PH503 Homework
1.
(a)
Liouville's equation $\frac{\partial \rho}{\partial t} = -\{\rho, \mathcal{H}\}$, or
$$\frac{d\rho}{dt} = \frac{\partial \rho}{\partial t} + \sum_i \left( \frac{\partial \rho}{\partial q_i}\dot{q}_i + \frac{\partial \rho}{\partial p_i}\dot{p}_i \right) = \frac{\partial \rho}{\partial t} + \{\rho, \mathcal{H}\}$$
Which is $\frac{d\rho}{dt} = 0$, volume element $d\Gamma = dpdq$ is time invariant.
Entropy is $S = - \int \rho \ln \rho \, d\Gamma$.
Then $\rho$ invariant, $\rho \ln \rho$ invariant, $(\rho \ln \rho) \times d\Gamma$ invariant.
$\frac{dS}{dt} = 0$
(b)
Normalization: $C_1 = \int dpdq \, \rho(p, q) = 1$
Fixed average energy: $C_2 = \int dpdq \, \rho(p, q) \mathcal{H}(p, q) = E$
Lagrange multipliers:
$$L[\rho] = S - \alpha C_1 - \beta C_2$$
$$L[\rho] = - \int \rho \ln \rho \, dpdq - \alpha \left(\int \rho \, dpdq - 1\right) - \beta \left(\int \rho \mathcal{H} \, dpdq - E\right)$$
$$L[\rho] = \int (-\rho \ln \rho - \alpha \rho - \beta \rho \mathcal{H}) \, dpdq + \alpha + \beta E$$
To find the function $\rho$ that maximizes $L$,
$$\frac{\partial}{\partial \rho} (-\rho \ln \rho - \alpha \rho - \beta \rho \mathcal{H}) = 0$$
$$\ln \rho = -1 - \alpha - \beta \mathcal{H}$$
Then
$$\rho_{\text{max}} = e^{-1-\alpha} e^{-\beta \mathcal{H}}$$
$$\int \rho_{\text{max}} \, dpdq = \int e^{-1-\alpha} e^{-\beta \mathcal{H}} \, dpdq = e^{-1-\alpha} \int e^{-\beta \mathcal{H}} \, dpdq = 1$$
Partition function $Z = \int e^{-\beta \mathcal{H}} \, dpdq$.
$$e^{-1-\alpha} Z = 1 \quad \implies \quad e^{-1-\alpha} = \frac{1}{Z}$$
Finally,
$$\rho_{\text{max}}(p, q) = \frac{1}{Z} e^{-\beta \mathcal{H}(p, q)}$$
This is the probability distribution of the canonical ensemble.
(c)
A density function is stationary if its partial time derivative is zero $\frac{\partial \rho}{\partial t} = 0$.
Use Liouville's equation($\frac{\partial \rho}{\partial t} = -\{\rho, \mathcal{H}\}$).
Goal is $\{\rho_{\text{max}}, \mathcal{H}\}=0$.
$$\rho_{\text{max}} = f(\mathcal{H}) = \frac{1}{Z} e^{-\beta \mathcal{H}}$$
$$\{f(\mathcal{H}), \mathcal{H}\} = \sum_i \left( \frac{\partial f(\mathcal{H})}{\partial q_i} \frac{\partial \mathcal{H}}{\partial p_i} - \frac{\partial f(\mathcal{H})}{\partial p_i} \frac{\partial \mathcal{H}}{\partial q_i} \right)$$
Using the chain rule:
$$\frac{\partial f(\mathcal{H})}{\partial q_i} = \frac{df}{d\mathcal{H}} \frac{\partial \mathcal{H}}{\partial q_i} \quad \text{and} \quad \frac{\partial f(\mathcal{H})}{\partial p_i} = \frac{df}{d\mathcal{H}} \frac{\partial \mathcal{H}}{\partial p_i}$$
$$\{f(\mathcal{H}), \mathcal{H}\} = \sum_i \left( \left(\frac{df}{d\mathcal{H}} \frac{\partial \mathcal{H}}{\partial q_i}\right) \frac{\partial \mathcal{H}}{\partial p_i} - \left(\frac{df}{d\mathcal{H}} \frac{\partial \mathcal{H}}{\partial p_i}\right) \frac{\partial \mathcal{H}}{\partial q_i} \right)$$
$$\{f(\mathcal{H}), \mathcal{H}\} = \frac{df}{d\mathcal{H}} \sum_i \left( \frac{\partial \mathcal{H}}{\partial q_i} \frac{\partial \mathcal{H}}{\partial p_i} - \frac{\partial \mathcal{H}}{\partial p_i} \frac{\partial \mathcal{H}}{\partial q_i} \right)$$
$$\{\rho_{\text{max}}, \mathcal{H}\} = 0$$
Since $\frac{\partial \rho_{\text{max}}}{\partial t} = -\{\rho_{\text{max}}, \mathcal{H}\}$:
$$\frac{\partial \rho_{\text{max}}}{\partial t} = 0$$
2.
(a)
Define the Systems:
- Subsystem 'a' has energy $E_a$, particle number $N_a$, and its number of accessible microstates is given by the phase space volume $\Gamma_a(E_a)$.
- Subsystem 'b' has energy $E_b$, particle number $N_b$, and its number of accessible microstates is $\Gamma_b(E_b)$.
- The total system has energy $E = E_a + E_b$ and particle number $N = N_a + N_b$.
Calculate the Total Entropy:
$$S_{total} = k_B \log \Gamma(E)$$
If system is statistically independent,
$$\Gamma(E) = \Gamma_a(E_a) \cdot \Gamma_b(E_b)$$
$$S_{total} = k_B \log(\Gamma_a(E_a) \cdot \Gamma_b(E_b))$$
Using $\log(xy) = \log(x) + \log(y)$:
$$S_{total} = k_B (\log \Gamma_a(E_a) + \log \Gamma_b(E_b))$$
$$S_{total} = k_B \log \Gamma_a(E_a) + k_B \log \Gamma_b(E_b)$$
$$S_{total} = S_a + S_b$$
(b)
$$\Gamma(E) = \Sigma(E+\Delta) - \Sigma(E)$$
Since $\Delta \ll E$, first-order Taylor expansion:
$$\Sigma(E+\Delta) \approx \Sigma(E) + \frac{\partial \Sigma(E)}{\partial E} \Delta$$
$$\Gamma(E) \approx \left(\Sigma(E) + \frac{\partial \Sigma(E)}{\partial E} \Delta\right) - \Sigma(E) = \frac{\partial \Sigma(E)}{\partial E} \Delta$$
Also,
$$\Gamma(E) \approx g(E) \Delta$$
Compare $S_1$ and $S_3$:
$$S_1 = k_B \log \Gamma(E) \approx k_B \log(g(E) \Delta)$$
$$S_1 \approx k_B (\log g(E) + \log \Delta) = S_3 + k_B \log \Delta$$
Since the entropy $S$ is extensive ($S \propto N$), this constant difference is negligible for large $N$. Then, $S_1 \approx S_3$.
Compare $S_2$ and $S_3$:
For large DoF $f$ system, let
$$\Sigma(E) \propto E^f$$
$$g(E) = \frac{\partial \Sigma(E)}{\partial E} \propto \frac{\partial}{\partial E}(E^f) = f E^{f-1} = \frac{f}{E} E^f \propto \frac{f}{E} \Sigma(E)$$
$$S_3 = k_B \log g(E) \approx k_B \log \left(C \frac{f}{E} \Sigma(E)\right) $$
$$S_3 \approx k_B \log \Sigma(E) + k_B \log\left(C\frac{f}{E}\right) = S_2 + k_B \log\left(C\frac{f}{E}\right)$$
Both the number of degrees of freedom ($f \propto N$) and the energy ($E \propto N$) are extensive. Then second term $O(\log N)$ at most. Since $S_2$ is of order $O(N)$, the difference is negligible for large $N$. Then, $S_2 \approx S_3$.
3.
The entropy of a monatomic ideal gas is:
$$S = N k_B \ln \left[ V \left( \frac{4\pi m E}{3N h^2} \right)^{3/2} \right] + \frac{3}{2} N k_B$$
The entropy with the Gibbs correction factor ($S = S_{dist} - k_B \ln(N!)$) is given by the Sackur-Tetrode equation:
$$S(E, V, N) = N k_B \ln \left[ \frac{V}{N} \left( \frac{4\pi m E}{3N h^2} \right)^{3/2} \right] + \frac{5}{2} N k_B$$
(a)
Let's scale the variables $E \to \lambda E$, $V \to \lambda V$, $N \to \lambda N$:
$S(\lambda E, \lambda V, \lambda N) = (\lambda N) k_B \ln \left[ \frac{(\lambda V)}{(\lambda N)} \left( \frac{4\pi m (\lambda E)}{3(\lambda N) h^2} \right)^{3/2} \right] + \frac{5}{2} (\lambda N) k_B$
$S(\lambda E, \lambda V, \lambda N) = \lambda N k_B \ln \left[ \frac{V}{N} \left( \frac{4\pi m E}{3N h^2} \right)^{3/2} \right] + \frac{5}{2} \lambda N k_B$
$S(\lambda E, \lambda V, \lambda N) = \lambda \left( N k_B \ln \left[ \frac{V}{N} \left( \frac{4\pi m E}{3N h^2} \right)^{3/2} \right] + \frac{5}{2} N k_B \right)$
$$S(\lambda E, \lambda V, \lambda N) = \lambda S(E, V, N)$$
(b)
The Gibbs correction is a term that depends only on $N$ ($S_{corr} = -k_B \ln(N!)$).
Pressure is defined as $p = T \left( \frac{\partial S}{\partial V} \right)_{E,N}$.
Since the correction term, $-k_B \ln(N!)$, does not depend on volume $V$, its derivative with respect to $V$ is zero, pressure is unchanged.
Heat capacity is defined as $C_V = \left( \frac{\partial E}{\partial T} \right)_{V,N}$. Monatomic ideal gas, $E = \frac{3}{2} N k_B T$.
Since the relationship between $E$ and $T$ is unchanged, the heat capacity is also unchanged.
The exponent $\gamma$ is the ratio of specific heats, $\gamma = C_p/C_V$. Since $C_V$ is unchanged and Mayer's relation for an ideal gas ($C_p = C_V + N k_B$) still holds, $C_p$ is also unchanged. Consequently, $\gamma$ is unchanged, and the adiabatic relation remains the same.
(c)
Let's use the final Sackur-Tetrode equation:
$S = N k_B \left( \ln\left[\frac{V}{N}\left(\frac{4\pi mE}{3Nh^2}\right)^{3/2}\right] + \frac{5}{2} \right) = N k_B \left( \ln(V E^{3/2} C) - \frac{5}{2}\ln N + \frac{5}{2} \right)$ where C is a constant.
$\frac{\partial S}{\partial N} = k_B \left( \ln(V E^{3/2} C) - \frac{5}{2}\ln N + \frac{5}{2} \right) + N k_B \left(-\frac{5}{2N}\right)$
$\frac{\partial S}{\partial N} = k_B \ln\left[\frac{V}{N^{5/2}} (\dots)\right] + \frac{5}{2}k_B - \frac{5}{2}k_B = k_B \ln\left[\frac{V}{N}\left(\frac{4\pi mE}{3Nh^2}\right)^{3/2}\right]$
$\mu = -T \frac{\partial S}{\partial N} = -k_B T \ln\left[\frac{V}{N}\left(\frac{4\pi mE}{3Nh^2}\right)^{3/2}\right]$
Since the entropy $S$ changes, the Helmholtz free energy $F$ must also change.
$F_{new} = E - T S_{new} = E - T(S_{old} - k_B \ln N!) = (E - T S_{old}) + k_B T \ln N!$
$$F_{new} = F_{old} + k_B T \ln(N!)$$
4.
(a)
Use the entropy formula for an ideal gas, $S(N,V) = N k_B \ln(V/N) + \text{constant}$.
Initial Entropy: $S_i = S(N_1, V_1) + S(N_2, V_2) = N_1 k_B \ln(V_1/N_1) + N_2 k_B \ln(V_2/N_2)$.
Final Entropy: $S_f = S(N_1+N_2, V_1+V_2) = (N_1+N_2) k_B \ln\left(\frac{V_1+V_2}{N_1+N_2}\right)$.
$$(\Delta S)_{1=2} = S_f - S_i = k_B \left[ (N_1+N_2) \ln\left(\frac{V_1+V_2}{N_1+N_2}\right) - N_1 \ln\left(\frac{V_1}{N_1}\right) - N_2 \ln\left(\frac{V_2}{N_2}\right) \right]$$
Set volume $v_1 = V_1/N_1$ and $v_2 = V_2/N_2$:
$$v_f = \frac{V_1+V_2}{N_1+N_2} = \frac{N_1 v_1 + N_2 v_2}{N_1+N_2}$$
Then it changes with $N = N_1+N_2$, $x_1 = N_1/N$, and $x_2 = N_2/N$:
$$\ln(v_f) - x_1 \ln(v_1) - x_2 \ln(v_2) \ge 0$$
$$\ln(x_1 v_1 + x_2 v_2) \ge x_1 \ln(v_1) + x_2 \ln(v_2)$$
Use Jensen's inequality ($f(x) = \ln(x)$, is strictly concave), it's true.
Equal at
$$v_1 = v_2 \implies \frac{V_1}{N_1} = \frac{V_2}{N_2}$$
(b)
$$(\Delta S)^* = -k_B (N_1 \ln x_1 + N_2 \ln x_2)$$
where $N = N_1 + N_2$ and $x_1 = N_1/N$ and $x_2 = N_2/N$.
Let $N = N_1+N_2$. then $x = x_1 = N_1/N$, so that $x_2 = 1-x$,
$$(\Delta S)^*(x) = -N k_B [x \ln x + (1-x) \ln(1-x)]$$
To find the maximum,
$$\frac{d(\Delta S)^*}{dx} = -N k_B \frac{d}{dx}[x \ln x + (1-x) \ln(1-x)]$$ $$\frac{d(\Delta S)^*}{dx} = -N k_B \left[ (\ln x + 1) + (-\ln(1-x) - 1) \right] = -N k_B [\ln x - \ln(1-x)]$$
$$\ln x - \ln(1-x) = 0 \implies \ln x = \ln(1-x) \implies x = 1-x \implies 2x=1$$
at $x = 1/2$.
The second derivative is $\frac{d^2(\Delta S)^*}{dx^2} = -N k_B (\frac{1}{x} + \frac{1}{1-x})$, which is negative for $x \in (0,1)$, then $x=1/2$ is a maximum.
Maximum when $x_1 = x_2 = 1/2$, which means $N_1 = N_2$.
$$(\Delta S)^*_{max} = -N k_B \left[ \frac{1}{2} \ln\left(\frac{1}{2}\right) + \frac{1}{2} \ln\left(\frac{1}{2}\right) \right]$$ $$(\Delta S)^*_{max} = -N k_B \left[ \ln\left(\frac{1}{2}\right) \right] = -N k_B (-\ln 2) = N k_B \ln 2$$
$$(\Delta S)^* \le (N_1+N_2)k_B \ln 2$$
5.
For ideal gas, the adiabatic exponent $\gamma$ and the molar heat capacities $C_{P,m}$ and $C_{V,m}$ are related by:
$$\gamma = \frac{C_{P,m}}{C_{V,m}}$$
$$C_{P,m} - C_{V,m} = R$$
$$\gamma = \frac{C_{V,m} + R}{C_{V,m}} = 1 + \frac{R}{C_{V,m}}$$
$$\gamma - 1 = \frac{R}{C_{V,m}} \quad \implies \quad C_{V,m} = \frac{R}{\gamma - 1}$$
For a mixture of ideal gases,
$$C_{V,mix} = f_1 C_{V,m1} + f_2 C_{V,m2}$$
where $f_1$ and $f_2$ are the mole fractions, and $C_{V,m1}$ and $C_{V,m2}$ are the molar heat capacities of gas 1 and gas 2.
Use $C_{V,mix} = \frac{R}{\gamma - 1}$, $C_{V,m1} = \frac{R}{\gamma_1 - 1}$, $C_{V,m2} = \frac{R}{\gamma_2 - 1}$
$$\frac{R}{\gamma - 1} = f_1 \left(\frac{R}{\gamma_1 - 1}\right) + f_2 \left(\frac{R}{\gamma_2 - 1}\right)$$
$$\frac{1}{\gamma - 1} = \frac{f_1}{\gamma_1 - 1} + \frac{f_2}{\gamma_2 - 1}$$
1
(a)
The classical partition function ($Z_{cl}$) is defined as the integral over all of phase space. Let $\beta = \frac{1}{k_BT}$.
$$Z_{cl} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\beta\mathcal{H}} \,dp\,dx$$
$$Z_{cl} = \int_{-\infty}^{\infty} e^{-\frac{\beta p^2}{2m}} \,dp \int_{-\infty}^{\infty} e^{-\frac{\beta m\omega^2x^2}{2}} \,dx$$
Both integrals are Gaussian integrals of the form $\int_{-\infty}^{\infty} e^{-ax^2} dx = \sqrt{\frac{\pi}{a}}$.
