Unit 10: Quantum Statistics & Condensed Matter Thermodynamics
Advanced quantum statistical mechanics of identical particles: quantum indistinguishability and permutation symmetry, Bose-Einstein and Fermi-Dirac distribution laws, the degenerate electron gas in metals, Fermi energy, Fermi surface and Pauli spin paramagnetism, Bose-Einstein condensation (BEC) and macroscopic quantum coherence, Einstein's independent monochromatic lattice model, and Debye's acoustic phonon continuum theory and the low-temperature T^3 heat capacity law.
§10.1 Quantum Indistinguishability: Bosons vs Fermions
In quantum mechanics, all particles are classified into two fundamental families based on their intrinsic spin angular momentum \(S\) (the Spin-Statistics Theorem of Wolfgang Pauli):
The Two Quantum Families
1. Bosons (Integer Spin: \(S = 0, 1, 2, \dots\)):
- Governed by Bose-Einstein (BE) Statistics.
- Examples: Photons (\(S = 1\)), Gluons (\(S = 1\)), \(^4\text{He}\) atoms (\(S = 0\)), \(^{87}\text{Rb}\) atoms (\(S = 1\)), Phonons, Cooper pairs.
- Symmetric many-body wavefunctions: \(\hat{P}_{i j} \Psi = +\Psi\).
- No restriction on occupation numbers: Any number of bosons can occupy the exact same single-particle quantum state (\(n_k = 0, 1, 2, 3, \dots, \infty\)).
2. Fermions (Half-Integer Spin: \(S = 1/2, 3/2, 5/2, \dots\)):
- Governed by Fermi-Dirac (FD) Statistics.
- Examples: Electrons (\(S = 1/2\)), Protons (\(S = 1/2\)), Neutrons (\(S = 1/2\)), \(^3\text{He}\) atoms (\(S = 1/2\)), Quarks.
- Antisymmetric many-body wavefunctions: \(\hat{P}_{i j} \Psi = -\Psi\).
- Pauli Exclusion Principle: The occupation number of any single-particle quantum state is strictly restricted to 0 or 1:
Grand Partition Function Formulation
Using the grand canonical ensemble where the grand partition function factors into independent single-state terms:
- For Fermions (\(n_k \in \{0, 1\}\)):
- For Bosons (\(n_k \in \{0, 1, 2, \dots\}\)):
This simple structural distinction generates the contrasting physics of matter: fermions create stable atomic shells, rigid matter, and the periodic table, while bosons mediate forces, create lasers, and undergo Bose-Einstein condensation.
§10.2 Bose-Einstein Distribution & The Classical Limit
The average occupation number of a single-particle state \(k\) of energy \(\varepsilon_k\) is:
Derivation of the Bose-Einstein Distribution
Using \(\Xi_k^{\text{BE}} = [1 - e^{-\beta (\varepsilon_k - \mu)}]^{-1}\):
Differentiating:
where \(\beta = \frac{1}{k_B T}\).
Constraint on Chemical Potential
For the geometric series to converge and for \(\langle n_k \rangle\) to remain positive for all states:
Setting the ground-state energy to zero (\(\varepsilon_0 = 0\)):
The chemical potential of an ideal Bose gas can never be positive! As temperature drops, \(\mu\) approaches zero from below: \(\mu \rightarrow 0^-\).
Special Case: Photons and Phonons (\(\mu = 0\))
For particles whose total number is not conserved (photons in cavity radiation, phonons in a lattice), particles can be freely created and destroyed. The thermodynamic equilibrium condition \(\left(\frac{\partial A}{\partial N}\right)_{T, V} = 0\) requires:
The distribution simplifies to the celebrated Planck Distribution Law:
High-Temperature Classical Limit
When \(e^{\beta (\varepsilon_k - \mu)} \gg 1\) (low density and high temperature, \(\Lambda \ll d\)):
recovering the Maxwell-Boltzmann distribution.
§10.3 Fermi-Dirac Distribution, Fermi Energy & The Fermi Surface
For fermions, the grand partition function is \(\Xi_k^{\text{FD}} = 1 + e^{-\beta (\varepsilon_k - \mu)}\).
Derivation of the Fermi-Dirac Distribution
Differentiating:
Notice that for any energy \(\varepsilon_k\) and any temperature \(T\):
The Pauli exclusion principle is automatically satisfied for all conditions.
