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Chapter 6 • Theory & Derivations

The Condensed State

Solids at low and high temperatures, Debye interpolation, thermal expansion and Grüneisen parameter, quantum Bose and Fermi liquids, and electronic band spectra of metals and dielectrics.

§6.1 Solids at Low and High Temperature

The thermal properties of solids reflect the quantum mechanical excitation of their crystal lattice vibrations and conduction electrons.

1. The High-Temperature Classical Limit

At high temperatures ($T \gg \Theta_D$), the thermal energy $k_B T$ is far larger than the quantum energy of any lattice vibration ($\hbar \omega_D$). All $3N$ normal modes of vibration are fully excited. According to the classical equipartition theorem, each harmonic mode contributes $k_B T$ of internal energy: $$U \to 3 N k_B T \implies C_V \to 3 N k_B = 3 R \approx 24.94\text{ J/(mol}\cdot\text{K)}$$ This is the classical Dulong-Petit value, which is identical for all non-magnetic monoatomic solids regardless of chemical identity.

2. The Low-Temperature Quantum Freezing

As temperature drops below the Debye temperature ($T \ll \Theta_D$), thermal energy becomes insufficient to excite high-frequency vibrational modes:
  • In **Insulators:** Lattice acoustic phonons dominate completely, yielding the universal cubic temperature dependence: $$C_V^{\text{insulator}}(T) = A T^3 = \frac{12\pi^4}{5} N k_B \left( \frac{T}{\Theta_D} \right)^3$$
  • In **Metals:** Both conduction electrons and acoustic phonons contribute: $$C_V^{\text{metal}}(T) = \gamma T + A T^3$$ Plotting $C_V / T$ versus $T^2$ yields a straight line whose $y$-intercept gives the electronic Sommerfeld constant $\gamma$ and whose slope gives the lattice coefficient $A$!

§6.2 Debye's Interpolation Formula and Universal Heat Capacity Scaling

Peter Debye synthesized the low- and high-temperature regimes into a single, universal dimensionless function that describes virtually all insulating crystals.

1. The Debye Function ($D(y)$)

The internal energy of a crystal is expressed in terms of the **Debye Function**: $$U(T) = 3 N k_B T \, D\left( \frac{\Theta_D}{T} \right)$$ where $y = \Theta_D / T$, and $D(y)$ is defined as: $$D(y) \equiv \frac{3}{y^3} \int_0^y \frac{x^3 dx}{e^x - 1}$$

2. The Universal Specific Heat Expression

Differentiating $U(T)$ gives the molar heat capacity: $$C_V(T) = 3 R \left[ 4 D(y) - \frac{3 y}{e^y - 1} \right] = 9 R \left( \frac{T}{\Theta_D} \right)^3 \int_0^{\Theta_D / T} \frac{x^4 e^x dx}{(e^x - 1)^2}$$ Notice that:
The reduced heat capacity $\frac{C_V}{3R}$ is a universal function of the single dimensionless ratio $\frac{T}{\Theta_D}$.
When experimental specific heat data for diverse solids (lead with $\Theta_D = 105\text{ K}$, copper with $\Theta_D = 343\text{ K}$, and diamond with $\Theta_D = 2230\text{ K}$) are plotted against $T / \Theta_D$, all data points collapse onto the exact same theoretical Debye master curve!

§6.3 Thermal Expansion of Solids and the Grüneisen Parameter

Why does a solid expand when heated? A perfectly harmonic crystal lattice ($V(x) = \frac{1}{2}c x^2$) **does not expand at all**! Thermal expansion is a direct consequence of **anharmonicity** in the interatomic potential.

1. Proof that Harmonic Oscillators Have Zero Thermal Expansion

Consider an atom displaced by $x$ from its equilibrium position. In a symmetric parabolic potential $V(x) = c x^2$: $$\langle x \rangle = \frac{\int_{-\infty}^\infty x e^{-\beta c x^2} dx}{\int_{-\infty}^\infty e^{-\beta c x^2} dx} = 0$$ Because the potential is strictly symmetric, the thermal average displacement $\langle x \rangle$ vanishes identically at all temperatures. A purely harmonic crystal never expands!