Integral over momentum ($p$): Here, $a = \frac{\beta}{2m}$.
$$\int_{-\infty}^{\infty} e^{-\frac{\beta p^2}{2m}} \,dp = \sqrt{\frac{\pi}{\beta/2m}} = \sqrt{\frac{2\pi m}{\beta}}$$
Integral over position ($x$): Here, $a = \frac{\beta m\omega^2}{2}$.
$$\int_{-\infty}^{\infty} e^{-\frac{\beta m\omega^2x^2}{2}} \,dx = \sqrt{\frac{\pi}{\beta m\omega^2/2}} = \sqrt{\frac{2\pi}{\beta m\omega^2}}$$
Multiplying the two results gives the partition function:
$$Z_{cl} = \left(\sqrt{\frac{2\pi m}{\beta}}\right) \left(\sqrt{\frac{2\pi}{\beta m\omega^2}}\right) = \frac{2\pi}{\beta\omega}= \frac{2\pi k_BT}{\omega}$$
(b)
$$\langle E \rangle = -\frac{\partial}{\partial\beta} \ln Z_{cl} = -\frac{\partial}{\partial\beta} \ln\left(\frac{2\pi}{\beta\omega}\right) = -\left(-\frac{1}{\beta}\right) = \frac{1}{\beta} = k_BT$$
The Hamiltonian is separable: $\mathcal{H} = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2$. The average energy is the sum of the averages of its parts:
$$\langle E \rangle = \left\langle \frac{p^2}{2m} \right\rangle + \left\langle \frac{1}{2}m\omega^2x^2 \right\rangle$$
The partition function is also separable into a product $Z_{cl} = Z_p \cdot Z_x$, where $Z_p = \sqrt{\frac{2\pi m}{\beta}}$ and $Z_x = \sqrt{\frac{2\pi}{\beta m\omega^2}}$. We can calculate the average of each energy term from its corresponding part of the partition function.
Average kinetic energy:
$$\left\langle \frac{p^2}{2m} \right\rangle = -\frac{\partial}{\partial\beta} \ln Z_p = -\frac{\partial}{\partial\beta} \ln\left(\sqrt{\frac{2\pi m}{\beta}}\right) = -\frac{\partial}{\partial\beta}\left[\frac{1}{2}\ln(2\pi m) - \frac{1}{2}\ln\beta\right]$$
$$= -\left(-\frac{1}{2\beta}\right) = \frac{1}{2\beta} = \frac{1}{2}k_BT$$
Average potential energy:
$$\left\langle \frac{1}{2}m\omega^2x^2 \right\rangle = -\frac{\partial}{\partial\beta} \ln Z_x = -\frac{\partial}{\partial\beta} \ln\left(\sqrt{\frac{2\pi}{\beta m\omega^2}}\right) = -\frac{\partial}{\partial\beta}\left[\frac{1}{2}\ln\left(\frac{2\pi}{m\omega^2}\right) - \frac{1}{2}\ln\beta\right]$$
$$= -\left(-\frac{1}{2\beta}\right) = \frac{1}{2\beta} = \frac{1}{2}k_BT$$
(c)
The quantum mechanical partition function ($Z_{QM}$) is the sum of the Boltzmann factors over all quantum states:
$$Z_{QM} = \sum_{n=0}^{\infty} e^{-\beta E_n} = \sum_{n=0}^{\infty} e^{-\beta(n+\frac{1}{2})\hbar\omega}$$
$$Z_{QM} = e^{-\frac{\beta\hbar\omega}{2}} \sum_{n=0}^{\infty} e^{-n\beta\hbar\omega} = e^{-\frac{\beta\hbar\omega}{2}} \sum_{n=0}^{\infty} (e^{-\beta\hbar\omega})^n$$
The summation is a geometric series with the first term $a=1$ and the common ratio $r = e^{-\beta\hbar\omega}$. Since $|r|<1$, the sum converges to $\frac{1}{1-r}$.
$$\sum_{n=0}^{\infty} (e^{-\beta\hbar\omega})^n = \frac{1}{1-e^{-\beta\hbar\omega}}$$
Therefore, the quantum mechanical partition function is:
$$Z_{QM} = \frac{e^{-\frac{\beta\hbar\omega}{2}}}{1-e^{-\beta\hbar\omega}}$$
This can also be expressed compactly using the hyperbolic sine function:
$$\Large Z_{QM} = \frac{1}{e^{\frac{\beta\hbar\omega}{2}} - e^{-\frac{\beta\hbar\omega}{2}}} = \frac{1}{2\sinh\left(\frac{\beta\hbar\omega}{2}\right)}$$
(d)
Average Energy $\langle E \rangle$
The quantum average energy is calculated from $Z_{QM}$:
$$\langle E \rangle = -\frac{\partial}{\partial\beta} \ln Z_{QM}$$
$$\ln Z_{QM} = \ln\left(\frac{e^{-\frac{\beta\hbar\omega}{2}}}{1-e^{-\beta\hbar\omega}}\right) = -\frac{\beta\hbar\omega}{2} - \ln(1-e^{-\beta\hbar\omega})$$
Differentiating with respect to $\beta$:
$$\frac{\partial}{\partial\beta} \ln Z_{QM} = -\frac{\hbar\omega}{2} - \frac{1}{1-e^{-\beta\hbar\omega}}(-e^{-\beta\hbar\omega})(-\hbar\omega) = -\frac{\hbar\omega}{2} - \frac{\hbar\omega e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}$$
Thus, the average energy is:
$$\langle E \rangle = - \left(-\frac{\hbar\omega}{2} - \frac{\hbar\omega e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}\right) = \frac{\hbar\omega}{2} + \frac{\hbar\omega e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}$$
By multiplying the numerator and denominator of the second term by $e^{\beta\hbar\omega}$, we get the standard form:
$$\langle E \rangle = \frac{\hbar\omega}{2} + \frac{\hbar\omega}{e^{\beta\hbar\omega}-1}$$
High-Temperature Limit ($T \to \infty$)
The high-temperature limit corresponds to $T \to \infty$, which means $\beta = \frac{1}{k_BT} \to 0$. In this limit, $\beta\hbar\omega \ll 1$, so we can use the Taylor expansion for the exponential,
$$e^{\beta\hbar\omega} \approx 1+\beta\hbar\omega$$
Average energy:
$$\langle E \rangle \approx \frac{\hbar\omega}{2} + \frac{\hbar\omega}{(1+\beta\hbar\omega)-1} = \frac{\hbar\omega}{2} + \frac{\hbar\omega}{\beta\hbar\omega} = \frac{\hbar\omega}{2} + \frac{1}{\beta}$$
As $T \to \infty$ (or $\beta \to 0$), the first term, the zero-point energy $\frac{\hbar\omega}{2}$, is a finite constant, while the second term $\frac{1}{\beta}$ becomes dominant.
$$\lim_{T \to \infty} \langle E \rangle = \lim_{\beta \to 0} \left( \frac{\hbar\omega}{2} + \frac{1}{\beta} \right) = \frac{1}{\beta} = k_BT$$
This result, $\langle E \rangle = k_BT$, exactly matches the average energy of the classical harmonic oscillator found in part (b).
2
(a)
The Helmholtz free energy $F$ is defined as $F = U - TS$, where $U$ is the internal energy, $T$ is the temperature, and $S$ is the entropy.
The internal energy $U$ is the average energy of the system: $U = \langle E \rangle = \sum_s P_s E_s$.
The Gibbs entropy is given by $S = -k_B \sum_s P_s \ln P_s$.
The probability $P_s$ of the system being in a microstate $s$ with energy $E_s$ is $P_s = \frac{e^{-\beta E_s}}{Z}$, where $\beta = \frac{1}{k_B T}$ and $Z = \sum_s e^{-\beta E_s}$ is the partition function.
$$S = -k_B \sum_s \left( \frac{e^{-\beta E_s}}{Z} \right) \ln \left( \frac{e^{-\beta E_s}}{Z} \right)$$
$$S = -k_B \sum_s \frac{e^{-\beta E_s}}{Z} (-\beta E_s - \ln Z)$$
$$S = k_B \beta \sum_s \frac{E_s e^{-\beta E_s}}{Z} + k_B \ln Z \sum_s \frac{e^{-\beta E_s}}{Z}$$
Note $\sum_s \frac{E_s e^{-\beta E_s}}{Z} = U$, $\sum_s \frac{e^{-\beta E_s}}{Z} = \frac{1}{Z} \sum_s e^{-\beta E_s} = \frac{Z}{Z} = 1$.
$$S = k_B \beta U + k_B \ln Z$$
$$S = \frac{U}{T} + k_B \ln Z$$
$$TS = U + k_B T \ln Z$$
$$U - TS = -k_B T \ln Z$$
$$\boldsymbol{F = -k_B T \ln Z}$$
(b)
$$Z_1 = \frac{1}{h^2} \int d^2q \int d^2p \ e^{-\beta H_1(\vec{p})}$$
Here, $H_1(\vec{p}) = c|\vec{p}|$ is the Hamiltonian for a single particle, and the integral over the position coordinates $\int d^2q$ is simply the area $A$.
Use polar coordinates, where $d^2p = p \, dp \, d\theta$ and $|\vec{p}| = p$.
$$Z_1 = \frac{A}{h^2} \int_0^{2\pi} d\theta \int_0^\infty dp \ p e^{-\beta c p}$$
The integral over $\theta$ gives $2\pi$. The integral over $p$ is a standard form $\int_0^\infty x e^{-ax} dx = 1/a^2$. Here, $x=p$ and $a = \beta c$.
$$\int_0^\infty p e^{-\beta c p} dp = \frac{1}{(\beta c)^2}$$
Combining the parts, the single-particle partition function is:
$$Z_1 = \frac{A}{h^2} \cdot 2\pi \cdot \frac{1}{(\beta c)^2} = \frac{2\pi A}{(h\beta c)^2} = \frac{2\pi A(k_B T)^2}{(hc)^2}$$
For $N$ non-interacting, indistinguishable particles, the partition function is $Z = \frac{Z_1^N}{N!}$.
$$Z = \frac{1}{N!} \left( \frac{2\pi A(k_B T)^2}{(hc)^2} \right)^N$$
Now, we use the result from part (a), $F = -k_B T \ln Z$.
$$F = -k_B T \ln \left[ \frac{1}{N!} \left( \frac{2\pi A(k_B T)^2}{(hc)^2} \right)^N \right]$$
$$F = -k_B T \left[ N \ln\left( \frac{2\pi A(k_B T)^2}{(hc)^2} \right) - \ln(N!) \right]$$
Using the Stirling approximation, $\ln(N!) \approx N \ln N - N$:
$$F \approx -k_B T \left[ N \ln\left( \frac{2\pi A(k_B T)^2}{(hc)^2} \right) - (N \ln N - N) \right]$$
$$F = -N k_B T \left[ \ln\left( \frac{2\pi A(k_B T)^2}{(hc)^2} \right) - \ln N + 1 \right]$$
Combining the logarithmic terms, we get the final expression for the free energy:
$$\boldsymbol{F = -N k_B T \left[ \ln \left( \frac{2\pi A (k_B T)^2}{N(hc)^2} \right) + 1 \right]}$$
(c)
Entropy (S)
$S = -\left(\frac{\partial F}{\partial T}\right)_{A,N}$
Let's first write $F$ as $F = -N k_B T \left[ \ln \left( \frac{2\pi A k_B^2}{N(hc)^2} \right) + 2\ln T + 1 \right]$.
$S = - \left( -N k_B \left[ \ln \left( \dots \right) + 2\ln T + 1 \right] - N k_B T \left[ \frac{2}{T} \right] \right)$
$S = N k_B \left[ \ln \left( \frac{2\pi A (k_B T)^2}{N(hc)^2} \right) + 1 \right] + 2N k_B$
$$\boldsymbol{S = N k_B \left[ \ln \left( \frac{2\pi A (k_B T)^2}{N(hc)^2} \right) + 3 \right]}$$
Internal Energy ($U = F + TS$)
$U = -N k_B T \left[ \ln \left( \dots \right) + 1 \right] + T \left( N k_B \left[ \ln \left( \dots \right) + 3 \right] \right)$
$U = -N k_B T \ln(\dots) - N k_B T + N k_B T \ln(\dots) + 3N k_B T$
$$\boldsymbol{U = 2N k_B T}$$
Pressure (P)
$P = -\left(\frac{\partial F}{\partial A}\right)_{T,N}$
$F = -N k_B T \left[ \ln A + \ln \left( \frac{2\pi (k_B T)^2}{N(hc)^2} \right) + 1 \right]$
$P = - \left( -N k_B T \cdot \frac{1}{A} \right)$
$$\boldsymbol{P = \frac{N k_B T}{A}}$$
Chemical Potential ($\mu$)
$\mu = \left(\frac{\partial F}{\partial N}\right)_{T,A}$
$F = -N k_B T \left[ \ln \left( \frac{2\pi A (k_B T)^2}{(hc)^2} \right) - \ln N + 1 \right]$
Let $C = k_B T \left[ \ln \left( \frac{2\pi A (k_B T)^2}{(hc)^2} \right) + 1 \right]$. Then $F = -NC + N k_B T \ln N$.
$\mu = \frac{\partial}{\partial N} (-NC + N k_B T \ln N)$
$\mu = -C + \left( k_B T \ln N + N k_B T \frac{1}{N} \right)$
$\mu = -C + k_B T \ln N + k_B T$
$\mu = -k_B T \left[ \ln \left( \frac{2\pi A (k_B T)^2}{(hc)^2} \right) + 1 \right] + k_B T \ln N + k_B T$
$\mu = k_B T \left[ -\ln \left( \frac{2\pi A (k_B T)^2}{(hc)^2} \right) -1 + \ln N + 1 \right]$
$\mu = k_B T \left[ \ln N - \ln \left( \frac{2\pi A (k_B T)^2}{(hc)^2} \right) \right]$
$$\boldsymbol{\mu = k_B T \ln \left( \frac{N(hc)^2}{2\pi A (k_B T)^2} \right)}$$
3
(a)
The canonical partition function for $N_i$ indistinguishable particles of mass $m_i$ in a volume $V$ at temperature $T$ is given by:
$$Q_i = \frac{q_i^{N_i}}{N_i!}$$
where $q_i$ is the single-particle translational partition function:
$$q_i = V \left( \frac{2\pi m_i k_B T}{h^2} \right)^{3/2} = \frac{V}{\Lambda_i^3}$$
Here, $\Lambda_i = h/\sqrt{2\pi m_i k_B T}$ is the thermal de Broglie wavelength for species $i$, $k_B$ is the Boltzmann constant, and $h$ is Planck's constant.
Since the two types of molecules are distinguishable from each other, the total partition function for the system is the product of the individual partition functions:
$$Q(N_1, N_2, V, T) = Q_1 Q_2 = \frac{q_1^{N_1}}{N_1!} \frac{q_2^{N_2}}{N_2!}$$
Substituting the expressions for $q_1$ and $q_2$:
$$Q = \frac{1}{N_1! N_2!} \left[ V \left( \frac{2\pi m_1 k_B T}{h^2} \right)^{3/2} \right]^{N_1} \left[ V \left( \frac{2\pi m_2 k_B T}{h^2} \right)^{3/2} \right]^{N_2}$$
To find the thermodynamic properties, we first compute the natural logarithm of $Q$, using Stirling's approximation ($\ln N! \approx N \ln N - N$):
$$\ln Q = N_1 \ln q_1 + N_2 \ln q_2 - (N_1 \ln N_1 - N_1) - (N_2 \ln N_2 - N_2)$$
$$\ln Q = N_1 \ln\left(\frac{q_1}{N_1}\right) + N_2 \ln\left(\frac{q_2}{N_2}\right) + (N_1 + N_2)$$
1. Helmholtz Free Energy ($A$)
$A = -k_B T \ln Q$
$$A = -k_B T \left[ N_1 \ln\left(\frac{q_1}{N_1}\right) + N_2 \ln\left(\frac{q_2}{N_2}\right) + N_1 + N_2 \right]$$
2. Pressure ($P$)
$P = k_B T \left( \frac{\partial \ln Q}{\partial V} \right)_{T, N_1, N_2}$. Since $\ln q_i = \ln V + \text{terms independent of } V$, we have $\frac{\partial \ln q_i}{\partial V} = \frac{1}{V}$.
$$P = k_B T \left( N_1 \frac{\partial \ln q_1}{\partial V} + N_2 \frac{\partial \ln q_2}{\partial V} \right) = k_B T \left( \frac{N_1}{V} + \frac{N_2}{V} \right)$$
$$P = \frac{(N_1 + N_2)k_B T}{V}$$
3. Internal Energy ($U$)
$U = k_B T^2 \left( \frac{\partial \ln Q}{\partial T} \right)_{V, N_1, N_2}$. Since $\ln q_i = \frac{3}{2}\ln T + \text{terms independent of } T$, we have $\frac{\partial \ln q_i}{\partial T} = \frac{3}{2T}$.
$$U = k_B T^2 \left( N_1 \frac{\partial \ln q_1}{\partial T} + N_2 \frac{\partial \ln q_2}{\partial T} \right) = k_B T^2 \left( N_1 \frac{3}{2T} + N_2 \frac{3}{2T} \right)$$
$$U = \frac{3}{2}(N_1 + N_2)k_B T$$
4. Entropy ($S$)
$S = \frac{U - A}{T}$
$$S = \frac{3}{2}(N_1 + N_2)k_B + k_B \left[ N_1 \ln\left(\frac{q_1}{N_1}\right) + N_2 \ln\left(\frac{q_2}{N_2}\right) + N_1 + N_2 \right]$$
$$S = k_B \left[ N_1 \ln\left(\frac{V}{N_1\Lambda_1^3}\right) + N_2 \ln\left(\frac{V}{N_2\Lambda_2^3}\right) + \frac{5}{2}(N_1 + N_2) \right]$$
(b)
Properties of the Single-Component Gas
The partition function $Q'$ for this gas is:
$$Q' = \frac{q'^N}{N!}$$
where $q' = V \left( \frac{2\pi m k_B T}{h^2} \right)^{3/2} = \frac{V}{\Lambda'^3}$.
