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### Quantum statistics

 Boltzmann Bose-Einstein Fermi-Dirac

Classical Boltzmann statistics assumes that the system is made up of distinguishable particles.
In the dice metaphor this corresponds to using coloured dice to ensure we can distinguish from . In general, this is appropriate in a classical system because we can, in principle (although with difficulty in practice), measure the position and trajectory of each particle and, by keeping track of them, distinguish them from one another.

The situation is different in quantum mechanics because of Heisenberg's uncertainty principle,
which puts a fundamental lower limit on the product of the uncertainties of position and momentum of any measurement. Therefore, an alternative statistical approach is needed for indistinguishable particles.
In the dice metaphor, we can achieve this by using identically coloured dice. Now there is no difference between and , and the number of microstates contributing to each macrostate is reduced accordingly.

For certain quantum particles, an additional constraint applies: Particles with half-integer spin (e.g. electrons) obey the Pauli exclusion principle;
they cannot occupy the exact same quantum state, i.e. have exactly the same set of quantum numbers. When playing dice, this is the equivalent of rejecting all those throws that result in the same number on several dice. Particles for which this applies are called fermions;
they obey the Fermi-Dirac statistics. Particles with integer spin can congregate in the same state and therefore aren't subject to this constraint. They are known as bosons
and are described by the Bose-Einstein statistics.

### Bose-Einstein statistics

Bose-Einstein statistics
describes indistinguishable, non-interacting quantum particles.
The derivation of the distribution function follows the same method used when deriving the Boltzmann distribution, but the statistical weight of the macrostates is different. Consider a situation where energy levels are close together. We can treat all states within a narrow range ${\rm d}E$ of energies as essentially degenerate. If there are $A_i$ levels within the range, then we need to slice up the range into $A_i$ "cells" using $A_i-1$ barriers between them. In addition, we need to distribute the $N_i$ particles in this energy range across those cells. In effect, we have a mixed distribution of barriers and particles across the size of the degenerate energy range. The number of microstates within $[E_i,E_i+{\rm d}E]$
is then $$\frac{(N_i+A_i-1)!}{N_i!(A_i-1)!}\qquad.$$ The numerator counts all objects (barriers and particles) to be distributed, while the denominator removes the duplicates due to the fact that we can swap any two particles without making a difference, neither will swapping any two barriers create a distinguishable microstate. However, swapping a barrier and a particle will change the size of two of the cells, resulting in a different microstate that contributes to the statistical weight of the macrostate. Up to this point, this description only deals with the situation in the energy range at $E_i$, but a similar situation applies to all the other energy ranges, too. Therefore, the statistical weight of a macrostate
is the product of all the individual weights of each of the energy ranges: $$\Omega=\prod_i{\frac{(N_i+A_i-1)!}{N_i!(A_i-1)!}}\approx\prod_i{\frac{(N_i+A_i)!}{N_i!(A_i)!}}\qquad.$$ If the system is large, the 1 is negligible compared to the number of particles.