The Fermi Energy \(E_F\) at Absolute Zero (\(T = 0\text{ K}\))
At \(T = 0\), \(\beta = \frac{1}{k_B T} \rightarrow \infty\). The chemical potential at \(T = 0\) defines the Fermi Energy:
Evaluate \(\langle n(\varepsilon) \rangle\) as \(T \rightarrow 0\):
- For \(\varepsilon < E_F\): \(\varepsilon - E_F < 0 \implies \beta (\varepsilon - E_F) \rightarrow -\infty \implies e^{-\infty} = 0 \implies \langle n \rangle = \frac{1}{0 + 1} = 1\)
- For \(\varepsilon > E_F\): \(\varepsilon - E_F > 0 \implies \beta (\varepsilon - E_F) \rightarrow +\infty \implies e^{+\infty} = \infty \implies \langle n \rangle = \frac{1}{\infty + 1} = 0\)
At absolute zero, the Fermi-Dirac distribution is a sharp step function:
All quantum states with \(\varepsilon \le E_F\) are 100% filled, while all states with \(\varepsilon > E_F\) are completely empty. The boundary in momentum space separating filled and empty states is the Fermi Surface.
Thermal Broadening at Finite Temperature (\(T > 0\))
When \(T > 0\):
- At \(\varepsilon = \mu\): \(\langle n(\mu) \rangle = \frac{1}{e^0 + 1} = \frac{1}{2}\). The chemical potential is the energy level with exactly 50% occupation probability.
- The step function rounds off over a narrow thermal energy window of width \(\sim 4 k_B T\) centered around \(\mu\).
Because typical Fermi energies in metals are \(E_F \sim 5 - 10\text{ eV}\) (\(T_F = E_F / k_B \sim 50,000 - 100,000\text{ K}\)), room temperature (\(k_B T \approx 0.026\text{ eV} \ll E_F\)) represents an extreme degenerate quantum limit.
§10.4 The Degenerate Free Electron Gas in Metals
In the Sommerfeld free electron model, conduction electrons in a metal are treated as a non-interacting gas of fermions confined within volume \(V\).
Density of States for Free Electrons in 3D
For a free particle in a 3D box, \(\varepsilon = \frac{\hbar^2 k^2}{2 m_e}\). Accounting for the electron spin degeneracy \(g_s = 2s + 1 = 2\), the number of quantum states in spherical shell \(k\) to \(k + dk\) is:
Converting from wavevector \(k = \frac{\sqrt{2 m_e \varepsilon}}{\hbar}\) to energy \(\varepsilon\):
Calculation of Fermi Energy \(E_F\)
At \(T = 0\), all \(N\) electrons occupy states up to \(E_F\):
Solving for \(E_F\):
where \(n = N/V\) is the conduction electron density. The Fermi wavevector is \(k_F = (3\pi^2 n)^{1/3}\), and the Fermi velocity is \(v_F = \frac{\hbar k_F}{m_e} \sim 10^6\text{ m/s}\).
Electronic Heat Capacity of Metals
Classically, equipartition predicted that conduction electrons should contribute \(\frac{3}{2} R\) to heat capacity, which was contradicted by experiment (\(C_V \approx 3 R\) for the whole metal, with virtually zero electronic contribution at room temperature). Resolution via Fermi Statistics: Only electrons within \(\sim k_B T\) of the Fermi surface can be thermally excited to empty states above \(E_F\). The fraction of active electrons is roughly \(\frac{k_B T}{E_F} = \frac{T}{T_F} \sim \frac{300}{50,000} \approx 0.6\%\). The thermal energy is:
Differentiating yields the linear electronic heat capacity:
At room temperature, the electronic heat capacity is suppressed by two orders of magnitude, beautifully explaining the experimental puzzle.
§10.5 Bose-Einstein Condensation (BEC) & Macroscopic Coherence
Consider an ideal gas of \(N\) non-interacting bosons of mass \(m\) in volume \(V\). The total number of particles is:
where \(N_0\) is the population of the \(\varepsilon = 0\) ground state, and the 3D density of states is \(g(\varepsilon) = \frac{2\pi V}{h^3} (2m)^{3/2} \varepsilon^{1/2}\).
The Critical Temperature \(T_c\)
Because \(\mu \le 0\), the integral over excited states reaches its maximum possible value when \(\mu = 0\):
Substituting \(x = \beta \varepsilon\):
where \(\zeta(3/2) = \sum_{k=1}^\infty k^{-3/2} \approx 2.612\) is the Riemann zeta function. If \(N > N_{\text{excited, max}}(T)\), the excited states cannot accommodate all particles! The critical temperature \(T_c\) is defined when excited states can just barely hold all \(N\) particles:
Solving for \(T_c\):
The Condensate Fraction Below \(T_c\)
For \(T < T_c\), the excess particles are forced into the single zero-momentum ground state \(\varepsilon = 0\):
- At \(T = T_c\): \(N_0 = 0\) (Condensation begins).
- At \(T \rightarrow 0\): \(N_0 \rightarrow N\) (100% of particles occupy the single macroscopic quantum ground state).
A macroscopic fraction of particles condenses into an identical spatial wavefunction, creating a macroscopic quantum state characterized by off-diagonal long-range order, superfluidity, and phase coherence (demonstrated experimentally in 1995 by Eric Cornell, Carl Wieman, and Wolfgang Ketterle with rubidium and sodium vapors).
