Boltzmann Statistics, Fermi-Dirac Statistics, and Bose-Einstein Statistics

Simplify the partition function by writing the total energy of a system as a sum of individual energies. The final equations depend on whether the individual particles of the system are fermions or bosons. They obey two different types of statistics, Fermi-Dirac or Bose-Einstein. Both under certain conditions can be reduced to Boltzmann statistics.

4-1 The Special Case of Boltzmann Statistics

All known particles fall into two clasees: those whose wave function must be symmetric under the operation of the interchange of two identical particles, and those whose wave function must be antisymmetric.

There are problems in which the Hamiltonian can be written as a sum of simpler Hamiltonians. The most obvious example is the case of a dilute gas. Another example is the decomposition of the Hamiltonian of a polyatomic molecule into various degrees of freedom.

<tex>H \approx H_

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+ H_

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+ H_

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+ H_

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</tex>

</center>

Canonical Partition Function

Denote individual energy states by \left

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, where the superscript denotes the particle and the subscript denotes the state.

<tex>Q(N, V, T) = \sum_

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e^

Unknown macro: {left ( epsilon_i^a + epsilon_j^b + epsilon_k^c + ... ) / k T}

</tex>

<tex>Q(N, V, T) = \sum_

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e^

Unknown macro: {left ( epsilon_i^a / k T right ) }

\sum_

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e^

Unknown macro: {left ( epsilon_j^b / k T right ) }

\sum_

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e^

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</tex>

<tex>q(V, T) = \sum_i e^{- \epsilon_i / kT}</tex>

<tex> \mbox

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</tex>

</center>

The last expression requires knowledge only of the energy values of an individual particle or quasi-particle. It is called a molecular partition function.

In a perfect crystal each atom is confined to one and only one lattice point. The particles are distinguishable.

Molecular Hamiltonian

Reduce an <tex>N</tex>-body problem further into individual degrees of freedom.

<tex>q_

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= q_

q_

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q_

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q_

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...</tex>

<tex>q_

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= \sum_i e^{- \epsilon_i^

Unknown macro: {trans}

/ kT }</tex>

</center>

An excellent approximation in most cases is obtained by carrying out the summation below in an unrestricted manner and dividing by <tex>N!</tex>. The number of molecular states available to a molecule at room temperature, say, is much greater than the number of molecules in the system for all but the most extreme densities. The indistinguishability of the particles has been included by dividing by <tex>N!</tex>. The condition is favored by large mass, high temperature, and low density. The results are valid for polyatomic molecules as well, since the translational energy states account for almost all of the nergy states available to any molecule.

<tex>Q(N, V, T) = \sum_

Unknown macro: {i, j, k, l...}

e^{-(\epsilon_i + \epsilon_j + \epsilon_k + \epsilon_l + ...)/kT}</tex>

<tex> \mbox

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</tex>

<tex>Q(N, V, T) = \frac

Unknown macro: {q(V, T)^N}

Unknown macro: {N!}

</tex>

<tex>q(V, T) = \sum_j e^{-\epsilon_j / kT } </tex>

</center>

Requirement of Boltzmann Statistics

The number of available molecular states is much greater than the number of particles in the system. The approximation becomes increasingly better at higher temperatures.

Probability of Molecules in _j_th Energy State

The total energy of the <tex>N</tex>-body system is below.

<tex>E = N \bar \epsilon</tex>

<tex>E = kT^2 \left ( \frac

Unknown macro: { partial ln Q }

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\right )_

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</tex>

<tex>E = N \sum_j \epsilon_j \frac{ e^{- \epsilon_j / k T}}

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</tex>

<tex>\bar \epsilon = \sum_j \epsilon_j \frac{ e^{- \epsilon_j / k T}}

</tex>

<tex>\mbox

Unknown macro: {Probability molecule in the jth energy state}

</tex>

<tex>\pi_j = \frac{e^{- \epsilon_j / k T}}{\sum_j e^{- \epsilon_j / k T}}</tex>

<tex>\pi_j = \frac{e^{- \epsilon_j / k T}}

Unknown macro: {q}

</tex>

</center>

The last expression is the probability that a molecule is in the _j_th vibrational state irrespective of the other degrees of freedom.

Consider the effect of symmetry requirements of N-body wave functions on the sum over states.

4-2 Fermi-Dirac and Bose-Einstein Statistics

The distribution function in the case of fermions is called Fermi-Dirac statistics, and the the distribution function in the case of bosons is called Bose-Einstein statistics. These are the only exact distributions. In the case of high temperature and/or low density, both of these go over into Boltzmann or classical distribution.

Treat the general case by means of the grand canonical ensemble. Let <tex>E_j(N, V)</tex> be the energy states available to a system containing <tex>N</tex> molecules. Let <tex>\epsilon_k</tex> be the molecular quantum states. Let <tex>n_k = n_k (E_j)</tex> be the number of molecules in the <tex>k</tex>th molecular state when the system itself is in the quantum state with energy <tex>E_j</tex>. A quantum state of the entire system is specified by the set <tex>

Unknown macro: {n_k}

</tex>. The energy of the system is expressed below.

<tex>E_j = \sum_k \epsilon_k n_k</tex>

<tex>Q(N, V, T) = \sum_n_k e^{- \beta \sum_i* \epsilon_i n_i}</tex>

</center>

Avoid the restriction that the sum of number of particles is equal to the total number of particles by using the grand canonical partition function.

