14.4 Quantum statistical mechanics

中文

Statistical mechanics is the physics oflarge systems ofparticles, which are usually identical. The systems are generally so large that only averages ofphysical quantities can be accurately dealt with. This section will give only the briefest introduction to this enormous and far ranging subject.

Density operator

Let a quantum system have a complete o.n. basis $|{\psi }_i⟩$ . Ifwe imagine the rest of the universe (taken in a somewhat restricted sense) to be spanned by an o.n. set $|{\theta }_a⟩$ , then the genera state of the combined system can be written

$$|\Psi ⟩=\sum _i\sum _a{c}_{ia}|{\psi }_i⟩|{\theta }_a⟩.$$

An operator $A$ acting on the system only acts on the vectors $|{\psi }_i⟩$ , hence

$$\begin{array}{rl}⟨A{⟩}_{\Psi }& =⟨\Psi |A|\Psi ⟩\\ & ={\sum }_i{\sum }_a{\sum }_j{\sum }_b\overline{c_{ia}}⟨{\theta }_a|⟨{\psi }_i|A|{\psi }_j⟩|{\theta }_b⟩c_{jb}\\ & ={\sum }_i{\sum }_a{\sum }_j{\sum }_b{A}_{ij}\overline{c_{ia}}{c}_{jb}{\delta }_{ab}\\ & ={\sum }_i{\sum }_j{A}_{ij}{\rho }_{ji}\end{array}$$

where

$$A_{ij}=⟨{\psi }_i|A|{\psi }_j⟩,{\rho }_{ji}=\sum _a{c}_{ja}\overline{c_{ia}}.$$

The operator $A$ can be written

$$A=\sum _i\sum _j{A}_{ij}|{\psi }_i⟩⟨{\psi }_j|.$$

Exercise

Verify that for any $|\varphi ⟩\in \mathcal{H}$

$$A|\varphi ⟩=\sum _i\sum _j{A}_{ij}|{\psi }_i⟩⟨{\psi }_j|\varphi ⟩.$$

Define the density operator $\rho$ as that having components ${\rho }_{ij}$ ,

$$\rho =\sum _i\sum _j{\rho }_{ij}|{\psi }_i⟩⟨{\psi }_j|,$$

which is hermitian since

$$\overline{{\rho }_{ji}}=\sum _a\overline{c_{ia}}{c}_{ja}={\rho }_{ij}.$$

A useful expression for the expectation value of $A$ is

$$⟨A⟩=\mathrm{tr}(A\rho )=\mathrm{tr}(\rho A)\text{\hspace{1em}(14.40)}$$

where the trace of an operator is given by

$$\mathrm{tr}B=\sum _i⟨{\psi }_i|B|{\psi }_i⟩=\sum _i{B}_{ii}.$$

Exercise

Show that the trace of an operator is independent of the o.n. basis $|{\psi }_i⟩$

Setting $A=I$ we have

$$⟨I⟩=⟨\Psi |\Psi ⟩=\Vert \Psi {\Vert }^2=1⟹\mathrm{tr}(\rho )=1,$$

and setting $A=|{\psi }_k⟩⟨{\psi }_k|$ gives

$$⟨A⟩=⟨\Psi |{\psi }_k⟩⟨{\psi }_k|\Psi ⟩=\Vert ⟨\Psi |{\psi }_k⟩{\Vert }^2\ge 0.$$

On the other hand

$$\begin{array}{c}\mathrm{tr}(A\rho )={\sum }_i⟨{\psi }_i|A\rho |{\psi }_i⟩={\sum }_i⟨{\psi }_i|{\psi }_k⟩⟨{\psi }_k|\rho |{\psi }_i⟩\\ ={\sum }_i{\delta }_{ik}⟨{\psi }_k|\rho |{\psi }_i⟩=⟨{\psi }_k|\rho |{\psi }_k⟩={\rho }_{kk}.\end{array}$$