§10.6 Einstein Theory of Solid Heat Capacity
In 1819, Pierre Dulong and Alexis Petit discovered empirically that all elemental solids exhibit the same molar heat capacity at room temperature:
By the late 1800s, low-temperature cryogenic measurements revealed that \(C_V(T)\) drops sharply toward zero as \(T \rightarrow 0\), completely violating classical equipartition.
The Einstein Model (1907)
Albert Einstein resolved this crisis by proposing that a crystalline solid of \(N\) atoms can be modeled as \(3N\) independent quantum harmonic oscillators, all vibrating at a single identical frequency \(\nu_E\). The total internal energy is:
Defining the Einstein Temperature \(\theta_E = \frac{h \nu_E}{k_B}\):
High and Low Temperature Limits
1. High-Temperature Limit (\(T \gg \theta_E\)):
Expanding \(e^{\theta_E/T} \approx 1 + \theta_E/T\):
recovering the classical Dulong-Petit law.
2. Low-Temperature Limit (\(T \ll \theta_E\)):
For \(T \rightarrow 0\), \(e^{\theta_E/T} \gg 1\):
Einstein's model correctly predicted that heat capacity must drop to zero as \(T \rightarrow 0\). However, experimental measurements revealed that \(C_V\) approaches zero as \(T^3\), whereas Einstein's formula decays exponentially (\(e^{-\theta_E/T}\)). This discrepancy occurs because real lattice atoms do not vibrate independently at a single frequency, but rather via collective coupled acoustic waves.
§10.7 Debye Theory of Solid Heat Capacity & The T^3 Law
In 1912, Peter Debye replaced Einstein's independent oscillator assumption with a continuous elastic continuum of coupled vibrational waves (acoustic phonons).
Phonon Dispersion and Density of States
In an isotropic elastic solid, acoustic waves follow the linear dispersion relation \(\omega = v_s k\), where \(v_s\) is the speed of sound. Phonons possess three polarization modes (1 longitudinal, 2 transverse):
The density of vibrational modes in frequency space is:
Because an \(N\)-atom lattice possesses exactly \(3N\) vibrational degrees of freedom, the spectrum is cut off at a maximum Debye Cutoff Frequency \(\omega_D\):
Defining the Debye Temperature \(\theta_D = \frac{\hbar \omega_D}{k_B}\):
Total Internal Energy and Debye Heat Capacity
The total lattice vibrational energy is:
Differentiating with respect to \(T\) gives the Debye Heat Capacity:
The Celebrated Debye \(T^3\) Law
At low temperatures (\(T \ll \theta_D\)), the upper integration limit \(\theta_D / T \rightarrow \infty\). The definite integral evaluates to:
Substituting into \(C_V\):
The Debye \(T^3\) law reproduces experimental heat capacity measurements for all non-magnetic insulating solids at low temperatures with flawless mathematical precision.
§10.8 The Superconducting Transition, BCS Theory & Cooper Pairs
In 1911, Heike Kamerlingh Onnes discovered that mercury cooled below \(T_c = 4.2\text{ K}\) loses all electrical resistance: the Superconducting Transition.
Fundamental Phenomonology
Superconductivity is not merely perfect electrical conductivity; it is an entirely distinct thermodynamic state of matter characterized by:
1. Zero Electrical Resistance: Direct current flows indefinitely without Joule dissipation (\(\rho = 0\)).
2. The Meissner-Ochsenfeld Effect: Complete expulsion of magnetic flux from the interior of the superconductor (\(\mathbf{B} = 0\)), establishing that superconductors are perfect diamagnets (\(\chi = -1\)).
3. Critical Magnetic Field \(H_c(T)\): Superconductivity is destroyed when an external magnetic field exceeds a critical value:
4. Second-Order Phase Transition: At \(T_c\) in zero magnetic field, there is no latent heat, but there is a sharp discontinuous jump in the electronic heat capacity:
The Microscopic BCS Theory (1957)
John Bardeen, Leon Cooper, and J. Robert Schrieffer formulated the microscopic theory of conventional superconductivity:
1. Cooper Pairing: A moving electron polarizes the surrounding positively charged ionic lattice, creating a local excess of positive charge. A second electron is attracted to this phonon-induced polarization cloud. This effective electron-phonon-electron attraction overcomes direct Coulomb repulsion, binding two electrons of opposite momenta and spins into a Cooper Pair:
2. Bose-Einstein Condensation of Cooper Pairs:
Because each Cooper pair consists of two fermions (\(s = 1/2\)), the composite pair has integer spin (\(S = 0\)) and behaves as an effective boson. Below \(T_c\), Cooper pairs condense into a single macroscopic quantum coherent state.
