Bose Systems

The **Bose-Einstein distribution function** governs the statistical occupancy of single-particle quantum states for any system of identical bosons (particles with integer intrinsic spin $s = 0, 1, 2, \dots$, possessing symmetric many-body wavefunctions).

1. Derivation via the Grand Canonical Ensemble

For bosons, any single-particle quantum state $i$ can be occupied by any number of particles $n_i \in \{0, 1, 2, 3, \dots\}$. The grand partition function for a single state $i$ is a geometric series: $$\Xi_i = \sum_{n_i=0}^\infty e^{-\beta(\epsilon_i - \mu)n_i} = \frac{1}{1 - e^{-\beta(\epsilon_i - \mu)}}$$ For this geometric series to converge, the ratio $e^{-\beta(\epsilon_i - \mu)}$ must be strictly less than $1$, requiring: $$\epsilon_i - \mu > 0 \implies \mu < \epsilon_0$$ The chemical potential of an ideal Bose gas must always be strictly less than the ground-state energy. The mean occupation number is: $$n(\epsilon_i) = \langle n_i \rangle = -\frac{1}{\beta}\frac{\partial \ln \Xi_i}{\partial \epsilon_i} = \frac{e^{-\beta(\epsilon_i - \mu)}}{1 - e^{-\beta(\epsilon_i - \mu)}} = \frac{1}{e^{(\epsilon_i - \mu)/(k_B T)} - 1}$$ This is the **Bose-Einstein Distribution Function**.

2. Fundamental Differences Between Quantum Statistics

• As $\epsilon \to \mu$, the Bose occupation number diverges: $n(\epsilon) \to +\infty$. Bosons enjoy occupying the same quantum state, leading to macroscopic condensation.
• In contrast, the Fermi-Dirac occupation can never exceed $1$: $f(\epsilon) \le 1$.
• In the classical dilute limit where $e^{(\epsilon - \mu)/k_B T} \gg 1$, both distributions converge to the classical Maxwell-Boltzmann exponential $e^{-(\epsilon - \mu)/k_B T}$.

Blackbody radiation is the thermal electromagnetic radiation within an enclosed cavity in thermodynamic equilibrium with its cavity walls.

1. Photons as a Bose Gas with Zero Chemical Potential

Photons are spin-1 massless bosons. In a cavity, photons are continuously absorbed and emitted by the cavity walls; their total number $N$ is not conserved. In thermal equilibrium, Helmholtz free energy $F$ is minimized with respect to photon number: $$\left(\frac{\partial F}{\partial N}\right)_{T, V} = \mu = 0$$ The chemical potential of a photon gas is **identically zero**: $\mu = 0$. The mean number of photons in a cavity mode of frequency $\nu$ is: $$\langle n_\nu \rangle = \frac{1}{e^{h\nu / k_B T} - 1}$$

2. Density of Electromagnetic Modes

Electromagnetic waves in a cavity of volume $V$ have wavevector $k = 2\pi \nu / c$. Because electromagnetic waves are transverse, there are $g = 2$ independent orthogonal polarization states for every wavevector: $$g(\nu) \, d\nu = 2 \times \frac{V}{(2\pi)^3} 4\pi k^2 dk = 2 \times \frac{4\pi V}{(2\pi)^3} \left(\frac{2\pi \nu}{c}\right)^2 \frac{2\pi}{c} d\nu = \frac{8\pi V}{c^3} \nu^2 d\nu$$

3. Planck's Spectral Radiation Formula

Multiplying mode density by average mode energy $\langle E_\nu \rangle = h\nu \langle n_\nu \rangle$: $$u(\nu) \, d\nu = \frac{1}{V} g(\nu) h\nu \langle n_\nu \rangle d\nu = \frac{8\pi h \nu^3}{c^3} \frac{d\nu}{e^{h\nu / k_B T} - 1}$$ Expressed in terms of wavelength $\lambda = c / \nu$ ($|d\nu| = \frac{c}{\lambda^2} d\lambda$): $$u(\lambda) \, d\lambda = \frac{8\pi h c}{\lambda^5} \frac{d\lambda}{e^{h c / (\lambda k_B T)} - 1}$$ This is **Planck's Law of Blackbody Radiation** (1900), which birthed quantum physics.

We now integrate Planck's law to derive the macroscopic thermodynamic properties of blackbody radiation.