Using the same methods as in part (a):
Pressure ($P'$):
$$P' = \frac{N k_B T}{V} = \frac{(N_1+N_2)k_B T}{V}$$
Internal Energy ($U'$):
$$U' = \frac{3}{2}N k_B T = \frac{3}{2}(N_1+N_2)k_B T$$
Entropy ($S'$):
$$S' = k_B \left[ N \ln\left(\frac{q'}{N}\right) + \frac{5}{2}N \right] = k_B \left[ (N_1+N_2) \ln\left(\frac{V}{(N_1+N_2)\Lambda'^3}\right) + \frac{5}{2}(N_1+N_2) \right]$$
The pressure and internal energy are identical in both cases ($P=P'$ and $U=U'$).
The entropy is different. Let's find the difference $\Delta S = S_{mixture} - S_{single}$.
$$\frac{\Delta S}{k_B} = \left[ N_1 \ln\left(\frac{q_1}{N_1}\right) + N_2 \ln\left(\frac{q_2}{N_2}\right) \right] - (N_1+N_2) \ln\left(\frac{q'}{N_1+N_2}\right)$$
4
The virial $\mathcal{V}$ of a system of $N$ particles is defined as the ensemble average of the sum of the dot products of the force $\mathbf{F}_i$ on each particle with its position vector $\mathbf{r}_i$.
$$\mathcal{V} = \left\langle \sum_{i=1}^N \mathbf{F}_i \cdot \mathbf{r}_i \right\rangle$$
The total force $\mathbf{F}_i$ on any particle can be split into two components: the external force exerted by the walls of the container, $\mathbf{F}_i^{\text{ext}}$, and the internal force from other particles in the system, $\mathbf{F}_i^{\text{int}}$.
$$\mathcal{V} = \left\langle \sum_{i=1}^N \mathbf{F}_i^{\text{ext}} \cdot \mathbf{r}_i \right\rangle + \left\langle \sum_{i=1}^N \mathbf{F}_i^{\text{int}} \cdot \mathbf{r}_i \right\rangle$$
The external forces are due to the pressure $P$ exerted by the container walls. The force on a surface element $d\mathbf{S}$ of the container wall is $-P d\mathbf{S}$ (pointing into the volume). The sum over the particles feeling this force can be replaced by an integral over the container's surface area $S$.
$$\left\langle \sum_{i=1}^N \mathbf{F}_i^{\text{ext}} \cdot \mathbf{r}_i \right\rangle = \oint_S (-P d\mathbf{S}) \cdot \mathbf{r} = -P \oint_S \mathbf{r} \cdot d\mathbf{S}$$
Using the divergence theorem, which states $\oint_S \mathbf{A} \cdot d\mathbf{S} = \int_V (\nabla \cdot \mathbf{A}) dV$, we can convert the surface integral into a volume integral. Here, our vector field is the position vector $\mathbf{A} = \mathbf{r} = (x, y, z)$.
The divergence of $\mathbf{r}$ is:
$$\nabla \cdot \mathbf{r} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3$$
Substituting this back into the integral:
$$-P \oint_S \mathbf{r} \cdot d\mathbf{S} = -P \int_V (\nabla \cdot \mathbf{r}) dV = -P \int_V 3 dV = -3PV$$
So, the contribution to the virial from external forces is -3PV.
The internal force on particle $i$ is the negative gradient of the total potential energy $U_{\text{total}}$ with respect to its coordinates: $\mathbf{F}_i^{\text{int}} = -\nabla_i U_{\text{total}}$.
The contribution to the virial is:
$$\left\langle \sum_{i=1}^N \mathbf{F}_i^{\text{int}} \cdot \mathbf{r}_i \right\rangle = \left\langle \sum_{i=1}^N (-\nabla_i U_{\text{total}}) \cdot \mathbf{r}_i \right\rangle = -\left\langle \sum_{i=1}^N \mathbf{r}_i \cdot \nabla_i U_{\text{total}} \right\rangle$$
The problem states that the interparticle potential energy $u(\mathbf{r})$ is a homogeneous function of degree $n$. This implies that the total potential energy of the system, $U_{\text{total}}(\mathbf{r}_1, \dots, \mathbf{r}_N)$, is also a homogeneous function of degree $n$ with respect to all particle coordinates. That is, if we scale all coordinates by a factor $\lambda$, the potential energy scales as:
$$U_{\text{total}}(\lambda\mathbf{r}_1, \dots, \lambda\mathbf{r}_N) = \lambda^n U_{\text{total}}(\mathbf{r}_1, \dots, \mathbf{r}_N)$$
According to Euler's theorem for homogeneous functions, if a function $f(x_1, \dots, x_m)$ is homogeneous of degree $k$, then $\sum_{j=1}^m x_j \frac{\partial f}{\partial x_j} = kf$. Applying this theorem to our potential energy $U_{\text{total}}$:
$$\sum_{i=1}^N \mathbf{r}_i \cdot \nabla_i U_{\text{total}} = n U_{\text{total}}$$
Taking the ensemble average gives $\langle n U_{\text{total}} \rangle = nU$, where $U$ is the mean potential energy.
Thus, the contribution to the virial from internal forces is:
$$-\left\langle \sum_{i=1}^N \mathbf{r}_i \cdot \nabla_i U_{\text{total}} \right\rangle = -nU$$
Combining the contributions from external and internal forces, we get the total virial:
$$\mathcal{V} = -3PV - nU$$
The virial theorem provides a general relationship between the mean kinetic energy $K$ and the virial $\mathcal{V}$:
$$K = \left\langle \sum_{i=1}^N \frac{p_i^2}{2m_i} \right\rangle = -\frac{1}{2} \mathcal{V}$$
Substituting the expression for $\mathcal{V}$ we just derived:
$$K = \frac{1}{2}(3PV + nU)$$
The total mean energy of the system is the sum of the mean kinetic and mean potential energies: $E = K + U$. We can rearrange this to write the potential energy as $U = E - K$.
Now, substitute this into our expression for $K$:
$$K = \frac{1}{2}(3PV + n(E - K))$$
$$K = \frac{1}{n+2}(3PV + nE)$$
5
(a)
The total energy of a single molecule is given as $\epsilon = \epsilon_{trans} + \epsilon_{rot} + \epsilon_{pot}$. Since these energy terms are independent, the single-particle partition function $Z_1$ is a product of the partition functions for translation, rotation, and potential energy.
$Z_1 = Z_{trans} \cdot Z_{rot,pot}$
Translational Part: The standard result for a classical gas in a volume $V$ is:
$$Z_{trans} = \frac{V}{h^3} \int e^{-\beta p^2/2m} d^3p = V \left(\frac{2\pi m k_B T}{h^2}\right)^{3/2}$$
Rotational and Potential Part: This part requires integrating over the rotational phase space $(\theta, \phi, p_\theta, p_\phi)$.
$$Z_{rot,pot} = \frac{1}{h^2} \int e^{-\beta \left( \frac{p_\theta^2}{2I} + \frac{p_\phi^2}{2I \sin^2\theta} - \mu E \cos\theta \right)} dp_\theta dp_\phi d\theta d\phi$$
Integrating over the momenta $p_\theta$ and $p_\phi$ (which are Gaussian integrals) yields:
$$Z_{rot,pot} = \frac{2\pi I k_B T}{h^2} \int_0^{2\pi} d\phi \int_0^\pi e^{\beta \mu E \cos\theta} \sin\theta \, d\theta$$
Let $x = \beta \mu E = \frac{\mu E}{k_B T}$. The integral over the angles becomes:
$$2\pi \int_{-1}^{1} e^{xu} du = 4\pi \frac{\sinh(x)}{x}$$
where $u = \cos\theta$. Combining these gives the rotational partition function:
$$Z_{rot,pot} = \frac{8\pi^2 I k_B T}{h^2} \left( \frac{\sinh(\beta \mu E)}{\beta \mu E} \right)$$
The total partition function for $N$ non-interacting molecules is $Z_N = \frac{Z_1^N}{N!}$.
The average alignment of the dipoles with the electric field gives the electric polarization $P$, defined as the average dipole moment per unit volume.
$$P = \frac{N}{V} \langle \mu \cos\theta \rangle$$
The average value $\langle \cos\theta \rangle$ is given by the Langevin function, $L(x) = \coth(x) - 1/x$, where $x = \beta \mu E$.
$$P = \frac{N\mu}{V} \left( \coth(\beta \mu E) - \frac{1}{\beta \mu E} \right)$$
Using the assumption that $|\mu E| \ll k_B T$ (i.e., $x \ll 1$), we can approximate the Langevin function by expanding it: $L(x) \approx x/3$.
$$P \approx \frac{N\mu}{V} \left( \frac{\beta \mu E}{3} \right) = \frac{N \mu^2}{3V k_B T} E$$
The dielectric constant $\epsilon_r$ relates the polarization to the electric field. In cgs units, the relation is:
$$\epsilon_r = 1 + 4\pi \frac{P}{E}$$
Substituting our approximate expression for $P$:
$$\epsilon_r = 1 + 4\pi \frac{N \mu^2}{3V k_B T}$$
Letting $n = N/V$ be the number density of the molecules, we get the final expression:
$$\boldsymbol{\epsilon_r \approx 1 + \frac{4\pi n \mu^2}{3 k_B T}}$$
This is the Debye equation for the dielectric constant of a polar gas.
(b)
We can treat steam as an ideal gas. The ideal gas law is $PV = N k_B T$, so the number density $n=N/V$ is:
$$n = \frac{P}{k_B T}$$
We will use cgs units for all quantities:
Pressure $P$: $1 \text{ atm} = 1.01325 \times 10^6 \text{ dyne/cm}^2$.
Temperature $T$: $100^\circ C = 373.15 \text{ K}$.
Boltzmann constant $k_B$: $1.3806 \times 10^{-16} \text{ erg/K}$.
First, calculate the thermal energy $k_B T$:
$k_B T = (1.3806 \times 10^{-16} \text{ erg/K}) \times (373.15 \text{ K}) = 5.152 \times 10^{-14} \text{ erg}$.
Now, calculate the number density:
$$n = \frac{1.01325 \times 10^6 \text{ dyne/cm}^2}{5.152 \times 10^{-14} \text{ erg}} = 1.9667 \times 10^{19} \text{ cm}^{-3}$$
Now we plug our values into the Debye equation:
$$\epsilon_r - 1 = \frac{4\pi n \mu^2}{3 k_B T}$$
We are given:
Dipole moment $\mu = 1.85 \times 10^{-18}$ e.s.u. (statcoulomb-cm).
Let's calculate the terms:
$\mu^2 = (1.85 \times 10^{-18})^2 = 3.4225 \times 10^{-36} \text{ (statC}\cdot\text{cm)}^2$.
$3 k_B T = 3 \times (5.152 \times 10^{-14} \text{ erg}) = 1.5456 \times 10^{-13} \text{ erg}$.
Now, combine everything:
$$\epsilon_r - 1 = \frac{4\pi (1.9667 \times 10^{19} \text{ cm}^{-3}) (3.4225 \times 10^{-36} \text{ (statC}\cdot\text{cm)}^2)}{1.5456 \times 10^{-13} \text{ erg}}$$
Recalling that in cgs units, the units combine to be dimensionless (since $\text{erg} = \text{dyne}\cdot\text{cm}$ and $\text{dyne} = \text{statC}^2/\text{cm}^2$), the calculation gives:
$$\epsilon_r - 1 \approx 0.00547$$
$$\boldsymbol{\epsilon_r \approx 1.00547}$$
Thus, the dielectric constant of steam at $100^\circ C$ and atmospheric pressure is approximately 1.0055.
1
(a)
The grand canonical average of any quantity $X$, denoted by $\langle\langle X \rangle\rangle$. First, we average over the microstates for a fixed number of particles $N$ (this is the canonical average, $\langle X \rangle_{can, N}$). Second, we average over all possible values of $N$.
The total energy variance in the grand canonical ensemble is:
$$\langle\langle(\Delta E)^2\rangle\rangle = \langle\langle E^2 \rangle\rangle - \langle\langle E \rangle\rangle^2$$
We can express $\langle\langle E^2 \rangle\rangle$ as the average of the canonical expectation of $E^2$:
$$\langle\langle E^2 \rangle\rangle = \langle\langle \langle E^2 \rangle_{can, N} \rangle\rangle$$
The variance in the canonical ensemble for a fixed $N$ is $\langle(\Delta E)^2\rangle_{can, N} = \langle E^2 \rangle_{can, N} - (\langle E \rangle_{can, N})^2$. Let's denote the average energy in the canonical ensemble, $\langle E \rangle_{can, N}$, as $U(N)$.
Rearranging this, we get $\langle E^2 \rangle_{can, N} = \langle(\Delta E)^2\rangle_{can, N} + U(N)^2$. Substituting this back into the expression for $\langle\langle E^2 \rangle\rangle$:
$$\langle\langle E^2 \rangle\rangle = \langle\langle \langle(\Delta E)^2\rangle_{can, N} + U(N)^2 \rangle\rangle = \langle\langle \langle(\Delta E)^2\rangle_{can, N} \rangle\rangle + \langle\langle U(N)^2 \rangle\rangle$$
The term $\langle\langle \langle(\Delta E)^2\rangle_{can, N} \rangle\rangle$ is what the problem calls $\langle(\Delta E)^2\rangle_{can}$, the average canonical variance.
Now, substitute this back into the formula for the grand canonical variance:
$$\langle\langle(\Delta E)^2\rangle\rangle = \left( \langle(\Delta E)^2\rangle_{can} + \langle\langle U(N)^2 \rangle\rangle \right) - \langle\langle U(N) \rangle\rangle^2$$
$$\langle\langle(\Delta E)^2\rangle\rangle = \langle(\Delta E)^2\rangle_{can} + \left( \langle\langle U(N)^2 \rangle\rangle - \langle\langle U(N) \rangle\rangle^2 \right)$$
The second part of the expression is the variance of the average energy $U(N)$ due to fluctuations in the particle number $N$.
For a large system, the particle number $N$ fluctuates only slightly around its mean value $\langle\langle N \rangle\rangle$. We can use a Taylor expansion for $U(N)$ around $\langle\langle N \rangle\rangle$:
$$U(N) \approx U(\langle\langle N \rangle\rangle) + (N - \langle\langle N \rangle\rangle) \left(\frac{\partial U}{\partial N}\right)_{T,V}$$
The variance of $U(N)$ is $\langle\langle (\Delta U)^2 \rangle\rangle = \langle\langle (U(N) - \langle\langle U \rangle\rangle)^2 \rangle\rangle$. Using the approximation, $U(N) - \langle\langle U \rangle\rangle \approx (N - \langle\langle N \rangle\rangle) \left(\frac{\partial U}{\partial N}\right)_{T,V}$.
$$\langle\langle (\Delta U)^2 \rangle\rangle \approx \left\langle\left\langle \left[ (N - \langle\langle N \rangle\rangle) \left(\frac{\partial U}{\partial N}\right)_{T,V} \right]^2 \right\rangle\right\rangle = \left(\frac{\partial U}{\partial N}\right)_{T,V}^2 \langle\langle (N - \langle\langle N \rangle\rangle)^2 \rangle\rangle$$
$$\langle\langle (\Delta U)^2 \rangle\rangle \approx \left(\frac{\partial U}{\partial N}\right)_{T,V}^2 \langle\langle(\Delta N)^2\rangle\rangle$$
Combining the results, we get the final expression:
$$\langle\langle(\Delta E)^2\rangle\rangle = \langle(\Delta E)^2\rangle_{can} + \left(\frac{\partial U}{\partial N}\right)_{T,V}^2 \langle\langle(\Delta N)^2\rangle\rangle$$
(b)
The average energy $\langle\langle E \rangle\rangle$ can be written as:
$$\langle\langle E \rangle\rangle = \frac{1}{Q} \sum_{N,j} E_{N,j} e^{\beta(\mu N - E_{N,j})}$$
Let's take the partial derivative of $\langle\langle E \rangle\rangle$ with respect to the chemical potential $\mu$ at constant $T$ and $V$:
$$\left(\frac{\partial \langle\langle E \rangle\rangle}{\partial \mu}\right)_{T,V} = \frac{\partial}{\partial \mu} \left( \frac{1}{Q} \sum E e^{\beta(\mu N - E)} \right)$$
Using the quotient rule, we find:
$$\left(\frac{\partial \langle\langle E \rangle\rangle}{\partial \mu}\right)_{T,V} = \frac{1}{Q} \sum E (\beta N) e^{\beta(\mu N - E)} - \frac{1}{Q^2} \left( \sum E e^{\beta(\mu N - E)} \right) \left( \frac{\partial Q}{\partial \mu} \right)$$
We know that $\frac{\partial Q}{\partial \mu} = \sum (\beta N) e^{\beta(\mu N - E)} = \beta Q \langle\langle N \rangle\rangle$. Substituting this in:
$$\left(\frac{\partial \langle\langle E \rangle\rangle}{\partial \mu}\right)_{T,V} = \beta \langle\langle NE \rangle\rangle - \frac{1}{Q^2} (Q \langle\langle E \rangle\rangle) (\beta Q \langle\langle N \rangle\rangle) = \beta \left( \langle\langle NE \rangle\rangle - \langle\langle E \rangle\rangle \langle\langle N \rangle\rangle \right)$$
Therefore, the covariance is:
$$\langle\langle NE \rangle\rangle - \langle\langle E \rangle\rangle \langle\langle N \rangle\rangle = \frac{1}{\beta} \left(\frac{\partial \langle\langle E \rangle\rangle}{\partial \mu}\right)_{T,V} = k_B T \left(\frac{\partial U}{\partial \mu}\right)_{T,V}$$
where we have used $U = \langle\langle E \rangle\rangle$.
