Chapter 52The Grand Canonical Distribution and Quantum Statistics — The Fermi and Bose Distributions

Example 52.1: Grand canonical distribution of an adsorption site (a miniature of Langmuir adsorption)

Suppose a solid surface has a single small “adsorption site” that can adsorb just one gas molecule. When the site is empty, the energy is $0$ and the number of particles is $N=0$. When one molecule is adsorbed at the site, the energy is $-\varepsilon_0$ ($\varepsilon_0\gt0$; the system is stabilized by adsorption, so the energy is negative) and the number of particles is $N=1$. Regard this site as a grand canonical system in contact with a gas (the supply source of the molecules to be adsorbed) at temperature $T$ and chemical potential $\mu$. (1) Find the ratio between the probability $P_{\text{empty}}$ that the site is empty and the probability $P_{\text{adsorbed}}$ that a molecule is adsorbed. (2) When $\mu$ is very large (the gas pressure is very high), which probability becomes larger? Read the answer off from the expression.

Solution. (1) Apply Eq. \eqref{eq:52-gc-prob} to the two states $(j,N)=(\text{empty},0)$ and $(j,N)=(\text{adsorbed},1)$. The empty state has $E_{j,N}=0,N=0$, so $P_{\text{empty}}\propto e^{-\beta(0-\mu\cdot0)}=e^{0}=1$. The adsorbed state has $E_{j,N}=-\varepsilon_0,N=1$, so $P_{\text{adsorbed}}\propto e^{-\beta(-\varepsilon_0-\mu\cdot1)}=e^{\beta(\varepsilon_0+\mu)}$. Therefore,

(2) The larger $\mu$ is (the higher the pressure and density of the gas; as we will see in Section 52.3, $\mu$ increases with them), the larger the exponent $\beta(\varepsilon_0+\mu)$ becomes, and $P_{\text{adsorbed}}/P_{\text{empty}}\to\infty$: the site is almost certainly filled by a molecule. Conversely, for $\mu\to-\infty$ (an extremely dilute gas) the ratio approaches $0$, and the site is almost certainly empty. This agrees perfectly with intuition (the higher the gas pressure, the more adsorption onto the surface proceeds), and it is the first concrete example showing that $\mu$ is a quantity that expresses “how easily particles can be pushed in.” (You will later notice that the statistics of this adsorption site is none other than the one-level version of the Fermi–Dirac distribution that we derive in Section 52.5.)

Example 52.2: Grand partition function of a classical ideal gas

Using the single-particle partition function of the monatomic ideal gas found in Section 51.1, $Z_1=V(2\pi mk_{\mathrm B}T/h^2)^{3/2}=V/\lambda_{\mathrm T}^3$ ($\lambda_{\mathrm T}$ is the thermal de Broglie wavelength), and the $N$-particle partition function $Z_N=Z_1^N/N!$ (same chapter), find the grand partition function $\Xi$ of the classical ideal gas.

Solution. Substituting $Z_N=Z_1^N/N!$ into Eq. \eqref{eq:52-Xi-ZN} (the form using the fugacity $z=e^{\beta\mu}$), we obtain

This is exactly the Taylor expansion (Maclaurin expansion) of the exponential function, $e^x=\sum_{N=0}^{\infty}x^N/N!$ (University Mathematics, Chapter 4), with $x=zZ_1$. Therefore,

so the grand partition function is obtained in a neat closed form (a single exponential function rather than an infinite series). Using this, from Eq. \eqref{eq:52-thm-J} we get $N=k_{\mathrm B}T\,\partial\ln\Xi/\partial\mu=k_{\mathrm B}T\cdot\beta zZ_1=zZ_1=e^{\beta\mu}Z_1$ (we used the fact that, when $\ln\Xi=zZ_1$ is differentiated with respect to $\mu$, the derivative of $z=e^{\beta\mu}$, $\partial z/\partial\mu=\beta z$, gives $\partial(zZ_1)/\partial\mu=\beta zZ_1$). We use this relation $N=e^{\beta\mu}Z_1$ in the next section to find the explicit form of the chemical potential.

Example 52.3: When does $\mu=0$ for nitrogen gas? — the limit of the classical ideal gas formula

Consider a gas of nitrogen molecules $\mathrm N_2$ (mass $m=28.0134\times1.6605\times10^{-27}\ \mathrm{kg}=4.652\times10^{-26}\ \mathrm{kg}$) at $T=300\ \mathrm K$ and $p=1\ \mathrm{atm}=101\,325\ \mathrm{Pa}$. If we consider only translational motion, then $n=p/(k_{\mathrm B}T)=2.446\times10^{25}\ \mathrm{m^{-3}}$ and $\lambda_{\mathrm T}=1.904\times10^{-11}\ \mathrm m$, and $\mu$ from Eq. \eqref{eq:52-mu-ideal} is $-0.403\ \mathrm{eV}$ ($-15.6\,k_{\mathrm B}T$) (this value agrees with the one obtained from the Sackur–Tetrode equation in Section 49.6). For this gas: (1) At what pressure, in $\mathrm{atm}$, does $\mu=0$ ($n\lambda_{\mathrm T}^3=1$) hold if the pressure is raised while $T=300\ \mathrm K$ is kept? (2) At what temperature, in $\mathrm K$, does $\mu=0$ hold if the gas is cooled while $p=1\ \mathrm{atm}$ is kept? (3) Find $n\lambda_{\mathrm T}^3$ at the boiling point of nitrogen, $77\ \mathrm K$, and consider why the classical ideal gas formula ceases to be usable.

Solution. First, at $T=300\ \mathrm K$ and $1\ \mathrm{atm}$, $z=n\lambda_{\mathrm T}^3=2.446\times10^{25}\times(1.904\times10^{-11})^3=1.69\times10^{-7}$.

(1) The number density at which $n\lambda_{\mathrm T}^3=1$ is $n=1/\lambda_{\mathrm T}^3=1/(1.904\times10^{-11}\ \mathrm m)^3=1.45\times10^{32}\ \mathrm{m^{-3}}$. The pressure is $p=nk_{\mathrm B}T=1.45\times10^{32}\times4.142\times10^{-21}\ \mathrm{Pa}=6.0\times10^{11}\ \mathrm{Pa}$, that is, about $5.9\times10^{6}\ \mathrm{atm}$ (about $600$ × 10,000 = six million times $1\ \mathrm{atm}$).

(2) At constant $p$, $n=p/(k_{\mathrm B}T)\propto T^{-1}$ and $\lambda_{\mathrm T}^3\propto T^{-3/2}$, so $z=n\lambda_{\mathrm T}^3\propto T^{-5/2}$. Solving $z=1.69\times10^{-7}\,(T/300\ \mathrm K)^{-5/2}=1$ gives $T=300\ \mathrm K\times(1.69\times10^{-7})^{2/5}=0.59\ \mathrm K$.

