ch4.tex

When we wish to decouple a system into non-interacting parts, different
statistics apply based on the underlying physics of the system. Three major
types of statistics exist with respect to molecular systems: Boltzmann,
Fermi-Dirac, and Bose-Einstein. To be exact Boltzmann statistics is the limit of
the other two at sufficiently high T.

\begin{itemize}
	\item Fermi-Dirac
		\begin{itemize}
			\item Occurs when the exchange of basis particles is antisymmetric
				with respect to the wave function.
			\item Example fermions are electrons which the extrapolation of the
				Fermi-Dirac statistics is the Pauli exclusion principle.
		\end{itemize}
	\item Bose-Einstein
		\begin{itemize}
			\item For systems where the exchange of basis particles is symmetric
				with respect to the wave function, Bose-Einstein statistics
				apply.
			\item Examples are protons and neutrons.
		\end{itemize}
	\item Boltzmann
		\begin{itemize}
			\item The high temperature limit of the other two.
			\item Ends up being the classical approximation of the quantum
				mechanical equations.
		\end{itemize}
\end{itemize}

\section{Boltzmann Statistics}\label{sec:Boltzmann Statistics}
\subsection{Hamiltonian Decoupling}
If a multiparticle system is dilute enough the interactions between particles is
negligible, and the system can be treated as a $N$ independent body problem.
Even in liquids and solids convenient ``basis particles'' can be chosen so the
system is modeled by a $N$ independent body system. In these cases where
interactions can be ignored, the Hamiltonian can be decoupled.
\begin{equation*}
	\mathcal{H} = \sum_{i=1}^{N}{\mathcal{H}_i}
\end{equation*}
In addition, for individual molecules if the individual modes of molecular
energy are decoupled,
\begin{equation*}
	\mathcal{H}_i = \sum_{m \in M}{\mathcal{H}_i^{m}},
\end{equation*}
where $M$ is the set of all decoupled energy modes.
\subsection{Partition Function Decoupling}
We now consider a $N$ distinguishable, independent, body system. Then,
\begin{align*}
	\mathcal{Q}(N,V,T) &= \sum_j{e^{-\beta E_j}} = \sum_{j,k,\dots}{e^{-\beta
	\cdot (\epsilon_j^a + \epsilon_k^b + \epsilon_l^c + \dots}} \\
					   &= q_a q_b q_c \dots
\end{align*}
Here we see the canonical partition function becomes $N$ individual particle
partition functions. Since the particles are identical, we have
\begin{align*}
	q(V, T) &= \sum_j{e^{-\beta \epsilon_j}}\\
	\mathcal{Q} &= q^{N}. 
\end{align*}
This equation is only true in a distinguishable system though which in molecular
systems in rarely the case. Including distinguishably leads to some problems.
Consider, two equivalent energy states,
\begin{align*}
	E_{tot} &= \epsilon_i^a + \epsilon_j^b + \epsilon_j^c + \dots \\
	E_{tot} &= \epsilon_j^a + \epsilon_i^b + \epsilon_j^c + \dots\quad .
\end{align*}
In fact, there are $N$ such equivalent arrangements. However, in the case that
all individual energies are different, there are $N!$ combinations. The problem,
thus, is that different energy combinations have differing numbers of equivalent
arrangements. This would imply that a simple divisor cannot be applied to reduce
the number of configurations of a distinguishable system to a indistinguishable
one.

Despite this fact, we can still often make the approximation of dividing the
distinguishable partition function by $N!$. At room temperature ($m=10^{-22}g$,
$a=10\text{cm}$), the number of energy states available to a particle in a box
is $O(10^{30})$. This number signifies that in all but extremely dense systems
at around room temperature or even colder is much greater than the number of
particles, so it is unlikely that any two particles adopt the same energy, so
\begin{equation*}
	\mathcal{Q} = \frac{q^{N}}{N!},
\end{equation*}
is an excellent approximation for all but `the lightest molecules at very low
temperatures.'
\subsection{Getting Molecular Properties}
Since we now are dealing with individual molecular partition functions, we can
get averages of molecular properties. The probability of being in state $j$ is
\begin{equation*}
	\pi_j = \frac{e^{-\beta \epsilon_j}}{q}.
\end{equation*}
Any mechanical properties is then,
\begin{equation*}
	\bar{M} = \sum_j{M_j \pi_j}.
\end{equation*}
The parallelism to the ensembles already mentioned is not an accident as
Boltzmann's original examination of statistical mechanics used a system of
molecules in thermal equilibrium as his start. In fact, we could from \textit{a
priori} principles derive the previous two equations.