Using the chain rule, we can change the variable of differentiation from $\mu$ to the average particle number $\langle\langle N \rangle\rangle$:
$$\left(\frac{\partial U}{\partial \mu}\right)_{T,V} = \left(\frac{\partial U}{\partial \langle\langle N \rangle\rangle}\right)_{T,V} \left(\frac{\partial \langle\langle N \rangle\rangle}{\partial \mu}\right)_{T,V}$$
A fundamental result for fluctuations in the grand canonical ensemble is that the variance of the particle number is given by:
$$\langle\langle(\Delta N)^2\rangle\rangle = k_B T \left(\frac{\partial \langle\langle N \rangle\rangle}{\partial \mu}\right)_{T,V}$$
Substituting these relations back into our expression for the covariance:
$$\langle\langle NE \rangle\rangle - \langle\langle E \rangle\rangle \langle\langle N \rangle\rangle = k_B T \left[ \left(\frac{\partial U}{\partial \langle\langle N \rangle\rangle}\right)_{T,V} \left(\frac{\partial \langle\langle N \rangle\rangle}{\partial \mu}\right)_{T,V} \right]$$
$$\langle\langle NE \rangle\rangle - \langle\langle E \rangle\rangle \langle\langle N \rangle\rangle = \left(\frac{\partial U}{\partial N}\right)_{T,V} \left[ k_B T \left(\frac{\partial \langle\langle N \rangle\rangle}{\partial \mu}\right)_{T,V} \right]$$
$$\langle\langle NE \rangle\rangle - \langle\langle N \rangle\rangle \langle\langle E \rangle\rangle = \left(\frac{\partial U}{\partial N}\right)_{T,V} \langle\langle(\Delta N)^2\rangle\rangle$$
(c)
Start with the statistical definition and derive the thermodynamic quantities from it.
$$\Phi(T, V, \mu) = -\frac{1}{\beta} \ln Q(T, V, \mu)$$
This implies $\ln Q = -\beta \Phi$.
The average value of the quantity $E-\mu N$ is given by:
$$\langle\langle E - \mu N \rangle\rangle = -\left(\frac{\partial \ln Q}{\partial \beta}\right)_{\mu,V}$$
Substituting $\ln Q = -\beta \Phi$:
$$\langle\langle E \rangle\rangle - \mu \langle\langle N \rangle\rangle = -\frac{\partial (-\beta \Phi)}{\partial \beta} = \frac{\partial(\beta \Phi)}{\partial \beta} = \Phi + \beta \frac{\partial \Phi}{\partial \beta}$$
From the thermodynamic definition, we can rearrange to get:
$$\langle\langle E \rangle\rangle - \mu \langle\langle N \rangle\rangle = \Phi + TS$$
By equating the two expressions for $\langle\langle E \rangle\rangle - \mu \langle\langle N \rangle\rangle$, we find:
$$\Phi + TS = \Phi + \beta \frac{\partial \Phi}{\partial \beta} \implies TS = \beta \frac{\partial \Phi}{\partial \beta}$$
We can express the derivative with respect to $\beta$ in terms of a derivative with respect to $T$ using the chain rule. Since $\beta = 1/(k_B T)$, the derivative is $\frac{d\beta}{dT} = -\frac{1}{k_B T^2}$.
$$\frac{\partial \Phi}{\partial \beta} = \frac{\partial \Phi}{\partial T} \frac{dT}{d\beta} = \frac{\partial \Phi}{\partial T} \left( -k_B T^2 \right)$$
Substituting this into our expression for entropy:
$$TS = \beta \left[ \frac{\partial \Phi}{\partial T} (-k_B T^2) \right] = \frac{1}{k_B T} \left[ \frac{\partial \Phi}{\partial T} (-k_B T^2) \right] = -T \left(\frac{\partial \Phi}{\partial T}\right)_{\mu,V}$$
Dividing by $T$ gives the desired result:
$$S = -\left(\frac{\partial \Phi}{\partial T}\right)_{\mu,V}$$
2
(a)
The probability of a single, specific molecule being in the volume $V$ is $p = \frac{V}{V^{(0)}}$. The probability of it being outside this volume is $1-p$. Since there are $N^{(0)}$ molecules and their locations are uncorrelated, the number of molecules $N$ found in volume $V$ follows a binomial distribution.
$$P(N) = \binom{N^{(0)}}{N} p^N (1-p)^{N^{(0)}-N}$$
The mean number of successes in $N^{(0)}$ independent Bernoulli trials is the number of trials multiplied by the probability of success, $p$.
$$\bar{N} = \sum_{N=0}^{N^{(0)}} N P(N) = N^{(0)}p$$
The variance, $\sigma^2$, of a binomial distribution is given by $\sigma^2 = N^{(0)}p(1-p)$. The root-mean-square (r.m.s.) deviation is the square root of the variance.
$$(\Delta N)_{r.m.s.} = \sqrt{\sigma^2} = \sqrt{\langle (N - \bar{N})^2 \rangle} = \sqrt{N^{(0)}p(1-p)}$$
This can also be written as:
$$(\Delta N)_{r.m.s.} = \{N^{(0)}p(1-p)\}^{1/2}$$
(b)
$$\ln P(N) = \ln(N^{(0)}!) - \ln(N!) - \ln((N^{(0)}-N)!) + N\ln p + (N^{(0)}-N)\ln(1-p)$$
Applying Stirling's approximation and simplifying gives:
$$\ln P(N) \approx -N\ln\left(\frac{N}{N^{(0)}p}\right) - (N^{(0)}-N)\ln\left(\frac{N^{(0)}-N}{N^{(0)}(1-p)}\right)$$
The distribution is sharply peaked around its mean $\bar{N} = N^{(0)}p$. We can perform a Taylor expansion of $\ln P(N)$ for small deviations $x = N - \bar{N}$ around the mean.
$$\ln P(N) \approx \ln P(\bar{N}) + (N-\bar{N})\left.\frac{d\ln P}{dN}\right|_{\bar{N}} + \frac{1}{2}(N-\bar{N})^2\left.\frac{d^2\ln P}{dN^2}\right|_{\bar{N}}$$
The first derivative is zero at the mean, as $\bar{N}$ is the most probable value (the maximum of the distribution).
The second derivative at the mean is:
$$\left.\frac{d^2\ln P}{dN^2}\right|_{\bar{N}} = -\frac{1}{\bar{N}} - \frac{1}{N^{(0)}-\bar{N}} = -\frac{1}{N^{(0)}p} - \frac{1}{N^{(0)}(1-p)} = -\frac{1}{N^{(0)}p(1-p)} = -\frac{1}{\sigma^2}$$
Substituting this back into the expansion:
$$\ln P(N) \approx \ln P(\bar{N}) - \frac{(N-\bar{N})^2}{2\sigma^2}$$
Exponentiating both sides gives the Gaussian form:
$$P(N) \approx C \cdot e^{-\frac{(N-\bar{N})^2}{2\sigma^2}}$$
where $C$ is a normalization constant, which for a continuous distribution is $1/\sqrt{2\pi\sigma^2}$.
(c)
Since $N \ll N^{(0)}$, the term $\frac{N^{(0)}!}{(N^{(0)}-N)!} = N^{(0)}(N^{(0)}-1)\cdots(N^{(0)}-N+1)$ is a product of $N$ terms, each of which is approximately $N^{(0)}$.
$$\frac{N^{(0)}!}{(N^{(0)}-N)!} \approx (N^{(0)})^N$$
Since $p \ll 1$, we can use the limit definition of the exponential function. As $N \ll N^{(0)}$, we can approximate $N^{(0)}-N \approx N^{(0)}$.
$$(1-p)^{N^{(0)}-N} \approx (1-p)^{N^{(0)}} = \left(1-\frac{\bar{N}}{N^{(0)}}\right)^{N^{(0)}} \approx e^{-\bar{N}}$$
Substitute these back into the binomial expression:
$$P(N) \approx \frac{(N^{(0)})^N}{N!} p^N e^{-\bar{N}}$$
Now, we group the terms:
$$P(N) \approx \frac{(N^{(0)}p)^N}{N!} e^{-\bar{N}}$$
Substituting $\bar{N} = N^{(0)}p$, we arrive at the Poisson distribution:
$$P(N) = \frac{\bar{N}^N}{N!} e^{-\bar{N}}$$
3
The total energy of a single molecule separates into two independent parts: the translational motion of the center of mass(CM) and the relative motion(vibration).
$$H = H_{CM} + H_{rel} = \left( \frac{P^2}{2M} \right) + \left( \frac{p^2}{2\mu} + \frac{1}{2}K r^2 \right)$$
Because the parts are independent, the single-particle partition function $Z_1$ is the product of the partition functions for each part:
$$Z_1 = Z_{CM} \cdot Z_{rel}$$
The partition function for a single particle is found by integrating over all possible positions and momenta (phase space):
$$Z_1 = \frac{1}{h^6} \int d^3\boldsymbol{R} \int d^3\boldsymbol{P} \int d^3\boldsymbol{r} \int d^3\boldsymbol{p} \ e^{-\beta H(\boldsymbol{R}, \boldsymbol{P}, \boldsymbol{r}, \boldsymbol{p})}$$
where $\beta = 1/(k_B T)$.
The center of mass behaves like a free particle of mass $M=2m$ in a box of volume $V$.
$$Z_{CM} = \frac{1}{h^3} \int_V d^3\boldsymbol{R} \int d^3\boldsymbol{P} \ e^{-\beta \frac{P^2}{2M}}$$
The position integral simply gives the volume $V$. The momentum integral is a standard Gaussian integral:
$$\int_{-\infty}^{\infty} e^{-a x^2} dx = \sqrt{\frac{\pi}{a}}$$
So, for the three momentum components ($P_x, P_y, P_z$):
$$\int d^3\boldsymbol{P} \ e^{-\beta \frac{P^2}{2M}} = \left( \int_{-\infty}^{\infty} e^{-\frac{\beta}{2M} P_x^2} dP_x \right)^3 = \left( \sqrt{\frac{2M\pi}{\beta}} \right)^3$$
Combining these results:
$$Z_{CM} = \frac{V}{h^3} (2\pi M k_B T)^{3/2}$$
Relative Motion Partition Function ($Z_{rel}$)
The relative motion is described by a 3D simple harmonic oscillator with reduced mass $\mu=m/2$.
$$Z_{rel} = \frac{1}{h^3} \int d^3\boldsymbol{r} \int d^3\boldsymbol{p} \ e^{-\beta \left( \frac{p^2}{2\mu} + \frac{1}{2}Kr^2 \right)}$$
This integral itself separates into kinetic and potential parts. The kinetic part is another Gaussian integral over momentum, just like the one for $Z_{CM}$:
$$\int d^3\boldsymbol{p} \ e^{-\beta \frac{p^2}{2\mu}} = \left( \sqrt{\frac{2\mu\pi}{\beta}} \right)^3$$
The potential part is a Gaussian integral over position:
$$\int d^3\boldsymbol{r} \ e^{-\beta \frac{Kr^2}{2}} = \left( \int_{-\infty}^{\infty} e^{-\frac{\beta K}{2} x^2} dx \right)^3 = \left( \sqrt{\frac{2\pi}{\beta K}} \right)^3$$
Combining these for $Z_{rel}$:
$$Z_{rel} = \frac{1}{h^3} \left( \frac{2\pi\mu}{\beta} \right)^{3/2} \left( \frac{2\pi}{\beta K} \right)^{3/2} = \left( \frac{2\pi}{h\beta} \right)^3 \sqrt{\frac{\mu^3}{K^3}} = \left( \frac{k_B T}{h\nu} \right)^3$$
where $\nu = \frac{1}{2\pi}\sqrt{K/\mu}$ is the classical vibrational frequency.
Total Partition Function ($Z_N$)
For a system of $N$ non-interacting, indistinguishable molecules, the total partition function is $Z_N = \frac{Z_1^N}{N!}$.
$$Z_N = \frac{1}{N!} \left[ \frac{V}{h^3} (2\pi M k_B T)^{3/2} \left( \frac{k_B T}{h\nu} \right)^3 \right]^N$$
Helmholtz Free Energy (F)
$$F = -k_B T \ln Z_N = -k_B T \left[ N \ln Z_1 - \ln N! \right]$$
Using Stirling's approximation ($\ln N! \approx N\ln N - N$):
$$F \approx -N k_B T \left[ \ln\left( \frac{Z_1}{N} \right) + 1 \right]$$
Substituting $Z_1$:
$$F \approx -N k_B T \left[ \ln\left( \frac{V}{N h^3} (2\pi M k_B T)^{3/2} \left( \frac{k_B T}{h\nu} \right)^3 \right) + 1 \right]$$
Let's look at the temperature dependence inside the logarithm: $T^{3/2} \cdot T^3 = T^{9/2}$.
$$F \approx -N k_B T \left[ \ln(\text{const}) + \frac{9}{2} \ln(T) \right]$$
Internal Energy (U)
The internal energy can be found from the partition function using the relation $U = -\frac{\partial}{\partial \beta} \ln Z_N$.
$$\ln Z_N = N \ln Z_1 - \ln N! = N \left[ \ln V + \frac{3}{2}\ln(2\pi M) - \frac{3}{2}\ln\beta + 3\ln(k_B) + 3\ln T + \dots \right]$$
Since $\beta = 1/(k_B T)$, we have $\ln T = -\ln(k_B\beta)$.
$$\ln Z_N = N \left[ \ln(\dots) - \frac{3}{2}\ln\beta - 3\ln\beta \right] = N \left[ \ln(\dots) - \frac{9}{2}\ln\beta \right]$$
Now, taking the derivative:
$$U = -\frac{\partial}{\partial \beta} \left[ N \ln(\dots) - \frac{9N}{2}\ln\beta \right] = - \left(-\frac{9N}{2\beta}\right) = \frac{9N}{2\beta}$$
$$U = \frac{9}{2} N k_B T$$
Mean Squared Separation $\langle r_{12}^2 \rangle$
We want to find the thermal average $\langle r_{12}^2 \rangle = \langle r^2 \rangle$. This value is related to the potential energy part of the Hamiltonian, $U_{pot} = \frac{1}{2}Kr^2$.
The average potential energy can be calculated from the relevant part of the partition function, which is the position integral in $Z_{rel}$:
$$Z_{pot} = \int d^3\boldsymbol{r} \ e^{-\beta \frac{Kr^2}{2}} = \left( \frac{2\pi}{\beta K} \right)^{3/2}$$
The average potential energy is:
$$\langle U_{pot} \rangle = -\frac{\partial}{\partial \beta} \ln Z_{pot} = -\frac{\partial}{\partial \beta} \left[ \frac{3}{2} (\ln(2\pi) - \ln\beta - \ln K) \right]$$
$$\langle U_{pot} \rangle = - \frac{3}{2} \left( -\frac{1}{\beta} \right) = \frac{3}{2\beta} = \frac{3}{2} k_B T$$
We also know from the definition of the average that $\langle U_{pot} \rangle = \left\langle \frac{1}{2}Kr^2 \right\rangle = \frac{1}{2}K\langle r^2 \rangle$. Equating the two expressions gives:
$$\frac{1}{2}K\langle r^2 \rangle = \frac{3}{2} k_B T$$
Solving for $\langle r^2 \rangle$:
$$\langle r_{12}^2 \rangle = \frac{3 k_B T}{K}$$
This confirms the result from the equipartition theorem, showing that the mean squared separation is directly proportional to temperature.
4
The grand partition function for a system of non-interacting identical particles is given by:
$$\mathcal{Z} = e^{z Z_1}$$
where $z = e^{\beta \mu}$ is the fugacity and $Z_1$ is the single-particle partition function.
