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Chapter 2 of Statistical Physics by David Tong. Sections 2.1–2.2 give ideal classical partition functions; section 2.5 gives the density virial expansion and Mayer f-function. No separately contributed chapter author is established.

Chapter 4 of Statistical Physics by David Tong. Section 4.4: reservoir constraints, Gibbs minimum, equations (4.173)–(4.175) and extensive free-energy relations. No separately contributed chapter author is established.

The Clapeyron equation connects the slope of a phase coexistence curve to the latent heat and the change in volume during a first-order phase transition. A steep boundary can indicate a large latent heat, a small change in volume, or both. This lets a measured boundary constrain caloric properties and predicts how changing pressure shifts a melting temperature.

For a pure substance in bulk phase coexistence, label the initial and final thermodynamic phases $\alpha$ and $\beta$. Both have the same absolute temperature $T$, pressure $p$, and chemical potential. With $s$, $h$, and $v$ denoting entropy, enthalpy, and volume on one common basis, define $\Delta s=s_{\beta }-s_{\alpha }$, $\Delta h=h_{\beta }-h_{\alpha }$, and $\Delta v=v_{\beta }-v_{\alpha }$. ThenThe derivative follows the phase coexistence curve; it is not a partial derivative within one thermodynamic phase. Use molar volume $v_m$ with molar latent heat $L_m$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or specific volume $v$ with latent heat $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Both give $dp/dT$ in $\mathrm{Pa}{\mathrm{K}}^{-1}$. For molar mass $M$, $L_m=ML$ and $\Delta v_m=M\Delta v$, so the predicted slope is independent of the chosen basis. This formulation assumes a simple compressible system, fixed composition, and negligible interface contributions; a curved interface can instead require the Young–Laplace equation.

For positive latent heat in the stated conversion direction, the sign of the slope is the sign of $\Delta v$. Converting liquid to gas normally increases volume and gives a positive slope. The negative melting-point slope of water illustrates the opposite sign. At a thermodynamic critical point, the coexisting gas and liquid merge: both $L$ and $\Delta v$ vanish, so their ratio needs a limiting analysis rather than substitution of zero into the Clapeyron equation.^[ref]

A short derivation uses the molar Gibbs free energy $g_m$, equal to the chemical potential for a pure substance. Equality persists along the boundary, so $d{g}_{m,\alpha }=d{g}_{m,\beta }$ with $d{g}_m=-s_{m}dT+v_{m}dp$. Therefore $-\Delta s_{m}dT+\Delta v_{m}dp=0$. Equality of $g_m=h_m-T{s}_m$ also gives $\Delta h_m=T\Delta s_m$. Eliminating $\Delta s_m$ yields the Clapeyron equation.

Consider a closed sample of $N$ ideal gas particles heated at constant pressure $p$, with only pressure-volume work and negligible changes of bulk kinetic energy or gravitational potential energy. In a quasistatic process, the work done by the gas isThe first law of thermodynamics, $\Delta U=Q-W$, then giveswhere enthalpy is $H=U+pV$. Constant pressure alone is insufficient if electrical or other work also transfers energy. ^[ref]

For a monatomic ideal gas in its classical regime, $\Delta U=\frac{3}{2}N{k}_B\Delta T$ and $Q=\frac{5}{2}N{k}_B\Delta T$. Thus part of the supplied heat raises the internal energy, and part moves the boundary against external pressure.

For liquid–gas phase coexistence, make two distinct approximations: the gas obeys the ideal gas law, $v_{m,g}=RT/p$, and its molar volume greatly exceeds the liquid molar volume, $v_{m,g}\gg v_{m,l}$. Here $R$ is the molar gas constant and $p$ is the saturation pressure of the pure substance. Substitution into the Clapeyron equation giveswhere $p_{\ast }>0$ is any fixed reference pressure, making the logarithm dimensionless. Constant latent heat is a further approximation, needed for the elementary Integrated Clausius-Clapeyron relation below.^[ref]

The neglected liquid molar volume has a transparent error estimate. If the gas is ideal but $ϵ=v_{m,l}/v_{m,g}$ is retained, thenThus ignoring the liquid molar volume underestimates this slope by a relative amount $ϵ$. This estimate does not bound errors from treating a real gas as ideal or from freezing $L_m(T)$. Near a thermodynamic critical point, the gas and liquid molar volumes approach each other and these dilute-vapor assumptions fail.