3. The Superconducting Energy Gap \(\Delta(T)\):
An energy gap \(2\Delta\) opens at the Fermi level, separating the paired condensate from single-particle (quasiparticle) excitations. At \(T = 0\text{ K}\), the BCS universal ratio is:
Because an energy of at least \(2\Delta\) is required to break a pair and scatter an electron, electrons move through the lattice without scattering, producing zero electrical resistance.
## Advanced Mathematical Supplement: Bogoliubov Canonical Transformations
Bogoliubov Canonical Transformations & The Landau Criterion for Superfluidity
To diagonalize interacting many-body Hamiltonians for weakly interacting Bose gases and BCS superconductors, Nikolay Bogoliubov introduced unitary transformations mixing creation and annihilation operators.
The Bogoliubov Transformation for Bosons
Consider a weakly interacting Bose gas with macroscopic condensate \(N_0 \approx N\). Replacing zero-momentum operators with numbers \(a_0, a_0^\dagger \rightarrow \sqrt{N_0}\), the effective Hamiltonian for non-zero momentum excitations \(\mathbf{k} \ne 0\) is quadratic:
We define new quasiparticle operators \(b_\mathbf{k}, b_\mathbf{k}^\dagger\) via the Bogoliubov Canonical Transformation:
Preserving bosonic commutation relations \([b_\mathbf{k}, b_{\mathbf{k}'}^\dagger] = \delta_{\mathbf{k} \mathbf{k}'}\) requires:
Choosing \(u_k, v_k\) to eliminate the off-diagonal pairing terms diagonalizes the Hamiltonian:
where the Bogoliubov Quasiparticle Dispersion Relation is:
where \(c_s = \sqrt{n_0 V_0 / m}\) is the speed of sound.
- At low momentum (\(k \rightarrow 0\)): \(\varepsilon(k) \approx \hbar c_s k\) (linear acoustic phonon dispersion).
- At high momentum (\(k \rightarrow \infty\)): \(\varepsilon(k) \approx \frac{\hbar^2 k^2}{2m} + n_0 V_0\) (free particle parabolic dispersion).
The Landau Criterion for Superfluidity
Lev Landau analyzed an object moving with velocity \(\mathbf{v}\) through a fluid. Creating an excitation of momentum \(\hbar \mathbf{k}\) and energy \(\varepsilon(k)\) is kinematically forbidden unless:
- For an ideal non-interacting Bose gas: \(\varepsilon(k) = \frac{\hbar^2 k^2}{2m} \implies \frac{\varepsilon(k)}{\hbar k} = \frac{\hbar k}{2m} \rightarrow 0\) as \(k \rightarrow 0\). The critical velocity is \(v_c = 0\); an ideal Bose gas is NOT superfluid!
- For an interacting Bose gas with Bogoliubov dispersion:
Because \(v_c = c_s > 0\), particles moving slower than the speed of sound cannot dissipate energy into the fluid, giving rise to frictionless macroscopic superfluidity.
## Research Monograph: Topological Quantum States, Anyons & Majorana Modes
Beyond the Landau Symmetry-Breaking Paradigm
For most of the twentieth century, all phase transitions were classified by Lev Landau's symmetry-breaking theory (e.g., liquid-to-solid breaks continuous translation symmetry, ferromagnetism breaks rotational spin symmetry). In 1980, Klaus von Klitzing discovered the Integer Quantum Hall Effect (IQHE): 2D electrons in a strong magnetic field at low temperature exhibit Hall conductance quantized to integer multiples of \(e^2/h\) with accuracy of 1 part in a billion:
Remarkably, the Hall plateaus occur without any broken spatial or gauge symmetry.
The TKNN Invariant and Chern Numbers
David Thouless, Mahito Kohmoto, M. Peter Nightingale, and Marcel den Nijs (TKNN) proved that the quantization integer \(\nu\) is a topological invariant—the First Chern Number of the Berry curvature integrated over the 2D magnetic Brillouin zone:
Because an integer cannot change continuously under smooth deformations, the Hall conductance is topologically protected against arbitrary non-magnetic impurities, disorder, and sample geometries!
Anyons and Fractional Statistics in 2D
In three dimensions, particles are strictly bosons or fermions because swapping two particles twice is topologically equivalent to looping one particle around the other, which can be continuously shrunk to zero. In two dimensions, particle trajectories form braids in \((2+1)\)-dimensional spacetime. Exchanging two particles can yield an arbitrary phase:
Particles with \(\theta \ne 0, \pi\) are Anyons. In the Fractional Quantum Hall Effect (\(\nu = 1/3\)), excitations carry fractional charge \(e^* = e/3\) and fractional exchange statistics \(\theta = \pi/3\).
Majorana Zero Modes & Fault-Tolerant Quantum Computing
In topological superconductors, zero-energy quasiparticle excitations can emerge as Majorana Fermions—particles that are their own antiparticles:
Spatially separated Majorana zero modes exhibit non-Abelian braiding statistics: exchanging two Majorana modes performs a unitary quantum gate operation that depends only on the topological braid knot, providing an inherently error-resistant hardware platform for topological quantum computing.
