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# 5.1: Statistical Mechanics of Noninteracting Quantum Systems

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## Bose and Fermi systems in the grand canonical ensemble

A noninteracting many-particle quantum Hamiltonian may be written as1

$\HH=\sum_\alpha \ve\ns_\alpha\, \Hn\ns_\alpha\ ,$

where $$\Hn\ns_\alpha$$ is the number of particles in the quantum state $$\alpha$$ with energy $$\ve\ns_\alpha$$. This form is called the second quantized representation of the Hamiltonian. The number eigenbasis is therefore also an energy eigenbasis. Any eigenstate of $$\HH$$ may be labeled by the integer eigenvalues of the $$\Hn\ns_\alpha$$ number operators, and written as $$\ket{n\ns_1\,,\,n\ns_2\,,\,\ldots}$$. We then have

$\Hn\ns_\alpha\,\ket{\Vn}=n\ns_\alpha\,\ket{\Vn}$

and

$\HH\,\ket{\Vn}=\sum_\alpha n\ns_\alpha\,\ve\ns_\alpha\,\ket{\Vn}\ .$

The eigenvalues $$n\ns_\alpha$$ take on different possible values depending on whether the constituent particles are bosons or fermions, viz.

$\begin{split} \hbox{ bosons}:\ &n\ns_\alpha\in\big\{0\,,\,1\,,\,2\,,\,3\,,\,\ldots\big\}\\ \hbox{ fermions}:\ &n\ns_\alpha\in\big\{0\,,\,1\big\}\ . \end{split}$

In other words, for bosons, the occupation numbers are nonnegative integers. For fermions, the occupation numbers are either 0 or 1 due to the Pauli principle, which says that at most one fermion can occupy any single particle quantum state. There is no Pauli principle for bosons.

The $$N$$-particle partition function $$Z\ns_N$$ is then

$Z\ns_N=\sum_{\{n\ns_\alpha\}}\,e^{-\beta\sum_\alpha n\ns_\alpha \ve\ns_\alpha}\,\delta\ns_{N\,,\,\sum_\alpha n\ns_\alpha}\ ,$

where the sum is over all allowed values of the set $$\{n\ns_\alpha\}$$, which depends on the statistics of the particles. Bosons satisfy Bose-Einstein (BE) statistics, in which $$n\ns_\alpha\in\{0\,,\,1\,,\,2\,,\,\ldots\}$$. Fermions satisfy Fermi-Dirac (FD) statistics, in which $$n\ns_\alpha\in\{0\,,\,1\}$$.

The OCE partition sum is difficult to perform, owing to the constraint $$\sum_\alpha n\ns_\alpha=N$$ on the total number of particles. This constraint is relaxed in the GCE, where

$\begin{split} \Xi&=\sum_N e^{\beta\mu N}\,Z\ns_N\\\ &=\sum_{\{n\ns_\alpha\}}e^{-\beta\sum_\alpha n\ns_\alpha \ve\ns_\alpha}\,e^{\beta\mu\sum_\alpha n\ns_\alpha }\\ &=\prod_\alpha\Bigg(\sum_{n\ns_\alpha} e^{-\beta(\ve\ns_\alpha-\mu)\,n\ns_\alpha}\Bigg)\ . \end{split}$

Note that the grand partition function $$\Xi$$ takes the form of a product over contributions from the individual single particle states.

We now perform the single particle sums:

\begin{aligned} \sum_{n=0}^\infty e^{-\beta(\ve-\mu)\,n}&={1\over 1-e^{-\beta(\ve-\mu)}} \qquad\hbox{ (bosons)}\\ \sum_{n=0}^1 e^{-\beta(\ve-\mu)\,n}&=1+e^{-\beta(\ve-\mu)} \qquad\hbox{ (fermions)}\ .\end{aligned}

Therefore we have

$\begin{split} \XBE&=\prod_\alpha {1\over 1-e^{-(\ve\ns_\alpha-\mu)/\kT}}\\ \OBE&=\kT\sum_\alpha\ln\!\Big(1-e^{-(\ve\ns_\alpha-\mu)/\kT}\Big) \end{split}$

and

$\begin{split} \XFD&=\prod_\alpha \Big(1+e^{-(\ve\ns_\alpha-\mu)/\kT}\Big)\\ \OFD&=-\kT\sum_\alpha\ln\!\Big(1+e^{-(\ve\ns_\alpha-\mu)/\kT}\Big). \end{split}$