<tex>\Theta (V, T, \mu) = \sum_

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^

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e^

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Q(N, V, T)</tex>

<tex>\Theta (V, T, \mu) = \sum_

^

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\lambda^N \sum_n_k * e^{- \beta \sum_i* \epsilon_i n_i}</tex>

<tex>\Theta (V, T, \mu) = \sum_

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^

\sum_n_k * \lambda^

Unknown macro: {sum n_i}

e^{- \beta \sum_i* \epsilon_j n_j}</tex>

<tex>\Theta (V, T, \mu) = \sum_

Unknown macro: {N = 0}

^

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\sum_n_k * \Pi_k \left ( \lambda e^{- \beta \epsilon_k} \right )^

</tex>

</center>

The next step is important and is allowed since we are summing over all values of <tex>N</tex> or since we are using the grand canonical partition function.

<tex>\Theta (V, T, \mu) = \sum_

Unknown macro: {n_1 = 0}

^{n_1 \mbox{max}} \sum_

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^{{n_2 \mbox{max}} ... \Pi_k \left ( \lambda e}{-\beta \epsilon_k} \right )^

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</tex>

<tex> \Theta (V, T, \mu) = \Pi_k \sum_

Unknown macro: {n_k = 0}

^{n_k \mbox

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\left ( \lambda e^{-\beta \epsilon_k} \right )^

</tex>

<tex>\mbox

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</tex>

<tex>\Theta_

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= \Pi_k \left (1 + \lambda e^{- \beta \epsilon_k} \right )</tex>

<tex>\mbox

Unknown macro: {bose-Einstein Statistics}

</tex>

<tex>\Theta_

= \Pi_k \sum_

Unknown macro: {n_k = 0}

^

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\left (\lambda e^{- \beta \epsilon_k} \right )^

Unknown macro: {n_k}

</tex>

<tex>\Theta_

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= \Pi_k \sum_

(1 - \lambda e^{- \beta \epsilon_k} \right )^{-1}</tex>

</center>

Below is a derivation of the average number of particles in the <tex>k</tex>th quantum state.

<tex>N = kT \left ( \frac

Unknown macro: {partial ln Theta}

Unknown macro: {partial mu}

\right )_

Unknown macro: {V, T}

</tex>

<tex>N = \lambda \left( \frac

Unknown macro: {partial lambda}

\right )_

Unknown macro: {V, T}

</tex>

<tex>N = \sum_k \frac{\lambda e^{\beta \epsilon_k}}{1 \pm \lambda e^{ \beta \epsilon_k}}</tex>

<tex>\bar n_k= \frac{ \lambda e^{\beta \epsilon_k}}{1 \pm \lambda e^{ \beta \epsilon_k}}</tex>

<tex>\bar E = N \bar \epsilon</tex>

<tex>\bar E = \sum_k \bar n_k \epsilon_k</tex>

<tex>\bar E = \sum_k \frac{\lambda \epsilon_k e^{- \beta \epsilon_k}}{1 \pm \lambda e^{- \beta \epsilon_k}}</tex>

</center>

Lastly, consider the following.

<tex>pV = kT \ln \Theta (V, T, \mu)</tex>

<tex>pV = \pm kT \sum_k \ln \left [1 \pm \lambda e^{-\beta \epsilon_k} \right]</tex>

</center>

The individual particles of the system are not independent becase of the symmetry requirements of the wave function.

Both statistics go over into Boltzmann

Both kinds of statistics should go over into Boltzmann or classical statistics in the limit of high temperature or low density, where the number of available quantum states is much greater than the number of particles. The average number of molecules in any state is very small.

<tex>\lambda \mbox

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</tex>

<tex>\bar n_k = \lambda e^{- \beta \epsilon_k}</tex>

<tex>\frac

Unknown macro: {bar n_k}

Unknown macro: {N}

= \frac{e^-

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}

</tex>

<tex>q = \sum_j e^

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</tex>

<tex>\bar E \right \sum_j \lambda \epsilon_j e^

</tex>

<tex>\bar \epsilon = \frac

Unknown macro: { bar E}

Unknown macro: {bar N}

\right \frac{\sum_j \epsilon_j e^{\beta \epsilon_j}}{\sum_j e^{\beta \epsilon_j}}</tex>

<tex>pV \right (\pm kT) \left (\pm \lambda \sum_j e^{- \beta \epsilon_j} \right )</tex>

<tex>pV = \lambda kT \sum_j e^{- \beta \epsilon_j}</tex>

<tex>pV = \lambda kT q</tex>

<tex>\beta p V = \ln \Theta</tex>

<tex>\beta p V = \lambda q</tex>

<tex>\lambda q = N</tex>

</center>

The penultimate equation above is the perfect gas law, and thus the formulas of Fermi-Dirac and Bose-Einstein statistics reduce to those of Boltzmann statistics in the classical limits.

<tex>\Theta = e^

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</tex>

<tex>\Theta = \sum_

Unknown macro: {N = 0}

^

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\frac

Unknown macro: {(lambda q)^N}

</tex>

<tex>Q(N, V, T) = \frac

Unknown macro: {q^N}

Unknown macro: {N!}

</tex>

</center>

Apply the limit of Boltzmann statistics to the simplest system, namely a monoatomic ideal gas.

Child pages

McQuarrie Chapter 4

4-1 The Special Case of Boltzmann Statistics

Canonical Partition Function

Molecular Hamiltonian

Requirement of Boltzmann Statistics

Probability of Molecules in _j_th Energy State

4-2 Fermi-Dirac and Bose-Einstein Statistics

Both statistics go over into Boltzmann