An extensive quantity measures how much of a thermodynamic system there is; an intensive quantity describes its state independently of that amount. The clean comparison is replication: take independent identical systems at the same thermodynamic equilibrium, with the same substance, composition and thermodynamic phase. Combining their accounting boundaries multiplies internal energy $U$, entropy $S$, volume $V$, mass $m$ and amount of substance $n$, while leaving temperature $T$, pressure $p$ and chemical potential $\mu$ unchanged.^[ref]

For $\lambda$ copies the ledger isAn integer $\lambda$ counts actual copies; a positive real $\lambda$ describes the corresponding bulk scaling law. Copies can remain separated: replication need not involve removing a partition, changing a container or performing any work. Distinguish $n$, the amount of substance measured in moles, from $N=n{N}_A$, the dimensionless count of specified entities, where $N_A$ is the Avogadro constant in ${\mathrm{mol}}^{-1}$. For example, $2\mathrm{mol}$ contains $1.204428152\times 10^{24}$ specified entities; tripling it gives $6\mathrm{mol}$ and $3.613284456\times 10^{24}$ entities. This uses the exact Avogadro constant $N_A=6.02214076\times 10^{23}{\mathrm{mol}}^{-1}$, which does not triple. In the differential $dU=TdS-pdV+\mu dn$, $\mu$ is a molar chemical potential in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$; the coefficient of $dN$ instead has units of energy per entity.

As a direct check, the ideal gas law $pV=nRT$ givesFor a monatomic ideal gas with only translational degrees of freedom active, internal energy of an ideal gas gives $U=3nRT/2$, so $U^{\prime }=\lambda U$ at unchanged temperature. Here $R$ is the molar gas constant. The scaling distinction lets a small sample supply an equation of state for a much larger amount: material state and inventory become separate questions.

Replication does not mean multiplying every linear dimension by $\lambda$: a geometrically similar container then has ${\lambda }^3$ times the volume. Nor does it mean combining samples at different states. Equal temperature establishes thermal equilibrium but does not imply equal pressure, composition or mass density. Density of a composite system makes this distinction concrete.

The Ising model is a spin model with an Ising spin ${\sigma }_v\in \{-1,1\}$ at every graph vertex $v$. For a finite graph with vertex set $V$ and edge set $E$, a uniform coupling $J$ and field parameter $h$ give the HamiltonianEach edge is counted once. Here $T>0$ is temperature, $k_B$ is the Boltzmann constant, and $\beta$ is inverse temperature; $J$ and $h$ have units of energy. For magnetic applications, $h$ includes the magnetic moment multiplying the applied magnetic field. The canonical ensemble assigns the Boltzmann distributionThe partition function $Z$ makes the probabilities sum to one. Positive $J$ rewards equal neighboring Ising spins, modeling ferromagnetism; negative $J$ rewards opposite neighbors, giving an antiferromagnetic Ising model. Positive $h$ rewards $+1$, while negative $h$ rewards $-1$.^[ref]

A single local rule can organize a whole graph: rewarding agreement on each edge makes all-equal Ising spin patterns favorable even though no edge directly couples distant vertices. The four-spin Ising ring makes this visible with only sixteen microstates. It also separates three different questions: whether neighbors agree, whether the mean spin magnetization is nonzero, and how large its thermal fluctuations are.

A motivating experiment is the University of Colorado Boulder's Curie-temperature demonstration. An iron rod suspended from a twisting wire on a stand is initially held near a permanent magnet by ferromagnetism. A gas burner supplied from a propane cylinder heats the rod; the readout is its motion away from the magnet, followed by return as it cools. Repeated cooling supplies a reversible comparison on the same sample. The source explicitly calls for heating the rod while keeping the magnet cool, since heating the magnet can reduce its strength. The inference is that warming the sample weakens its magnetic attraction across the Curie temperature. This is a qualitative force readout, with no calibrated temperature or magnetization trace supplied: it motivates competition between alignment and thermal fluctuations, but does not identify $J$, measure individual Ising spins, or realize a four-spin Ising ring.^[ref]

Ernst Ising's 1925 paper examined a linear chain of two-orientation magnetic elements with neighboring interactions. In its calculation for a large chain, the zero-field mean magnetization vanishes and the model has no spontaneous magnetization. The inspected English translation states this immediately after equation (8) in §2. This historical result concerns the chain; it does not establish the behavior of a two-dimensional Ising model.^[ref]

Define the instantaneous spin magnetization $m=\frac{1}{4}{\sum }_i{\sigma }_i$. Global reversal sends $m$ to $-m$ without changing the zero-field energy. Pairing every microstate with its reversed partner proves $⟨m⟩=0$ at every positive temperature, even when almost all probability lies on the two fully aligned patterns. A vanishing mean does not imply weak alignment: the limiting probability distribution is half at $m=+1$ and half at $m=-1$.