2. Anharmonicity and Thermal Expansion

The real interatomic potential (such as the Lennard-Jones potential) is asymmetric, resisting compression more strongly than expansion. Expanding about the minimum: $$V(x) = c x^2 - g x^3 - f x^4, \quad (g, f > 0)$$ Treating the cubic anharmonic term $-g x^3$ as a perturbation: $$e^{-\beta V(x)} \approx e^{-\beta c x^2} (1 + \beta g x^3)$$ $$\langle x \rangle = \frac{\int_{-\infty}^\infty x e^{-\beta c x^2} (1 + \beta g x^3) dx}{\int_{-\infty}^\infty e^{-\beta c x^2} dx} = \frac{\beta g \int_{-\infty}^\infty x^4 e^{-\beta c x^2} dx}{\int_{-\infty}^\infty e^{-\beta c x^2} dx} = \frac{3 g}{4 c^2} k_B T$$ The average interatomic spacing increases linearly with temperature: $\langle x \rangle \propto T$.

3. The Grüneisen Parameter and Equation of State

As a solid expands ($V$ increases), phonon vibrational frequencies shift downward: $$\gamma_G \equiv -\frac{d\ln \omega_D}{d\ln V} = -\frac{V}{\omega_D}\frac{d\omega_D}{dV}$$ where $\gamma_G$ is the dimensionless **Grüneisen Parameter** (typically $\sim 1$ to $2$). The volumetric thermal expansion coefficient $\beta_V = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P$ satisfies the **Mie-Grüneisen Equation**: $$\beta_V = \frac{\gamma_G C_V}{V K_T}$$ where $K_T$ is the isothermal bulk modulus. Thermal expansion is directly proportional to lattice heat capacity: as $T \to 0$, $\beta_V \propto T^3$, vanishing at absolute zero in agreement with the Third Law of Thermodynamics.

§6.4 Quantum Liquids with Bose-Type Spectrum

Liquid Helium-4 ($^4\text{He}$) is the prototypical strongly interacting **quantum Bose liquid**.

1. The Elementary Excitation Spectrum

Lev Landau (1941) proposed that elementary excitations in Liquid $^4\text{He}$ are single quasiparticles whose energy $\epsilon(p)$ depends continuously on momentum $p$:
  1. Phonon Branch ($p/\hbar < 0.6\text{ Å}^{-1}$): Longitudinal sound waves with linear dispersion: $$\epsilon(p) = c_s p$$ where $c_s \approx 238\text{ m/s}$ is the speed of first sound.
  2. Maxon Peak ($p/\hbar \approx 1.1\text{ Å}^{-1}$): A local maximum around $\epsilon / k_B \approx 14\text{ K}$.
  3. Roton Minimum ($p/\hbar \approx 1.92\text{ Å}^{-1}$): A parabolic dip around characteristic momentum $p_0$: $$\epsilon(p) \approx \Delta + \frac{(p - p_0)^2}{2\mu}$$ with energy gap $\Delta / k_B \approx 8.65\text{ K}$, $p_0 / \hbar \approx 1.92\text{ Å}^{-1}$, and effective mass $\mu \approx 0.16 m_4$.

2. Bogoliubov's Microscopic Theory

Nikolay Bogoliubov (1947) derived this excitation spectrum microscopically for a weakly interacting Bose gas using canonical transformation: $$E(k) = \sqrt{\epsilon_k^2 + 2 n U_0 \epsilon_k}$$ where $\epsilon_k = \frac{\hbar^2 k^2}{2m}$ is the free-particle kinetic energy, and $U_0 = \frac{4\pi \hbar^2 a}{m}$ is the repulsive contact interaction.
  • At low wavevector ($k \to 0$): $E(k) \approx \sqrt{2 n U_0 \frac{\hbar^2 k^2}{2m}} = \hbar c_s k$, reproducing linear sound waves with sound speed $c_s = \sqrt{n U_0 / m}$.
  • At high wavevector ($k \to \infty$): $E(k) \to \epsilon_k + n U_0$, recovering quadratic single-particle behavior.