Hence all diagonal elements of the density matrix are positive, ${\rho }_{kk}=⟨{\psi }_k|\rho |{\psi }_k⟩\ge 0.\mathrm{As}-$ suming $\rho$ is a complete operator, select $|{\psi }_i⟩$ to be eigenvectors, $\rho |{\psi }_i⟩=w_i|{\psi }_i⟩$ , so that $\rho$ is diagonalized

$$\rho =\sum _i{w}_i|{\psi }_i⟩⟨{\psi }_i|.\text{\hspace{1em}(14.41)}$$

We then have

$$\sum _i{w}_i=1,w_i\ge 0.$$

The interpretation of the density operator $\rho ,$ , or its related state $|\Psi ⟩$ , is as a mixed state of the system, with the $i$th eigenstate $|{\psi }_i⟩$ having probability $w_i.\mathrm{A}$ pure state occurs when there exists $k$ such that $w_k=1$ and $w_i=0$ for all $i\ne k$ . In this case ${\rho }^2=\rho$ and the density operator is idempotent – it acts as a projection operator into the one-dimensional subspace spanned by the associated eigenstate ${\psi }_k$

Exercise

Show the converse: $i\cdot {\rho }^2=\rho$ then all $w_i=1\mathrm{or}0.$ , and there exists $k$ such tha $\rho =|{\psi }_k⟩⟨{\psi }_k|$

Exercise

Show that the probability of finding the system in a state $|\chi ⟩$ is $\mathrm{tr}\rho |\chi ⟩⟨\chi |$

Example 14.11

Consider a beam of photons in the $z$-direction. Let $|{\psi }_1⟩$ be the state of a photon polarized in the $x$-direction, and $|{\psi }_2⟩$ be the state of a photon polarized in the $y$-direction. The general state is a linear sum of these two,

$$|\psi ⟩=a|{\psi }_1⟩+b|{\psi }_2⟩\text{ where }|a{|}^2+|b{|}^2=1.$$

The pure state represented by this vector has density operator $\rho =|\psi ⟩⟨\psi |$ , having compo nents

$${\rho }_{ij}=⟨{\psi }_i|\rho |{\psi }_j⟩=⟨{\psi }_i|\psi ⟩\overline{⟨{\psi }_j|\psi ⟩}=\left(\begin{array}{cc}a\overline{a}& a\overline{b}\\ b\overline{a}& b\overline{b}\end{array}\right).$$

For example, the pure states corresponding to $45^{\circ }$ -polarization $\left(a=b=\frac{1}{\sqrt{2}}\right)$ and

$135^{\circ }$ -polarization $\left(a=-b=-\frac{1}{\sqrt{2}}\right)$ have respective density operators

$$\rho =\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right)\text{ and }\rho =\left(\begin{array}{cc}\frac{1}{2}& -\frac{1}{2}\\ -\frac{1}{2}& \frac{1}{2}\end{array}\right).$$

A half–half mixture of $45^{\circ }\cdot$ - and 135◦-polarized photons is indistinguishable from an equa mixture of $x$-polarized photons and $y$-polarized photons, since

$$\begin{array}{l}\left(\frac{1}{\sqrt{2}}|{\psi }_1⟩+\frac{1}{\sqrt{2}}|{\psi }_2⟩\right)\left(\frac{1}{\sqrt{2}}⟨{\psi }_1|+\frac{1}{\sqrt{2}}⟨{\psi }_2|\right)+\left(\frac{1}{\sqrt{2}}|{\psi }_1⟩-\frac{1}{\sqrt{2}}|{\psi }_2⟩\right)\left(\frac{1}{\sqrt{2}}⟨{\psi }_1|-\frac{1}{\sqrt{2}}⟨{\psi }_2|\right)\\ =\frac{1}{2}|{\psi }_1⟩⟨{\psi }_1|+\frac{1}{2}|{\psi }_2⟩⟨{\psi }_2|.\end{array}$$

From Schrödinger’s equation Eq. (14.13),

$$i\hslash \frac{\mathrm{d}}{\mathrm{d}t}|{\psi }_i⟩=H|{\psi }_i⟩,-i\hslash \frac{\mathrm{d}}{\mathrm{d}t}⟨{\psi }_i|=⟨{\psi }_i|H.$$

It follows that the density operator satisfies the evolution equation

$$\frac{\mathrm{d}\rho }{\mathrm{d}t}=\frac{-i}{\hslash }[H,\rho ].\text{\hspace{1em}(14.42)}$$

From the solution Eq. (14.14), Eq. (14.15) of the Schrödinger equation, the solution of Eq. (14.42) is

$$\rho (t)={\mathrm{e}}^{(-i/\hslash )Ht}\rho (0){\mathrm{e}}^{(i/\hslash )Ht}.\text{\hspace{1em}(14.43)}$$