(3) At $T=77\ \mathrm K$ ($1\ \mathrm{atm}$), $z=1.69\times10^{-7}\times(77/300)^{-5/2}\approx5\times10^{-6}$, which is far smaller than $1$. That is, nitrogen liquefies because of intermolecular forces long before $\mu=0$ is approached, while the quantum-mechanical extent $\lambda_{\mathrm T}\approx3.8\times10^{-11}\ \mathrm m$ is still far shorter than the intermolecular spacing $(k_{\mathrm B}T/p)^{1/3}\approx2.2\times10^{-9}\ \mathrm m$. The reason the classical ideal gas formula cannot be used for nitrogen gas is not quantum effects but intermolecular forces; quantum effects ($n\lambda_{\mathrm T}^3\sim1$) become essential for helium, which is light and has weak intermolecular forces, and for particles of small mass such as electrons (Section 52.7).

Caution. Here we ignored the internal motion of the molecule (rotation and vibration) and the binding energy, and considered the chemical potential of translational motion only. For a diatomic molecule, the translational $Z_1=V/\lambda_{\mathrm T}^3$ is multiplied by the internal partition function $z_{\mathrm{int}}$, so $\mu=k_{\mathrm B}T\ln(n\lambda_{\mathrm T}^3/z_{\mathrm{int}})$. For the rotation of $\mathrm N_2$ at room temperature, $z_{\mathrm{rot}}=T/(2\theta_{\mathrm{rot}})\approx50$ (the rotational temperature is $\theta_{\mathrm{rot}}=2.88\ \mathrm K$, and the factor 2 is the symmetry number of a molecule whose two atoms are identical), so $\mu$ is lowered further by $k_{\mathrm B}T\ln50\approx0.10\ \mathrm{eV}$. Also, the value of $\mu$ itself depends on the choice of the origin of energy (whether or not the binding energy is included).

Example 52.4: Re-deriving the equation of state of an ideal gas from $J=-pV$

Starting from the grand partition function of the ideal gas $\Xi=\exp(zZ_1)$ (Eq. \eqref{eq:52-Xi-idealgas}) and $Z_1=V/\lambda_{\mathrm T}^3$, derive the equation of state $pV=Nk_{\mathrm B}T$ from each of $p=-(\partial J/\partial V)_{T,\mu}$ and $J=-pV$.

Solution. Since $\ln\Xi=zZ_1=e^{\beta\mu}V/\lambda_{\mathrm T}^3$, we have $J=-k_{\mathrm B}T\ln\Xi=-k_{\mathrm B}T\,e^{\beta\mu}V/\lambda_{\mathrm T}^3$. When $T,\mu$ are fixed, both $\lambda_{\mathrm T}$ and $e^{\beta\mu}$ are constants, so $J$ is proportional to $V$. Therefore

On the other hand, the mean number of particles is $N=zZ_1=e^{\beta\mu}V/\lambda_{\mathrm T}^3$, so $e^{\beta\mu}/\lambda_{\mathrm T}^3=N/V$. Substituting this into $p$ gives $p=k_{\mathrm B}T\,N/V$, that is, $pV=Nk_{\mathrm B}T$. As another route, from $J=-pV$ we have $pV=-J=k_{\mathrm B}T\ln\Xi=k_{\mathrm B}T\,zZ_1=k_{\mathrm B}T\,N$, which gives the same equation directly. By either route, the equation of state of high school physics has been reproduced (by a route different from the derivation from the canonical distribution in Chapter 51).

Example 52.5: The particle-number distribution of an ideal gas is a Poisson distribution

For a subsystem of an ideal gas (grand partition function $\Xi=e^{zZ_1}$), find the probability $P(N)$ that the number of particles is exactly $N$, and verify $\langle N\rangle$ and $\langle(\Delta N)^2\rangle$. Also find the relative fluctuation of the number of molecules contained in a $1\ \mu\mathrm m^3$ region of air at $300\ \mathrm K$ and $1\ \mathrm{atm}$.

Solution. Adding up the probabilities of all the states with $N$ particles gives $P(N)=\sum_je^{-\beta(E_{j,N}-\mu N)}/\Xi=z^NZ_N/\Xi$. Using $Z_N=Z_1^N/N!$ and $\Xi=e^{zZ_1}$, we have

This is the Poisson distribution. We have $\sum_NP(N)=e^{-\bar N}\sum_N\bar N^N/N!=e^{-\bar N}e^{\bar N}=1$, so it is normalized. The mean is $\sum_NNP(N)=\bar N$, and the variance is $\langle N^2\rangle-\bar N^2=\bar N$ (using $\langle N(N-1)\rangle=\bar N^2$). This agrees with $\langle(\Delta N)^2\rangle=\langle N\rangle$ found above. A volume of air of $1\ \mu\mathrm m^3=10^{-18}\ \mathrm m^3$ contains, from $n=2.446\times10^{25}\ \mathrm{m^{-3}}$ of Example 52.3, $\bar N=2.4\times10^7$ molecules, and the relative fluctuation is $1/\sqrt{\bar N}=2.0\times10^{-4}$ ($0.02\,\%$).

Example 52.6: Hydrogen across a semipermeable membrane — what is equal when $\mu$ is equal?

A thin film of palladium (Pd) can be used as a semipermeable membrane that passes only hydrogen $\mathrm H_2$ and does not pass nitrogen $\mathrm N_2$. On the left side of the membrane is pure hydrogen at a pressure of $2.0\ \mathrm{atm}$; on the right side is a mixture of hydrogen and nitrogen (total pressure $3.0\ \mathrm{atm}$); the temperature is the same on both sides. After enough time has passed for the flow of hydrogen to stop, what are the partial pressures of hydrogen and nitrogen on the right side? Treat the gases as ideal gases.

Solution. Since only hydrogen can pass through the membrane, the equilibrium condition is that the chemical potential of hydrogen is equal on the two sides, $\mu_{\mathrm H_2}^{\text{left}}=\mu_{\mathrm H_2}^{\text{right}}$ (nitrogen does not pass through the membrane, so there is no need to equalize $\mu$ for nitrogen). The chemical potential of hydrogen in an ideal gas (even in a mixture, each component behaves independently of the other components, because the molecules exert no forces on one another) is determined by the partial pressure $p_{\mathrm H_2}$ of that component: $\mu_{\mathrm H_2}=\mu^\circ(T)+k_{\mathrm B}T\ln(p_{\mathrm H_2}/p^\circ)$. Since $\mu^\circ(T)$ is common to the two sides, equal $\mu$ is equivalent to equal partial pressures of hydrogen. Therefore the partial pressure of hydrogen on the right side is the same as on the left, $2.0\ \mathrm{atm}$, and the partial pressure of nitrogen is $3.0-2.0=1.0\ \mathrm{atm}$. The total pressures, $2.0\ \mathrm{atm}$ on the left and $3.0\ \mathrm{atm}$ on the right, are not equal, but there is no contradiction: this is because nitrogen cannot pass through the membrane and pushes against the wall on the right side. The purification of hydrogen (extracting pure hydrogen from hydrogen containing impurities) is carried out on this principle.

Example 52.7: The number of ways to place two particles in three single-particle states

In how many ways can two particles be placed in three single-particle states $a,b,c$? Count the ways for each of the following cases, and write out the sets of occupation numbers $(n_a,n_b,n_c)$: (1) distinguishable particles, (2) indistinguishable bosons, (3) indistinguishable fermions.