1. Total Energy Density and the Stefan-Boltzmann Law

The total electromagnetic energy density in the cavity is: $$u(T) = \int_0^\infty u(\nu) \, d\nu = \frac{8\pi h}{c^3} \int_0^\infty \frac{\nu^3 d\nu}{e^{h\nu / k_B T} - 1}$$ Substituting $x = \frac{h\nu}{k_B T}$: $$u(T) = \frac{8\pi h}{c^3} \left(\frac{k_B T}{h}\right)^4 \int_0^\infty \frac{x^3 dx}{e^x - 1}$$ The standard Bose-Einstein definite integral is $\int_0^\infty \frac{x^3 dx}{e^x - 1} = \frac{\pi^4}{15}$. Therefore: $$u(T) = \left( \frac{8\pi^5 k_B^4}{15 c^3 h^3} \right) T^4 = a T^4$$ where $a = 7.5657 \times 10^{-16}\text{ J/(m}^3\cdot\text{K}^4)$ is the radiation constant. The radiant emissive power (flux density) from an ideal blackbody surface is: $$J = \frac{c}{4} u(T) = \sigma T^4$$ where $\sigma = \frac{a c}{4} = 5.6704 \times 10^{-8}\text{ W/(m}^2\cdot\text{K}^4)$ is the **Stefan-Boltzmann constant**.

2. Wien's Displacement Law

Maximizing $u(\lambda)$ with respect to $\lambda$ yields $\frac{d}{d\lambda} u(\lambda) = 0$, leading to: $$\frac{hc}{\lambda_{\max} k_B T} \left( 1 - e^{-hc/(\lambda_{\max} k_B T)} \right) = 5$$ Solving numerically gives $x = \frac{hc}{\lambda_{\max} k_B T} \approx 4.9651$, establishing **Wien's Displacement Law**: $$\lambda_{\max} T = \frac{h c}{4.9651 k_B} = 2.8978 \times 10^{-3}\text{ m}\cdot\text{K}$$

3. Radiation Pressure and Entropy of the Photon Gas

Because photons are relativistic ($E = p c$), the radiation pressure is: $$P = \frac{1}{3} u(T) = \frac{1}{3} a T^4$$ The total Helmholtz free energy is: $$F = U - TS = -PV = -\frac{1}{3} a V T^4$$ The entropy of the photon gas is: $$S = -\left(\frac{\partial F}{\partial T}\right)_V = \frac{4}{3} a V T^3$$

In a crystalline solid, atomic nuclei do not vibrate independently; their motion is coupled by interatomic electrostatic forces, forming collective vibrational wave modes called **phonons**.

1. Phonons as Quasiparticles

A phonon is a quantum of crystal lattice vibration possessing energy $E = \hbar \omega$ and crystal momentum $\mathbf{p} = \hbar \mathbf{k}$. Like photons:

• Phonons are bosons with spin 0.
• Phonons are created and destroyed thermally without number conservation; their chemical potential is zero: $\mu = 0$.
• Mean phonon occupancy is given by Planck's distribution: $$\langle n_\omega \rangle = \frac{1}{e^{\hbar \omega / k_B T} - 1}$$

2. Acoustic and Optical Phonon Branches

For a 3D crystal lattice with $N$ unit cells and $p$ atoms per basis:

• Total degrees of freedom: $3pN$.
• $3$ **Acoustic Branches:** At long wavelengths ($k \to 0$), atoms in a cell move in phase; frequency is linear: $\omega = v_s k$ (sound waves).
• $3(p - 1)$ **Optical Branches:** Adjacent basis atoms vibrate out of phase, creating oscillating electric dipoles that couple to light.

Peter Debye (1912) recognized that low-temperature heat capacity is dominated by long-wavelength acoustic phonons, which can be treated as an elastic continuous medium.

1. The Debye Density of States and Debye Cutoff Frequency

In an isotropic elastic solid with 1 longitudinal and 2 transverse sound speeds: $$g(\omega) = \frac{V}{2\pi^2} \left( \frac{1}{v_L^3} + \frac{2}{v_T^3} \right) \omega^2 = \frac{3V}{2\pi^2 v_s^3} \omega^2$$ Because a crystal of $N$ atoms has exactly $3N$ vibrational modes, the spectrum must terminate at a maximum **Debye Cutoff Frequency** $\omega_D$: $$\int_0^{\omega_D} g(\omega) \, d\omega = 3N \implies \frac{V}{2\pi^2 v_s^3} \omega_D^3 = 3N \implies \omega_D = v_s \left( \frac{6\pi^2 N}{V} \right)^{1/3}$$ We define the **Debye Temperature** $\Theta_D$: $$\Theta_D \equiv \frac{\hbar \omega_D}{k_B}$$