The rest of the derivation follows the familiar pattern. First, take the logarithm of the statistical weight: $$\ln\Omega=\sum_i\ln(N_i+A_i)!-\sum_i\ln N_i!-\sum_i\ln A_i!\qquad.$$ Some of the sums cancel out after applying Stirling's formula: $$\qquad=\sum_i(N_i+A_i)\ln(N_i+A_i)\color{red}{\cancel{\color{black}{-\sum_i(N_i+A_i)}}}-\sum_iN_i\ln N_i\color{red}{\cancel{\color{black}{+\sum_i N_i}}}-\sum_iA_i\ln A_i\color{red}{\cancel{\color{black}{+\sum_iA_i}}}\qquad.$$ To find the most probable macrostate, we need to calculate the total differential of $\ln{\Omega}$ and set it to zero: $${\rm d}\ln\Omega=\sum_i{\frac{{\rm d}}{{\rm d}N_i}(N_i+A_i)\ln(N_i+A_i){\rm d}N_i}-\sum_i{\frac{{\rm d}}{{\rm d}N_i}N_i\ln N_i{\rm d}N_i}-\sum_i{\color{red}{\cancel{\color{black}{\frac{{\rm d}}{{\rm d}N_i}A_i\ln A_i}}}{\rm d}N_i}\qquad.$$ The first sum needs an application of the product rule: $$\qquad=\color{red}{\cancel{\color{black}{\sum_i\frac{N_i+A_i}{N_i+A_i}{\rm d}N_i}}}+\sum_i\ln(N_i+A_i){\rm d}N_i\color{red}{\cancel{\color{black}{-\sum_i\frac{N_i}{N_i}{\rm d}N_i}}}-\sum_i\ln N_i{\rm d}N_i\qquad.$$ Now the first and third term cancel, and the other two can be combined: $$\qquad=\sum_i\ln\frac{N_i+A_i}{N_i}{\rm d}N_i=\sum_i\ln\left(\frac{A_i}{N_i}+1\right){\rm d}N_i\overset{!}{=}0\qquad.$$ At the same time, the total number of particles and the total energy of the system are maintained: $$N=\sum_iN_i\quad\Rightarrow\quad{\rm d}N=\sum_i{\rm d}N_i\overset{!}{=}0$$ $$E=\sum_iN_iE_i\quad\Rightarrow\quad{\rm d}E=\sum_iE_i{\rm d}N_i\overset{!}{=}0\qquad.$$ The three equations are combined using Lagrange multipliers:
$$\sum_i\left[\ln\left(\frac{A_i}{N_i}+1\right)+a+bE_i\right]{\rm d}N_i\overset{!}{=}0\qquad,$$ leading to a population of a specific level, taking into account the extent of degeneracy: $$\frac{N_i}{A_i}=\frac{1}{\exp{\left(-a-\frac{E_i}{k_BT}\right)}-1}\qquad.$$ Apart from the degeneracy, the difference from the Boltzmann distribution lies in the -1 in the denominator.

### Fermi-Dirac statistics

The Fermi-Dirac statistics for indistinguishable particles with half-integer spin
can be derived in a very similar way. Since fermions cannot be in the exact same quantum state (energy level), all the cells in the degenerate energy range can only either be empty or occupied by a single particle. This means we don't need to distribute barriers since they are in fixed locations - each cell is the same size. The contribution of each energy range to the statistical weight is then the number of combinations of the degenerate cells, divided by the number of indistinguishable combinations due to either swapping two empty cells or two occupied cells. As before, the total statistical weight is the product of these contributions from each energy range: $$\Omega=\prod_i\frac{A_i!}{N_i!(A_i-N_i)!}\quad.$$ Again, we take the logarithm, $$\ln\Omega=\sum_i\ln(A_i!)-\sum_i\ln(N_i!)-\sum_i\ln[(A_i-N_i)!]\qquad,$$ apply Stirling's formula, $$\qquad=\sum_iA_i\ln A_i\color{red}{\cancel{\color{black}{-\sum_iA_i}}}-\sum_iN_i\ln N_i\color{red}{\cancel{\color{black}{+\sum_iN_i}}}-\sum_i(A_i-N_i)\ln(A_i-Ni)\color{red}{\cancel{\color{black}{+\sum_i(A_i-N_i)}}}\qquad,$$ and compare with Bose-Einstein:
$$\ln\Omega_{BE}=\color{red}{-}\sum_iA_i\ln A_i-\sum_iN_i\ln N_i\color{red}{+}\sum_i(A_i\color{red}{+}N_i)\ln(A_i\color{red}{+}N_i)$$ The formula is very similar; only the four highlighted signs are different for the two quantum statistics. By carrying through these sign changes, we have for the population of a particular energy range: $$\frac{N_i}{A_i}=\frac{1}{\exp{\left(-a-\frac{E_i}{k_BT}\right)}\color{red}{+}1}$$ The only difference is the +1 in the denominator.