Solution. (1) Particle 1 and particle 2 can each independently go into any of the three states, so there are $3\times3=9$ ways. (2) Since the particles are indistinguishable, exchanging the two particles gives the same state. Counting by the sets of occupation numbers $(n_a,n_b,n_c)$, these are sets of nonnegative integers satisfying $n_a+n_b+n_c=2$, so there are $6$ ways: $(2,0,0),(0,2,0),(0,0,2),(1,1,0),(1,0,1),(0,1,1)$ ($3$ ways in which $2$ particles are in the same state and $3$ ways in which they are in different states). (3) Fermions cannot have $2$ particles in the same state, so removing from (2) the three ways with $n_i=2$ leaves $(1,1,0),(1,0,1),(0,1,1)$, that is, $3$ ways.

In summary, the numbers of ways are in the ratio $9:6:3$. The number of sets in which both particles are in the same state ($n_i=2$) is $3$ for distinguishable particles and also $3$ for bosons, but it is $0$ for fermions. For distinguishable particles there are $6$ ways in which they go into different states (two ways each, by exchanging particles 1 and 2), but for indistinguishable particles these merge into $3$ ways.

Example 52.8: Eight particles in a box: the energy price of the Pauli exclusion principle

Consider a particle of mass $m$ confined in a cubic box of side $L$. The energy of a single-particle state is determined by three positive integers $(n_x,n_y,n_z)$: $\varepsilon=\varepsilon_0(n_x^2+n_y^2+n_z^2)$, with $\varepsilon_0=\dfrac{\pi^2\hbar^2}{2mL^2}$ (this is the one-dimensional box result $E_n=n^2\varepsilon_0$ of University Mathematics, Section 45.3, summed over the three directions). Find the lowest energy (the energy at absolute zero), in units of $\varepsilon_0$, of (1) $8$ fermions of spin $\tfrac12$ (each $(n_x,n_y,n_z)$ has two spin states) and (2) $8$ bosons of spin $0$. (3) Obtain the numerical values for electrons in a box with $L=1\ \mathrm{nm}$.

Solution. First we list the single-particle states in order of increasing energy. In order of increasing $n_x^2+n_y^2+n_z^2$: $(1,1,1)$ gives $3$; $(2,1,1)$ and its permutations (three arrangements in all) give $6$; $(2,2,1)$ and its permutations (three arrangements in all) give $9$; $(3,1,1)$ and its permutations (three arrangements in all) give $11$; $(2,2,2)$ (one arrangement) gives $12$.

(1) Each $(n_x,n_y,n_z)$ can hold up to $2$ fermions, one with spin up and one with spin down. The orbital of energy $3\varepsilon_0$ (one orbital) holds $2$ particles and the orbitals of energy $6\varepsilon_0$ (three orbitals) hold $6$ particles, so a total of $2+6=8$ particles are packed in. The total energy is $2\times3\varepsilon_0+6\times6\varepsilon_0=6\varepsilon_0+36\varepsilon_0=42\,\varepsilon_0$.

(2) All eight bosons can go into the lowest state $(1,1,1)$, so the total energy is $8\times3\varepsilon_0=24\,\varepsilon_0$.

(3) For an electron with $L=1\ \mathrm{nm}$, $\varepsilon_0=\pi^2(1.0546\times10^{-34})^2/(2\times9.109\times10^{-31}\times10^{-18})\ \mathrm J=6.02\times10^{-20}\ \mathrm J=0.376\ \mathrm{eV}$, so (1) gives $42\times0.376=15.8\ \mathrm{eV}$ and (2) gives $24\times0.376=9.02\ \mathrm{eV}$. Because of the exclusion principle, fermions are forced to have a higher energy than the same number of bosons, by $18\,\varepsilon_0$ (in this case $6.8\ \mathrm{eV}$). This difference is the origin of the enormous energy and pressure of the degenerate electron gas that we will see in Section 52.8.

Example 52.9: How “blurred” is the Fermi distribution at room temperature?

At $T=300\ \mathrm K$ ($k_{\mathrm B}T=0.02585\ \mathrm{eV}$), find the value of the Fermi distribution $f(\varepsilon)$ for each of $\varepsilon-\mu=-0.10,\,-0.05,\,+0.05,\,+0.10,\,+0.20\ \mathrm{eV}$. Also find the width in $\varepsilon$ over which the distribution changes from $f=0.9$ to $f=0.1$. How does it compare with the $E_{\mathrm F}\approx5\ \mathrm{eV}$ of a metal?

Solution. We compute $x=(\varepsilon-\mu)/k_{\mathrm B}T$ and substitute it into $f=1/(e^x+1)$. For $\varepsilon-\mu=-0.10\ \mathrm{eV}$, $x=-3.868$ and $e^{x}=0.02087$, so $f=1/(1.02087)=0.9795$. For $-0.05\ \mathrm{eV}$, $x=-1.934$ and $e^{x}=0.1445$, so $f=0.8737$. For $+0.05\ \mathrm{eV}$, $x=+1.934$ and $e^x=6.919$, so $f=0.1263$ ($f(-0.05)+f(+0.05)=1$ confirms the symmetry). For $+0.10\ \mathrm{eV}$, $x=3.868$ and $e^x=47.9$, so $f=0.0205$. For $+0.20\ \mathrm{eV}$, $x=7.736$ and $e^x=2290$, so $f=4.4\times10^{-4}$.

The $x$ for which $f=0.9$ is, from $e^x=1/9$, $x=-\ln9=-2.197$, and for $f=0.1$ it is $x=+2.197$, so the width is $4.394\,k_{\mathrm B}T=4.394\times0.02585\ \mathrm{eV}=0.114\ \mathrm{eV}$. Relative to $E_{\mathrm F}\approx5\ \mathrm{eV}$ this is only $0.114/5\approx2\,\%$, so the Fermi distribution at room temperature is merely the corner of the $T=0$ step function slightly rounded.

Example 52.10: The values of the three distributions at the same $x$, and the constraint on $\mu$ for the Bose distribution

(1) For $x=(\varepsilon-\mu)/k_{\mathrm B}T=0.5$, find the mean occupation numbers of the Fermi, Boltzmann, and Bose distributions, together with the respective fluctuations $\langle(\Delta n_\varepsilon)^2\rangle$. (2) Find the value of $x$ at which $\langle n_\varepsilon\rangle=1$ in the Bose distribution. Is there an $x$ at which $\langle n_\varepsilon\rangle=1$ in the Fermi distribution? (3) Why can we not set $\mu=+0.01\,k_{\mathrm B}T$ in a system of bosons whose lowest energy is $\varepsilon_0=0$?

Solution. (1) $e^{0.5}=1.6487$. Fermi: $\langle n_\varepsilon\rangle=1/(1.6487+1)=1/2.6487=0.3775$, and the fluctuation is $0.3775\times(1-0.3775)=0.2350$. Boltzmann: $\langle n_\varepsilon\rangle=e^{-0.5}=0.6065$, and the fluctuation is $0.6065$. Bose: $\langle n_\varepsilon\rangle=1/(1.6487-1)=1/0.6487=1.5415$, and the fluctuation is $1.5415\times(1+1.5415)=3.918$. Even at the same $x$, the values increase in the order Fermi $\lt$ Boltzmann $\lt$ Bose, and the differences in the fluctuations are larger still.