Hence for any function $\begin{array}{r}f(\rho )={\sum }_{i}f(w_i)|{\psi }_i⟩⟨{\psi }_i|\end{array}$ the trace is constant,

$$\mathrm{tr}(f(\rho ))=\mathrm{tr}({\mathrm{e}}^{(-i/\hslash )Ht}f(\rho (0)){\mathrm{e}}^{(i/\hslash )Ht})=\mathrm{tr}(f(\rho (0)))=\text{ const. }$$

A mixed state is said to be stationary if $\mathrm{d}\rho /\mathrm{d}t=0$ . From Eq. (14.42) this implies $[H,\rho ]=0$ , and for any pair of energy eigenvectors

$$H|E_j⟩=E_j|E_j⟩,H|E_k⟩=E_k|E_k⟩$$

we have

$$0=⟨E_j|\rho H-H\rho |E_k⟩=(E_k-E_j)⟨E_j|\rho |E_k⟩.$$

Hence, if $E_j\ne E_k$ then $⟨E_j|\rho |E_k⟩={\rho }_{jk}=0$ , and if $H$ has no degenerate energy levels then

$$\rho =\sum _i{w}_i|E_i⟩⟨E_i|,$$

which is equivalent to the assertion that the density operator is a function of the Hamiltonian, $\rho =\rho (H)$ . If $H$ has degenerate energy levels then $\rho$ and $H$ can be simultaneously diagonalized, and it is possible to treat this as a limiting case of non-degenerate levels. It is reasonable therefore to assume that $\rho =\rho (H)$ in all cases.

Ensembles

An ensemble of physical systems is another way of talking about the density operator. Essentially, we consider a large number of copies of the same system, within certain con straints, to represent a statistical system of particles. Each member of the ensemble is a possible state of the system; it is an eigenstate of the Hamiltonian and the density operator tells us its probability within the ensemble.

One of the simplest examples is the microcanonical ensemble, where $\rho$ is constant for energy values in a narrow range, $E<E_k<E+\Delta E$ , and $w_j=0$ for all energy values $E_j$ outside this range. For those energy values within the allowed range, we set $w_k=w=1/s$ where $s$ is the number of energy values in the range $(E,E+\Delta E)$ . Let $j(E)$ be the number of states with energy $E_k<E$ , then

$$s=j(E+\Delta E)-j(E)=\Sigma (E)\Delta E,$$

where

$$\Sigma (E)=\frac{\mathrm{d}j(E)}{\mathrm{d}E}=\text{ the density of states. }$$

For the microcanonical ensemble all $w_k=0$ for $E_k\le E$ or $E_k\ge E+\Delta E$ , while

$$w_k=\frac{1}{\Sigma (E)\Delta E}\text{ if }E<E_k<E+\Delta E.$$

The canonical ensemble can be thought of as a system embedded in a heat reservoir consisting of the external world. Let $H$ be the Hamiltonian of the system and $H_R$ that of the reservoir. The total Hamiltonian of the universe is $H_U=H_R+H$ . Suppose the system is in the eigenstate $|{\psi }_m⟩$ of energy $E_m$

$$H|{\psi }_m⟩=E_m|{\psi }_m⟩$$

and let $|\Psi ⟩=|\theta ⟩|{\psi }_m⟩$ be the total state of the universe. If we assume the universe to be in a microcanonical ensemble, then

$$H_U|\Psi ⟩=E_U|\Psi ⟩\text{ where }E<E_U<E+\Delta E.$$

Using the decomposition $H_U=H_R+H$ we have

$$H_R|\theta ⟩|{\psi }_m⟩+|\theta ⟩E_m|{\psi }_m⟩=E_U|\theta ⟩|{\psi }_m⟩,$$

whence

$$H_R|\theta ⟩=(E_U-E_m)|\theta ⟩.$$

Thus $|\theta ⟩$ is an eigenstate of $H_R$ with energy $E_U-E_m.\mathrm{If}{\Sigma }_R(E_R)$ is the density of states in the reservoir, then

$$w_m{\Sigma }_U(E_U)\Delta E={\Sigma }_R(E_U-E_m)\Delta E,$$

whence

$$w_m=\frac{{\Sigma }_R(E_U-E_m)}{{\Sigma }_U(E_U)}.$$

For $E_m\ll E_U$ , as expected of a system in a much larger reservoir,

$$\ln w_m=const.-\beta E_m,$$

most commonly written in the form

$$w_m=\frac{1}{Z}{\mathrm{e}}^{-\beta E_m}\text{ where }Z=\sum _{m=0}^{\infty }{\mathrm{e}}^{-\beta E_m},\text{\hspace{1em}(14.44)}$$

