The basic theory of statistical mechanics has been covered. The partition function and thermodynamic equations have been connected. Steps are below.

• solve Schrodinger equation for values of energy levels
• use an ensemble that is most convenient
• find the partition function
• plug into the thermodynamic framework

The next few lectures involve applying theory to examples, such as gases, solids, and solutions. There is a distinction between distinguishable and indistinguishable particles.

Partition function for non-interacting particles

Evaluate the partition function, <math>Q</math>, for an ideal gas and a mixture of non-interacting atoms. Consider lattice vibrations and monoatomic and diatomic gases. There needs to be some interaction between the particles in order for the system to equilibrate. Assume the atoms are weakly interacting and exchange energy to equilibrate, but that otherwise the system consists of non-interacting particles. The Hamiltonian can be de-coupled. Consider the sum of Hamiltonians with an ideal gas.

<br>

<math>\hat H = \hat H_a + \hat H_b + \hat H_c +...</math>

<br>

<math>\hat H \Psi_

= \epsilon^a + \epsilon^b + \epsilon^c + ...</math>
<br>

</center>

Hydrogen atoms

Consider two atoms of hydrogen at rest that are far away enough from each other that the wave packets are non-interacting. The solution of one is the same as the other. There is no need to solve the many-body wavefunction. Below is the equation that would be solved for <math>10^

</math> particles, a graph of two non-interacting wave packets, and the equation that is applicable to single particle states and solved for a single particle spectrum.
<center>

<br>

<math>\hat H \Psi_{\mbox{total}} = E_{\mbox{total}} \Psi_{\mbox{total}}</math>

<br>

</center>

Consider two atoms that are close together and form a molecule. The wavefunctions can't be divided into two Hamiltonians. They can't be separate.

Comments

The term <math>\epsilon_l^a</math> is the energy of particle <math>a</math> in single particle state <math>l</math>. Each particle can be in a different state as allowed by the spectra. The state of the whole syste is defined by <math>\nu</math>. The state of the whole system gives state energy.

<center>

<math>\nu = \left

</math>
<br>

<math>E_

} = \epsilon_

^c+\epsilon_

^d</math>
<br>

<math>E_

= \epsilon_

^a + \epsilon_

^c+\epsilon_

^d</math>
<br>

<math>Q = \sum_

e^{-\beta E_{\nu}}</math>
<br>

<math>Q = \sum_l \sum_m \sum_n e^

)</math>

<br>

<math>Q = \sum_l e^

</math>

<br>

<math>Q=q^a q^b q^c</math>

<br>

<math>Q = q^N</math>

</center>

Distinguishable versus Indistinguishable Particles

The derivation above considered <math>N</math> non-interacting, distinguishable particles with the same energy spectrum. The result only holds for distinguishable particles. There is a summation of all possible states. The localization of the particles on the surgace results in distinguishability. Any ensemble is ok to choose. The term <math>q</math> represents the mini-partition function. This is the energy spectrum of one.

<p>
</p>

Consider distinguishable particles. A label is on the particles and they are tracked. They are localized in space and treated as distinguishable. The model that is applied to the syste is not related to property or how the system behaves. Fermions can be either distinguishable or indsitnguishable. There could be different cases. Considering atoms in a lattice to be distinguishable particles works well in lattice dynamics. The atoms are labeled by position. Labels can be on atoms in crystals and polymer chains. However, generally the sum cannot be performed where indicies are unrestricted. The unrestricted sum over all possible states cannot be performed.

<center>

<br>

<math>Q \ne \sum_l \sum_m \sum_n e^

</math>

<br>

</center>

Consider two different states below. They are labeled as two different states. If the particles are indistinguishable, there is no difference between the states. One can't tell which is which. Both can't be included in the summation; this would result in overcounting. There is a localization criteria for indistinguishability. There are two distinct states if particles can be labelled but the same state if the particles are indistinguishable.

<center>

<table>

<tr>

<td>
<math>1s</math>
</td>

<td>
<math>2s</math>
</td>

<td>
<math>1s</math>
</td>

<td>
<math>1s</math>
</td>

</tr>

<tr>

<td>
<math>2s</math>
</td>

<td>
<math>1s</math>
</td>

<td>
<math>1s</math>
</td>

<td>
<math>1s</math>
</td>

</tr>

</table>

</center>

Example from research

It was shown that electrons are localized to a lattice in a material. There is associated configurational entropy. The phase diagram looks different at a result. The diagram looks much more like what is calculated. The localization of the particles results in a physical effect in the phase diagram.

Energy and delocalization

Consider the Schrodinger equation and look at the lowest energy levels. There is a pattern in considering the delocalization of ideal gas particles. The higher the energy the more uniform the distribution and more delocalized the particles.

Fermions and bosons

Fermions and bosons are two kinds of particles. Fermions are half spin particles. They are electrons or protons or any molecule with sum of spins a multiple of one-half. Some states are forbidden. Two fermions can't be in the same state. The indices of the summation depend on each other. There is a restriction on the summation since no two fermions can occupy the same state. There would be an overcounting problem with an unrestricted sum.

<p>
</p>

Bosons are of integer spin. Photons, phonons, and molecules that can be invented with integer spin are <math>N</math>-spin particles. With bosons, there is not the restriction found with fermions. There is an unlimited number of particles in the same state.

Unrestricted summation

Consider an unrestricted summation over all states. Consider the case when states are different with some permutation. Below are permutations of states and the number of times the states are counted. Each should only be counted once. There are a lot of states like the second set of states below, while the first states are extremely rare. Below is the partition function under conditions of low density of particles, a large number of states, <math>N</math>, and particles that aren't too heavy.

<center>

<br>

<math>\epsilon_l + \epsilon_m + \epsilon_m + \epsilon_m + ... \leftarrow \rightarrow \epsilon_m + \epsilon_l + \epsilon_m + \epsilon_m + ...</math>

<br>

<math> \mbox

N \mbox

</math>
<br>

<math>Q = \frac

</math>
</center>