2. Total Lattice Energy and Debye Specific Heat

The total vibrational internal energy is: $$U(T) = \int_0^{\omega_D} \hbar \omega \langle n_\omega \rangle g(\omega) d\omega = 9 N k_B T \left( \frac{T}{\Theta_D} \right)^3 \int_0^{\Theta_D/T} \frac{x^3 dx}{e^x - 1}$$ Differentiating with respect to $T$ yields the **Debye Heat Capacity**: $$C_V(T) = 9 N k_B \left( \frac{T}{\Theta_D} \right)^3 \int_0^{\Theta_D/T} \frac{x^4 e^x}{(e^x - 1)^2} dx$$

3. The Low-Temperature Debye $T^3$ Law

At low temperatures ($T \ll \Theta_D$), the upper integration limit extends to infinity: $\int_0^\infty \frac{x^4 e^x dx}{(e^x - 1)^2} = \frac{4\pi^4}{15}$. $$C_V(T) = 9 N k_B \left( \frac{T}{\Theta_D} \right)^3 \frac{4\pi^4}{15} = \frac{12\pi^4}{5} N k_B \left( \frac{T}{\Theta_D} \right)^3$$ This is the famous **Debye $T^3$ Law**. It perfectly matches experimental calorimeter data for all non-magnetic insulating solids at low temperatures.

Satyendra Nath Bose (1924) and Albert Einstein (1925) predicted that an ideal gas of massive bosons undergoes a spectacular phase transition at low temperatures, with a macroscopic fraction of particles collapsing into the identical zero-momentum ground state.

1. Maximum Capacity of Excited States

For a 3D gas of $N$ non-relativistic bosons of mass $m$ in volume $V$, the density of states is $g(\epsilon) = \frac{2\pi V}{h^3} (2m)^{3/2} \epsilon^{1/2}$. The total number of particles in excited states ($\epsilon > 0$) is: $$N_{\text{exc}} = \int_0^\infty \frac{g(\epsilon) d\epsilon}{e^{(\epsilon - \mu)/k_B T} - 1}$$ Because $\mu \le 0$, the maximum number of bosons that excited states can hold occurs when $\mu \to 0^-$: $$N_{\text{exc}}^{\max}(T) = \frac{2\pi V (2m)^{3/2}}{h^3} (k_B T)^{3/2} \int_0^\infty \frac{x^{1/2} dx}{e^x - 1} = V \left( \frac{m k_B T}{2\pi \hbar^2} \right)^{3/2} \zeta(3/2)$$ where $\zeta(3/2) \approx 2.6124$ is the Riemann zeta function.

2. The Critical Condensation Temperature ($T_c$)

When the temperature falls below a critical value $T_c$, the maximum capacity of excited states becomes strictly less than the total number of particles: $N_{\text{exc}}^{\max}(T) < N$. Setting $N_{\text{exc}}^{\max}(T_c) = N$ defines the **Bose-Einstein Transition Temperature**: $$T_c = \frac{2\pi \hbar^2}{m k_B} \left( \frac{N/V}{\zeta(3/2)} \right)^{2/3} = \frac{2\pi \hbar^2}{m k_B} \left( \frac{n}{2.6124} \right)^{2/3}$$

3. Macroscopic Ground-State Condensation Fraction

For $T < T_c$, all surplus particles must condense into the single ground state $\epsilon_0 = 0$: $$N_0(T) = N - N_{\text{exc}}(T) = N \left[ 1 - \left( \frac{T}{T_c} \right)^{3/2} \right]$$ Below $T_c$, a macroscopic fraction of the gas forms a single giant macroscopic quantum wavepacket, experimentally observed in 1995 in trapped Rubidium-87 atoms by Cornell, Wieman, and Ketterle (Nobel Prize 2001).

When Helium-4 ($^4\text{He}$, a spin-0 boson) is cooled below $T_\lambda = 2.17\text{ K}$ at saturated vapor pressure, it undergoes the **Lambda Transition** from normal Liquid He-I to superfluid **Liquid He-II**.