where the last identity follows from ${\sum }_m{w}_m=1$ . The density operator for the canonica ensemble is thus

$$\rho =\frac{1}{Z}{\mathrm{e}}^{-\beta H}\text{\hspace{1em}(14.45)}$$

where, by the identity $\mathrm{tr}\rho =1$

$$Z=\mathrm{tr}{\mathrm{e}}^{-\beta H}=\sum _{m=0}^{\infty }{\mathrm{e}}^{-\beta E_m},\text{\hspace{1em}(14.46)}$$

known as the canonical partition function. The average energy is

$$U=⟨E⟩=\mathrm{tr}(\rho H)=\frac{1}{Z}\sum _k{E}_k{\mathrm{e}}^{-\beta E_k}=-\frac{\partial \ln Z}{\partial \beta }.\text{\hspace{1em}(14.47)}$$

Example 14.12

Consider a linear harmonic oscillator having Hamiltonian given by Eq. (14.19). The energy eigenvalues are

$$E_m=\frac{1}{2}\hslash \omega +m\hslash \omega$$

and the partition function is

$$\begin{array}{l}Z={\sum }_{m=0}^{\infty }{\mathrm{e}}^{-\beta E_m}\\ ={\mathrm{e}}^{-\frac{1}{2}\beta \hslash \omega }{\sum }_{m=0}^{\infty }{\left({\mathrm{e}}^{-\beta \hslash \omega }\right)}^m.\\ =\frac{{\mathrm{e}}^{\beta \hslash \omega /2}}{{\mathrm{e}}^{\beta \hslash \omega }-1}.\end{array}$$

From Eq. (14.47) the average energy is

$$\begin{array}{c}U=-\frac{\partial \ln Z}{\partial \beta }\\ =\frac{1}{2}\hslash \omega +\frac{\hslash \omega }{{\mathrm{e}}^{\beta \hslash \omega }-1}.\end{array}$$

As $\beta \to 0$ we have $U\approx {\beta }^{-1}$ . This is the classical limit $U=kT$ , where $T$ is the temperature, and is an indication of the identity $\beta =1/kT.\mathrm{As}\beta \to \infty$ we arrive at the low temperature limit,

$$U\to \frac{1}{2}\hslash \omega +\hslash \omega {\mathbf{e}}^{-\beta \hslash \omega }\approx \frac{1}{2}\hslash \omega .$$

The entropy is defined as

$$S=-k\mathrm{tr}(\rho \ln \rho )=-k\sum _i{w}_i\ln w_i.\text{\hspace{1em}(14.48)}$$

For a pure state, $w_i=1\mathrm{or}0.$ , we have $S=0$ . This is interpreted as a state of maximum order. For a completely random state, $w_i=const.=1/N$ where $N$ is the total number of states in the ensemble (assumed finite here), the entropy is

$$S=k\ln N.$$

This state of maximal disorder corresponds to a maximum value of $S,$ as may be seen by using the method of Lagrange multipliers: the maximum of $S$ occurs where $\mathrm{d}S=0$ subject to the constraint ${\sum }_i{w}_i=1$ 2

$$\begin{array}{c}\mathrm{d}S=0⟹\mathrm{d}{\sum }_i{w}_i\ln w_i-\lambda {\sum }_i\mathrm{d}{w}_i=0\\ ⟹{\sum }_i(1+\ln w_i-\lambda )\mathrm{d}{w}_i=0.\end{array}$$

Since $\mathrm{d}{w}_i$ is arbitrary the Lagrange multiplier is $\lambda =1+\ln w_i$ , so that

$$w_1=w_2=\cdots ={\mathrm{e}}^{\lambda -1}.$$

Exercise

Two systems may be said to be independent iftheir combined density operator is $\rho ={\rho }_1{\rho }_2={\rho }_2{\rho }_1$ . Show that the entropy has the additive property for independent systems, $S=S_1+S_2$