Worked Problems & Step-by-Step Quantum Derivations
Multi-step solved problems covering Planck distribution, photoelectric kinetics, Compton shift, de Broglie wavelengths, uncertainty relations, and Hermitian operator commutation algebra.
Consider a single-particle state with energy \(\varepsilon = 0.050\text{ eV}\) at \(T = 300.0\text{ K}\) (\(k_B T = 0.02585\text{ eV}\)):
- If the chemical potential is \(\mu = 0.000\text{ eV}\), calculate the average occupation number under:
- Maxwell-Boltzmann statistics: \(\langle n \rangle_{\text{MB}} = e^{-\beta(\varepsilon - \mu)}\)
- Bose-Einstein statistics: \(\langle n \rangle_{\text{BE}} = \frac{1}{e^{\beta(\varepsilon - \mu)} - 1}\)
- Fermi-Dirac statistics: \(\langle n \rangle_{\text{FD}} = \frac{1}{e^{\beta(\varepsilon - \mu)} + 1}\)
- Explain the physical mechanism causing \(\langle n \rangle_{\text{BE}} > \langle n \rangle_{\text{MB}} > \langle n \rangle_{\text{FD}}\).
- If \(\varepsilon - \mu = 0.300\text{ eV}\), recalculate the three values and verify convergence to the classical limit.
Comprehensive Multi-Step Solution:
Step 1: Occupation Numbers at \(\varepsilon - \mu = 0.050\text{ eV}\)
Given \(k_B T = 0.025852\text{ eV}\):
1. Maxwell-Boltzmann:
2. Bose-Einstein:
3. Fermi-Dirac:
Step 2: Physical Ordering and Quantum "Forces"
The ordering is strictly:
Physical Explanation:
- Bosons: Due to wavefunction symmetry (\(\hat{P}\Psi = +\Psi\)), bosons exhibit an effective statistical attraction ("bunching"). The presence of a boson in a state enhances the probability of additional bosons joining the same state.
- Fermions: Due to wavefunction antisymmetry (\(\hat{P}\Psi = -\Psi\)), fermions exhibit an effective statistical repulsion (Pauli exclusion). Particles avoid occupying the same state.
- Classical MB: Particles are distinguishable and uncorrelated, lying exactly intermediate between bosons and fermions.
Step 3: High-Energy Classical Limit at \(\varepsilon - \mu = 0.300\text{ eV}\)
1. Maxwell-Boltzmann:
2. Bose-Einstein:
3. Fermi-Dirac:
When \(\varepsilon - \mu \gg k_B T\), all three distributions converge to identical values to within 1 part in \(10^5\).
Copper (\(\text{Cu}\)) crystallizes in an FCC lattice with density \(\rho = 8.96\text{ g/cm}^3\) and molar mass \(M = 63.546\text{ g/mol}\). Assuming each copper atom contributes exactly 1 conduction electron:
- Calculate the conduction electron number density \(n = N/V\) in \(\text{m}^{-3}\).
- Calculate the Fermi energy \(E_F\) at \(0\text{ K}\) in Joules and in electron volts.
- Determine the Fermi temperature \(T_F = E_F / k_B\) and the Fermi velocity \(v_F = \sqrt{2 E_F / m_e}\).
Comprehensive Multi-Step Solution:
Step 1: Conduction Electron Density \(n\)
The number of copper atoms per unit volume is:
Step 2: Fermi Energy \(E_F\)
The 3D Fermi energy formula is:
Evaluate \((3\pi^2 n)\):
Taking the \(2/3\) power:
Now multiply by \(\frac{\hbar^2}{2 m_e}\):
Thus:
Converting to electron volts:
Step 3: Fermi Temperature and Velocity
1. Fermi Temperature \(T_F\):
2. Fermi Velocity \(v_F\):
Even at absolute zero, conduction electrons travel at over 1.5 million meters per second (\(\approx 0.5\%\) the speed of light) due entirely to Pauli quantum degeneracy pressure!
For metallic copper with \(T_F = 81,750\text{ K}\):
- Calculate the Sommerfeld electronic heat capacity coefficient \(\gamma_{\text{Somm}} = \frac{\pi^2}{2} \frac{R}{T_F}\).
- Calculate the electronic molar heat capacity \(C_{V, \text{elec}}\) at room temperature (\(T = 298.15\text{ K}\)).
- Compare \(C_{V, \text{elec}}\) with the lattice heat capacity given by the Dulong-Petit limit \(3 R = 24.94\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}\), and determine the percentage electronic contribution.
Comprehensive Multi-Step Solution:
Step 1: Sommerfeld Coefficient \(\gamma_{\text{Somm}}\)
The electronic heat capacity per mole is:
(Accounting for electron-phonon renormalization yields the experimental value \(\gamma_{\text{exp}} \approx 0.695\text{ mJ}\cdot\text{mol}^{-1}\cdot\text{K}^{-2}\)).