The energy of a single atom is the sum of its kinetic and magnetic potential energies, which are independent: $\epsilon = \epsilon_{kin} + \epsilon_{mag}$. Therefore, the single-particle partition function $Z_1$ is a product of the translational and magnetic partition functions:
$$Z_1 = Z_{trans} \cdot Z_{mag}$$
For a particle of mass $m$ in a volume $V$, the translational partition function is:
$$Z_{trans} = \frac{V}{\lambda_{th}^3}$$
where $\lambda_{th} = \frac{h}{\sqrt{2\pi m k_B T}}$ is the thermal de Broglie wavelength.
The problem states the atom has two magnetic energy states: $-\mu_B H$ and $+\mu_B H$. The partition function is the sum over the Boltzmann factors for these states:
$$Z_{mag} = \sum_{i} e^{-\beta \epsilon_{mag,i}} = e^{-\beta(-\mu_B H)} + e^{-\beta(+\mu_B H)}$$
$$Z_{mag} = e^{\beta \mu_B H} + e^{-\beta \mu_B H} = 2 \cosh(\beta \mu_B H)$$
The full single-particle partition function is:
$$Z_1 = \frac{V}{\lambda_{th}^3} \left[ 2 \cosh(\beta \mu_B H) \right]$$
Finally, the grand partition function is:
$$\mathcal{Z}(T, V, \mu, H) = \exp\left(z \frac{V}{\lambda_{th}^3} \left[ 2 \cosh(\beta \mu_B H) \right] \right)$$
Magnetization is found from the grand potential, $\Omega = -k_B T \ln \mathcal{Z}$, using the thermodynamic relation $M = -\left(\frac{\partial \Omega}{\partial H}\right)_{T, V, \mu}$.
First, let's find the grand potential:
$$\Omega = -k_B T \ln(\mathcal{Z}) = -k_B T \left( z \frac{V}{\lambda_{th}^3} \left[ 2 \cosh(\beta \mu_B H) \right] \right)$$
The average number of particles in the system is $\langle N \rangle = -\left(\frac{\partial \Omega}{\partial \mu}\right)_{T, V, H} = z \frac{\partial (-\Omega)}{\partial z} = z Z_1$. So we can write:
$$\Omega = -k_B T \langle N \rangle$$
Note that $\langle N \rangle$ itself depends on $H$: $\langle N \rangle(H) = z \frac{V}{\lambda_{th}^3} 2 \cosh(\beta \mu_B H)$.
Now, we differentiate $\Omega$ with respect to $H$:
$$M = -\frac{\partial}{\partial H} \left[ -k_B T z \frac{V}{\lambda_{th}^3} 2 \cosh(\beta \mu_B H) \right]$$
$$M = k_B T z \frac{V}{\lambda_{th}^3} \cdot 2 \frac{\partial}{\partial H} \left[ \cosh(\beta \mu_B H) \right]$$
$$M = k_B T z \frac{V}{\lambda_{th}^3} \cdot 2 \left[ \sinh(\beta \mu_B H) \cdot (\beta \mu_B) \right]$$
Substituting $\beta = 1/(k_B T)$:
$$M = z \frac{V}{\lambda_{th}^3} \cdot 2 \mu_B \sinh(\beta \mu_B H)$$
We can express this in terms of the average particle number $\langle N \rangle$:
$$M = \left( z \frac{V}{\lambda_{th}^3} 2 \cosh(\beta \mu_B H) \right) \mu_B \frac{\sinh(\beta \mu_B H)}{\cosh(\beta \mu_B H)}$$
$$M = \langle N \rangle \mu_B \tanh(\beta \mu_B H)$$
The process occurs at constant temperature and volume. The first law of thermodynamics including magnetic work is $dU = dQ + dW$, where the work done on the system is $dW = -P dV + M dH$. Since volume is constant ($dV=0$), we have $dU = dQ + M dH$.
The heat absorbed by the system is $dQ = dU - M dH$. The total heat given off by the system when the field is reduced from $H$ to 0 is:
$$Q_{out} = - \int_{H \to 0} dQ = - \int_{H}^{0} (dU - M dH') = \int_{0}^{H} (dU - M dH')$$
Since the process is isothermal, $dU = \left(\frac{\partial U}{\partial H'}\right)_{T,V,\mu} dH'$.
$$Q_{out} = \int_{0}^{H} \left(\frac{\partial U}{\partial H'}\right)_{T,V,\mu} dH' - \int_{0}^{H} M dH' = U(H) - U(0) + \int_0^H M dH'$$
This expression is incorrect. The heat given off is $Q_{out} = -\Delta Q = - ( \Delta U - W)$, where $W = \int M dH$ is the work done by the system.
$$Q_{out} = W - \Delta U = \int_H^0 M dH' - (U(0) - U(H)) = U(H) - U(0) - \int_H^0 M dH'$$
$$Q_{out} = U(H) - U(0) + \int_0^H M(H') dH'$$
Let's find the total energy $U$. The average energy per particle is $\langle \epsilon \rangle = \langle \epsilon_{trans} \rangle + \langle \epsilon_{mag} \rangle$.
$\langle \epsilon_{trans} \rangle = \frac{3}{2} k_B T$
$\langle \epsilon_{mag} \rangle = -\frac{\partial}{\partial \beta} \ln(Z_{mag}) = -\frac{\partial}{\partial \beta} \ln(2\cosh(\beta\mu_B H)) = -\mu_B H \tanh(\beta \mu_B H)$
So the total energy is $U(H) = \langle N \rangle_H \left[ \frac{3}{2} k_B T - \mu_B H \tanh(\beta \mu_B H) \right]$.
When $H=0$, $U(0) = \langle N \rangle_0 \left[ \frac{3}{2} k_B T \right]$.
Let's define $\langle N \rangle_0 = \langle N \rangle(H=0) = z \frac{V}{\lambda_{th}^3} \cdot 2$. Then $\langle N \rangle_H = \langle N \rangle_0 \cosh(\beta \mu_B H)$.
Now we compute the integral part:
$$\int_0^H M(H') dH' = \int_0^H \langle N \rangle_{H'} \mu_B \tanh(\beta \mu_B H') dH'$$
$$= \int_0^H \langle N \rangle_0 \cosh(\beta \mu_B H') \mu_B \tanh(\beta \mu_B H') dH'$$
$$= \langle N \rangle_0 \mu_B \int_0^H \sinh(\beta \mu_B H') dH'$$
$$= \langle N \rangle_0 \mu_B \left[ \frac{\cosh(\beta \mu_B H')}{\beta \mu_B} \right]_0^H = \langle N \rangle_0 k_B T \left( \cosh(\beta \mu_B H) - 1 \right)$$
Now we assemble $Q_{out}$:
$$Q_{out} = \overbrace{\langle N \rangle_0 \cosh(\beta \mu_B H)\left[\frac{3}{2}k_B T - \mu_B H \tanh(\beta \mu_B H)\right]}^{U(H)} - \overbrace{\langle N \rangle_0 \frac{3}{2}k_B T}^{U(0)} + \overbrace{\langle N \rangle_0 k_B T(\cosh(\beta \mu_B H)-1)}^{\int M dH'}$$
$$Q_{out} = \langle N \rangle_0 \left[ \frac{3}{2}k_B T \cosh(\beta\mu_B H) - \mu_B H \sinh(\beta\mu_B H) - \frac{3}{2}k_B T + k_B T \cosh(\beta\mu_B H) - k_B T \right]$$
Grouping terms:
$$Q_{out} = \langle N \rangle_0 \left[ \left(\frac{3}{2}k_B T + k_B T\right)(\cosh(\beta\mu_B H)-1) - \mu_B H \sinh(\beta\mu_B H) \right]$$
$$Q_{out} = \langle N \rangle_0 \left[ \frac{5}{2}k_B T (\cosh(\beta\mu_B H)-1) - \mu_B H \sinh(\beta\mu_B H) \right]$$
This is the total heat given off by the system. $\langle N \rangle_0$ is the average number of atoms that would be in the system at zero magnetic field, which can be treated as a constant for the process.
5
The fundamental principle we'll use is that at equilibrium, the chemical potential of the molecules in the gaseous phase ($\mu_g$) must be equal to the chemical potential of the molecules in the adsorbed phase ($\mu_a$).
We can model the gas as a single-component ideal gas. The chemical potential for an ideal gas is given by:
$$\mu_g = k_B T \ln \left( \frac{n_g \lambda^3}{Z_{int}} \right)$$
where:
$k_B$ is the Boltzmann constant.
$T$ is the temperature.
$n_g = P_g / (k_B T)$ is the number density of the gas.
$P_g$ is the pressure of the gas.
$\lambda = h / \sqrt{2\pi m k_B T}$ is the thermal de Broglie wavelength.
$Z_{int}$ is the internal partition function of a gas molecule (for rotations, vibrations, etc.), which depends on temperature.
Substituting the expression for $n_g$, we get:
$$\mu_g = k_B T \ln \left( \frac{P_g \lambda^3}{k_B T Z_{int}(T)} \right)$$
For the adsorbed phase, we use the Langmuir model, which assumes:
- There is a fixed number of identical adsorption sites, $N_{sites}$.
- Each site can be occupied by at most one molecule.
- The adsorbed molecules do not interact with each other.
Let $N_a$ be the number of adsorbed molecules. The equilibrium fraction of occupied sites is $\theta = N_a / N_{sites}$.
The canonical partition function, $Z_a$, for the $N_a$ adsorbed molecules is the product of the combinatorial ways to arrange the molecules on the sites and the partition function of a single molecule raised to the power of $N_a$.
$$Z_a = \binom{N_{sites}}{N_a} [q(T)]^{N_a}$$
where:
$\binom{N_{sites}}{N_a} = \frac{N_{sites}!}{N_a!(N_{sites}-N_a)!}$ is the number of ways to arrange $N_a$ molecules on $N_{sites}$ sites.
$q(T)$ is the partition function for a single adsorbed molecule. This includes a term for the binding energy to the site ($-\epsilon_0$) and vibrational modes, so $q(T) = e^{\epsilon_0 / k_B T} Z_{vib}(T)$.
The Helmholtz free energy is $F_a = -k_B T \ln Z_a$. Using Stirling's approximation ($\ln N! \approx N \ln N - N$), we can find $\ln Z_a$:
$$\ln Z_a \approx N_{sites} \ln N_{sites} - N_a \ln N_a - (N_{sites}-N_a)\ln(N_{sites}-N_a) + N_a \ln q(T)$$
The chemical potential is the derivative of the free energy with respect to the number of adsorbed particles, $\mu_a = (\frac{\partial F_a}{\partial N_a})_{T, N_{sites}}$:
$$\mu_a = -k_B T \frac{\partial}{\partial N_a} \ln Z_a$$
$$\mu_a = -k_B T [-\ln N_a - 1 + \ln(N_{sites}-N_a) + 1 + \ln q(T)]$$
$$\mu_a = k_B T \ln \left( \frac{N_a}{N_{sites}-N_a} \right) - k_B T \ln q(T)$$
By dividing the numerator and denominator by $N_{sites}$ and using $\theta = N_a / N_{sites}$, we get:
$$\mu_a = k_B T \ln \left( \frac{\theta}{1-\theta} \right) - k_B T \ln q(T)$$
Now we set the chemical potentials equal: $\mu_g = \mu_a$.
$$k_B T \ln \left( \frac{P_g \lambda^3}{k_B T Z_{int}(T)} \right) = k_B T \ln \left( \frac{\theta}{1-\theta} \right) - k_B T \ln q(T)$$
We can cancel $k_B T$ from all terms and combine the logarithms on the right side:
$$\ln \left( \frac{P_g \lambda^3}{k_B T Z_{int}(T)} \right) = \ln \left( \frac{\theta}{(1-\theta)q(T)} \right)$$
Exponentiating both sides gives:
$$\frac{P_g \lambda^3}{k_B T Z_{int}(T)} = \frac{\theta}{(1-\theta)q(T)}$$
Finally, we solve for the gas pressure, $P_g$:
$$P_g = \frac{\theta}{1-\theta} \left[ \frac{k_B T Z_{int}(T)}{\lambda^3 q(T)} \right]$$
$\lambda^3 \propto T^{-3/2}$ is a function of temperature.
Therefore, we can define this entire term as "a certain function of temperature," call it $K(T)$.
$$P_g = \frac{\theta}{1-\theta} \times K(T)$$
Problem 1
Density Matrix: The density matrix is: $$\rho = \frac{1}{Z} e^{-\beta H}$$
For a non-interacting system, $H = \sum_{i=1}^N h_i$ where $h_i$ is the single-particle Hamiltonian.
The matrix element is: $$\langle x'_1, x'_2, \ldots, x'_N | \rho | x_1, x_2, \ldots, x_N \rangle = \frac{1}{Z} \langle x'_1, x'_2, \ldots, x'_N | e^{-\beta \sum_i h_i} | x_1, x_2, \ldots, x_N \rangle$$
Since the states are unsymmetrized products:
$$= \frac{1}{Z} \prod_{i=1}^N \langle x'i | e^{-\beta h_i} | x_i \rangle = \frac{1}{Z} \prod{i=1}^N \rho_1(x'_i, x_i)$$
where $\rho_1(x', x) = \langle x' | e^{-\beta h} | x \rangle$ is the single-particle density matrix.
Partition Function: $$Z = \text{Tr}(\rho) = \int dx_1 dx_2 \cdots dx_N \langle x_1, x_2, \ldots, x_N | e^{-\beta H} | x_1, x_2, \ldots, x_N \rangle$$
$$= \int dx_1 dx_2 \cdots dx_N \prod_{i=1}^N \langle x_i | e^{-\beta h_i} | x_i \rangle$$
$$= \prod_{i=1}^N \left(\int dx_i \langle x_i | e^{-\beta h} | x_i \rangle\right) = (Z_1)^N$$
where $Z_1 = \int dx \langle x | e^{-\beta h} | x \rangle$ is the single-particle partition function.
Problem 2
The canonical ensemble average of any operator $A$ is: $$\langle A \rangle = \frac{1}{Z} \text{Tr}(A e^{-\beta H})$$
where $Z = \text{Tr}(e^{-\beta H})$ and $\beta = 1/(k_B T)$.
Average energy $$\langle H \rangle = \frac{1}{Z} \text{Tr}(H e^{-\beta H}) = -\frac{\partial \ln Z}{\partial \beta}$$
Second moment $$\langle H^2 \rangle = \frac{1}{Z} \text{Tr}(H^2 e^{-\beta H})$$
Note that: $$\frac{\partial}{\partial \beta}\left(\frac{1}{Z} \text{Tr}(H e^{-\beta H})\right) = -\frac{1}{Z^2}\frac{\partial Z}{\partial \beta}\text{Tr}(H e^{-\beta H}) - \frac{1}{Z}\text{Tr}(H^2 e^{-\beta H})$$
$$\frac{\partial \langle H \rangle}{\partial \beta} = -\frac{1}{Z^2}\frac{\partial Z}{\partial \beta}\text{Tr}(H e^{-\beta H}) - \langle H^2 \rangle$$
$$= -\langle H \rangle \frac{1}{Z}\frac{\partial Z}{\partial \beta} - \langle H^2 \rangle = \langle H \rangle^2 - \langle H^2 \rangle$$
Therefore: $$\langle H^2 \rangle - \langle H \rangle^2 = -\frac{\partial \langle H \rangle}{\partial \beta}$$
Relate to heat capacity
The heat capacity at constant volume is: $$C_v = \frac{\partial \langle H \rangle}{\partial T}$$
Using $\beta = 1/(k_B T)$, we have $\frac{\partial}{\partial T} = -\frac{\beta^2}{k_B}\frac{\partial}{\partial \beta}$:
$$C_v = -\frac{\beta^2}{k_B}\frac{\partial \langle H \rangle}{\partial \beta} = \frac{\beta^2}{k_B}(\langle H^2 \rangle - \langle H \rangle^2)$$
Therefore: $$\boxed{\langle H^2 \rangle - \langle H \rangle^2 = k_B T^2 C_v}$$
Problem 3
(a) Derive the average occupation number
Each single-particle state k can have occupation number $n_k \in {0, 1, 2, \ldots, \ell}$.
The grand canonical partition function for state k is: $$\Xi_k = \sum_{n_k=0}^{\ell} e^{-\beta(E_{n_k} - \mu n_k)} = \sum_{n_k=0}^{\ell} (z e^{-\beta \epsilon(k)})^{n_k}$$
where $z = e^{\beta \mu}$ is the fugacity, and the energy of $n_k$ particles in state k is $E_{n_k} = n_k \epsilon(k)$.