1. The Two-Fluid Model (Tisza and Landau)

Liquid He-II behaves hydrodynamically as an interpenetrating mixture of two components: $$\rho = \rho_n + \rho_s$$

• Superfluid Component ($\rho_s$): Fraction of atoms in the macroscopic quantum condensate. It possesses **strictly zero viscosity** ($\eta_s = 0$) and **zero entropy** ($s_s = 0$). It can flow with zero friction through sub-micron capillaries!
• Normal Fluid Component ($\rho_n$): Thermal gas of elementary quasiparticle excitations (phonons and rotons). It has normal viscosity and carries all the entropy of the liquid.

As $T \to 0\text{ K}$, $\rho_n / \rho \to 0$ and $\rho_s / \rho \to 1$.

2. Landau's Criterion for Superfluidity and the Roton Spectrum

Lev Landau (1941) asked: *At what speed will an object moving through He-II experience frictional drag by creating elementary excitations?* By energy and momentum conservation, creation of an excitation with energy $\epsilon(p)$ and momentum $p$ requires: $$v > v_c = \min_p \left( \frac{\epsilon(p)}{p} \right)$$ For He-II, the excitation spectrum exhibits a local minimum at $p_0 / \hbar \approx 1.92\text{ Å}^{-1}$ with energy gap $\Delta / k_B \approx 8.6\text{ K}$ called the **Roton minimum**: $$\epsilon(p) \approx \Delta + \frac{(p - p_0)^2}{2\mu}$$ The minimum ratio $\epsilon(p)/p$ is: $$v_c = \frac{\Delta}{p_0} \approx 58\text{ m/s}$$ For any flow velocity $v < v_c$, creating excitations is kinematically forbidden by quantum mechanics: the liquid flows with **strictly zero dissipation**!

We now explore the thermodynamic behavior of the degenerate Bose gas across the BEC transition.

1. Pressure Below $T_c$

In the condensed phase ($T < T_c$), the chemical potential is locked at zero: $\mu = 0$. The grand potential $\Phi_G = -PV$ is given solely by excited states: $$P(T) = k_B T \left( \frac{m k_B T}{2\pi \hbar^2} \right)^{3/2} \zeta(5/2) \propto T^{5/2}$$ where $\zeta(5/2) \approx 1.3415$. Notice that:

Below $T_c$, the pressure depends strictly on temperature $T$ and is completely independent of volume $V$! $$\left(\frac{\partial P}{\partial V}\right)_T = 0$$

The isothermal compressibility $\kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T \to \infty$ diverges. Compressing the gas simply transfers particles from excited states into the zero-volume condensate without changing pressure.

2. Heat Capacity Across the Transition

The heat capacity is: $$C_V(T) = \begin{cases} \frac{15}{4} k_B \zeta(5/2) \left( \frac{m k_B T}{2\pi \hbar^2} \right)^{3/2} V \propto T^{3/2} & \text{for } T < T_c \\ \frac{3}{2} N k_B \left[ 1 + 0.231 \left( \frac{T_c}{T} \right)^{3/2} + \dots \right] & \text{for } T > T_c \end{cases}$$ At $T = T_c$, $C_V$ reaches a sharp peak with value $C_V(T_c) \approx 1.925 N k_B$, exhibiting a cusp that marks a continuous third-order thermodynamic transition in the ideal gas (and a famous logarithmic $\lambda$-peak in interacting Liquid Helium).

The thermal properties of a diatomic gas (such as $N_2, O_2, CO$) depend on the quantum excitation of rotational and vibrational molecular levels.

1. Molecular Rotational Partition Function

Treating the diatomic molecule as a rigid rotor with moment of inertia $I = \mu_m r_0^2$: $$E_J = \frac{\hbar^2}{2I} J(J + 1) = k_B \Theta_{\text{rot}} J(J + 1), \quad J = 0, 1, 2, \dots$$ where $\Theta_{\text{rot}} \equiv \frac{\hbar^2}{2 I k_B}$ is the **characteristic rotational temperature** (typically $2\text{ K}$ to $85\text{ K}$). The degeneracy of each rotational level is $g_J = 2J + 1$: $$z_{\text{rot}} = \sum_{J=0}^\infty (2J + 1) e^{-J(J+1) \Theta_{\text{rot}} / T}$$ For $T \gg \Theta_{\text{rot}}$ (room temperature for most gases), the sum can be approximated by an integral: $$z_{\text{rot}} \approx \int_0^\infty (2J + 1) e^{-J(J+1) \Theta_{\text{rot}} / T} dJ = \frac{T}{\sigma \Theta_{\text{rot}}}$$ where $\sigma$ is the symmetry number ($\sigma = 2$ for homonuclear $N_2$, $\sigma = 1$ for heteronuclear $CO$). The rotational heat capacity in this high-temperature limit is $C_V^{\text{rot}} = R$.