(2) Bose: from $1/(e^x-1)=1$ we get $e^x=2$, so $x=\ln2=0.693$. Fermi: setting $1/(e^x+1)=1$ gives $e^x=0$, which does not occur for any finite $x$ ($\langle n_\varepsilon\rangle\to1$ only as $x\to-\infty$; by the exclusion principle it never exceeds $1$).

(3) For the lowest state ($\varepsilon_0=0$), substituting $\mu=0.01\,k_{\mathrm B}T$ into $\langle n_0\rangle=1/(e^{(0-\mu)/k_{\mathrm B}T}-1)$ gives $1/(e^{-0.01}-1)=1/(-0.00995)\approx-100.5$, which is negative, so the number of particles would be negative. We must have $\mu\lt\varepsilon_0=0$ (Eq. \eqref{eq:52-mu-bound}).

Example 52.11: Occupation numbers of light: visible light is “particle-like,” radio waves are “wave-like”

Photons obey the Bose distribution with $\mu=0$, $\langle n_\varepsilon\rangle=1/(e^{h\nu/k_{\mathrm B}T}-1)$. (1) Find the mean number of photons in the mode of wavelength $500\ \mathrm{nm}$ in the thermal radiation from the surface of the Sun ($T=5800\ \mathrm K$). (2) Find the mean number of photons in the mode of frequency $10\ \mathrm{GHz}$ (microwave) in a cavity at room temperature, $T=300\ \mathrm K$.

Solution. (1) $\nu=c/\lambda=2.998\times10^8/(5.00\times10^{-7})=5.996\times10^{14}\ \mathrm{Hz}$, and $h\nu=6.626\times10^{-34}\times5.996\times10^{14}=3.973\times10^{-19}\ \mathrm J=2.480\ \mathrm{eV}$. Then $x=h\nu/k_{\mathrm B}T=3.973\times10^{-19}/(1.381\times10^{-23}\times5800)=4.961$, so $\langle n_\varepsilon\rangle=1/(e^{4.961}-1)=1/141.9=7.05\times10^{-3}$. (2) $h\nu=6.626\times10^{-34}\times10^{10}=6.626\times10^{-24}\ \mathrm J=4.14\times10^{-5}\ \mathrm{eV}$ and $x=6.626\times10^{-24}/(1.381\times10^{-23}\times300)=1.600\times10^{-3}$, so by $e^x-1\approx x$ we get $\langle n_\varepsilon\rangle\approx1/x=625$ (equal to $k_{\mathrm B}T/h\nu$).

For visible light, the mean is only $0.007$ photons per mode, so the modes are almost empty and the photons can be counted one by one as “particles.” For microwaves, a single mode contains several hundred photons, and the radiation can be treated as a continuous “wave” made up of a great many photons, using the theory of classical electromagnetic waves (the Rayleigh–Jeans law, $\langle n_\varepsilon\rangle\,h\nu\approx k_{\mathrm B}T$). The answer to “Is light a particle or a wave?” is decided by whether the occupation number is sufficiently smaller or sufficiently larger than $1$.

Example 52.12: Helium-4 gas and electrons in a metal — which is classical?

(1) For helium-4 gas at $T=300\ \mathrm K$ and $p=1\ \mathrm{atm}$ (mass $m=4.0026\times1.6605\times10^{-27}\ \mathrm{kg}=6.646\times10^{-27}\ \mathrm{kg}$, $g_s=1$), find $n\lambda_{\mathrm T}^3$ and the relative deviation of the pressure due to quantum statistics. (2) For the conduction electrons of copper (density $8.96\ \mathrm{g/cm^3}$, molar mass $63.55\ \mathrm{g/mol}$; one electron per atom, $g_s=2$), find $n\lambda_{\mathrm T}^3/g_s$ at $T=300\ \mathrm K$, and estimate the temperature $T_{\mathrm d}$ required for the electrons to behave classically (the temperature at which $n\lambda_{\mathrm T}^3/g_s=1$).

Solution. (1) $n=p/k_{\mathrm B}T=2.446\times10^{25}\ \mathrm{m^{-3}}$. $\lambda_{\mathrm T}=h/\sqrt{2\pi mk_{\mathrm B}T}=6.626\times10^{-34}/\sqrt{2\pi\times6.646\times10^{-27}\times4.142\times10^{-21}}=5.04\times10^{-11}\ \mathrm m$. $n\lambda_{\mathrm T}^3=2.446\times10^{25}\times(5.04\times10^{-11})^3=3.13\times10^{-6}$. The deviation is $n\lambda_{\mathrm T}^3/2^{5/2}=3.13\times10^{-6}/5.657=5.5\times10^{-7}$. Since helium-4 is a boson, the pressure is lower than the classical value by $5.5\times10^{-7}$ ($0.00006\,\%$), but this is far smaller than the effect of the intermolecular forces of a real gas and is entirely negligible.

(2) The number density of copper atoms is $\dfrac{8.96\times10^{3}\ \mathrm{kg/m^3}}{63.55\times10^{-3}\ \mathrm{kg/mol}}\times6.022\times10^{23}\ \mathrm{mol^{-1}}=8.49\times10^{28}\ \mathrm{m^{-3}}$ (the number density of electrons is the same). From the electron mass $m_{\mathrm e}=9.109\times10^{-31}\ \mathrm{kg}$ we get $\lambda_{\mathrm T}=6.626\times10^{-34}/\sqrt{2\pi\times9.109\times10^{-31}\times4.142\times10^{-21}}=4.30\times10^{-9}\ \mathrm m=4.3\ \mathrm{nm}$. $n\lambda_{\mathrm T}^3=8.49\times10^{28}\times(4.30\times10^{-9})^3=6.8\times10^{3}$, and dividing by $g_s=2$ gives $n\lambda_{\mathrm T}^3/g_s=3.4\times10^3\gg1$. The thermal de Broglie wavelength of the electron, $4.3\ \mathrm{nm}$, is far longer than the distance between electrons, $n^{-1/3}=0.23\ \mathrm{nm}$, so the waves overlap completely (the gas is strongly degenerate). The temperature $T_{\mathrm d}$ is obtained by solving $n\lambda_{\mathrm T}^3/g_s=\dfrac{n}{g_s}\left(\dfrac{h^2}{2\pi mk_{\mathrm B}T}\right)^{3/2}=1$ for $T$, which gives

(Here $h^2/(2\pi m_{\mathrm e}k_{\mathrm B})=5.56\times10^{-15}\ \mathrm{m^2\,K}$. Also $(4.25\times10^{28})^{1/3}=3.49\times10^{9}\ \mathrm{m^{-1}}$, its square is $(4.25\times10^{28})^{2/3}=1.22\times10^{19}\ \mathrm{m^{-2}}$, and $5.56\times10^{-15}\times1.22\times10^{19}=6.8\times10^{4}\ \mathrm K$.) For the electrons in copper to behave like a classical gas requires about $7\times10^4\ \mathrm K$ (more than 10 times the temperature of the Sun's surface), and at room temperature they are completely quantum (Fermi degenerate). Compared with the molecules of air (Example 52.3, $n\lambda_{\mathrm T}^3=1.7\times10^{-7}$), the two differ by as much as a factor of $10^{10}$. This is why, in Section 52.8, the electrons in a metal must be treated as a Fermi gas rather than as a classical ideal gas.