This is a geometric series: $$\Xi_k = \sum_{n_k=0}^{\ell} (z e^{-\beta \epsilon(k)})^{n_k} = \frac{1 - (z e^{-\beta \epsilon(k)})^{\ell+1}}{1 - z e^{-\beta \epsilon(k)}}$$
The average occupation number is: $$\langle n_k \rangle = \frac{1}{\Xi_k} \sum_{n_k=0}^{\ell} n_k (z e^{-\beta \epsilon(k)})^{n_k} = z \frac{\partial \ln \Xi_k}{\partial z}$$
$$\ln \Xi_k = \ln[1 - (z e^{-\beta \epsilon(k)})^{\ell+1}] - \ln[1 - z e^{-\beta \epsilon(k)}]$$
$$\frac{\partial \ln \Xi_k}{\partial z} = -\frac{(\ell+1)(z e^{-\beta \epsilon(k)})^{\ell} e^{-\beta \epsilon(k)}}{1 - (z e^{-\beta \epsilon(k)})^{\ell+1}} + \frac{e^{-\beta \epsilon(k)}}{1 - z e^{-\beta \epsilon(k)}}$$
Therefore: $$\langle n_k \rangle = \frac{z e^{-\beta \epsilon(k)}}{1 - z e^{-\beta \epsilon(k)}} - \frac{(\ell+1)(z e^{-\beta \epsilon(k)})^{\ell+1}}{1 - (z e^{-\beta \epsilon(k)})^{\ell+1}}$$
$$= \boxed{\frac{1}{z^{-1}e^{\beta\epsilon(k)} - 1} - \frac{\ell + 1}{(z^{-1}e^{\beta\epsilon(k)})^{\ell+1} - 1}}$$
(b) Check limiting cases
Case ℓ = 1 (Fermi-Dirac): $$\langle n_k \rangle = \frac{1}{z^{-1}e^{\beta\epsilon(k)} - 1} - \frac{2}{(z^{-1}e^{\beta\epsilon(k)})^{2} - 1}$$
$$= \frac{1}{z^{-1}e^{\beta\epsilon(k)} - 1} - \frac{2}{(z^{-1}e^{\beta\epsilon(k)} - 1)(z^{-1}e^{\beta\epsilon(k)} + 1)}$$
$$= \frac{z^{-1}e^{\beta\epsilon(k)} + 1 - 2}{(z^{-1}e^{\beta\epsilon(k)} - 1)(z^{-1}e^{\beta\epsilon(k)} + 1)} = \frac{z^{-1}e^{\beta\epsilon(k)} - 1}{(z^{-1}e^{\beta\epsilon(k)} - 1)(z^{-1}e^{\beta\epsilon(k)} + 1)}$$
$$= \frac{1}{z^{-1}e^{\beta\epsilon(k)} + 1} = \frac{1}{e^{\beta(\epsilon(k) - \mu)} + 1}$$
This is the Fermi-Dirac distribution.
Case ℓ → ∞ (Bose-Einstein): As $\ell \to \infty$, if $|z e^{-\beta \epsilon(k)}| < 1$ (which must hold for convergence): $$(z^{-1}e^{\beta\epsilon(k)})^{\ell+1} \to \infty$$
So the second term vanishes: $$\langle n_k \rangle \to \frac{1}{z^{-1}e^{\beta\epsilon(k)} - 1} = \frac{1}{e^{\beta(\epsilon(k) - \mu)} - 1}$$
This is the Bose-Einstein distribution.
Problem 4
Part 1: Derive the partition function
For N indistinguishable particles, the quantum partition function is: $$Q_N(V,T) = \frac{1}{N!} \sum_{\text{states}} e^{-\beta E_n}$$
In the classical limit (high temperature), we use the correspondence between quantum states and classical phase space: $$Q_N(V,T) = \frac{1}{N! h^{3N}} \int e^{-\beta H} d^{3N}p , d^{3N}r$$
For an ideal gas with weak interactions, $H = K + U$ where:
- $K = \sum_{i=1}^N \frac{p_i^2}{2m}$ (kinetic energy)
- $U = \sum_{i<j} v(r_{ij})$ (interaction potential)
The momentum integrals separate: $$\int e^{-\beta \sum_i p_i^2/2m} d^{3N}p = \left(\int e^{-\beta p^2/2m} d^3p\right)^N = \left(2\pi m k_B T\right)^{3N/2}$$
Using $h = 2\pi\hbar$ and defining the thermal wavelength $\lambda = h/\sqrt{2\pi m k_B T}$: $$\left(2\pi m k_B T\right)^{3N/2} = \frac{(2\pi\hbar)^{3N}}{2^{3N}\lambda^{3N}} = \frac{h^{3N}}{2^{3N}}$$
Therefore: $$Q_N(V,T) = \frac{1}{N! h^{3N}} \cdot \frac{h^{3N}}{2^{3N}} \int e^{-\beta U} d^{3N}r$$
$$\boxed{Q_N(V,T) = \frac{1}{N! 2^{3N}} Z_N(V,T)}$$
where $Z_N(V,T) = \int \exp\left\{-\beta \sum_{i<j} v(r_{ij})\right\} d^{3N}r$
Part 2: First-order correction to equation of state
For weak interactions, expand: $$e^{-\beta \sum_{i<j} v(r_{ij})} \approx 1 - \beta \sum_{i<j} v(r_{ij}) + O(\beta^2)$$
The configurational integral becomes: $$Z_N \approx \int \left[1 - \beta \sum_{i<j} v(r_{ij})\right] d^{3N}r$$
$$= V^N - \beta \sum_{i<j} \int v(r_{ij}) d^{3N}r$$
For each pair $(i,j)$, integrating over all other coordinates gives $V^{N-2}$: $$\int v(r_{ij}) d^{3N}r = V^{N-2} \int v(r) d^3r$$
The number of pairs is $\binom{N}{2} = \frac{N(N-1)}{2}$:
$$Z_N \approx V^N \left[1 - \beta \frac{N(N-1)}{2V^2} \int v(r) d^3r\right]$$
For large N: $$Z_N \approx V^N \left[1 - \beta \frac{N^2}{2V^2} \int v(r) d^3r\right]$$
Define the second virial coefficient: $$B_2(T) = -\frac{1}{2}\int \left(e^{-\beta v(r)} - 1\right) d^3r \approx -\frac{\beta}{2}\int v(r) d^3r$$
Then: $$\ln Z_N \approx N \ln V - \beta \frac{N^2}{V} B_2(T)$$
The Helmholtz free energy is: $$F = -k_B T \ln Q_N = -k_B T \left[\ln Z_N - \ln(N!) - 3N \ln 2\right]$$
Using Stirling's approximation $\ln N! \approx N \ln N - N$: $$F \approx -Nk_B T \ln\left(\frac{eV}{N \cdot 2^3}\right) + Nk_B T \frac{N}{V} B_2(T)$$
The pressure is: $$P = -\frac{\partial F}{\partial V}\bigg|_{T,N} = \frac{Nk_B T}{V} - \frac{N^2 k_B T}{V^2} B_2(T)$$
$$\boxed{PV = Nk_B T\left(1 + \frac{N}{V}B_2(T) + \ldots\right)}$$
This is the virial expansion to first order, where $B_2(T)$ is the second virial coefficient.
Problem 5
Let ${|E_m\rangle}$ be the complete orthonormal set of energy eigenstates with $H|E_m\rangle = E_m|E_m\rangle$.
The exact partition function is: $$Q(\beta) = \text{Tr}(e^{-\beta H}) = \sum_m e^{-\beta E_m}$$
Expand trial states in energy eigenbasis
Any trial state $|\phi_n\rangle$ can be expanded as: $$|\phi_n\rangle = \sum_m c_{nm} |E_m\rangle$$
where $\sum_m |c_{nm}|^2 = 1$ (normalization).
Calculate expectation value $$\langle \phi_n | H | \phi_n \rangle = \sum_m |c_{nm}|^2 E_m$$
Apply convexity of exponential
Since $e^{-\beta x}$ is a convex function for $\beta > 0$, by Jensen's inequality: $$e^{-\beta \sum_m |c_{nm}|^2 E_m} \leq \sum_m |c_{nm}|^2 e^{-\beta E_m}$$
Therefore: $$e^{-\beta \langle \phi_n | H | \phi_n \rangle} \leq \sum_m |c_{nm}|^2 e^{-\beta E_m}$$
Sum over all trial states
$$\sum_n e^{-\beta \langle \phi_n | H | \phi_n \rangle} \leq \sum_n \sum_m |c_{nm}|^2 e^{-\beta E_m} = \sum_m \left(\sum_n |c_{nm}|^2\right) e^{-\beta E_m}$$
Since ${\phi_n}$ is orthonormal (but not necessarily complete): $$\sum_n |c_{nm}|^2 = \sum_n |\langle E_m | \phi_n \rangle|^2 \leq 1$$
with equality when ${\phi_n}$ is complete (by completeness relation).
Therefore: $$\sum_n e^{-\beta \langle \phi_n | H | \phi_n \rangle} \leq \sum_m e^{-\beta E_m} = Q(\beta)$$
$$\boxed{Q(\beta) \geq \sum_n \exp{-\beta \langle \phi_n | H | \phi_n \rangle}}$$
Equality condition:
Equality holds if and only if:
- $\sum_n |c_{nm}|^2 = 1$ for all $m$ (completeness)
- $e^{-\beta \langle \phi_n | H | \phi_n \rangle} = \sum_m |c_{nm}|^2 e^{-\beta E_m}$ with equality in Jensen's inequality
Condition 2 requires $|\phi_n\rangle$ to be eigenstates (so all weight is on one eigenvalue). Combined with condition 1, we need ${\phi_n}$ to be a complete orthonormal set of eigenfunctions of $H$.
Physical significance: This inequality provides a variational bound for the partition function, useful for approximation methods like the variational principle in quantum mechanics.
Problem 1
For an ideal quantum gas, we work in the Grand Canonical Ensemble. The partition function $\mathcal{Z}$ for non-interacting particles is the product of the partition functions for each single-particle energy state $k$ with energy $\epsilon_k$.
$$\mathcal{Z} = \prod_k \mathcal{Z}_k$$
The grand potential $\Omega$ (which is equal to $-PV$) is given by:
$$PV = k_B T \ln \mathcal{Z} = k_B T \sum_k \ln \mathcal{Z}_k$$
For Fermions (exclusion principle, $n_k \in \{0, 1\}$) and Bosons ($n_k \in \{0, 1, 2, \dots\}$), the sum over states yields:
$$\ln \mathcal{Z} = \pm \sum_k \ln \left( 1 \pm z e^{-\beta \epsilon_k} \right)$$
Thus, the pressure $P$ is:
$$\frac{P}{k_B T} = \pm \frac{1}{V} \sum_k \ln \left( 1 \pm z e^{-\beta \epsilon_k} \right) \quad \text{--- (Eq. 1)}$$
The total number of particles $N$ is determined by $N = z \frac{\partial}{\partial z} (\ln \mathcal{Z})$:
$$N = \sum_k \frac{1}{z^{-1}e^{\beta \epsilon_k} \pm 1} \quad \text{--- (Eq. 2)}$$
For a large volume $V$, the energy levels are closely spaced. We convert the sum over states $\sum_k$ into an integral over energy $\epsilon$ using the density of states $g(\epsilon)$.
For a free particle in 3 dimensions ($\epsilon = p^2/2m$), the density of states (including spin factor $g_s$, assumed to be 1 here) is:
$$g(\epsilon) d\epsilon = \frac{V}{h^3} 4\pi p^2 dp = \frac{2\pi V (2m)^{3/2}}{h^3} \epsilon^{1/2} d\epsilon$$
Let's define a constant $C$ to simplify notation:
$$C = \frac{2\pi (2m)^{3/2}}{h^3}$$
Substituting the integral form into (Eq. 2):
$$N = \int_0^\infty \frac{C V \epsilon^{1/2}}{z^{-1}e^{\beta \epsilon} \pm 1} d\epsilon$$
Divide by $V$ to get density $n$:
$$n = \frac{N}{V} = C \int_0^\infty \frac{\epsilon^{1/2}}{z^{-1}e^{\beta \epsilon} \pm 1} d\epsilon$$
Change variable: $x = \beta \epsilon \implies \epsilon = k_B T x, \quad d\epsilon = k_B T dx$.
$$n = C (k_B T)^{3/2} \int_0^\infty \frac{x^{1/2}}{z^{-1}e^x \pm 1} dx$$
Using the definition of the Thermal de Broglie wavelength $\lambda = \frac{h}{\sqrt{2\pi m k_B T}}$, one can show that $C (k_B T)^{3/2} = \frac{1}{\lambda^3} \cdot \frac{2}{\sqrt{\pi}}$.
The integral is related to the Polylogarithm (or Bose/Fermi integrals):
$$\int_0^\infty \frac{x^{s-1}}{z^{-1}e^x \pm 1} dx = \Gamma(s) \text{Li}_s^{\pm}(z)$$
For density, $s = 3/2$, and $\Gamma(3/2) = \frac{\sqrt{\pi}}{2}$. Thus:
$$n = \frac{1}{\lambda^3} \text{Li}_{3/2}^{\pm}(z)$$
Substituting the integral form into (Eq. 1):
$$\frac{P}{k_B T} = \pm C \int_0^\infty \epsilon^{1/2} \ln \left( 1 \pm z e^{-\beta \epsilon} \right) d\epsilon$$
We perform Integration by Parts:
- $u = \ln(1 \pm z e^{-\beta \epsilon}) \implies du = \frac{\mp \beta z e^{-\beta \epsilon}}{1 \pm z e^{-\beta \epsilon}} d\epsilon = \frac{-\beta}{z^{-1}e^{\beta \epsilon} \pm 1} d\epsilon$
- $dv = \epsilon^{1/2} d\epsilon \implies v = \frac{2}{3} \epsilon^{3/2}$
$$\frac{P}{k_B T} = \pm C \left( \left[ \frac{2}{3}\epsilon^{3/2} \ln(\dots) \right]_0^\infty - \int_0^\infty \frac{2}{3}\epsilon^{3/2} \left( \frac{-\beta}{z^{-1}e^{\beta \epsilon} \pm 1} \right) d\epsilon \right)$$
The boundary term vanishes. We are left with:
$$\frac{P}{k_B T} = \frac{2}{3} C \beta \int_0^\infty \frac{\epsilon^{3/2}}{z^{-1}e^{\beta \epsilon} \pm 1} d\epsilon$$
Using the same variable change $x = \beta \epsilon$:
$$\frac{P}{k_B T} = \frac{2}{3} C (k_B T)^{3/2} \int_0^\infty \frac{x^{3/2}}{z^{-1}e^x \pm 1} dx$$
Using the integral identity with $s=5/2$ and $\Gamma(5/2) = \frac{3}{2}\frac{\sqrt{\pi}}{2}$:
$$\frac{P}{k_B T} = \frac{1}{\lambda^3} \text{Li}_{5/2}^{\pm}(z)$$
Now we have the two key equations (where $+$ is Bose, $-$ is Fermi inside the Li function definition):
1. $n \lambda^3 = \text{Li}_{3/2}^{\pm}(z) = z \pm \frac{z^2}{2^{3/2}} + \frac{z^3}{3^{3/2}} \pm \dots$
2. $\frac{P \lambda^3}{k_B T} = \text{Li}_{5/2}^{\pm}(z) = z \pm \frac{z^2}{2^{5/2}} + \frac{z^3}{3^{5/2}} \pm \dots$
We assume $z \ll 1$ (High T, Low Density).
$$n \lambda^3 \approx z \pm \frac{z^2}{2^{3/2}}$$
(Upper sign $+$ for Bose, Lower sign $-$ for Fermi).
We need to solve for $z$ in terms of $n \lambda^3$. Let $y = n \lambda^3$.
$$y = z \pm \frac{z^2}{2\sqrt{2}}$$
To first order, $z \approx y$.
Substitute $z = y$ into the small second-order term:
$$y \approx z \pm \frac{y^2}{2\sqrt{2}} \implies z \approx y \mp \frac{y^2}{2\sqrt{2}}$$
Restoring variables:
$$z \approx n\lambda^3 \mp \frac{(n\lambda^3)^2}{2^{3/2}}$$
$$\frac{P \lambda^3}{k_B T} \approx z \pm \frac{z^2}{2^{5/2}}$$
Substitute our expression for $z$:
$$\frac{P \lambda^3}{k_B T} \approx \left( n\lambda^3 \mp \frac{(n\lambda^3)^2}{2^{3/2}} \right) \pm \frac{1}{2^{5/2}} \left( n\lambda^3 \mp \dots \right)^2$$
Keep terms only up to order $(n\lambda^3)^2$:
$$\frac{P \lambda^3}{k_B T} \approx n\lambda^3 \mp \frac{(n\lambda^3)^2}{2^{3/2}} \pm \frac{(n\lambda^3)^2}{2^{5/2}}$$
Divide by $\lambda^3$:
$$\frac{P}{k_B T} \approx n \mp \frac{n^2 \lambda^3}{2^{3/2}} \pm \frac{n^2 \lambda^3}{2^{5/2}}$$
Factor out $\frac{n^2 \lambda^3}{2^{3/2}}$:
$$\frac{P}{k_B T} \approx n \mp \frac{n^2 \lambda^3}{2^{3/2}} \left( 1 - \frac{1}{2} \right) = n \mp \frac{n^2 \lambda^3}{2^{5/2}}$$
Since $2^{5/2} = 4\sqrt{2}$, we arrive at the final Equations of State:
For Ideal Bose Gas:
$$PV = N k_B T \left( 1 - \frac{1}{4\sqrt{2}} \frac{N \lambda^3}{V} \right)$$
For Ideal Fermi Gas:
$$PV = N k_B T \left( 1 + \frac{1}{4\sqrt{2}} \frac{N \lambda^3}{V} \right)$$
Problem 2
To evaluate integrals of the form $I = \int_0^\infty \frac{H(\epsilon)}{e^{\beta(\epsilon-\mu)} + 1} d\epsilon$ at low temperatures (large $\beta$), we use the Sommerfeld expansion up to the order of $T^4$:
$$I \approx \int_0^\mu H(\epsilon) d\epsilon + \frac{\pi^2}{6}(k_B T)^2 H'(\mu) + \frac{7\pi^4}{360}(k_B T)^4 H'''(\mu)$$
We will define the Fermi temperature $T_F$ and Fermi energy $\epsilon_F$ such that $\epsilon_F = k_B T_F$. The expansion parameter is the small ratio $\frac{T}{T_F}$.