The Ising ring magnetization fluctuations measure this distinction through variance. In the $d=0$ class, both patterns have $m^2=1$. Among the $d=2$ patterns, eight have $m^2=1/4$ and four have $m^2=0$; the two $d=4$ patterns also have $m^2=0$. ConsequentlyIt equals $6/17$ at $K=(\ln 2)/4$, approaches $1/4$ for four independent fair Ising spins at $K\to 0$, and approaches $1$ as $K\to \infty$. Unlike the energy variance, the spin magnetization variance remains large in the low-temperature symmetric mixture.

The same thermal fluctuation controls the field response. Allow $h$ to vary before setting it to zero; differentiating the Boltzmann distribution inserts $\beta {\sum }_i{\sigma }_i=4\beta m$, givingThis is a spin-normalized magnetic susceptibility with respect to the energy-valued parameter $h$; conversion to physical magnetization and magnetic field requires the magnetic moment and sample volume factors.

For finite $N$, $Z$ is a finite positive sum of exponential functions. Its free energy $-k_{B}T\log Z$ is a real analytic function for $T>0$, so the smooth heat capacity peak here is not a critical phase transition. Spontaneous magnetization concerns the order of limits: first a thermodynamic limit, then removal of a symmetry-breaking field. The four-spin Ising ring alone establishes neither that limit nor any two-dimensional Ising model result.^[ref]

Latent heat is the heat absorbed or released in a first-order phase transition through its entropy jump. For an equilibrium conversion from thermodynamic phase $\alpha$ to thermodynamic phase $\beta$ at fixed pressure, evaluate both thermodynamic phases at the same temperature and pressure on their phase coexistence curve:Use one normalization throughout: $L$, $S$ and $H$ may all be per mole, giving $L$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or all per unit mass, giving $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Here $S$ is entropy and $H$ is enthalpy on that basis. Equal Gibbs free energies of the coexisting thermodynamic phases give $\Delta H=T_{\mathrm{tr}}\Delta S$. Identifying this enthalpy difference with supplied heat assumes a closed system and an equilibrium conversion with only pressure-volume work. ^[ref]

For a pure substance with positive latent heat per unit mass $L$, let $x$ be the phase fraction of the higher-enthalpy thermodynamic phase, measured by mass. While both thermodynamic phases coexist at the fixed pressure,Adding heat changes $x$ rather than the temperature; this creates the horizontal portion of a heating curve. Once $x=1$, additional heat raises the temperature according to specific heat capacity at constant pressure. Reversing the conversion releases heat. The plateau represents changing enthalpy, rather than a finite single-thermodynamic phase heat capacity satisfying $Q=C_p\Delta T$. ^[ref]

For a constructed fixed-pressure model, take $m=1\mathrm{kg}$, initial temperature $280\mathrm{K}$, transition temperature $T_{\mathrm{tr}}=300\mathrm{K}$, solid specific heat capacity at constant pressure $c_s=2\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, latent heat $L=100\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$, and liquid specific heat capacity at constant pressure $c_l=3\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. These are stipulated model parameters, not empirical properties of a named material. Heating the solid to $300\mathrm{K}$ takes $40\mathrm{kJ}$ of sensible heat; converting it fully to liquid takes another $100\mathrm{kJ}$ of heat; heating the liquid to $320\mathrm{K}$ takes $60\mathrm{kJ}$. The total is $200\mathrm{kJ}$. After a total heat input of $90\mathrm{kJ}$, $x=(90-40)/100=1/2$ and the temperature is still $300\mathrm{K}$. The corresponding per-unit-mass entropy jump is $L/T_{\mathrm{tr}}=\frac{1}{3}\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$.

A continuous phase transition has no latent heat. Its heat capacity can have a jump or a singularity without a finite interval in which added heat converts coexisting thermodynamic phases at a constant temperature. A heating plateau therefore requires the stated phase coexistence conditions; it is not a feature of every phase transition. ^[ref]

An ice bath can absorb heat while its temperature barely changes: the heat converts ice into water. The latent heat of fusion quantifies this melting budget. For a pure substance at its equilibrium melting temperature $T_m(p)$ and specified pressure $p$, define it here per unit mass:Here $h_l$ and $h_s$ are the specific enthalpies of the coexisting bulk liquid and solid, and $H$ is enthalpy. For ordinary melting, $L>0$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$; an amount $m$ has a total fusion enthalpy $mL$ in joules. A molar fusion enthalpy instead has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and equals $ML$, where $M=m/n$ is mass divided by amount of substance in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Always check the normalization before inserting a tabulated value into an energy balance.^{[ref] [ref]}