§6.5 Quantum Liquids with Fermi-Type Spectrum (Landau Fermi Liquid Theory)

Liquid Helium-3 ($^3\text{He}$, a spin-1/2 fermion) and conduction electrons in heavy fermion metals represent strongly interacting **quantum Fermi liquids**.

1. Landau Fermi Liquid Theory

Lev Landau (1956) recognized that strong repulsive interactions do not destroy the sharpness of the Fermi surface. Instead, there is a continuous one-to-one correspondence between the eigenstates of a non-interacting Fermi gas and the low-energy excitations of the interacting system, called **quasiparticles**.

2. Quasiparticles and Effective Mass ($m^*$)

A quasiparticle is an individual fermion dressed by a surrounding cloud of interactions with other particles. Near the Fermi surface, the quasiparticle energy is: $$\epsilon(p) \approx \epsilon_F + v_F^* (p - p_F) = \epsilon_F + \frac{p_F}{m^*} (p - p_F)$$ where $m^*$ is the **quasiparticle effective mass**: $$\frac{m^*}{m} = 1 + \frac{1}{3} F_1^s$$ In Liquid $^3\text{He}$ at ambient pressure, $m^* \approx 3.0 m_3$; under high pressure ($30\text{ bar}$), $m^* \approx 6.0 m_3$.

3. Quasiparticle Lifetime Divergence

By Pauli exclusion, scattering between quasiparticles near the Fermi surface requires both initial and final states to lie within energy $k_B T$ of $\epsilon_F$. The scattering rate scales as $\Gamma \propto (k_B T)^2 + (\epsilon - \epsilon_F)^2$. Therefore, the quasiparticle lifetime diverges at low temperatures: $$\tau \propto \frac{1}{T^2} \to \infty$$ As $T \to 0$, quasiparticles become infinitely well-defined quantum excitations, explaining why free-electron models work remarkably well in metals!

§6.6 The Electronic Spectra of Metals

In a crystalline metal, conduction electrons move in the periodic electrostatic potential created by the ionic lattice: $V(\mathbf{r} + \mathbf{R}) = V(\mathbf{r})$.

1. Bloch's Theorem and Energy Bands

According to Bloch's Theorem, single-electron wavefunctions are plane waves modulated by the lattice periodicity: $$\psi_{n\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}} u_{n\mathbf{k}}(\mathbf{r}), \quad u_{n\mathbf{k}}(\mathbf{r} + \mathbf{R}) = u_{n\mathbf{k}}(\mathbf{r})$$ The energy spectrum separates into continuous **energy bands** $\epsilon_n(\mathbf{k})$ separated by **bandgaps**.

2. The Electronic Hallmark of a Metal

A material is an electrical conductor (metal) if and only if:
The highest occupied energy level—the Fermi Energy $\epsilon_F$—lies inside a partially filled electronic energy band.
Because empty quantum states are available immediately above $\epsilon_F$ with infinitesimal energy spacing $d\epsilon \to 0$:
  • An applied electric field $\mathbf{E}$ accelerates electrons into adjacent unoccupied states, producing a net electrical current (Drude-Sommerfeld conductivity $\sigma = \frac{n e^2 \tau}{m^*}$).
  • The density of states at the Fermi level $g(\epsilon_F) > 0$ is finite, yielding linear electronic heat capacity $C_V = \gamma T$ and Pauli paramagnetism $\chi = \mu_B^2 g(\epsilon_F)$.

§6.7 The Electronic Spectra of Solid Dielectrics

In solid dielectrics (insulators and semiconductors), the Fermi level lies within a forbidden energy gap separating filled and empty electronic bands.

1. Band Structure of Insulators and Semiconductors

In an intrinsic dielectric at $T = 0\text{ K}$:
  • The **Valence Band** is $100\%$ completely filled with electrons.
  • The **Conduction Band** is completely empty.
  • They are separated by a finite **Bandgap** $E_g$: $$E_g = \epsilon_c - \epsilon_v$$ For semiconductors, $E_g \lesssim 3\text{ eV}$ (Silicon: $1.12\text{ eV}$, Gallium Arsenide: $1.42\text{ eV}$). For insulators, $E_g \gtrsim 4\text{ eV}$ (Diamond: $5.47\text{ eV}$, Silicon Dioxide: $9.0\text{ eV}$).