\section{Fermi-Dirac and Bose-Einstein Statistics}%
\label{sec:Fermi-Dirac and Bose-Einstein Statistics}
\subsection{Common Derivation}
Derivation of the exact distribution of states in an independent system obeying
either Fermi-Dirac or Bose-Einstein statistics is most readily done in the grand
canonical ensemble. We let $E_j(N, V)$ be the energy available to a $N$ body
system, $\epsilon_k$ the molecular quantum states, and $n_k$ the occupancy
number of a molecular quantum state. Thus it follows,
\begin{align*}
	E_j &= \sum_k{\epsilon_k n_k}\\
	N &= \sum_k{n_k}.
\end{align*}

From here, we take the $\Xi$ and find its form on a molecular basis.
\begin{equation*}
	\Xi = \sum_N{\lambda^{N}} \sum_j{e^{-\beta E_j}} =
	\sum_N{\lambda^{N}}\sum_{n_k \in S}{e^{-\beta \cdot \sum_i{\epsilon_i n_i}}},
\end{equation*}
where $S$ is the set of all occupancies numbers that obeys $\sum_k{n_k} = N$.
Some more manipulation leads to,
\begin{align*}
	\Xi &= \sum_{N=0}^{\infty} \sum_{n_k \in S}{\lambda^{\sum_i{n_i}} e^{-\beta
	\cdot \sum_i{\epsilon_i n_i}}}\\
		&= \sum_{N=0}^{\infty} \sum_{n_k \in S} \prod_k{(\lambda e^{-\beta
		\epsilon_k})^{n_k}}.
\end{align*}
We have moved $\lambda$ into the second summation which just distributes
$\lambda^{N}$ to each term. We then isolated $n_k$ by using a series product.
From here, we must notice that since $N \to \infty$, every distribution of $n_k
\in \mathbb{N}$ compatible with the occupancy number requirements of the given
statistics will be summed over. Thus each $n_k$ will vary over all its possible values in
combination with every other occupancy number. The knowledge allows us to write
the previous equation as,
\begin{align*}
	\Xi &= \sum_{n_1}^{\max(n_1)} \sum_{n_2}^{\max(n_2)} \sum_{n_3}^{\max(n_3)}
	\cdots \prod_k{(\lambda e^{-\beta \epsilon_k})^{n_k}}\\
	\text{or}\\
	\Xi &= \sum_{n_1}^{\max(n_1)}{(\lambda e^{-\beta \epsilon_1})^{n_1}}
	\sum_{n_2}^{\max(n_2)}{(\lambda e^{-\beta \epsilon_2})^{n_2}}\cdots\quad.
\end{align*}
This is simple the product of $N$ terms so we have,
\begin{equation*}
	\Xi = \prod_k \sum_{n_k}^{\max(n_k)}{(\lambda e^{-\beta \epsilon_k})^{n_k}}
\end{equation*}

\subsection{Fermi-Dirac}
For Fermi-Dirac statistics each state is either uniquely occupied or not
occupied at all. Therefore, the summation above expands to two terms for
occupied and unoccupied states.
\begin{equation*}
	\Xi = \prod_k (1 + \lambda e^{-\beta \epsilon_k}).
\end{equation*}
\subsection{Bose-Einstein}
In the case of Bose-Einstein statistics each state can be occupied any number of
times without restriction. This becomes,
\begin{align*}
	\Xi &= \prod_k \sum_{n_k}^{\infty}{(\lambda e^{-\beta \epsilon_k})^{n_k}}\\
		&= \prod_k \sum_{i_k}^{\infty}{x^{i_k}}\\
		&= \prod_k (1 + x)^{-1} \quad \text{for } x < 1\\
		&= \prod_k (1 + \lambda e^{-\beta \epsilon_k})^{-1}
\end{align*}
The last two lines uses the result from the geometric series. One way to write
this result is,
\begin{equation*}
	\Xi_{BE}^{FD} = \prod_k{(1 + \lambda e^{-\beta \epsilon_k})^{\pm 1}}.
\end{equation*}
From here we note that,
\begin{equation*}
	\ln\Xi = \sum_k{\ln{(1 \pm \lambda e^{-\beta \epsilon_k})}}.
\end{equation*}
Thus $\bar{E}$ can be calculated by first noting,
\begin{align*}
	\bar{n}_k &= \frac{\lambda e^{-\beta \epsilon_k}}{1 \pm \lambda e^{-\beta
	\epsilon_k}}\\
	\text{then,}\\
	\bar{E} &= \sum_k{\bar{n}_k \epsilon_k} = \sum_k{\frac{\epsilon_k
		\lambda e^{-\beta \epsilon_k}}{1 \pm \lambda e^{-\beta
		\epsilon_k}}}.
\end{align*}

By allowing $T \to \infty$ or $\frac{N}{V} \to 0$, $\lambda \to 0$ which means
$\bar{n}_k \to 0$. This can be shown to make both Fermi-Dirac and Bose-Einstein
statistics go to the Boltzmann limit.