If the Hamiltonian depends on a parameter $a$, we define ‘generalized force’ $A$ conjugate to $a$ by

$$A=⟨-\frac{\partial H}{\partial a}⟩=-\mathrm{tr}\left(\rho \frac{\partial H}{\partial a}\right).\text{\hspace{1em}(14.49)}$$

For example, for a gas in a volume $V$, the pressure $p$ is defined as

$$p=⟨-\frac{\partial H}{\partial V}⟩=-\mathrm{tr}\left(\rho \frac{\partial H}{\partial V}\right).$$

If $H|{\psi }_k⟩=E_k|{\psi }_k⟩$ where $E_k=E_k(a)$ and $|{\psi }_k⟩=|{\psi }_k(a)⟩$ , then

$$\frac{\partial H}{\partial a}|{\psi }_k⟩+H\frac{\partial }{\partial a}|{\psi }_k⟩=\frac{\partial E_k}{\partial a}|{\psi }_k⟩+E_k\frac{\partial }{\partial a}|{\psi }_k⟩$$

so that

$$\begin{array}{l}A=-⟨\frac{\partial H}{\partial a}⟩=-\mathrm{tr}\left(\rho \frac{\partial H}{\partial a}\right)\\ =-{\sum }_k{w}_k⟨{\psi }_k|\frac{\partial H}{\partial a}|{\psi }_k⟩\\ =-{\sum }_k{w}_k(⟨{\psi }_k|\frac{\partial E_k}{\partial a}|{\psi }_k⟩+⟨{\psi }_k|E_k-H|\frac{\partial }{\partial a}{\psi }_k⟩)\\ =-{\sum }_k{w}_k(\frac{\partial E_k}{\partial a}\Vert {\psi }_k{\Vert }^2+⟨{\psi }_k|E_k-E_k|\frac{\partial }{\partial a}{\psi }_k⟩)\\ =-{\sum }_k{w}_k\frac{\partial E_k}{\partial a}.\end{array}$$

The total work done under a change of parameter $\mathrm{d}a$ is defined to be

$$\mathrm{d}W=-\mathrm{d}U=-\sum _k{w}_k\frac{\partial E_k}{\partial a}=A\mathrm{d}a.$$

For a change in volume this gives the classical formula $dW=p\mathrm{d}V$

For the canonical ensemble we have, by Eq. (14.44),

$$A=-\sum _k{w}_k\frac{\partial E_k}{\partial a}=\frac{1}{\beta }\frac{\partial \ln Z}{\partial a},\text{\hspace{1em}(14.50)}$$

and as the entropy is given by

$$S=-k\sum _k{w}_k\ln w_k=k\sum _k{w}_k(\ln Z+\beta E_k)=k(\ln Z+\beta U)$$

we have

$$\mathrm{d}S=k(\mathrm{d}\ln Z+\beta \mathrm{d}U)=\frac{1}{T}(\mathrm{d}U+A\mathrm{d}a)\text{\hspace{1em}(14.51)}$$

where

$$\beta =\frac{1}{kT}.$$

This relation forms the basic connection between statistical mechanics and thermodynamics (see Section 16.4); the quantity $T$ is known as the temperature of the system.

Systems of identical particles

For a system of $N$ identical particles, bosons or fermions, let $h$ be the Hamiltonian of each individual particle, having eigenstates

$$h|{\phi }_a⟩={\epsilon }_a|{\phi }_a⟩(a=0,1,\dots ).$$

The Hamiltonian of the entire system $i\begin{array}{r}H:{\mathcal{H}}^N\to {\mathcal{H}}^N\end{array}$ given by

$$H=h_1+h_2+\cdots +h_N$$

where

$$h_i|{\psi }_1⟩\dots |{\psi }_i⟩\dots |{\psi }_N⟩=|{\psi }_1⟩\dots h|{\psi }_i⟩\dots |{\psi }_N⟩.$$