2. Thermal Carrier Generation and Chemical Potential

At finite temperature $T > 0$, thermal fluctuations promote electrons across the bandgap into the conduction band, leaving empty states (**holes**) in the valence band. The electron density $n$ and hole density $p$ are: $$n = N_c e^{-(\epsilon_c - \mu)/k_B T}, \quad p = N_v e^{-(\mu - \epsilon_v)/k_B T}$$ where $N_c = 2\left(\frac{m_e^* k_B T}{2\pi \hbar^2}\right)^{3/2}$ and $N_v = 2\left(\frac{m_h^* k_B T}{2\pi \hbar^2}\right)^{3/2}$ are the effective densities of states. Equating $n = p = n_i$ (intrinsic semiconductor): $$n_i = \sqrt{N_c N_v} \exp\left( -\frac{E_g}{2 k_B T} \right)$$ $$\mu = \frac{\epsilon_c + \epsilon_v}{2} + \frac{3}{4} k_B T \ln\left( \frac{m_h^*}{m_e^*} \right)$$ The Fermi level in an intrinsic dielectric sits essentially at the mid-gap! Electrical conductivity grows exponentially with temperature: $\sigma(T) \propto e^{-E_g / (2 k_B T)}$, in stark contrast to metals where resistance increases with temperature due to phonon scattering.

📝 Chapter Worked Examples & Exercises

Complete derivations & analytical proofs
Medium Example 6.1: Debye Temperature and Maximum Lattice Frequency in Diamond

For diamond, the atomic number density is $n = 1.76 \times 10^{29}\text{ atoms/m}^3$ and the average sound speed is $v_s = 1.20 \times 10^4\text{ m/s}$. (a) Calculate the Debye cutoff frequency $\omega_D$ and Debye temperature $\Theta_D$. (b) Explain why diamond feels remarkably cold to the touch and has a high thermal conductivity at room temperature ($300\text{ K}$).

Step 1: Calculate the Debye cutoff frequency
$$\omega_D = v_s (6\pi^2 n)^{1/3} = (1.20 \times 10^4\text{ m/s}) [6\pi^2 (1.76 \times 10^{29}\text{ m}^{-3})]^{1/3}$$ $$6\pi^2 (1.76 \times 10^{29}) = 1.042 \times 10^{31} \implies (1.042 \times 10^{31})^{1/3} = 2.184 \times 10^{10}\text{ m}^{-1}$$ $$\omega_D = (1.20 \times 10^4) \times (2.184 \times 10^{10}) = 2.621 \times 10^{14}\text{ rad/s}$$

This is the maximum lattice vibrational frequency in diamond.

Step 2: Determine the Debye temperature
$$\Theta_D = \frac{\hbar \omega_D}{k_B} = \frac{(1.055 \times 10^{-34}\text{ J}\cdot\text{s})(2.621 \times 10^{14}\text{ s}^{-1})}{1.381 \times 10^{-23}\text{ J/K}} = 2,002\text{ K}$$

Because $\Theta_D \approx 2,000\text{ K}$ is far above room temperature ($300\text{ K}$), diamond is deeply in the quantum regime at room temperature.

Step 3: Analyze thermal properties at 300 K
$$\frac{T}{\Theta_D} = \frac{300}{2002} \approx 0.15 \ll 1$$ $$C_V \approx \frac{12\pi^4}{5} R \left(\frac{300}{2002}\right)^3 \approx 6.1\text{ J/(mol}\cdot\text{K)} \ll 3R (24.9\text{ J/(mol}\cdot\text{K)})$$

Because phonon modes are largely frozen, phonon-phonon Umklapp scattering is exceptionally rare, giving diamond an extraordinarily large phonon mean free path and the highest room-temperature thermal conductivity of any bulk solid ($k \approx 2200\text{ W/(m}\cdot\text{K)}$).