2. Molecular Vibrational Partition Function

Treated as a 1D harmonic oscillator with vibrational frequency $\omega$: $$\Theta_{\text{vib}} \equiv \frac{\hbar \omega}{k_B}$$ For typical molecules, $\Theta_{\text{vib}} \sim 1,000\text{ K}$ to $3,000\text{ K}$ (e.g., $N_2$: $\Theta_{\text{vib}} = 3,374\text{ K}$). $$z_{\text{vib}} = \frac{e^{-\Theta_{\text{vib}} / 2T}}{1 - e^{-\Theta_{\text{vib}} / T}}$$ $$C_V^{\text{vib}} = R \left( \frac{\Theta_{\text{vib}}}{T} \right)^2 \frac{e^{\Theta_{\text{vib}}/T}}{(e^{\Theta_{\text{vib}}/T} - 1)^2}$$ At room temperature ($300\text{ K} \ll \Theta_{\text{vib}}$), vibrational modes are frozen into their quantum ground state.

In a homonuclear diatomic molecule like Hydrogen ($H_2$), quantum statistics of the two identical atomic nuclei imposes strict selection rules on molecular rotation!

1. Nuclear Spin States of Hydrogen ($H_2$)

Each proton has nuclear spin $I = 1/2$ (fermion). The two proton spins couple to give total nuclear spin $I_{\text{tot}} \in \{0, 1\}$:

• Para-Hydrogen ($I_{\text{tot}} = 0$, Nuclear Singlet): $$\chi_{\text{para}} = \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle) \quad (1\text{ antisymmetric nuclear spin state})$$
• Ortho-Hydrogen ($I_{\text{tot}} = 1$, Nuclear Triplet): $$\chi_{\text{ortho}} \in \left\{ |\uparrow\uparrow\rangle, \, \frac{|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle}{\sqrt{2}}, \, |\downarrow\downarrow\rangle \right\} \quad (3\text{ symmetric nuclear spin states})$$

2. Total Wavefunction Symmetry Requirement

Because protons are fermions, the total molecular wavefunction must be **antisymmetric** under the exchange of the two protons: $$\Psi_{\text{total}} = \psi_{\text{trans}} \times \psi_{\text{vib}} \times \psi_{\text{rot}} \times \chi_{\text{spin}}$$ The spatial exchange of the two nuclei is equivalent to an inversion: $\mathbf{r} \to -\mathbf{r}$, which multiplies the rotational spherical harmonic $Y_{JM}(\theta, \phi)$ by $(-1)^J$:

• Even $J$ ($J = 0, 2, 4, \dots$): Symmetric rotational state ($(-1)^J = +1$).
• Odd $J$ ($J = 1, 3, 5, \dots$): Antisymmetric rotational state ($(-1)^J = -1$).

To make $\Psi_{\text{total}}$ antisymmetric:

1. Para-Hydrogen: Antisymmetric spin state $\implies$ Must have **EVEN rotational levels only** ($J = 0, 2, 4, \dots$). Ground state energy $E_0 = 0$ ($J=0$).
2. Ortho-Hydrogen: Symmetric spin state $\implies$ Must have **ODD rotational levels only** ($J = 1, 3, 5, \dots$). Lowest state is $J=1$ with energy $E_1 = 2 k_B \Theta_{\text{rot}}$.

3. The High-Temperature 3:1 Equilibrium Ratio

At room temperature ($T \gg \Theta_{\text{rot}} \approx 85\text{ K}$), all nuclear spin states are equally populated. The equilibrium ratio is given by their nuclear spin statistical weights: $$\frac{\text{Ortho}}{\text{Para}} = \frac{g_{\text{ortho}}}{g_{\text{para}}} = \frac{3}{1} = 75\% \text{ Ortho}, \quad 25\% \text{ Para}$$ When hydrogen is cooled to liquid temperature ($20\text{ K}$) without a catalyst, conversion from ortho ($J=1$) to para ($J=0$) is extraordinarily slow (taking weeks) because it requires an electron-nuclear spin-flip. Catalysts (activated carbon, ferric oxide) are added to speed conversion and prevent boil-off in liquid hydrogen rocket propellant storage!