itemize} 
\newpage
	\textbf{Discussion} \\
		For high temperatures, the equation $c_V(T) = N \kb$ corresponds to the experimental
		results. However, for temperatures $\rightarrow 0$ the deviation increases. This comes
		from the assumption of no interaction between the atoms. That fact becomes important
		when the thermodynamic oscillation frequency is in the range of sound waves.
		
		The collapse of this model at temperatures near absolute zero led to the {\it Debye modell}. \\
	
	\textbf{Appendix}
	  \begin{itemize} 
	    \item[a)] The Hamiltonian is given by:
 							  \begin{align}
								  \HOp = \sum\limits_{i=1}^{N} \hbar \omega \left(\hat{a}_i^+ \hat{a}_i + \frac{1}{2}\right) 
								  		 = \sum\limits_{i=1}^{N} \hbar \omega \left(\hat{N}_i+ \frac{1}{2}\right) 
								  		 = \sum\limits_{i=1}^{N} \hbar \omega \left(n_i+ \frac{1}{2}\right) \ \ ,
								\label{eq:Hamiltonian}
							  \end{align} 
	    					where the last step uses, that the system state is an eigenstate. With \eqref{eq:Hamiltonian},
	    					we get the partition sum:
 							  \begin{align}
								  Z(\beta, N) &= \tx{Tr} \ e^{-\beta \HOp} 
								  		= \sum\limits_{j=0}^{\infty} e^{-\beta \hbar \omega \sum_{i=0}^N n_i + \frac{1}{2}} 
								  		= \prod_{i=1}^N \sum\limits_{j=0}^{\infty} e^{-\beta \hbar \omega \left(n_i + \frac{1}{2}\right)} \notag \\
								  	 &= \prod_{i=1}^N \frac{e^{-\frac{\beta \hbar \omega}{2}}}{1-e^{-\beta \hbar \omega}} 
								  		= \left(\frac{e^{-\frac{\beta \hbar \omega}{2}}}{1-e^{-\beta \hbar \omega}}\right)^N \ \ .
								\label{eq:PartialSum}
							  \end{align} 
								We compute the free energy:
 							  \begin{align*}
								  F(\beta, N) = - \frac{1}{\beta} \ln Z(\beta, N) 
								  						= \frac{\hbar \omega N}{2} + \frac{N}{\beta} \ln \left(1 - e^{-\beta \hbar \omega} \right)
							  \end{align*} 
								and the average energy:
 							  \begin{align*}
								  \langle E \rangle &= - \frac{\partial}{\partial \beta} \ln Z(\beta, N) \\
								  									&= N \frac{\partial}{\partial \beta} \frac{\beta \hbar \omega}{2} 
								  									 + N \frac{\partial}{\partial \beta} \ln \left(1 - e^{-\beta \hbar \omega} \right) \\
								  									&= \frac{1}{2} N \hbar \omega 
								  									 + \frac{N \hbar \omega e^{-\beta \hbar \omega}}{1-e^{-\beta \hbar \omega}} 
								  									 = \frac{1}{2} N \hbar \omega + \frac{N \hbar \omega}{e^{\beta \hbar \omega} -1} \ \ .
							  \end{align*} 								
								With that, we compute the specific heat (with $\beta = \frac{1}{\kb T}$):
 							  \begin{align*}
								  c_V(T) &= \frac{\partial \langle E \rangle}{\partial T} \\
								   			 &= \frac{\partial}{\partial T} \frac{N \hbar \omega}{\exp \left(\frac{\hbar \omega}{\kb T}-1 \right)} \\
								   			 &= -N \hbar \omega \cdot \frac{\exp(\frac{\hbar \omega}{\kb T})(-\frac{\hbar \omega}{\kb T^2})}
								   			 														 	 {\left(\exp \left(\frac{\hbar \omega}{\kb T} \right) -1\right)^2} \\
								   			 &= \frac{N \hbar^2 \omega^2 \cdot \exp\left(\frac{\hbar \omega}
								   			 				 {\kb T}\right)}{\kb T^2 \cdot \exp \left(\frac{\hbar \omega}{\kb T} - 1\right)^2} \ \ .
							  \end{align*} 																
	    					As the energy, that has to be plotted, is given by:
 							  \begin{align*}
								  E(T,N) = F(T,N) + T \cdot S(T,N) \ \ ,
							  \end{align*} 			
	    					we need to calculate $S(T,N)$:
 							  \begin{align*}
								  S(T,N) &= - \frac{\partial F(\beta, N)}{\partial T} \\
								  			 &= - N \kb \ln \left(1 - \exp \left(-\frac{\hbar \omega}{\kb T} \right) \right)
								  			  - \frac{N \kb T \cdot (-\frac{\hbar \omega}{\kb T^2})}{1 - \exp \left(-\frac{\hbar \omega}{\kb T}\right)} \\
								  			 &= \frac{N \hbar \omega}{T} \frac{1}{1 - \exp\left(-\frac{\hbar \omega}{\kb T}\right)}
								  			  - N \kb \cdot \ln\left(1 - \exp \left(-\frac{\hbar \omega}{\kb T}\right)\right) \ \ .
							  \end{align*} 				    					
	    					This leads to:
 							  \begin{align*}
								  E(T,N) = \frac{1}{2} N \hbar \omega + \frac{N \hbar \omega}{1 - \exp \left(-\frac{\hbar \omega}{\kb T}\right)} \ \ .
							  \end{align*} 	
			\item[b)] We consider the two cases $T \rightarrow 0$ and $T \rightarrow \infty$ for $c_V(T)$.
 							  \begin{align*}
									\lim_{T \rightarrow 0} c_V(T) 
									&= \lim_{T \rightarrow 0} \frac{N \hbar^2 \omega^2 \cdot \exp\left(\frac{\hbar \omega}
								   			 											  {\kb T}\right)}{\kb T^2 \cdot \exp \left(\frac{\hbar \omega}{\kb T} - 1\right)^2} \\
								 	&\approx \lim_{T \rightarrow 0} \frac{N \hbar^2 \omega^2}
								   			 											  {\kb T^2 \cdot \exp \left(\frac{\hbar \omega}{\kb T}\right)} \\
								  &= 0 \\
								  \lim_{T \rightarrow \infty} c_V(T) 
								  &= \lim_{T \rightarrow 0} \frac{N \hbar^2 \omega^2 \cdot \exp\left(\frac{\hbar \omega}
								   			 											  {\kb T}\right)}{\kb T^2 \cdot \exp \left(\frac{\hbar \omega}{\kb T} - 1\right)^2} \\
								  &\approx \lim_{T \rightarrow 0} \frac{N\hbar^2 \omega^2}{\kb T^2} \cdot \left(\frac{\kb T}{\hbar \omega}\right)^2 \\
								  &= N \kb
							  \end{align*} 	
			
	  \end{itemize} 
	
	
\end{document}