Example 52.13: Density of states and level spacing of the electrons in a 1 cm cube of copper

For copper of volume $V=1\ \mathrm{cm}^3$, find the density of states $g(E_{\mathrm F})$ (in units of $\mathrm{eV^{-1}}$) at the Fermi energy $E_{\mathrm F}=7.04\ \mathrm{eV}$, and the level spacing $1/g(E_{\mathrm F})$ there, from the electron number density $n=8.49\times10^{28}\ \mathrm{m^{-3}}$ (the expression $g(\varepsilon)=\tfrac32N\varepsilon^{1/2}/E_{\mathrm F}^{3/2}$ for $g$ in terms of $N,E_{\mathrm F}$ will be derived in Section 52.8.2).

Solution. $N=nV=8.49\times10^{28}\times10^{-6}=8.49\times10^{22}$. $g(E_{\mathrm F})=\tfrac32N/E_{\mathrm F}=1.5\times8.49\times10^{22}/7.04\ \mathrm{eV}=1.81\times10^{22}\ \mathrm{eV^{-1}}$. Therefore the level spacing is $1/g=5.5\times10^{-23}\ \mathrm{eV}$, and as many as $5\times10^{20}$ levels fit within $k_{\mathrm B}T=0.026\ \mathrm{eV}$. The measure of the spacing of the lowest levels, $\varepsilon_0=\pi^2\hbar^2/(2mL^2)=6.0\times10^{-34}\ \mathrm J=3.8\times10^{-15}\ \mathrm{eV}$ ($L=1\ \mathrm{cm}$, electron mass), is also far smaller than $k_{\mathrm B}T$. In fact, if we count the states of the cubical box one by one in the range $\varepsilon/\varepsilon_0\le10^5$, the count reaches $99.3\,\%$ of the value given by the formula $N(\varepsilon)$, and it approaches $100\,\%$ as the range is widened.

Example 52.14: Fermi energy of copper (about $7\ \mathrm{eV}$)

Assuming that copper (density $8.96\ \mathrm{g/cm^3}$, molar mass $63.55\ \mathrm{g/mol}$) contributes one conduction electron per atom, find (1) the number density $n$ of electrons, (2) $k_{\mathrm F}$, (3) $E_{\mathrm F}$ (in $\mathrm{eV}$), (4) $T_{\mathrm F}$, and (5) $v_{\mathrm F}$. (6) Find the ratio of the Fermi energy to $k_{\mathrm B}T$ at $T=300\ \mathrm K$.

Solution. (1) The number of atoms per unit volume is $n=\dfrac{8.96\times10^{3}\ \mathrm{kg/m^3}}{63.55\times10^{-3}\ \mathrm{kg/mol}}\times6.022\times10^{23}\ \mathrm{mol^{-1}}=8.49\times10^{28}\ \mathrm{m^{-3}}$.

(2) $k_{\mathrm F}=(3\pi^2n)^{1/3}=(3\times9.870\times8.49\times10^{28})^{1/3}\ \mathrm{m^{-1}}=(2.514\times10^{30})^{1/3}=1.36\times10^{10}\ \mathrm{m^{-1}}$.

(3) $E_{\mathrm F}=\dfrac{\hbar^2k_{\mathrm F}^2}{2m}=\dfrac{(1.0546\times10^{-34})^2\times(1.36\times10^{10})^2}{2\times9.109\times10^{-31}}\ \mathrm J=1.13\times10^{-18}\ \mathrm J=7.04\ \mathrm{eV}$ ($1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm J$). (For copper, the value is usually quoted as $7.0\ \mathrm{eV}$.)

(4) $T_{\mathrm F}=E_{\mathrm F}/k_{\mathrm B}=1.13\times10^{-18}/1.381\times10^{-23}=8.2\times10^4\ \mathrm K$.

(5) $v_{\mathrm F}=\hbar k_{\mathrm F}/m=1.0546\times10^{-34}\times1.36\times10^{10}/9.109\times10^{-31}=1.57\times10^6\ \mathrm{m/s}$ (about $0.5\,\%$ of the speed of light).

(6) Since $k_{\mathrm B}T=0.02585\ \mathrm{eV}$, we have $E_{\mathrm F}/k_{\mathrm B}T=7.04/0.02585=272$ and $T/T_{\mathrm F}=300/8.2\times10^4=0.0037$. Room temperature is only $0.4\,\%$ of the Fermi temperature. Even at $T=0$ the electrons move with a mean kinetic energy as large as $\tfrac35E_{\mathrm F}=4.2\ \mathrm{eV}$ (classically equivalent to $T\approx3\times10^4\ \mathrm K$), and the thermal energy at room temperature changes only a small part of it.

Example 52.15: Degeneracy pressure of the electron gas, and the bulk modulus of copper

Find the pressure $p=\tfrac25nE_{\mathrm F}$ of the electron gas of copper at $T=0$. Compare it with the pressure $nk_{\mathrm B}T$ that a classical ideal gas of the same density would exert at $300\ \mathrm K$. Further, find the bulk modulus $K=-V\,\partial p/\partial V$ and compare it with the measured value (about $140\ \mathrm{GPa}$).

Solution. $p=\tfrac25\times8.49\times10^{28}\times7.04\times1.602\times10^{-19}\ \mathrm{Pa}=3.83\times10^{10}\ \mathrm{Pa}=38\ \mathrm{GPa}$ (about $3.8\times10^5$ atmospheres). For a classical ideal gas, $nk_{\mathrm B}T=8.49\times10^{28}\times4.14\times10^{-21}=3.5\times10^8\ \mathrm{Pa}=0.35\ \mathrm{GPa}$, so the pressure of the electron gas is about $110$ times larger (the ratio is $\tfrac25E_{\mathrm F}/k_{\mathrm B}T=0.4\times272=109$). Since $p\propto n^{5/3}\propto V^{-5/3}$, the bulk modulus is $-V\,\partial p/\partial V=\tfrac53p=\tfrac23nE_{\mathrm F}=64\ \mathrm{GPa}$. This is a little under half of the measured value of about $140\ \mathrm{GPa}$, so the order of magnitude is right for a free-electron approximation. It shows that the degeneracy pressure of the electrons accounts for an important part of the reason why metals are hard to compress.

Example 52.16: How large is the electronic specific heat of copper at room temperature compared with the classical value?

Using Theorem 52.5, find the electronic specific heat of $1\ \mathrm{mol}$ of copper ($N=N_{\mathrm A}$ electrons, $T_{\mathrm F}=8.2\times10^4\ \mathrm K$) at $T=300\ \mathrm K$, and compare it with the classical value $\tfrac32R$ and with the Dulong–Petit lattice specific heat $3R$. Also estimate the fraction of electrons that are excited by heat.

Solution. $T/T_{\mathrm F}=300/(8.2\times10^4)=3.66\times10^{-3}$. With $R=8.314\ \mathrm{J/(mol\,K)}$,

This is $0.150/12.47=1.2\,\%$ of the classical value $\tfrac32R=12.47\ \mathrm{J/(mol\,K)}$ and $0.6\,\%$ of the lattice specific heat $3R=24.9\ \mathrm{J/(mol\,K)}$. The fraction of electrons excited by heat is only $\sim\tfrac32T/T_{\mathrm F}=0.55\,\%$. This is why the specific heat of a metal at room temperature is nearly equal to $3R$ in spite of the presence of the electrons.