Hard Example 6.2: Intrinsic Carrier Concentration and Fermi Level in Silicon

Silicon has an indirect bandgap $E_g = 1.12\text{ eV}$ at $T = 300\text{ K}$. The effective masses of electrons and holes are $m_e^ = 1.08 m_0$ and $m_h^ = 0.56 m_0$. (a) Calculate the intrinsic carrier density $n_i$ at $300\text{ K}$. (b) Determine the offset of the Fermi level from the mid-gap position.

Step 1: Compute effective densities of states N_c and N_v
$$N_c = 2\left(\frac{2\pi m_e^* k_B T}{h^2}\right)^{3/2} = 2.81 \times 10^{25}\text{ m}^{-3}$$ $$N_v = 2\left(\frac{2\pi m_h^* k_B T}{h^2}\right)^{3/2} = 1.04 \times 10^{25}\text{ m}^{-3}$$

These are the effective quantum densities of states of the conduction and valence bands.

Step 2: Calculate intrinsic carrier concentration n_i
$$n_i = \sqrt{N_c N_v} e^{-E_g / (2 k_B T)}$$ $$\sqrt{N_c N_v} = \sqrt{(2.81 \times 10^{25})(1.04 \times 10^{25})} = 1.71 \times 10^{25}\text{ m}^{-3}$$ $$\frac{E_g}{2 k_B T} = \frac{1.12\text{ eV}}{2 (0.02585\text{ eV})} = 21.663$$ $$n_i = (1.71 \times 10^{25}) \times e^{-21.663} = (1.71 \times 10^{25})(3.91 \times 10^{-10}) = 6.69 \times 10^{15}\text{ m}^{-3} = 6.69 \times 10^9\text{ cm}^{-3}$$

At room temperature, only roughly one atom in ten trillion is thermally ionized.

Step 3: Calculate Fermi level offset from mid-gap
$$\Delta \mu = \mu - \frac{\epsilon_c + \epsilon_v}{2} = \frac{3}{4} k_B T \ln\left(\frac{m_h^*}{m_e^*}\right) = \frac{3}{4} (0.02585\text{ eV}) \ln\left(\frac{0.56}{1.08}\right) = -0.0127\text{ eV} = -12.7\text{ meV}$$

Because holes are lighter than electrons ($m_h^ < m_e^$), the Fermi level is shifted downward by $12.7\text{ meV}$ below the exact center of the bandgap.

Hard Example 6.3: Grüneisen Parameter and Thermal Expansion of Aluminum

For Aluminum, the molar heat capacity at $T = 300\text{ K}$ is $C_V = 24.2\text{ J/(mol}\cdot\text{K)}$, molar volume is $V_m = 1.00 \times 10^{-5}\text{ m}^3\text{/mol}$, isothermal bulk modulus is $K_T = 76\text{ GPa}$, and the linear thermal expansion coefficient is $\alpha_L = 23.1 \times 10^{-6}\text{ K}^{-1}$. (a) Calculate the volumetric thermal expansion coefficient $\beta_V = 3\alpha_L$. (b) Use the Mie-Grüneisen relation to calculate the Grüneisen parameter $\gamma_G$.

Step 1: Compute volumetric thermal expansion coefficient
$$\beta_V = 3 \alpha_L = 3 \times (23.1 \times 10^{-6}\text{ K}^{-1}) = 6.93 \times 10^{-5}\text{ K}^{-1}$$

For an isotropic cubic crystal, volumetric expansion is three times linear expansion.

Step 2: Solve Mie-Grüneisen equation for gamma_G
$$\beta_V = \frac{\gamma_G C_V}{V_m K_T} \implies \gamma_G = \frac{\beta_V V_m K_T}{C_V}$$ $$\gamma_G = \frac{(6.93 \times 10^{-5}\text{ K}^{-1}) \times (1.00 \times 10^{-5}\text{ m}^3\text{/mol}) \times (76 \times 10^9\text{ Pa})}{24.2\text{ J/(mol}\cdot\text{K)}}$$ $$\gamma_G = \frac{52.668}{24.2} \approx 2.18$$

The Grüneisen parameter of Aluminum is $2.18$, which is characteristic of anharmonic acoustic phonon shifts in fcc metals.