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5.1: Statistical Mechanics of Noninteracting Quantum Systems

  • Page ID
    18569
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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 \(g(\ve)\).