The eigenstates of the total Hamiltonian,

$$H|{\Phi }_k⟩=E_k|{\Phi }_k⟩,$$

are linear combinations of state vectors

$$|{\psi }_1⟩|{\psi }_2⟩\dots |{\psi }_N⟩$$

such that

$$E_k={\epsilon }_0+{\epsilon }_1+\cdots +{\epsilon }_N.$$

If $n_0$ particles are in state $|{\phi }_0,n_1$ particles in state $|{\phi }_1$ , etc. then the energy eigenstates are determined by the set of occupation numbers $(n_0,n_1,n_2,\dots )$ such that

$$E_k=\sum _{a=0}^{\infty }{n}_a{\epsilon }_a.$$

If we are looking for eigenstates that are simultaneously eigenstates of the permutation operators $P_{:}$ , then they must be symmetric states $P|{\Psi }_k⟩=|{\Psi }_k⟩$ for bosons, and antisym metric states $P|{\Psi }_k⟩=(-1{)}^P|{\Psi }_k⟩$ in the case of fermions. Let $S$ be the symmetrization operator and $A$ the antisymmetrization operator

$$S=\frac{1}{N!}\sum _{P}P,A=\frac{1}{N!}\sum _P(-1{)}^{P}P.$$

Both are hermitian and idempotent

$$S^{\ast }=\frac{1}{N!}\sum _P{P}^{\ast }=\frac{1}{N!}\sum _P{P}^{-1}=S,A^{\ast }=A\text{ since }(-1{)}^{P^{\ast }}=(-1{)}^P$$ $$S^2=S,A^2=A,AS=SA=0.$$

Thus $S$ and $A$ are orthogonal projection operators, and for any state $|\Psi ⟩$

$$PS|\Psi ⟩=S|\Psi ⟩,PA|\Psi ⟩=(-1{)}^{P}A|\Psi ⟩$$

for all permutations $P$. For bosons the eigenstates are of the form $S|{\phi }_{a_1}⟩|{\phi }_{a_2}⟩\dots |{\phi }_{a_N}⟩$ , while for fermions they are $A|{\phi }_{a_1}⟩|{\phi }_{a_2}⟩\dots |{\phi }_{a_N}⟩$ . In either case the state of the system is completely determined by the occupation numbers $n_0,n_1,\dots$ . For bosons the occupation numbers run from 0 to $N_{:}$ , while the Pauli exclusion principle implies that fermionic occupation numbers only take on values 0 or 1. Thus, for the canonical distribution

$$w(n_0,n_1,\dots )=\frac{1}{Z}{\mathrm{e}}^{-\beta {\sum }_a{n}_a{\epsilon }_a},\sum _a{n}_a=N,$$

where

$$Z_{\mathrm{Bose}}=\sum _{n_0=0}^N\sum _{n_1=0}^N\dots {\mathrm{e}}^{-\beta {\sum }_a{n}_a{\epsilon }_a},$$ $$Z_{\text{ Fermi }}=\sum _{n_0=0}^1\sum _{n_1=0}^1\dots {\mathrm{e}}^{-\beta {\sum }_a{n}_a{\epsilon }_a}.$$

The constraint ${\sum }_a{n}_a=N$ makes these sums quite difficult to calculate directly.

In the classical version where particles are distinguishable, all ways of realizing a configuration are counted separately,

$$\begin{array}{l}{Z}_{\text{Boltzmann}}={\sum }_{n_0=0}^N{\sum }_{n_1=0}^N\dots {\mathrm{e}}^{-\beta {\sum }_a{n}_a{\epsilon }_a}\frac{N!}{n_0!n_1!\dots }\\ ={\left({\mathrm{e}}^{-\beta {\epsilon }_0}+{\mathrm{e}}^{-\beta {\epsilon }_1}+\dots \right)}^N\\ =(Z_1{)}^N,\end{array}$$

where $Z_1$ is the one-particle partition function

$$Z_1=\sum _a{\mathrm{e}}^{-\beta {\epsilon }_a}.$$

The average energy is

$$U=⟨E⟩=-\frac{\partial \ln Z}{\partial \beta }=\frac{N{\sum }_a{\epsilon }_a{\mathrm{e}}^{-\beta {\epsilon }_a}}{{\sum }_a{\mathrm{e}}^{-\beta {\epsilon }_a}}=N⟨\epsilon ⟩.$$

It is generally accepted that $Z_{\mathrm{Boltzmann}}$ should be divided by $N$!, discounting all possible permutations of particles, in order to avoid the Gibbs paradox.