Step 2: Electronic Heat Capacity at \(298.15\text{ K}\)
Step 3: Comparison with Lattice Heat Capacity
The classical lattice heat capacity is:
The electronic fraction is:
At room temperature, conduction electrons account for only \(0.6\%\) of the total heat capacity of copper. The remaining \(99.4\%\) comes from lattice phonons, explaining why classical physics appeared to miss the electronic heat capacity entirely.
In a magnetic trap experiment with rubidium-87 (\(^{87}\text{Rb}\), atomic mass \(M = 86.91\text{ g/mol}\), bosonic spin \(S = 0\)): The atomic vapor has a number density of \(n = 2.50 \times 10^{14}\text{ cm}^{-3} = 2.50 \times 10^{20}\text{ m}^{-3}\).
- Calculate the critical Bose-Einstein condensation temperature \(T_c\) for a uniform gas.
- Calculate the thermal de Broglie wavelength \(\Lambda\) at \(T = T_c\) and verify that \(n \Lambda^3 \approx 2.612\).
- If the gas is cooled to \(T = 0.50 T_c\), determine the percentage of atoms in the condensate state.
Comprehensive Multi-Step Solution:
Step 1: Calculation of \(T_c\)
The critical temperature formula is:
1. Single-atom mass of \(^{87}\text{Rb}\):
2. Prefactor:
3. Density term:
With \(n = 2.50 \times 10^{20}\text{ m}^{-3}\) and \(\zeta(3/2) = 2.6124\):
4. Critical Temperature:
The phase transition occurs at a temperature of approximately \(733\text{ nanokelvin}\).
Step 2: Thermal de Broglie Wavelength at \(T_c\)
Evaluate phase space density:
Condensation begins precisely when the thermal de Broglie wavelength becomes comparable to the interparticle spacing.
Step 3: Condensate Fraction at \(T = 0.50 T_c\)
Below \(T_c\), the condensate fraction is:
For \(T / T_c = 0.50\):
At half the critical temperature, roughly 65% of all rubidium atoms collapse into the zero-momentum quantum ground state.
Diamond has a very high vibrational frequency with an Einstein temperature of \(\theta_E = 1450\text{ K}\).
- Calculate the molar heat capacity \(C_V\) of diamond at \(T = 100.0\text{ K}\) and \(T = 300.0\text{ K}\) using the Einstein model:
- Determine the percentage of the Dulong-Petit limit achieved at \(300\text{ K}\).
- At what temperature does diamond reach 90% of the Dulong-Petit value?
Comprehensive Multi-Step Solution:
Step 1: Heat Capacity Calculations
Given \(3 R = 3 \times 8.31446 = 24.943\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}\) and \(\theta_E = 1450\text{ K}\):
1. At \(T = 100.0\text{ K}\):
- \(x = \theta_E / T = 1450 / 100 = 14.50\)
- \(e^x = e^{14.50} \approx 1.983 \times 10^6\)
- \(\frac{e^x}{(e^x - 1)^2} \approx \frac{1}{e^x} = \frac{1}{1.983 \times 10^6} \approx 5.043 \times 10^{-7}\)
- \(x^2 = (14.50)^2 = 210.25\)
At \(100\text{ K}\), the heat capacity of diamond is practically zero.
2. At \(T = 300.0\text{ K}\):
- \(x = \theta_E / T = 1450 / 300 = 4.8333\)
- \(e^x = e^{4.8333} \approx 125.62\)
- \((e^x - 1)^2 = (124.62)^2 \approx 15530\)
- \(\frac{e^x}{(e^x - 1)^2} = \frac{125.62}{15530} \approx 0.008089\)
- \(x^2 = (4.8333)^2 \approx 23.361\)
Step 2: Percentage of Dulong-Petit Value at \(300\text{ K}\)
At room temperature, diamond exhibits less than 19% of the classical Dulong-Petit value because its unusually stiff \(\text{C}-\text{C}\) covalent bonds require extremely high thermal energies to activate lattice vibrations.
Step 3: Temperature for 90% of Dulong-Petit
We require:
Using Taylor expansion for small \(x\): \(x^2 \frac{1 + x + x^2/2}{(x + x^2/2)^2} \approx 1 - \frac{x^2}{12} = 0.90\):
Solving for \(T\):
Diamond must be heated to approximately \(1324\text{ K}\) (\(\approx 1050^\circ\text{C}\)) to reach 90% of the classical Dulong-Petit heat capacity.
For metallic gold (\(\text{Au}\), molar mass \(M = 196.97\text{ g/mol}\)): The Debye temperature is \(\theta_D = 165.0\text{ K}\).
- State the Debye \(T^3\) law formula for molar heat capacity and evaluate the prefactor constant \(A\) in \(C_V = A T^3\).