$n_x^2$$n_y^2$$n_z^2$$n^2 1113 2116 1216 1126 2219 2129 1229 31111 13111 11311 22212 32114 31214 23114 21314 13214 12314 particle in a box standing wave$$\psi=\sin\left(\frac{n\pi x}{a}\right)$$quantum number, n=1,2,3\dots energy eigenvalue$$E_i=\frac{n^2a^2}{8ma^2}$$in three dimensions:$$n^2=n_x^2+n_y^2+n_z^2$$degeneracy, g_i phase space$$n^2=n_x^2+n_y^2+n_z^2=\frac{8ma^2E}{h^2}N=\frac{1}{8}\cdot\frac{4}{3}\pi r^3=\frac{4\pi}{3}\left(\frac{2ma^2E}{h^2}\right)^{\frac{3}{2}}\frac{{\rm d}N}{{\rm d}E}=\frac{2\pi a^3}{h^3}(2m)^{\frac{3}{2}}\sqrt{E}=g(E)$$density of states, g(E), bundle of states statistical weight$$\Omega_i=\frac{g_i^{N_i}}{N_i!}$$### Comparison of the particle statistics To summarise, the three particle statistics are summarised below:$$\begin{eqnarray} \textbf{Boltzmann}&:&\frac{N_i}{g_i}=\frac{1}{\exp{\left(-\frac{E_i-\mu}{k_BT}\right)}}\\ \textbf{Bose-Einstein}&:&\frac{N_i}{g_i}=\frac{1}{\exp{\left(-\frac{E_i-\mu}{k_BT}\right)}\color{red}{-}1}\\ \textbf{Fermi-Dirac}&:&\frac{N_i}{g_i}=\frac{1}{\exp{\left(-\frac{E_i-\mu}{k_BT}\right)}\color{red}{+}1} \end{eqnarray}$$chemical potential, \mu=ak_BT The figure provides a decision matrix to determine which statistics is appropriate along with some classes of systems where each would be applicable. ### Partitition functions partition function distribution over energy levels translation vibration rotation electronic states total energy$$E=E_{\textrm{trans}}+E_{\textrm{vib}}+E_{\textrm{rot}}+E_{\textrm{el}}$$total partition function$$z=\sum_i\exp{\left(-\frac{E_i}{k_BT}\right)} =\sum_i\exp{\left(-\frac{E_{i,\textrm{trans}}}{k_BT}\right)}\exp{\left(-\frac{E_{i,\textrm{vib}}}{k_BT}\right)}\exp{\left(-\frac{E_{i,\textrm{rot}}}{k_BT}\right)}\exp{\left(-\frac{E_{i,\textrm{el}}}{k_BT}\right)}$$independent contributions$$z=\left(\sum_iexp{\left(-\frac{E_{i,\textrm{trans}}}{k_BT}\right)}\right)\left(\sum_iexp{\left(-\frac{E_{i,\textrm{vib}}}{k_BT}\right)}\right)\left(\sum_iexp{\left(-\frac{E_{i,\textrm{rot}}}{k_BT}\right)}\right)\left(\sum_iexp{\left(-\frac{E_{i,\textrm{el}}}{k_BT}\right)}\right) =z_{\textrm{trans}}z_{\textrm{vib}}z_{\textrm{rot}}z_{\textrm{el}}\ln{z}=\ln{z_{\textrm{trans}}}+\ln{z_{\textrm{vib}}}+\ln{z_{\textrm{rot}}}+\ln{z_{\textrm{el}}}$$quantum numbers translational partition function particle in a box$$E_{\textrm{el}}=\frac{p^2h^2}{8ma^2}z_{\textrm{el}}=\sum_{p=1}^{\infty}\exp{\left(-\frac{p^2h^2}{8ma^2k_BT}\right)}$$vibrational partition function harmonic oscillator$$E_{\textrm{vib}}=(m+\frac{1}{2})\hbar\omega$$zero-point energy$$z_{\textrm{vib}}=\sum_{m=1}^{\infty}\exp{\left(-\frac{(m+\frac{1}{2})\hbar\omega}{k_BT}\right)}$$rotational partition function rigid rotor$$E_{\textrm{rot}}=\frac{j(j+1)h^2}{8\pi^2I}$$moment of inertia, I degeneracy, g_j=2j+1$$z_{\textrm{rot}}=\sum_{j=0}^{\infty}(2j+1)\exp{\left(-\frac{j(j+1)h^2}{8\pi^2Ik_BT}\right)}$$electronic partition function Schrödinger equation hydrogen atom$$E_{\textrm{el}}=-\frac{\mu Z^2e^4}{8\epsilon_0^2h^2n^2}$$reduced mass$$z_{\textrm{el}}=\sum_{n=1}^{\infty}\exp{\left(-\frac{\mu Z^2e^4}{8\epsilon_0^2h^2n^2k_BT}\right)}$\$

Now that we have statistics for both distinguishable and indistinguishable particles, we need to deal with the assumption made so far that the particles do not interact. This is achieved by using ensemble statistics.