In the quantum case, it is easier to consider an even larger distribution, wherein the number of particles is no longer fixed. Assuming an open system, allowing exchange of particles between the system and reservoir, an argument similar to that used to arrive at the canonical ensemble gives

$$w_{mN}=\frac{1}{Z_g}{\mathrm{e}}^{\alpha N-\beta E_m},$$

where

$$Z_g=\sum _{N=0}^{\infty }\sum _m{\mathrm{e}}^{\alpha N-\beta E_m}.$$

This is known as the partition function for the grand canonical ensemble. In terms of the density operator,

$$\rho =\frac{1}{Z_g}{\mathrm{e}}^{-\beta H+\alpha N},Z_g=\mathrm{tr}{\mathrm{e}}^{-\beta H+\alpha N}.$$

For a system of identical particles,

$$w(n_0,n_1,\dots )=\frac{1}{Z}{\mathrm{e}}^{\alpha {\sum }_a{n}_a-\beta {\sum }_a{n}_a{\epsilon }_a},$$

where ${\sum }_a{n}_a=N$ is no longer fixed. We can therefore write

$$\begin{array}{l}{Z}_{\text{Bose}}={\sum }_{n_0=0}^{\infty }{\sum }_{n_1=0}^{\infty }\dots {\mathrm{e}}^{\alpha {\sum }_a{n}_a-\beta {\sum }_a{n}_a{\epsilon }_a}\\ ={\sum }_{n_0=0}^{\infty }{\mathrm{e}}^{(\alpha n_0-\beta {\epsilon }_0)n_0}{\sum }_{n_1=0}^{\infty }{\mathrm{e}}^{(\alpha n_1-\beta {\epsilon }_1)n_1}\dots \\ =\frac{1}{1-{\mathrm{e}}^{\alpha -\beta {\epsilon }_0}}\frac{1}{1-{\mathrm{e}}^{\alpha -\beta {\epsilon }_1}}\dots \end{array}$$

and

$$\ln Z_{\text{ Bose }}=-\sum _{a=0}^{\infty }\ln \left(1-\lambda {\mathrm{e}}^{-\beta {\epsilon }_a}\right),\lambda ={\mathrm{e}}^{\alpha }.$$

Similarly

$$Z_{\text{ Fermi }}=\sum _{n_0=0}^1\sum _{n_1=0}^1\dots {\mathrm{e}}^{\alpha {\sum }_a{n}_a-\beta {\sum }_a{n}_a{\epsilon }_a}\dots$$

results in

$$\ln Z_{\text{ Fermi }}=\sum _{a=0}^{\infty }\ln \left(1+\lambda e^{-\beta {\epsilon }_a}\right),\lambda =e^{\alpha }.$$

Summarizing, we have

$$\ln Z=\pm \sum _{a=0}^{\infty }\ln \left(1\pm \lambda {\mathrm{e}}^{-\beta {\epsilon }_a}\right)\text{\hspace{1em}(14.52)}$$

where the sign occurs for fermions, the sign for bosons.

The average occupation numbers are

$$\begin{array}{l}⟨n_a⟩={\sum }_{n_0}{\sum }_{n_1}\dots n_{a}w(n_0,n_1,\dots )\\ =\frac{1}{Z}{\sum }_{n_0}{\sum }_{n_1}\dots n_a{\mathrm{e}}^{{\sum }_a(\alpha -\beta {\epsilon }_a)n_a}\\ =-\frac{1}{\beta }\frac{\partial }{\partial {\epsilon }_a}\ln Z.\end{array}$$

Using Eq. (14.52),

$$⟨n_a⟩=-\frac{1}{\beta }\frac{\partial }{\partial {\epsilon }_a}\left(\pm \ln \left(1\pm \lambda {\mathrm{e}}^{-\beta {\epsilon }_a}\right)\right)=\frac{1}{{\lambda }^{-1}{\mathrm{e}}^{\beta {\epsilon }_a}\pm 1},\text{\hspace{1em}(14.53)}$$

where the sign applies to Fermi particles, and the to Bose. The parameter $\lambda$ is often written $\lambda =e^{\beta \mu }$ , where $\mu$ is known as the chemical potential. It can be shown, using the method of steepest descent (see [10]) that the formulae Eq. (14.53) are also valid for the canonical ensemble. In this case the total particle number is fixed so that the chemica potential may be found from

$$\sum _a⟨n_a⟩=\frac{1}{Z}\sum _{n_0}\sum _{n_1}\dots \left(\sum _a{n}_a\right){\mathrm{e}}^{{\sum }_b\left(\alpha -\beta {\epsilon }_b\right)n_b}=N\sum _{n_0}\sum _{n_1}\dots w\left(n_0,n_1,\dots \right)=N.$$

That is,

$$N=\sum _a\frac{1}{{\mathrm{e}}^{\beta ({\epsilon }_a-\mu )}\pm 1}$$

and

$$U=⟨E⟩=-\frac{\partial Z}{\partial \beta }=\sum _a\frac{{\epsilon }_a}{{\mathrm{e}}^{\beta ({\epsilon }_a-\mu )}\pm 1}=\sum _a{\epsilon }_a⟨n_a⟩.$$

Application to perfect gases, black body radiation and other systems may be found in any standard book on statistical mechanics [1, 10–12].