Example 52.17: The temperature below which the electronic specific heat exceeds the lattice specific heat

The specific heat of a metal at low temperatures is expressed as the sum $C=\gamma T+\beta T^3$ of the electronic contribution $\gamma T$ and the contribution $\beta T^3$ of lattice vibrations (phonons; Debye's $T^3$ law, derived in Chapter 53). For copper, compute $\gamma$ (per $1\ \mathrm{mol}$) in the free-electron approximation, and compute $\beta=\dfrac{12\pi^4}5\dfrac R{\theta_{\mathrm D}^3}$ from the Debye temperature $\theta_{\mathrm D}=343\ \mathrm K$, and find the temperature $T^*$ at which $\gamma T=\beta T^3$. What happens if we use the measured $\gamma$ (about $0.69\ \mathrm{mJ/(mol\,K^2)}$)?

Solution. In the free-electron approximation, $\gamma=\dfrac{\pi^2}2\dfrac R{T_{\mathrm F}}=4.935\times8.314/(8.2\times10^4)\ \mathrm{J/(mol\,K^2)}=5.0\times10^{-4}\ \mathrm{J/(mol\,K^2)}=0.50\ \mathrm{mJ/(mol\,K^2)}$. Also, $\beta=\dfrac{12\pi^4}5\times\dfrac{8.314}{343^3}=233.8\times\dfrac{8.314}{4.036\times10^7}=4.82\times10^{-5}\ \mathrm{J/(mol\,K^4)}$. From $\gamma T=\beta T^3$ we get $T^*=\sqrt{\gamma/\beta}=\sqrt{5.0\times10^{-4}/4.82\times10^{-5}}\ \mathrm K=3.2\ \mathrm K$. Using the measured $\gamma\approx0.69\times10^{-3}$ gives $T^*=3.8\ \mathrm K$, so in both cases the temperature is a few $\mathrm K$. Below about $4\ \mathrm K$ the electronic contribution is the larger, and if $C/T=\gamma+\beta T^2$ is plotted against $T^2$, the intercept with the vertical axis is $\gamma$ (the standard way to determine $\gamma$ experimentally). The fact that the measured $\gamma$ is about $1.4$ times larger than that of the free-electron approximation is interpreted as follows: because of electron–electron interactions and interactions with the lattice, the electrons in a crystal behave as if they had an effective mass heavier than that of free electrons ($\gamma\propto g(E_{\mathrm F})\propto m$).

Example 52.18: Estimating the condensation temperature of liquid helium-4

The density of liquid helium-4 near the $\lambda$ point is about $0.145\ \mathrm{g/cm^3}$. Regarding the helium-4 atoms as an ideal Bose gas of particles of mass $m=6.646\times10^{-27}\ \mathrm{kg}$, estimate the condensation temperature $T_{\mathrm c}$. Compare it with the measured $\lambda$ point, $2.17\ \mathrm K$.

Solution. The number density is $n=\dfrac{0.145\times10^{3}\ \mathrm{kg/m^3}}{6.646\times10^{-27}\ \mathrm{kg}}=2.18\times10^{28}\ \mathrm{m^{-3}}$. Substitute into Eq. \eqref{eq:52-Tc}. We have $n/\zeta(\tfrac32)=2.18\times10^{28}/2.612=8.35\times10^{27}\ \mathrm{m^{-3}}$, and its $2/3$ power is $(8.35\times10^{27})^{2/3}=4.12\times10^{18}\ \mathrm{m^{-2}}$. Also, $\dfrac{h^2}{2\pi mk_{\mathrm B}}=\dfrac{(6.626\times10^{-34})^2}{2\pi\times6.646\times10^{-27}\times1.381\times10^{-23}}=7.61\times10^{-19}\ \mathrm{m^2\,K}$. Therefore

This is somewhat higher than the measured $\lambda$ point, $2.17\ \mathrm K$, but both the order of magnitude and the value are close. The discrepancy arises because in liquid helium the interactions between atoms cannot be neglected, so the ideal-gas assumption does not hold. What matters is that “the statistics of bosons alone can predict almost the right temperature”; indeed, the isotope helium-3 (a fermion) has no such $\lambda$ transition.

Example 52.19: Condensation temperature and condensate fraction of a cold atomic gas

Find $T_{\mathrm c}$ for a uniform ideal Bose gas of $^{87}$Rb atoms ($m=86.909\times1.6605\times10^{-27}=1.4432\times10^{-25}\ \mathrm{kg}$) with number density $n=1.0\times10^{20}\ \mathrm{m^{-3}}$ ($10^{14}\ \mathrm{cm^{-3}}$). Also find the fraction condensed at $T=0.5\,T_{\mathrm c}$ and the temperature at which half of the atoms are condensed.

Solution. $\dfrac{h^2}{2\pi mk_{\mathrm B}}=\dfrac{(6.626\times10^{-34})^2}{2\pi\times1.4432\times10^{-25}\times1.381\times10^{-23}}=3.505\times10^{-20}\ \mathrm{m^2\,K}$ and $\left(\dfrac n{2.612}\right)^{2/3}=(3.828\times10^{19})^{2/3}=1.136\times10^{13}\ \mathrm{m^{-2}}$. Therefore $T_{\mathrm c}=3.505\times10^{-20}\times1.136\times10^{13}=3.98\times10^{-7}\ \mathrm K=0.40\ \mu\mathrm K$ (microkelvin). At $T=0.5\,T_{\mathrm c}$, Eq. \eqref{eq:52-N0} gives $N_0/N=1-0.5^{3/2}=1-0.354=0.646$. The temperature at which half of the atoms are condensed follows from $1-(T/T_{\mathrm c})^{3/2}=\tfrac12$: $T/T_{\mathrm c}=2^{-2/3}=0.630$, that is, $T=0.25\ \mu\mathrm K$. This is an extremely low temperature, $10^{-9}$ times room temperature ($300\ \mathrm K$), and it has to be realized artificially. Also, the thermal de Broglie wavelength at the transition point is $\lambda_{\mathrm T}=1.377\,n^{-1/3}=1.377\times2.154\times10^{-7}\ \mathrm m=0.30\ \mu\mathrm m$, which is nearly equal to the interparticle distance of $0.22\ \mu\mathrm m$.

Exercise 52.1: The Langmuir adsorption isotherm for an adsorption site

The adsorption site of Example 52.1 (empty: $E=0,N=0$; adsorbed: $E=-\varepsilon_0,N=1$) is in contact with a monatomic ideal gas of molecules of mass $m$ (pressure $p$, temperature $T$). (1) Using the chemical potential of the gas, $\mu=k_{\mathrm B}T\ln(p\lambda_{\mathrm T}^3/k_{\mathrm B}T)$, show that the probability that the site is occupied (the coverage) $\theta$ can be written as $\theta=p/(p+p_0)$ with $p_0=\dfrac{k_{\mathrm B}T}{\lambda_{\mathrm T}^3}e^{-\varepsilon_0/k_{\mathrm B}T}$. (2) For argon ($m=39.948\ \mathrm u$) gas with $T=300\ \mathrm K$ and $\varepsilon_0=0.40\ \mathrm{eV}$, find $p_0$, find $\theta$ at $1\ \mathrm{atm}$, and find the pressure at which $\theta=0.9$.