We can combine these expressions into one, writing

$\Omega(T,V,\mu)=\pm\kT\,\sum_\alpha\ln\!\Big(1\mp e^{-(\ve\ns_\alpha-\mu)/\kT}\Big), \label{Oqsm}$

where we take the upper sign for Bose-Einstein statistics and the lower sign for Fermi-Dirac statistics. Note that the average occupancy of single particle state $$\alpha$$ is

$\langle \Hn\ns_\alpha\rangle={\pz\Omega\over\pz\ve\ns_\alpha}={1\over e^{(\ve\ns_\alpha-\mu)/\kT}\mp 1}, \label{benum}$

and the total particle number is then

$N(T,V,\mu)=\sum_\alpha {1\over e^{(\ve\ns_\alpha-\mu)/\kT}\mp 1}. \label{Ntot}$

We will henceforth write $$n\ns_\alpha(\mu,T)=\langle \Hn\ns_\alpha\rangle$$ for the thermodynamic average of this occupancy.

## Quantum statistics and the Maxwell-Boltzmann limit

Consider a system composed of $$N$$ noninteracting particles. The Hamiltonian is

$\HH=\sum_{j=1}^N\> \Hh\ns_j\ .$

The single particle Hamiltonian $$\Hh$$ has eigenstates $$\tket{\alpha}$$ with corresponding energy eigenvalues $$\ve\ns_\alpha$$. What is the partition function? Is it

$Z\ \ {\mathop{\hbox to 10pt{=}}\limits^{\!\!\!{ ?}}}\sum_{\alpha\ns_1}\ \cdots\ \sum_{\alpha\ns_N} e^{-\beta \big(\ve\ns_{\alpha\ns_1}+ \ \ve\ns_{\alpha\ns_2} + \ \ldots\ +\ \ve\ns_{\alpha\ns_N}\big)}=\zeta^N\ ,$

where $$\zeta$$ is the single particle partition function,

$\zeta=\sum_\alpha e^{-\beta \ve\ns_\alpha}\ .$

For systems where the individual particles are distinguishable, such as spins on a lattice which have fixed positions, this is indeed correct. But for particles free to move in a gas, this equation is wrong. The reason is that for indistinguishable particles the many particle quantum mechanical states are specified by a collection of occupation numbers $$n\ns_\alpha$$, which tell us how many particles are in the single-particle state $$\sket{\alpha}$$. The energy is

$E=\sum_{\alpha} n\ns_\alpha \,\ve\ns_\alpha$

and the total number of particles is

$N=\sum_\alpha n\ns_\alpha\ .$

That is, each collection of occupation numbers $$\{n\ns_\alpha\}$$ labels a unique many particle state $$\ket{\{n\ns_\alpha\}}$$. In the product $$\zeta^N$$, the collection $$\{n\ns_\alpha\}$$ occurs many times. We have therefore overcounted the contribution to $$Z\ns_N$$ due to this state. By what factor have we overcounted? It is easy to see that the overcounting factor is

$\hbox{ degree of overcounting}\quad=\quad {N!\over\prod_\alpha n\ns_\alpha !}\qquad, \nonumber$

which is the number of ways we can rearrange the labels $$\alpha\ns_j$$ to arrive at the same collection $$\{n\ns_\alpha\}$$. This follows from the multinomial theorem,

$\Bigg(\sum_{\alpha=1}^K x\ns_\alpha\Bigg)^{\!\!N}=\sum_{n\ns_1}\sum_{n\ns_2}\>\cdots\>\sum_{n\ns_K}{N!\over n\ns_1!\,n\ns_2!\cdots n\ns_K!} \ x_1^{n\ns_1}\,x_2^{n\ns_2}\cdots x_K^{n\ns_K}\>\delta\ns_{N,n\ns_1\,+\,\ldots\,+\,n\ns_K}\ .$

Thus, the correct expression for $$Z\ns_N$$ is

$\begin{split} Z\ns_N&=\sum_{\{n\ns_\alpha\}}\,e^{-\beta\sum_\alpha n\ns_\alpha \ve\ns_\alpha}\,\delta\ns_{N,\sum_\alpha n\ns_\alpha}\\ &=\sum_{\alpha\ns_1}\sum_{\alpha\ns_2}\>\cdots\>\sum_{\alpha\ns_N}\left({\prod_\alpha n\ns_\alpha!\over N!}\right)\> e^{-\beta(\ve_{\alpha\ns_1}+\ \ve_{\alpha\ns_2} + \ \ldots\ +\ \ve_{\alpha\ns_N})}\ . \end{split}$

In the high temperature limit, almost all the $$n\ns_\alpha$$ are either $$0$$ or $$1$$, hence

$Z\ns_N\approx{\zeta^N\over N!}\ .$

This is the classical Maxwell-Boltzmann limit of quantum statistical mechanics. We now see the origin of the $$1/N!$$ term which is so important in the thermodynamics of entropy of mixing.