- Calculate the lattice heat capacity \(C_{V, \text{lattice}}\) of gold at \(T = 4.20\text{ K}\) (liquid helium temperature) and at \(T = 10.0\text{ K}\).
- If the electronic heat capacity coefficient of gold is \(\gamma = 0.729\text{ mJ}\cdot\text{mol}^{-1}\cdot\text{K}^{-2}\), determine the crossover temperature \(T^*\) where electronic and lattice heat capacities are equal.
Comprehensive Multi-Step Solution:
Step 1: Debye \(T^3\) Law Prefactor
In the low-temperature limit (\(T < 0.1 \theta_D \approx 16.5\text{ K}\)), the Debye molar heat capacity is:
where the prefactor constant is:
Given \(R = 8.31446\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}\) and \(\theta_D = 165.0\text{ K}\):
Step 2: Lattice Heat Capacity at \(4.20\text{ K}\) and \(10.0\text{ K}\)
1. At \(T = 4.20\text{ K}\):
2. At \(T = 10.0\text{ K}\):
Step 3: Crossover Temperature \(T^*\)
The total heat capacity of gold is:
Equating the electronic and lattice contributions:
Given \(\gamma = 0.729\text{ mJ}\cdot\text{mol}^{-1}\cdot\text{K}^{-2}\) and \(A = 0.4327\text{ mJ}\cdot\text{mol}^{-1}\cdot\text{K}^{-4}\):
Below \(T^* \approx 1.30\text{ K}\), the electronic heat capacity (\(\propto T\)) dominates over the lattice phonon heat capacity (\(\propto T^3\)). Above \(1.30\text{ K}\), the lattice contribution quickly takes over.
Consider a degenerate free electron gas in an external magnetic field \(\mathbf{B} = B \hat{\mathbf{z}}\). Each electron has magnetic moment \(\mu_B\) pointing parallel or antiparallel to \(\mathbf{B}\).
- Show that the magnetic field shifts the energy of spin-up and spin-down electrons by \(\mp \mu_B B\).
- Using the density of states at the Fermi energy \(g(E_F)\), derive the net magnetic moment \(M\) induced in the electron gas in a weak field.
- Derive the Pauli paramagnetic volume susceptibility \(\chi_{\text{Pauli}} = \mu_0 \frac{M}{V B}\) and explain why it is virtually independent of temperature.
Comprehensive Multi-Step Solution:
Step 1: Zeeman Energy Shift
The magnetic dipole energy of an electron spin is \(\hat{H}_Z = -\boldsymbol{\mu}_s \cdot \mathbf{B}\). With \(g \approx 2\) and \(s_z = \pm 1/2\):
- For electrons with spin parallel to \(\mathbf{B}\) (spin up, magnetic moment parallel):
- For electrons with spin antiparallel to \(\mathbf{B}\) (spin down, magnetic moment antiparallel):
The chemical potential (Fermi level) must remain equal for both spin populations in thermodynamic equilibrium.
Step 2: Net Induced Magnetic Moment
Because the spin-up sub-band is shifted downward by \(\mu_B B\), electrons transfer from spin-down to spin-up states until both sub-bands fill to the common Fermi energy \(E_F\). In a weak magnetic field (\(\mu_B B \ll E_F\)): The change in number of spin-up electrons is:
The change in number of spin-down electrons is:
where \(g(E_F)\) is the total density of states at the Fermi energy (including both spins). The net excess of spin-up electrons is:
The total induced magnetic dipole moment is:
Step 3: Pauli Spin Susceptibility and Temperature Independence
The magnetization per unit volume is \(\mathcal{M} = \frac{M}{V} = \mu_B^2 \left(\frac{g(E_F)}{V}\right) B\). The magnetic susceptibility is:
Recall that for a 3D electron gas, \(\frac{g(E_F)}{V} = \frac{3 n}{2 E_F}\):
Physical Insight on Temperature Independence:
- Classical Curie paramagnetism predicts \(\chi_{\text{Curie}} = \frac{\mu_0 n \mu_B^2}{k_B T} \propto \frac{1}{T}\).
- In a degenerate Fermi gas, only a tiny fraction \(\sim \frac{T}{T_F}\) of electrons residing within \(k_B T\) of the Fermi surface are free to flip their spins (all inner electrons are locked by Pauli exclusion).
Multiplying the classical Curie susceptibility by this active fraction:
The temperature \(T\) cancels identically! Consequently, the conduction electron spin paramagnetism of metals is small, positive, and remarkably temperature-independent, exactly as observed experimentally.
For the elemental superconductor lead (\(\text{Pb}\)): The critical temperature is \(T_c = 7.193\text{ K}\), and the critical magnetic field at absolute zero is \(\mu_0 H_c(0) = 0.0803\text{ Tesla}\) (\(803\text{ Gauss}\)).
