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Microstates and Degeneracy

A microstate is a full specification of all degrees of freedom of a system. A system may be conveniently defined as having $N$ degrees of freedom confined to a volume $V$. In general, microscopic degrees of freedom are quantum numbers. The index $\nu$ in Eq. 1 runs over all unique combination of quantum number values. There is a spectrum of energy eigenstates for any system given by

\begin{displaymath}
\mathscr{H}\left\vert\nu\right> = E_\nu\left\vert\nu\right>,
\end{displaymath} (2)

where $\mathscr{H}$ is the Hamiltonian, $\left\vert\nu\right>$ is shorthand (``ket'' notation) for the system wavefunction in state $\nu$, and $E_\nu$ is the energy of state $\nu$. In contrast to model systems usually considered in elementary quantum mechanics, the number of distinct microstates of systems of $\sim 10^{23}$ particles that have the same energy $E$ is very large, and this set of eigenstates is in practice impossible to obtain explicitly. This is indeed why we must instead treat this set statistically. We refer to the number of states that satisfy a given energy as the degeneracy of the energy level, denoted $\Omega(E,N,V)$:

\begin{displaymath}
\Omega(N,V,E) = \begin{array}{l}
\mbox{number of microstate...
...}\\
\mbox{energy between $E$ and $E+\delta E$.}
\end{array}\end{displaymath}

The many ``equivalent'' states numbering $\Omega(N,V,E)$ is called a microcanonical ensemble.

The Ising spin lattice is a simple statistical mechanical model with discrete energy levels which we can now introduce to gain some understanding of what it means to say $\Omega$ is ``large.'' Imagine a linear array of $N$ spins, each pointing either ``up'' or ``down.''

portrait A 1-D Ising system.


Let us suppose that the Hamiltonian of this system is given by

\begin{displaymath}
\mathscr{H} = h\sum_{i=1}^{N}s_i
\end{displaymath} (3)

where $s_i$ is -1 if spin $i$ is ``down'' and +1 if spin $i$ is ``up,'' and $h$ is some unit of energy. The ground state, the state with the lowest energy, has all spins down, so $\Omega_0 = 1$. The next state up has one spin up, but there are $N$ possible microstates that have this energy: $\Omega_1 = N$. The next state up has two spins, and there are $N(N-1)/2$ such microstates: $\Omega_2 =
N(N-1)/2$. For $m$ spins flipped, there are $\Omega_m = N!/(N-m)!m!$ distinct microstates. Thus we see that working with $\Omega$ for statistical mechanical systems means working with enormous numbers.

Although quantum mechanics tells us that atomic systems have discrete energy levels, when systems contain very large numbers of atoms, these energy levels become so closely spaced relative to their span that they may effectively be considered a continuum. We can thus pass into a classical (as opposed to quantum mechanical) representation, where the microstate for a system of $N$ particles is specified by a point in a $6N$-dimensional phase space:

\begin{displaymath}
\left(r^N,p^N\right) \equiv \left({\bf r}_1,{\bf r}_2,...,{\bf r}_N;{\bf p}_1,...,{\bf p}_N\right).
\end{displaymath} (4)

We can denote the number of states in a microcanonical ensemble for a classical system using the Dirac delta function:
\begin{displaymath}
\Omega(N,V,E) = \frac{1}{h^{3N}N!}\int\int d {\bf r}^N d {\b...
...elta\left[\mathscr{H}\left({\bf r}^N,{\bf p}^N\right)-E\right]
\end{displaymath} (5)

The microcanonical ensemble represents a hyperdimensional surface in the phase space dimensioned by $N$ particles with positions limited by the extent of $V$. The factorial in Eq. 5, $N!$, takes into account that the particles are indistingishable; that is, ordering particle labels is not important. $h$ is Planck's constant; note that it has units of (length)(momentum). Think of it as a quantum-mechanically-required ``mesh discretization'' for continuous space (it arises due to the Heisenberg uncertainty relation). It also nondimensionalizes the partition function. We will encounter it again in the next section, but we will also see why these ``prefactors'' are not essential ingredients of most molecular simulations.

You may wonder why there seem to be two viewpoints of statistical mechanics, quantum and classical. First, there really aren't two viewpoints: the classical picture is an approximation of the more general quantum mechanical picture. But statistical mechanics as a discipline was first formalized by Gibbs and Boltzmann before quantum mechanics was widely accepted, so it dealt necessarily with systems of classical particles obeying Newtonian equations of motion; that is, on classical mechanics. There appears to be a general concensus that it is easier to introduce statistical mechanical concepts using the ``sum-over-states'' notation of quantum statistical mechanics, rather than the apparently more cumbersome (and anyway approximate) ``integral-over-phase-space'' notation of classical statistical mechanics.


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