Finally, starting with the expressions for the grand partition function for Bose-Einstein or Fermi-Dirac particles, and working in the low density limit where $$n\ns_\alpha(\mu,T)\ll 1$$ , we have $$\ve\ns_\alpha-\mu\gg\kT$$, and consequently

$\begin{split} \Omega\ns_{\ssr{BE}/\ssr{FD}}&=\pm\kT\,\sum_\alpha\ln\!\Big(1\mp e^{-(\ve\ns_\alpha-\mu)/\kT}\Big)\\ &\longrightarrow-\kT\sum_\alpha e^{-(\ve\ns_\alpha-\mu)/\kT}\equiv \Omega\ns_\ssr{MB}\ . \end{split}$

This is the Maxwell-Boltzmann limit of quantum statistical mechanics. The occupation number average in the Maxwell-Boltzmann limit is then

$\langle \Hn\ns_\alpha\rangle=e^{-(\ve\ns_\alpha-\mu)/\kT}\ .$

## Single particle density of states

The single particle density of states per unit volume $$g(\ve)$$ is defined as

$g(\ve)={1\over V}\sum_\alpha \delta(\ve-\ve\ns_\alpha)\ .$

We can then write

$\Omega(T,V,\mu)=\pm V\kT\!\!\int\limits_{-\infty}^\infty\!\!\!d\ve\,g(\ve)\,\ln\!\Big(1\mp e^{-(\ve-\mu)/\kT}\Big) \ .$

For particles with a dispersion $$\ve(\Bk)$$, with $$\Bp=\hbar\Bk$$, we have

$\begin{split} g(\ve)&=\Sg\!\int\!\!{d^d\!k\over (2\pi)^d}\>\delta(\ve-\ve(\Bk)\big)\\ &={\Sg\,\Omega\ns_d\over (2\pi)^d}\,{k^{d-1}\over {d\ve/dk}}\ . \end{split}$

where $$\Sg=2S\!+\!1$$ is the spin degeneracy, and where we assume that $$\ve(\Bk)$$ is both isotropic and a monotonically increasing function of $$k$$. Thus, we have

$g(\ve)={\Sg\,\Omega\ns_d\over (2\pi)^d}\,{k^{d-1}\over {d\ve/dk}}=\begin{cases} {\Sg\over\pi}\,{dk\over d\ve} & d=1 \\ &\\ {\Sg\over 2\pi}\,k\,{dk\over d\ve} & d=2 \\ &\\ {\Sg\over 2\pi^2}\,k^2\,{dk\over d\ve} & d=3\ .\end{cases}$

In order to obtain $$g(\ve)$$ as a function of the energy $$\ve$$ one must invert the dispersion relation $$\ve=\ve(k)$$ to obtain $$k=k(\ve)$$.

Note that we can equivalently write

$g(\ve)\,d\ve = \Sg\>{d^d\!k\over (2\pi)^d} = {\Sg\,\Omega\ns_d\over (2\pi)^d}\>k^{d-1}\,dk$

to derive $$g(\ve)$$.

For a spin-$$S$$ particle with ballistic dispersion $$\ve(\Bk)=\hbar^2\Bk^2/2m$$, we have

$g(\ve)={2S\!+\!1\over \RGamma(d/2)}\bigg({m\over 2\pi\hbar^2}\bigg)^{\!d/2}\ve^{{d\over 2}-1}\ \RTheta(\ve) , \label{BDOS}$

where $$\RTheta(\ve)$$ is the step function, which takes the value $$0$$ for $$\ve<0$$ and $$1$$ for $$\ve\ge 0$$. The appearance of $$\RTheta(\ve)$$ simply says that all the single particle energy eigenvalues are nonnegative. Note that we are assuming a box of volume $$V$$ but we are ignoring the quantization of kinetic energy, and assuming that the difference between successive quantized single particle energy eigenvalues is negligible so that $$g(\ve)$$ can be replaced by the average in the above expression. Note that

$n(\ve,T,\mu)={1\over e^{(\ve-\mu)/\kT}\mp 1}.$

This result holds true independent of the form of $$g(\ve)$$. The average total number of particles is then

$N(T,V,\mu)=V\!\!\!\int\limits_{-\infty}^\infty\!\!\!d\ve\,g(\ve)\,{1\over e^{(\ve-\mu)/\kT}\mp 1}, \label{numeqn}$

which does depend on $$g(\ve)$$.