The number density $n = N/V$ is fixed. Using the density of states $g(\epsilon) = C\epsilon^{1/2}$ (where $H(\epsilon) = \epsilon^{1/2}$), we set up the equation for $N$:
$$N = \int_0^\infty \frac{C\epsilon^{1/2}}{e^{\beta(\epsilon-\mu)} + 1} d\epsilon$$
Applying the expansion (using $H'(\mu) = \frac{1}{2}\mu^{-1/2}$ and $H'''(\mu) = \frac{3}{8}\mu^{-5/2}$):
$$N = C \left[ \frac{2}{3}\mu^{3/2} + \frac{\pi^2}{6}(k_B T)^2 \frac{1}{2}\mu^{-1/2} + \frac{7\pi^4}{360}(k_B T)^4 \frac{3}{8}\mu^{-5/2} \right]$$
At $T=0$, $\mu = \epsilon_F$, so $N = \frac{2}{3}C\epsilon_F^{3/2}$. Equating the two expressions for $N$ and simplifying:
$$\epsilon_F^{3/2} = \mu^{3/2} \left[ 1 + \frac{\pi^2}{8}\left(\frac{k_B T}{\mu}\right)^2 + \frac{7\pi^4}{640}\left(\frac{k_B T}{\mu}\right)^4 \right]$$
To solve for $\mu$, we assume an expansion of the form $\mu \approx \epsilon_F [1 + A(\frac{T}{T_F})^2 + B(\frac{T}{T_F})^4]$. Inverting the series carefully (retaining terms up to $T^4$) yields:
$$\mu = \epsilon_F \left[ 1 - \frac{\pi^2}{12}\left(\frac{T}{T_F}\right)^2 - \frac{\pi^4}{80}\left(\frac{T}{T_F}\right)^4 \right]$$
Since $z = e^{\beta \mu}$, $\ln z = \frac{\mu}{k_B T}$.
$$\ln z = \frac{\epsilon_F}{k_B T} \left[ 1 - \frac{\pi^2}{12}\left(\frac{T}{T_F}\right)^2 - \frac{\pi^4}{80}\left(\frac{T}{T_F}\right)^4 \right]$$
The internal energy is $E = \int_0^\infty \epsilon g(\epsilon) f(\epsilon) d\epsilon$. Here $H(\epsilon) = \epsilon^{3/2}$.
Derivatives: $H'(\mu) = \frac{3}{2}\mu^{1/2}$, $H'''(\mu) = -\frac{3}{8}\mu^{-3/2}$.
Applying the expansion:
$$E = C \left[ \frac{2}{5}\mu^{5/2} + \frac{\pi^2}{6}(k_B T)^2 \frac{3}{2}\mu^{1/2} - \frac{7\pi^4}{360}(k_B T)^4 \frac{3}{8}\mu^{-3/2} \right]$$
Factor out the zero-temperature energy $E_0 = \frac{3}{5}N\epsilon_F$. Note that $\frac{2}{5}C\mu^{5/2} = E_0 (\frac{\mu}{\epsilon_F})^{5/2}$.
$$E = E_0 \left( \frac{\mu}{\epsilon_F} \right)^{5/2} \left[ 1 + \frac{5\pi^2}{8}\left(\frac{k_B T}{\mu}\right)^2 - \frac{7\pi^4}{384}\left(\frac{k_B T}{\mu}\right)^4 \right]$$
Now, substitute the expansion for $\mu$ into this equation and keep terms up to $T^4$. This involves Taylor expanding $(\mu/\epsilon_F)^{5/2}$ and the terms in the bracket.
After algebraic simplification, the terms combine to:
$$E = \frac{3}{5} N \epsilon_F \left[ 1 + \frac{5\pi^2}{12}\left(\frac{T}{T_F}\right)^2 - \frac{\pi^4}{16}\left(\frac{T}{T_F}\right)^4 \right]$$
For a non-relativistic ideal quantum gas, the relationship $PV = \frac{2}{3}E$ holds exactly.
Using the result for $E$:
$$P = \frac{2}{5} n \epsilon_F \left[ 1 + \frac{5\pi^2}{12}\left(\frac{T}{T_F}\right)^2 - \frac{\pi^4}{16}\left(\frac{T}{T_F}\right)^4 \right]$$
The heat capacity is defined as $C_V = \left(\frac{\partial E}{\partial T}\right)_V$.
Differentiating the expression for $E$ with respect to $T$:
$$C_V = \frac{3}{5} N \epsilon_F \left[ \frac{5\pi^2}{12} \cdot 2 \frac{T}{T_F^2} - \frac{\pi^4}{16} \cdot 4 \frac{T^3}{T_F^4} \right]$$
Substituting $\epsilon_F = k_B T_F$:
$$C_V = \frac{\pi^2}{2} N k_B \left(\frac{T}{T_F}\right) \left[ 1 - \frac{3\pi^2}{10}\left(\frac{T}{T_F}\right)^2 \right]$$
The most efficient way to find $F$ is using the thermodynamic relation:
$$F = E - TS$$
However, using the Euler relation (or Gibbs-Duhem) $G = \mu N$ and $G = F + PV$, we have:
$$F = \mu N - PV$$
Substitute the expressions for $\mu$ and $P$:
1. $\mu N = N \epsilon_F \left[ 1 - \frac{\pi^2}{12}(\frac{T}{T_F})^2 - \frac{\pi^4}{80}(\frac{T}{T_F})^4 \right]$
2. $PV = \frac{2}{3}E = \frac{2}{5} N \epsilon_F \left[ 1 + \frac{5\pi^2}{12}(\frac{T}{T_F})^2 - \frac{\pi^4}{16}(\frac{T}{T_F})^4 \right]$
Subtracting ($F = \mu N - \frac{2}{3}E$):
$$F = \frac{3}{5} N \epsilon_F \left[ 1 - \frac{5\pi^2}{12}\left(\frac{T}{T_F}\right)^2 + \frac{\pi^4}{48}\left(\frac{T}{T_F}\right)^4 \right]$$
$\mu \approx \epsilon_F \left[ 1 - \frac{\pi^2}{12}\left(\frac{T}{T_F}\right)^2 - \frac{\pi^4}{80}\left(\frac{T}{T_F}\right)^4 \right]$
$E \approx \frac{3}{5} N \epsilon_F \left[ 1 + \frac{5\pi^2}{12}\left(\frac{T}{T_F}\right)^2 - \frac{\pi^4}{16}\left(\frac{T}{T_F}\right)^4 \right]$
$P \approx \frac{2}{5} n \epsilon_F \left[ 1 + \frac{5\pi^2}{12}\left(\frac{T}{T_F}\right)^2 - \frac{\pi^4}{16}\left(\frac{T}{T_F}\right)^4 \right]$
$C_V \approx \frac{\pi^2}{2} N k_B \left(\frac{T}{T_F}\right) \left[ 1 - \frac{3\pi^2}{10}\left(\frac{T}{T_F}\right)^2 \right]$
$F \approx \frac{3}{5} N \epsilon_F \left[ 1 - \frac{5\pi^2}{12}\left(\frac{T}{T_F}\right)^2 + \frac{\pi^4}{48}\left(\frac{T}{T_F}\right)^4 \right]$
Problem 3
(a)
The definition of isothermal compressibility is:
$$\kappa_T = -\frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_T$$
For a fixed number of particles $N$, the specific volume $v = V/N = 1/n$. Therefore, we can rewrite the derivative in terms of particle density $n$:
$$V = \frac{N}{n} \implies dV = -\frac{N}{n^2} dn$$
$$\kappa_T = -\frac{1}{V} \frac{dV}{dP} = -\frac{n}{N} \left( -\frac{N}{n^2} \right) \left( \frac{\partial n}{\partial P} \right)_T = \frac{1}{n} \left( \frac{\partial n}{\partial P} \right)_T$$
To find $(\frac{\partial n}{\partial P})_T$, we use the standard parametric equations for an ideal Fermi gas (where $z = e^{\beta\mu}$ is the fugacity and $\lambda$ is the thermal de Broglie wavelength):
$$\frac{P}{kT} = \frac{g}{\lambda^3} f_{5/2}(z)$$
$$n = \frac{N}{V} = \frac{g}{\lambda^3} f_{3/2}(z)$$
At constant temperature $T$, $\lambda$ is constant. Thus, $P$ and $n$ only vary with $z$. We use the recurrence relation for Fermi-Dirac functions: $z \frac{d}{dz} f_\nu(z) = f_{\nu-1}(z)$.
Differentiation with respect to $z$:
$$\left( \frac{\partial P}{\partial z} \right)_T = \frac{kT g}{\lambda^3} \frac{1}{z} f_{3/2}(z)$$
$$\left( \frac{\partial n}{\partial z} \right)_T = \frac{g}{\lambda^3} \frac{1}{z} f_{1/2}(z)$$
Now we compute the ratio:
$$\left( \frac{\partial n}{\partial P} \right)_T = \frac{(\partial n / \partial z)_T}{(\partial P / \partial z)_T} = \frac{\frac{g}{\lambda^3 z} f_{1/2}(z)}{\frac{kT g}{\lambda^3 z} f_{3/2}(z)} = \frac{1}{kT} \frac{f_{1/2}(z)}{f_{3/2}(z)}$$
Substituting this back into the expression for $\kappa_T$:
$$\kappa_T = \frac{1}{n} \left[ \frac{1}{kT} \frac{f_{1/2}(z)}{f_{3/2}(z)} \right]$$
$$\mathbf{\kappa_T = \frac{1}{nkT} \frac{f_{1/2}(z)}{f_{3/2}(z)}}$$
Thermodynamics relates isothermal and adiabatic compressibility via the heat capacity ratio $\gamma = C_P / C_V$:
$$\kappa_S = \frac{\kappa_T}{\gamma}$$
From Problem 8.3, the heat capacity ratio is given by:
$$\gamma = \frac{5}{3} \frac{f_{5/2}(z) f_{1/2}(z)}{\{f_{3/2}(z)\}^2}$$
Substituting $\kappa_T$ and $\gamma$:
$$\kappa_S = \left[ \frac{1}{nkT} \frac{f_{1/2}(z)}{f_{3/2}(z)} \right] \times \left[ \frac{3 \{f_{3/2}(z)\}^2}{5 f_{5/2}(z) f_{1/2}(z)} \right]$$
Canceling terms ($f_{1/2}$ cancels out, one $f_{3/2}$ cancels out):
$$\mathbf{\kappa_S = \frac{3}{5nkT} \frac{f_{3/2}(z)}{f_{5/2}(z)}}$$
We use the Sommerfeld expansion approximations for the Fermi functions at large $z$:
$$f_\nu(z) \approx \frac{(\ln z)^\nu}{\Gamma(\nu+1)} \left[ 1 + \nu(\nu-1) \frac{\pi^2}{6} (\ln z)^{-2} \right]$$
Also, for a Fermi gas with fixed density $n$, the fugacity is related to the Fermi energy $\epsilon_F$ by:
$$kT \ln z \approx \epsilon_F \left[ 1 - \frac{\pi^2}{12} \left( \frac{kT}{\epsilon_F} \right)^2 \right]$$
Which implies:
$$\frac{1}{kT \ln z} \approx \frac{1}{\epsilon_F} \left[ 1 + \frac{\pi^2}{12} \left( \frac{kT}{\epsilon_F} \right)^2 \right]$$
For $\kappa_T$:
We need the ratio $f_{1/2}/f_{3/2}$.
$$f_{1/2} \approx \frac{(\ln z)^{1/2}}{\Gamma(3/2)} \left[ 1 - \frac{\pi^2}{24} (\ln z)^{-2} \right]$$
$$f_{3/2} \approx \frac{(\ln z)^{3/2}}{\Gamma(5/2)} \left[ 1 + \frac{\pi^2}{8} (\ln z)^{-2} \right]$$
The ratio of the Gamma functions is $\Gamma(5/2)/\Gamma(3/2) = 3/2$.
$$\frac{f_{1/2}}{f_{3/2}} \approx \frac{3}{2 \ln z} \frac{\left[ 1 - \frac{\pi^2}{24} (\ln z)^{-2} \right]}{\left[ 1 + \frac{\pi^2}{8} (\ln z)^{-2} \right]} \approx \frac{3}{2 \ln z} \left[ 1 - \left( \frac{\pi^2}{24} + \frac{\pi^2}{8} \right) (\ln z)^{-2} \right]$$
$$\approx \frac{3}{2 \ln z} \left[ 1 - \frac{\pi^2}{6} (\ln z)^{-2} \right]$$
Substitute this into $\kappa_T$ and use the expansion for $\ln z$:
$$\kappa_T = \frac{1}{nkT} \frac{3}{2 \ln z} \left[ 1 - \frac{\pi^2}{6} \left( \frac{kT}{\epsilon_F} \right)^2 \right]$$
$$\kappa_T \approx \frac{3}{2n} \underbrace{\frac{1}{kT \ln z}}_{\approx \frac{1}{\epsilon_F}[1+\frac{\pi^2}{12}(\dots)]} \left[ 1 - \frac{\pi^2}{6} \left( \frac{kT}{\epsilon_F} \right)^2 \right]$$
$$\kappa_T \approx \frac{3}{2n\epsilon_F} \left[ 1 + \left( \frac{1}{12} - \frac{2}{12} \right)\pi^2 \left( \frac{kT}{\epsilon_F} \right)^2 \right]$$
$$\mathbf{\kappa_T \simeq \frac{3}{2n\epsilon_F} \left[ 1 - \frac{\pi^2}{12} \left( \frac{kT}{\epsilon_F} \right)^2 \right]}$$
For $\kappa_S$:
We need the ratio $f_{3/2}/f_{5/2}$.
$$\frac{f_{3/2}}{f_{5/2}} \approx \frac{\Gamma(7/2)}{\Gamma(5/2) \ln z} \frac{[1 + \frac{\pi^2}{8}(\ln z)^{-2}]}{[1 + \frac{5\pi^2}{8}(\ln z)^{-2}]} \approx \frac{5}{2 \ln z} \left[ 1 - \frac{\pi^2}{2} (\ln z)^{-2} \right]$$
Substitute into $\kappa_S$:
$$\kappa_S \approx \frac{3}{5nkT} \frac{5}{2 \ln z} \left[ 1 - \frac{\pi^2}{2} \left(\frac{kT}{\epsilon_F}\right)^2 \right] = \frac{3}{2n} \frac{1}{kT \ln z} \left[ 1 - \frac{\pi^2}{2} (\dots) \right]$$
Using the $\ln z$ expansion again:
$$\kappa_S \approx \frac{3}{2n\epsilon_F} \left[ 1 + \frac{\pi^2}{12} (\dots) \right] \left[ 1 - \frac{\pi^2}{2} (\dots) \right]$$
$$\text{Correction: } \frac{1}{12} - \frac{6}{12} = -\frac{5}{12}$$
$$\mathbf{\kappa_S \simeq \frac{3}{2n\epsilon_F} \left[ 1 - \frac{5\pi^2}{12} \left( \frac{kT}{\epsilon_F} \right)^2 \right]}$$
(b)
We start with the provided thermodynamic relation:
$$C_P - C_V = T V \kappa_T \left( \frac{\partial P}{\partial T} \right)_V^2$$
First, determine $(\frac{\partial P}{\partial T})_V$. For an ideal quantum gas, the internal energy density is related to pressure by $P = \frac{2}{3} \frac{E}{V}$.
$$\left( \frac{\partial P}{\partial T} \right)_V = \frac{\partial}{\partial T} \left( \frac{2}{3} \frac{E}{V} \right)_V = \frac{2}{3V} \left( \frac{\partial E}{\partial T} \right)_V = \frac{2}{3V} C_V$$
Substitute this back into the relation:
$$C_P - C_V = T V \kappa_T \left( \frac{2 C_V}{3V} \right)^2 = T V \kappa_T \frac{4 C_V^2}{9 V^2} = \frac{4 T C_V^2}{9 V} \kappa_T$$
Now, substitute the expression for $\kappa_T = \frac{1}{nkT} \frac{f_{1/2}(z)}{f_{3/2}(z)}$ derived in part (a). Note that $V = N/n$:
$$C_P - C_V = \frac{4 T C_V^2}{9 (N/n)} \cdot \frac{1}{n k T} \frac{f_{1/2}(z)}{f_{3/2}(z)}$$
$$C_P - C_V = \frac{4 C_V^2}{9 N k} \frac{f_{1/2}(z)}{f_{3/2}(z)}$$
Dividing both sides by $C_V$:
$$\mathbf{\frac{C_P - C_V}{C_V} = \frac{4 C_V}{9 N k} \frac{f_{1/2}(z)}{f_{3/2}(z)}}$$
Low Temperature Limit:
At low temperatures ($kT \ll \epsilon_F$), the heat capacity of a Fermi gas is:
$$C_V \simeq \frac{\pi^2}{2} Nk \frac{kT}{\epsilon_F}$$
The ratio of Fermi functions (as used in part a) is:
$$\frac{f_{1/2}(z)}{f_{3/2}(z)} \simeq \frac{3}{2} \frac{kT}{\epsilon_F}$$
Substituting these into the derived equation:
$$\frac{C_P - C_V}{C_V} \simeq \frac{4}{9Nk} \left( \frac{\pi^2}{2} Nk \frac{kT}{\epsilon_F} \right) \left( \frac{3}{2} \frac{kT}{\epsilon_F} \right)$$
$$= \frac{4}{9Nk} \cdot \frac{3 \pi^2}{4} Nk \left( \frac{kT}{\epsilon_F} \right)^2$$
$$= \frac{12 \pi^2}{36} \left( \frac{kT}{\epsilon_F} \right)^2$$
$$\mathbf{ \simeq \frac{\pi^2}{3} \left( \frac{kT}{\epsilon_F} \right)^2 }$$
(c)
We check if the ratio of the compressibilities matches the adiabatic index $\gamma$ from Problem 8.3.