Hint: Substitute $e^{\beta\mu}=p\lambda_{\mathrm T}^3/k_{\mathrm B}T$ into the result of Example 52.1, $\theta=1/(1+e^{-\beta(\varepsilon_0+\mu)})$.

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Exercise 52.2: Distribution of the number of particles in an ideal gas

Consider a subsystem of an ideal gas with grand partition function $\Xi=e^{zZ_1}$, and let the mean number of particles be $\bar N=4$. (1) Find the probability $P(N)$ that the number of particles is exactly $N$. (2) Find $P(0)$, $P(4)$, and $P(N\ge8)$. (3) Show that the variance equals the mean by computing $\langle N(N-1)\rangle$. (4) In air at room temperature and $1\ \mathrm{atm}$, the volume for which $\bar N=4$ corresponds to a cube of what side length, in $\mathrm{nm}$?

Hint: $P(N)=z^NZ_N/\Xi$ and $Z_N=Z_1^N/N!$. For (3), rewrite $\sum_NN(N-1)\bar N^N/N!$ as a sum over $N\ge2$. For (4), use $n=p/k_{\mathrm B}T$.

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Exercise 52.3: Putting two particles into three states with energies $0,\varepsilon,2\varepsilon$

Two particles are placed in three single-particle states with energies $0,\varepsilon,2\varepsilon$. In the canonical distribution at temperature $T$, write the partition function $Z$ in terms of $x=e^{-\varepsilon/k_{\mathrm B}T}$ for (1) distinguishable particles, (2) bosons, and (3) fermions. (4) Find the mean energy at $k_{\mathrm B}T=\varepsilon$, and also state its limit as $T\to0$. (5) At $k_{\mathrm B}T=\varepsilon$, compare the probability that the two particles are in the same state for the cases (1) and (2).

Hint: The microstates are those of Example 52.7. The energy is the sum of the energies of the two particles, and the mean energy is $-\partial\ln Z/\partial\beta$.

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Exercise 52.4: Occupation probability of electrons in the conduction band of a semiconductor

In an intrinsic semiconductor, the chemical potential $\mu$ lies almost exactly at the middle of the band gap (an approximation that assumes the densities of states of electrons and holes are nearly equal). At $T=300\ \mathrm K$, for (1) silicon (band gap $E_{\mathrm g}=1.12\ \mathrm{eV}$) and (2) germanium ($E_{\mathrm g}=0.66\ \mathrm{eV}$), find from the Fermi distribution the probability that an electron occupies the state at the bottom of the conduction band, and verify that it can be approximated by the Boltzmann factor $e^{-E_{\mathrm g}/2k_{\mathrm B}T}$. Also find the probability that the state at the top of the valence band is empty (that is, that there is a hole).

Hint: $\varepsilon-\mu=E_{\mathrm g}/2$. The probability of a hole is $1-f(\mu-E_{\mathrm g}/2)$; use the symmetry $f(\mu+\delta)+f(\mu-\delta)=1$.

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Exercise 52.5: Occupation probability of a donor level — counting the spin degeneracy

The level $E_{\mathrm d}$ of a donor (an impurity that supplies one electron) in an n-type semiconductor can hold at most one electron (there are two possibilities, spin up or spin down; a second electron cannot enter because of the repulsion between electrons). Hence this level has three states: “empty,” “one spin-up electron,” and “one spin-down electron.” (1) Write the grand partition function $\Xi_{\mathrm d}$ of this level and find the mean occupation number $\langle n_{\mathrm d}\rangle$. (2) For $T=300\ \mathrm K$ and $E_{\mathrm d}-\mu=0.030\ \mathrm{eV}$, find $\langle n_{\mathrm d}\rangle$ and compare it with the value of the ordinary Fermi distribution $1/(e^x+1)$ ($x=(E_{\mathrm d}-\mu)/k_{\mathrm B}T$), which does not take the spin degeneracy into account. (3) State the reason for the difference.

Hint: Add up $\Xi_{\mathrm d}=\sum e^{-\beta(E-\mu N)}$ over the three states. $k_{\mathrm B}T=0.02585\ \mathrm{eV}$ (at $300\ \mathrm K$).

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Exercise 52.6: Two-dimensional electron gas

Consider electrons (mass $m$, $2$ spin states) confined to a square plane of side $L$ (area $L^2$). (1) Following the counting of states in wavenumber space (Section 52.8), show that the density of states $g(\varepsilon)$ is a constant, $mL^2/(\pi\hbar^2)$, independent of $\varepsilon$. (2) Find the Fermi energy $E_{\mathrm F}$ for an areal density $n_2=N/L^2$. (3) At finite temperature, compute $N=\int_0^\infty g f\dd\varepsilon$ exactly and show that $\mu=k_{\mathrm B}T\ln\bigl(e^{E_{\mathrm F}/k_{\mathrm B}T}-1\bigr)$. (4) For the two-dimensional electron gas formed at a semiconductor interface ($n_2=3\times10^{11}\ \mathrm{cm^{-2}}$, effective mass $m^*=0.067\,m_{\mathrm e}$), find $E_{\mathrm F}$ and $T_{\mathrm F}$.

Hint: (1) Divide the area of the quarter disk ($1/4$ of a circle) in the first quadrant by the area $(\pi/L)^2$ per lattice point. (3) $\int_0^\infty\dfrac{\dd\varepsilon}{e^{(\varepsilon-\mu)/k_{\mathrm B}T}+1}=k_{\mathrm B}T\ln(1+e^{\mu/k_{\mathrm B}T})$ (substitute $u=e^{-\varepsilon/k_{\mathrm B}T}$).

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Exercise 52.7: Electronic specific heat of aluminum

Regarding aluminum (density $2.70\ \mathrm{g/cm^3}$, molar mass $26.98\ \mathrm{g/mol}$, $3$ valence electrons) as a free electron gas, find (1) the electron number density $n$, (2) $E_{\mathrm F}$ and $T_{\mathrm F}$, and (3) the electronic specific heat coefficient $\gamma$ per $1\ \mathrm{mol}$ of atoms, together with the ratio of the electronic specific heat at $300\ \mathrm K$ to the classical value $\tfrac32\times3R$.

Hint: $1\ \mathrm{mol}$ of atoms supplies $N=3N_{\mathrm A}$ electrons. $\gamma=\tfrac{\pi^2}2Nk_{\mathrm B}/T_{\mathrm F}$.

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Exercise 52.8: Comparing $T_{\mathrm F}$ of a Fermi gas with $T_{\mathrm c}$ of a Bose gas

For the same mass $m$ and the same number density $n$, find numerically the ratio $T_{\mathrm c}/T_{\mathrm F}$, where $T_{\mathrm F}$ is the Fermi temperature of (a) an ideal Fermi gas of spin $\tfrac12$ ($g_s=2$) and $T_{\mathrm c}$ is the condensation temperature of (b) an ideal Bose gas of spin $0$. Also, for mass $m=3.016\ \mathrm u$ (helium-3) and $n=1.6\times10^{28}\ \mathrm{m^{-3}}$, what is $T_{\mathrm F}$ in $\mathrm K$? What can we conclude from this comparison?