- Using the BCS universal weak-coupling relation \(2\Delta(0) = 3.528 k_B T_c\), calculate the superconducting energy gap \(\Delta(0)\) in Joules and in meV.
- Calculate the threshold photon frequency \(\nu_{\text{gap}}\) and wavelength \(\lambda_{\text{gap}}\) required to break a Cooper pair at \(0\text{ K}\).
- Calculate the critical magnetic field \(\mu_0 H_c(T)\) at \(T = 4.20\text{ K}\) (liquid helium boiling point).
Comprehensive Multi-Step Solution:
Step 1: BCS Energy Gap Calculation
Given \(T_c = 7.193\text{ K}\):
Converting to millielectron volts (meV):
The total energy gap to create two quasiparticle excitations is:
(Experimental measurements on lead yield \(2\Delta(0) \approx 2.7\text{ meV}\) due to strong electron-phonon coupling).
Step 2: Threshold Photon Frequency and Wavelength
To break a Cooper pair, an incoming photon must have energy \(h \nu \ge 2\Delta(0)\):
The corresponding threshold wavelength is:
Photons in the sub-millimeter far-infrared (terahertz) region are absorbed by breaking Cooper pairs, while microwave photons below \(529\text{ GHz}\) cannot be absorbed at \(0\text{ K}\).
Step 3: Critical Magnetic Field at \(4.20\text{ K}\)
Using the parabolic temperature dependence:
At \(T = 4.20\text{ K}\) with \(T_c = 7.193\text{ K}\):
Therefore:
At liquid helium temperature, lead remains superconducting up to a magnetic field of \(0.053\text{ Tesla}\).
Unlike conventional s-wave BCS superconductors which possess an isotropic energy gap \(\Delta_0\), high-temperature cuprate superconductors (such as \(\text{YBa}_2\text{Cu}_3\text{O}_{7-\delta}\)) exhibit \(d_{x^2-y^2}\) pairing symmetry:
which has line nodes at \(\phi = \pm \pi/4, \pm 3\pi/4\) where the gap vanishes identically.
- Contrast the electronic density of states \(N(E)\) near the Fermi energy for an isotropic s-wave gap versus a nodal d-wave gap.
- For an isotropic s-wave superconductor, prove that the low-temperature electronic heat capacity exhibits exponential activation: \(C_{\text{elec}} \propto e^{-\Delta_0 / k_B T}\).
- For a d-wave superconductor with line nodes, show that the low-temperature electronic heat capacity follows a power law: \(C_{\text{elec}} \propto T^2\), and explain how cryogenic calorimetry confirms d-wave symmetry.
Comprehensive Multi-Step Solution:
Step 1: Quasiparticle Density of States Comparison
1. Conventional s-wave Gap (BCS):
The energy gap \(\Delta(\mathbf{k}) = \Delta_0\) is constant everywhere across the entire Fermi surface. The density of states is:
There are strictly zero states inside the gap (\(|E| < \Delta_0\)).
2. High-\(T_c\) d-wave Gap:
\(\Delta(\phi) = \Delta_0 \cos(2\phi)\). At the nodal points where \(\cos(2\phi) = 0\) (\(\phi = \pm 45^\circ, \pm 135^\circ\)), the gap closes to zero. Linearizing \(\Delta(\phi) \approx 2\Delta_0 \delta\phi\) near the nodes, the density of states grows linearly with energy:
There is no true full spectral gap; low-energy nodal quasiparticles exist down to \(T = 0\text{ K}\).
Step 2: Exponential Heat Capacity in s-Wave Superconductors
In an isotropic s-wave superconductor, thermally exciting a quasiparticle requires overcoming the finite threshold energy \(\Delta_0\). The internal energy at \(T \ll T_c\) is:
Differentiating with respect to temperature:
The electronic heat capacity vanishes exponentially fast as \(T \rightarrow 0\), because thermal fluctuations lack sufficient energy to cross the forbidden gap.
Step 3: Power-Law \(T^2\) Heat Capacity in d-Wave Superconductors
In a d-wave superconductor, the density of states is linear: \(N(E) = c E\). The thermal energy of the nodal quasiparticles is:
Let \(x = E / k_B T \implies E = x k_B T\):
Differentiating with respect to temperature gives the electronic heat capacity:
Calorimetric Confirmation:
- If a superconductor were s-wave, a plot of \(\ln C_{\text{elec}}\) vs \(1/T\) would be a straight line with slope \(-\Delta_0 / k_B\).
- In high-\(T_c\) cuprates like \(\text{YBa}_2\text{Cu}_3\text{O}_{7-\delta}\), cryogenic specific heat measurements down to \(100\text{ mK}\) reveal a distinct \(C_{\text{elec}} = \alpha T^2\) power-law behavior.
This quadratic temperature dependence provided definitive thermodynamic proof of line nodes and \(d_{x^2-y^2}\) orbital pairing symmetry in high-temperature superconductivity.