Problems

Problem 14.17

Show that the correctly normalized fermion states are

$$\frac{1}{\sqrt{N!}}\sum _P(-1{)}^{P}P|{\phi }_{a_1}⟩|{\phi }_{a_2}⟩\dots |{\phi }_{a_N}⟩$$

and normalized boson states are

$$\frac{1}{\sqrt{N!}\sqrt{n_0!}\sqrt{n_1!}\dots }\sum _{P}P|{\phi }_{a_1}⟩|{\phi }_{a_2}⟩\dots |{\phi }_{a_N}⟩.$$

Problem 14.18

Calculate the canonical partition function, mean energy $U$ and entropy $S,$ for a system having just two energy levels 0 and $E$. If $E=E(a)$ for a parameter ${}^{a,}$ calculate the force $A$ and verify the thermodynamic relation $\begin{array}{r}\mathrm{d}S=\frac{1}{T}(\mathrm{d}U+A\mathrm{d}a)\end{array}$

Problem 14.19

let $\rho ={\mathsf{e}}^{-\beta H}$ be the unnormalized canonical distribution. For a free particle of mass $m$ in one dimension show that its position representation form $\rho (x,x^{\prime };\beta )=⟨x|\rho |x^{\prime }⟩$ satisfies the diffusion equation

$$\frac{\partial \rho (x,x^{\prime };\beta )}{\partial \beta }=\frac{{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x^2}\rho (x,x^{\prime };\beta )$$

with ‘initial’ condition $\rho (x,x^{\prime };0)=\delta (x-x^{\prime })$ . Verify that the solution is

$$\rho (x,x^{\prime };\beta )={\left(\frac{m}{2\pi {\hslash }^2\beta }\right)}^{1/2}{\mathrm{e}}^{-m(x-x^{\prime }{)}^2/2{\hslash }^2\beta }.$$

Problem 14.20

A solid can be regarded as being made up of $3N$ independent quantum oscillators of angular frequency $\omega .$ Show that the canonical partition function is given by

$$Z={\left(\frac{{\mathrm{e}}^{-\beta \hslash \omega /2}}{1-{\mathrm{e}}^{-\beta \hslash \omega }}\right)}^{3N},$$

and the specific heat is given by

$$C_V=\frac{\mathrm{d}U}{\mathrm{d}T}=3Nk{\left(\frac{T_0}{T}\right)}^2\frac{{\mathrm{e}}^{T_0/T}}{{\left({\mathrm{e}}^{T_0/T}-1\right)}^2}\text{ where }k{T}_0=\hslash \omega .$$

Show that the high temperature limit $T\gg T_0$ is the classical value $C_V=3Nk.$

Problem 14.21

Show that the average occupation numbers for the classical distribution, $Z_{Boltzmann}$ are given by

$$⟨n_a⟩=-\frac{1}{\beta }\frac{\partial }{\partial {\epsilon }_a}\ln Z_{\mathrm{Boltzmann}}=\lambda {\mathrm{e}}^{-\beta {\epsilon }_a}.$$

Hence show that

$$⟨n_a{⟩}_{\text{ Fermi }}<⟨n_a{⟩}_{\text{ Boltzmann }}<⟨n_a{⟩}_{\text{ Bose }}$$

and that all three types agree approximately for low occupation numbers $⟨n_a⟩\ll 1$

Problem 14.22

A spin system consists of $N$ particles of magnetic moment $\mu$ in a magnetic field $B$. When $n$ particles have spin up, $N-n$ spin down, the energy is $E_n=n\mu B-(N-n)\mu B=(2n-N)\mu B$ . Show that the canonical partition function is

$$Z=\frac{\sinh ((N+1)\beta \mu B)}{\sinh \beta \mu B}.$$

Evaluate the mean energy $U$ and entropy $S,$ sketching their dependence on the variable $x=\beta \mu B$