Hint: $T_{\mathrm F}=\dfrac{\hbar^2}{2mk_{\mathrm B}}(3\pi^2n)^{2/3}$ and $T_{\mathrm c}=\dfrac{2\pi\hbar^2}{mk_{\mathrm B}}(n/\zeta(\tfrac32))^{2/3}$. Taking the ratio, $m,n$ cancel.

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Exercise 52.9: When the reservoir is an ideal gas, check $\beta$ and $\mu$

Suppose that the reservoir $R$ is an ideal gas consisting of $N_R$ monatomic molecules (volume $V_R$, energy $E_R$). Its entropy is given by the Sackur–Tetrode equation (Section 49.6), $S_R=N_Rk_{\mathrm B}\Bigl[\ln\Bigl\{\dfrac{V_R}{N_R}\Bigl(\dfrac{4\pi mE_R}{3h^2N_R}\Bigr)^{3/2}\Bigr\}+\dfrac52\Bigr]$. (1) Compute the partial derivatives $\bigl(\partial\ln\Omega_R/\partial E_R\bigr)_{N_R}$ and $\bigl(\partial\ln\Omega_R/\partial N_R\bigr)_{E_R}$ of $\ln\Omega_R=S_R/k_{\mathrm B}$. (2) Using $E_R=\tfrac32N_Rk_{\mathrm B}T$, show that the former equals $\beta=1/k_{\mathrm B}T$ and the latter equals $-\beta\mu$ ($\mu=k_{\mathrm B}T\ln(n\lambda_{\mathrm T}^3)$), and verify that $P_{j,N}\propto\exp\bigl[-\beta(E_{j,N}-\mu N)\bigr]$ follows from the Taylor expansion of Section 52.1. (3) In a state with $E_{j,N}=0$, by what factor does the number of states of the reservoir change when exactly one particle is handed over to the system $s$? Find it for helium-4 gas at $T=300\ \mathrm K$ and $p=1\ \mathrm{atm}$.

Hint: Expand $S_R/k_{\mathrm B}=N_R\bigl[\ln V_R+\tfrac32\ln E_R-\tfrac52\ln N_R+\tfrac32\ln\tfrac{4\pi m}{3h^2}+\tfrac52\bigr]$ and then differentiate (Section 49.6). For (3), use the first-order term in $N$, $-N(\partial\ln\Omega_R/\partial N)$.

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Exercise 52.10: Grand partition function of a system of fermions with two orbitals

Fermions are placed in two single-particle states with energies $0$ and $\varepsilon$. The system is in contact with a reservoir at temperature $T$ and chemical potential $\mu$. (1) Write out all the microstates (the pairs of occupation numbers $(n_1,n_2)$) and express the grand partition function $\Xi=\sum e^{-\beta(E-\mu N)}$ in terms of $z=e^{\beta\mu}$ and $x=e^{-\beta\varepsilon}$. Verify that it factorizes as $\Xi=(1+z)(1+zx)$. (2) Compute $\langle N\rangle=k_{\mathrm B}T\,\partial\ln\Xi/\partial\mu$ and show that it equals $f(0)+f(\varepsilon)$ ($f$ is the Fermi distribution). (3) For $k_{\mathrm B}T=\varepsilon$ and $\mu=\varepsilon/2$, find $\langle N\rangle$ and the probability that both states are occupied.

Hint: (2) $\langle N\rangle=z\,\partial\ln\Xi/\partial z$. The logarithm of a product is the sum of the logarithms.

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Exercise 52.11: $\Xi$, $\mu$, and pressure of a two-dimensional ideal gas

Consider a two-dimensional ideal gas in which monatomic molecules of mass $m$ move only on a plane of area $S$ (for example, adsorbed molecules moving about on a solid surface). The single-particle partition function is $Z_1=S/\lambda_{\mathrm T}^2$ ($\lambda_{\mathrm T}=h/\sqrt{2\pi mk_{\mathrm B}T}$ is the same thermal de Broglie wavelength as in three dimensions), and $Z_N=Z_1^N/N!$. (1) Find the grand partition function $\Xi$. (2) From $N=k_{\mathrm B}T\,\partial\ln\Xi/\partial\mu$, derive the relation $\mu=k_{\mathrm B}T\ln(n_2\lambda_{\mathrm T}^2)$ between the areal density $n_2=N/S$ and $\mu$. (3) In two dimensions the grand potential is $J=-\Pi S$ ($\Pi$ is the surface pressure, a force per unit length, with the unit $\mathrm{N/m}$). Show that $\Pi=n_2k_{\mathrm B}T$. (4) For argon ($m=39.948\ \mathrm u$), $T=300\ \mathrm K$, and $n_2=10^{16}\ \mathrm{m^{-2}}$, find $\mu$ (in $\mathrm{eV}$) and $\Pi$.

Hint: Follow the same procedure as in Examples 52.2 and 52.4. $Z_1=S/\lambda_{\mathrm T}^2$ is obtained from the three-dimensional $V/\lambda_{\mathrm T}^3$ by the replacements $V\to S$ and $\lambda_{\mathrm T}^3\to\lambda_{\mathrm T}^2$.

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Exercise 52.12: How the Fermi distribution becomes a step function at low temperature

For the Fermi distribution $f(\varepsilon)=1/\bigl(e^{(\varepsilon-\mu)/k_{\mathrm B}T}+1\bigr)$: (1) Show, from the limits of the exponential function, that the limit $T\to0$ ($\beta\to\infty$) is a step function, treating the cases $\varepsilon\lt\mu$ and $\varepsilon\gt\mu$ separately. What is the value at $\varepsilon=\mu$? (2) For the two cases $k_{\mathrm B}T/\mu=0.1$ and $0.01$, find the values of $f$ at $\varepsilon/\mu=0.9,\ 0.99,\ 1.01,\ 1.1$ and summarize them in a table. Describe how the distribution approaches a step.

Hint: (1) If $\varepsilon\lt\mu$, then $(\varepsilon-\mu)/k_{\mathrm B}T=-(\mu-\varepsilon)\beta\to-\infty$. (2) Compute $x=(\varepsilon-\mu)/k_{\mathrm B}T=(\varepsilon/\mu-1)/(k_{\mathrm B}T/\mu)$.

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Exercise 52.13: Why does condensation not occur in a two-dimensional ideal Bose gas?

Consider an ideal gas of bosons of mass $m$ and spin $0$ confined to a plane of area $A$. (1) Using the density of states $g(\varepsilon)=Am/(2\pi\hbar^2)$ (constant for $\varepsilon\gt0$), express the number density $n=N/A$ in terms of $z=e^{\mu/k_{\mathrm B}T}$ and the thermal de Broglie wavelength $\lambda_{\mathrm T}=h/\sqrt{2\pi mk_{\mathrm B}T}$. (2) Solve this for $z$. (3) Explain, by comparison with the three-dimensional case (Section 52.10), that at finite temperatures $\mu\lt0$ continues to hold and Bose–Einstein condensation does not occur. (4) Find $z$ and $\mu/k_{\mathrm B}T$ when $n\lambda_{\mathrm T}^2=1$.

Hint: $\displaystyle\int_0^\infty\dfrac{\dd x}{z^{-1}e^{x}-1}=-\ln(1-z)$. $\lambda_{\mathrm T}^2=2\pi\hbar^2/(mk_{\mathrm B}T)$.

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