From part (a):
$$\kappa_T = \frac{1}{nkT} \frac{f_{1/2}}{f_{3/2}}$$
$$\kappa_S = \frac{3}{5nkT} \frac{f_{3/2}}{f_{5/2}}$$
Taking the ratio:
$$\gamma = \frac{\kappa_T}{\kappa_S} = \frac{ \frac{1}{nkT} \frac{f_{1/2}}{f_{3/2}} }{ \frac{3}{5nkT} \frac{f_{3/2}}{f_{5/2}} }$$
$$\gamma = \frac{f_{1/2}}{f_{3/2}} \cdot \frac{5 f_{5/2}}{3 f_{3/2}}$$
$$\mathbf{\gamma = \frac{5}{3} \frac{f_{5/2}(z) f_{1/2}(z)}{\{f_{3/2}(z)\}^2}}$$
This precisely matches the expression for $\gamma$ given in Problem 8.3. Thus, the results are consistent.
Problem 4
In two dimensions, the density of states $g(\varepsilon)$ is constant. For a gas in an area $A$:
$$g(\varepsilon) d\varepsilon = \frac{2\pi m A}{h^2} d\varepsilon$$
(assuming the same spin degeneracy factor $g_s$ for both, or $g_s=1$ for simplicity as it cancels out).
The particle number $N$ for ideal quantum gases in 2D is given by the integrals:
$$N_F = \int_0^\infty \frac{g(\varepsilon)}{z_F^{-1} e^{\beta \varepsilon} + 1} d\varepsilon = \frac{A}{\lambda^2} \ln(1+z_F)$$
$$N_B = \int_0^\infty \frac{g(\varepsilon)}{z_B^{-1} e^{\beta \varepsilon} - 1} d\varepsilon = -\frac{A}{\lambda^2} \ln(1-z_B)$$
where $\lambda = \frac{h}{\sqrt{2\pi m k_B T}}$ is the thermal de Broglie wavelength.
Since we are comparing the two systems at the same $N$, $T$, and $A$, we equate $N_F = N_B$:
$$\frac{A}{\lambda^2} \ln(1+z_F) = -\frac{A}{\lambda^2} \ln(1-z_B)$$
$$\ln(1+z_F) = \ln(1-z_B)^{-1}$$
$$1+z_F = \frac{1}{1-z_B}$$
Rearranging this yields the relation given in the hint:
$$(1+z_F)(1-z_B) = 1 \quad \implies \quad \mathbf{z_B = \frac{z_F}{1+z_F}}$$
The internal energy $E$ in 2D is given by $\int \varepsilon \cdot n(\varepsilon) d\varepsilon$.
$$E_F = \frac{A}{\lambda^2} k_B T f_2(z_F)$$
$$E_B = \frac{A}{\lambda^2} k_B T g_2(z_B)$$
where the functions are defined as:
$$f_2(z) = \int_0^z \frac{\ln(1+t)}{t} dt \quad \text{and} \quad g_2(z) = \int_0^z \frac{-\ln(1-t)}{t} dt$$
We need to verify the identity:
$$f_2(z_F) - g_2(z_B) = \frac{1}{2} \ln^2(1+z_F)$$
Let's verify this by differentiation. Let $y(z_F) = f_2(z_F) - g_2\left(\frac{z_F}{1+z_F}\right)$. We take the derivative with respect to $z_F$.
1. $\frac{d}{dz_F} f_2(z_F) = \frac{\ln(1+z_F)}{z_F}$
2. For the second term, let $u = \frac{z_F}{1+z_F}$. Then $\frac{du}{dz_F} = \frac{(1+z_F) - z_F}{(1+z_F)^2} = \frac{1}{(1+z_F)^2}$.
Also note that $1-u = 1 - \frac{z_F}{1+z_F} = \frac{1}{1+z_F}$.
$$\frac{d}{dz_F} g_2(u) = \left( \frac{-\ln(1-u)}{u} \right) \frac{du}{dz_F}$$
$$= \frac{-\ln(\frac{1}{1+z_F})}{\frac{z_F}{1+z_F}} \cdot \frac{1}{(1+z_F)^2} = \frac{\ln(1+z_F) (1+z_F)}{z_F} \cdot \frac{1}{(1+z_F)^2} = \frac{\ln(1+z_F)}{z_F(1+z_F)}$$
Now subtract the derivatives:
$$y'(z_F) = \frac{\ln(1+z_F)}{z_F} - \frac{\ln(1+z_F)}{z_F(1+z_F)}$$
$$y'(z_F) = \frac{\ln(1+z_F)}{z_F} \left( 1 - \frac{1}{1+z_F} \right) = \frac{\ln(1+z_F)}{z_F} \left( \frac{z_F}{1+z_F} \right)$$
$$y'(z_F) = \frac{\ln(1+z_F)}{1+z_F}$$
Integrate $y'(z_F)$ with respect to $z_F$:
$$\int \frac{\ln(1+z_F)}{1+z_F} dz_F = \frac{1}{2} \ln^2(1+z_F)$$
Since $f_2(0)=0$ and $g_2(0)=0$, the integration constant is 0. Thus, we have proven:
$$\mathbf{f_2(z_F) = g_2(z_B) + \frac{1}{2} \ln^2(1+z_F)}$$
Using the energy expressions:
$$E_F(N, T) = \frac{A}{\lambda^2} k_B T f_2(z_F)$$
$$E_B(N, T) = \frac{A}{\lambda^2} k_B T g_2(z_B)$$
Substitute the relation $f_2(z_F) = g_2(z_B) + \frac{1}{2} \ln^2(1+z_F)$ into the expression for $E_F$:
$$E_F(N, T) = \frac{A}{\lambda^2} k_B T \left[ g_2(z_B) + \frac{1}{2} \ln^2(1+z_F) \right]$$
$$E_F(N, T) = \underbrace{\frac{A}{\lambda^2} k_B T g_2(z_B)}_{E_B(N, T)} + \frac{A}{\lambda^2} k_B T \cdot \frac{1}{2} \ln^2(1+z_F)$$
Now, recall the particle number equation: $N = \frac{A}{\lambda^2} \ln(1+z_F)$.
This implies $\ln(1+z_F) = \frac{N \lambda^2}{A}$.
Substitute this into the second term:
$$\text{Difference} = \frac{A k_B T}{2 \lambda^2} \left( \frac{N \lambda^2}{A} \right)^2 = \frac{A k_B T}{2 \lambda^2} \frac{N^2 \lambda^4}{A^2} = \frac{N^2}{2A} k_B T \lambda^2$$
Recall that $\lambda^2 = \frac{h^2}{2\pi m k_B T}$. Therefore, $k_B T \lambda^2 = \frac{h^2}{2\pi m}$, which is a constant independent of temperature.
$$\text{Difference} = \frac{N^2}{2A} \left( \frac{h^2}{2\pi m} \right) = \text{Constant (independent of T)}$$
Thus:
$$\mathbf{E_F(N, T) = E_B(N, T) + E_0}$$
(Where $E_0$ is actually the ground state energy of the Fermi gas at $T=0$).
The specific heat at constant volume is defined as:
$$C_V = \left( \frac{\partial E}{\partial T} \right)_{N, V}$$
Differentiating the energy relation with respect to $T$:
$$C_{V, F} = \frac{\partial}{\partial T} (E_B(N, T) + \text{const})$$
$$C_{V, F} = \frac{\partial E_B}{\partial T}$$
$$\mathbf{C_{V, F}(N, T) = C_{V, B}(N, T)}$$
Conclusion: The specific heat of an ideal Fermi gas is identical to that of an ideal Bose gas in two dimensions for all $N$ and $T$.
Problem 5
We start with Equation (8.2.20) from the text, which gives the paramagnetic susceptibility $\chi$ in terms of the chemical potential $\mu_0$ of a fictitious spinless Fermi gas:
$$\chi = \frac{2 n \mu^{*2}}{\left. \frac{\partial \mu_0(xN)}{\partial x} \right|_{x=1/2}} \quad \dots (1)$$
Here, $x$ is a fraction of the total number of particles $N$. Let the number of particles in this fictitious system be $N' = xN$. The density of this fictitious system is $n' = N'/V = x n$.
For a spinless ideal Fermi gas (degeneracy $g=1$), the relationship between particle density and fugacity $z' = \exp(\mu_0/kT)$ is:
$$n' = \frac{1}{\lambda^3} f_{3/2}(z')$$
Substituting $n' = xn$:
$$xn = \frac{1}{\lambda^3} f_{3/2}(z') \quad \dots (2)$$
We need to evaluate the denominator of Eq. (1): $\frac{\partial \mu_0}{\partial x}$.
Since $\mu_0 = kT \ln z'$, we have $\frac{\partial \mu_0}{\partial x} = \frac{kT}{z'} \frac{\partial z'}{\partial x}$.
Differentiating Eq. (2) with respect to $x$:
$$n = \frac{1}{\lambda^3} \frac{d f_{3/2}(z')}{d z'} \frac{\partial z'}{\partial x}$$
Using the recurrence relation $z \frac{d f_\nu(z)}{dz} = f_{\nu-1}(z)$, we get $\frac{d f_{3/2}}{dz'} = \frac{1}{z'} f_{1/2}(z')$.
$$n = \frac{1}{\lambda^3} \frac{1}{z'} f_{1/2}(z') \frac{\partial z'}{\partial x}$$
Solving for $\frac{\partial z'}{\partial x}$:
$$\frac{\partial z'}{\partial x} = \frac{n \lambda^3 z'}{f_{1/2}(z')}$$
Now substitute this into the expression for $\frac{\partial \mu_0}{\partial x}$:
$$\frac{\partial \mu_0}{\partial x} = \frac{kT}{z'} \left( \frac{n \lambda^3 z'}{f_{1/2}(z')} \right) = \frac{n \lambda^3 kT}{f_{1/2}(z')}$$
We must evaluate this derivative at $x = 1/2$.
At $x = 1/2$, the density of the fictitious spinless system is $n/2$.
From Eq. (2): $n/2 = \frac{1}{\lambda^3} f_{3/2}(z')$.
Now, consider the actual spin-1/2 Fermi gas ($g=2$). Its density equation is:
$$n = \frac{2}{\lambda^3} f_{3/2}(z) \implies \frac{n}{2} = \frac{1}{\lambda^3} f_{3/2}(z)$$
Comparing the fictitious system at $x=1/2$ with the real system, we see that their fugacities are identical: $z' = z$.
So, the denominator at $x=1/2$ becomes:
$$\left. \frac{\partial \mu_0}{\partial x} \right|_{x=1/2} = \frac{n \lambda^3 kT}{f_{1/2}(z)}$$
Substitute this back into Eq. (1):
$$\chi = \frac{2 n \mu^{*2}}{\left( \frac{n \lambda^3 kT}{f_{1/2}(z)} \right)} = \frac{2 \mu^{*2} f_{1/2}(z)}{\lambda^3 kT}$$
Finally, use the relation $n = \frac{2}{\lambda^3} f_{3/2}(z)$ to substitute for $\lambda^3$.
$$\frac{1}{\lambda^3} = \frac{n}{2 f_{3/2}(z)}$$
$$\chi = \frac{2 \mu^{*2} f_{1/2}(z)}{kT} \cdot \left( \frac{n}{2 f_{3/2}(z)} \right)$$
Simplifying gives the desired result:
$$\mathbf{\chi = \frac{n \mu^{*2}}{kT} \frac{f_{1/2}(z)}{f_{3/2}(z)}}$$
At low temperatures ($T \rightarrow 0$), the fugacity $z$ is large ($z \gg 1$). We use the Sommerfeld asymptotic expansions:
$$f_{1/2}(z) \approx \frac{(\ln z)^{1/2}}{\Gamma(3/2)} \left[ 1 - \frac{\pi^2}{24} (\ln z)^{-2} \right]$$
$$f_{3/2}(z) \approx \frac{(\ln z)^{3/2}}{\Gamma(5/2)} \left[ 1 + \frac{\pi^2}{8} (\ln z)^{-2} \right]$$
The ratio is:
$$\frac{f_{1/2}(z)}{f_{3/2}(z)} \approx \frac{\Gamma(5/2)}{\Gamma(3/2) \ln z} \frac{1 - \frac{\pi^2}{24}(\ln z)^{-2}}{1 + \frac{\pi^2}{8}(\ln z)^{-2}}$$
Since $\Gamma(5/2)/\Gamma(3/2) = 3/2$:
$$\frac{f_{1/2}}{f_{3/2}} \approx \frac{3}{2 \ln z} \left[ 1 - \left( \frac{\pi^2}{24} + \frac{\pi^2}{8} \right) (\ln z)^{-2} \right] = \frac{3}{2 \ln z} \left[ 1 - \frac{\pi^2}{6} (\ln z)^{-2} \right]$$
Approximating $\ln z \approx \varepsilon_F / kT$ for the correction term, and using the standard expansion $\frac{1}{\ln z} \approx \frac{kT}{\varepsilon_F} [1 + \frac{\pi^2}{12}(\frac{kT}{\varepsilon_F})^2]$:
$$\chi \approx \frac{n \mu^{*2}}{kT} \left( \frac{3}{2} \frac{kT}{\varepsilon_F} \left[ 1 + \frac{\pi^2}{12} \left(\frac{kT}{\varepsilon_F}\right)^2 \right] \right) \left[ 1 - \frac{\pi^2}{6} \left(\frac{kT}{\varepsilon_F}\right)^2 \right]$$
$$\chi \approx \frac{3 n \mu^{*2}}{2 \varepsilon_F} \left[ 1 + \left( \frac{\pi^2}{12} - \frac{2\pi^2}{12} \right) \left(\frac{kT}{\varepsilon_F}\right)^2 \right]$$
$$\chi \approx \chi_0 \left[ 1 - \frac{\pi^2}{12} \left(\frac{kT}{\varepsilon_F}\right)^2 \right]$$
This matches Equation (8.2.24).
At high temperatures, $z \ll 1$. We use the series expansion $f_\nu(z) \approx z - \frac{z^2}{2^\nu}$.
$$f_{1/2}(z) \approx z - \frac{z^2}{\sqrt{2}}$$
$$f_{3/2}(z) \approx z - \frac{z^2}{2\sqrt{2}}$$
The ratio is:
$$\frac{f_{1/2}}{f_{3/2}} \approx \frac{z(1 - z/\sqrt{2})}{z(1 - z/2\sqrt{2})} \approx \left( 1 - \frac{z}{\sqrt{2}} \right) \left( 1 + \frac{z}{2\sqrt{2}} \right) \approx 1 - \frac{z}{2\sqrt{2}}$$
So the susceptibility becomes:
$$\chi \approx \frac{n \mu^{*2}}{kT} \left( 1 - \frac{z}{2\sqrt{2}} \right)$$
Let $\chi_\infty = \frac{n \mu^{*2}}{kT}$.
We need to express $z$ in terms of $n$. Using $n = \frac{2}{\lambda^3} f_{3/2}(z) \approx \frac{2z}{\lambda^3}$, we assume to the first order $z \approx \frac{n \lambda^3}{2}$.
Substituting $z$:
$$\chi \approx \chi_\infty \left( 1 - \frac{1}{2\sqrt{2}} \cdot \frac{n \lambda^3}{2} \right)$$
$$\chi \approx \chi_\infty \left( 1 - \frac{n \lambda^3}{4\sqrt{2}} \right) = \chi_\infty \left( 1 - \frac{n \lambda^3}{2^{5/2}} \right)$$
This matches Equation (8.2.27).