For a closed system converting completely from solid to liquid at fixed pressure, with only pressure-volume work and negligible changes in bulk kinetic energy and potential energy, the first law of thermodynamics gives $Q_p=\Delta H=mL$. Both endpoints are at $T_m(p)$: warming initially colder ice or subsequently warming the resulting water requires additional sensible heat. Since $H=U+pV$, the corresponding specific internal energy jump is $\Delta u=L-p\Delta v$, where $v=V/m$; $L$ includes the work associated with the change in volume. Heat is transferred energy, whereas enthalpy is a state function. ^[ref]

Let $x=m_l/m$ be the liquid phase fraction by mass, rather than by volume. Neglecting interfacial contributions to enthalpy, equilibrium phase coexistence at fixed pressure givesIn the model above, reaching $T_m$ takes $2.10\mathrm{kJ}$ of sensible heat; the plateau then absorbs $33.40\mathrm{kJ}$, ending at a cumulative $35.50\mathrm{kJ}$. At $18.80\mathrm{kJ}$, half the mass is liquid and the temperature is still $T_m$. The final warming takes another $4.18\mathrm{kJ}$. The rounded $L$ and $c_l$ are inspired by the introductory ice-melting calorimetry values in Case Western's demonstration; constant properties and exact equilibrium throughout are stipulated model assumptions. ^[ref]

Within one thermodynamic phase, specific heat capacity at constant pressure is $c_p=(\partial h/\partial T{)}_p$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; a sample's heat capacity at constant pressure is $C_p=m{c}_p$, in $\mathrm{J}{\mathrm{K}}^{-1}$. The plateau is a finite enthalpy change at zero temperature change through varying $x$, so neither a finite single-phase $c_p$ nor $Q=m{c}_p\Delta T$ represents it. The heating curve needs separate sensible heat intervals and a fusion contribution. ^[ref]

The plateau assumes a pure substance, a stable bulk solid-liquid phase coexistence pair, fixed pressure, and sufficiently slow, uniform heating to remain near thermodynamic equilibrium. The total mass and composition must remain fixed, with negligible evaporation and no other phase transition. A fast heater can produce temperature gradients, so a local thermometer need not read the equilibrium $T_m$. Neither $T_m$ nor $L$ is universal across pressures or compositions: common-pressure melting-point slope describes the pressure dependence of $T_m$, while mixtures can melt over a temperature range as their coexisting compositions change. Their liquidus must be consulted instead of assuming a pure-substance plateau. ^[ref]

This makes ice useful as a temperature buffer: under these conditions it can absorb $m_{s}L$ before the remaining solid mass $m_s$ disappears, after which the water warms. If a stipulated constant net input power $P$ goes entirely into melting, that buffer lasts $m_{s}L/P$. Reversing the equilibrium conversion releases the same fusion enthalpy during freezing. The beautiful thermodynamic connection is $L=T_m(s_l-s_s)$, with $s$ the specific entropy and $T_m$ absolute temperature: equal Gibbs free energies at phase coexistence allow different enthalpies and entropies. The nearly stationary thermometer hides a finite, measurable change of state. ^[ref]

A normal Fermi liquid is an interacting fermion phase continuously connected to a Fermi gas without broken symmetry. Its Landau quasiparticles are dressed particle/hole excitations near the Fermi surface, with the same conserved quantum numbers, finite quasiparticle residue and a decay width small compared with their excitation energy. Their reference dispersion already includes zero-temperature interaction renormalization; “bare quasiparticle” here does not mean a free atom.^[ref]

The rate assumes $k_{B}T,|\xi |\ll E_F$, $\xi =ϵ-\mu$, regular scattering amplitudes and density of states, equilibrium Fermi-Dirac distributions, and two-body collisions conserving energy and momentum. Pauli exclusion principle restricts the available phase space. It describes a regular three-dimensional normal liquid, not a one-dimensional interacting electron liquid. On the surface, ${\tau }^{-1}\propto T^2$; at $T=0$ and $\xi >0$, it is proportional to ${\xi }^2$. Hence excitations become increasingly sharp. The quoted asymmetric occupation factor is for particle out-scattering: a hole decay or full spectral width also needs the complementary in-scattering process; it does not imply exponentially stable deep holes.

Symmetry $f_{ab}=f_{ba}$ cancels the quadratic functional's factor one-half, giving the thermal Landau quasiparticle energy shiftHere $a=(\mathbf{k},\lambda )$, and the derivative is with respect to an occupation, not particle density. Replacing the homogeneous momentum sum by an energy integral defines a smooth angularly averaged kernel $H_a(ϵ)$. With $\delta n=n_F(ϵ;\mu ,T)-\Theta ({\mu }_0-ϵ)$,Fixed density gives $\delta \mu =-({\pi }^2/6)(k_{B}T{)}^2{D}^{\prime }({\mu }_0)/D({\mu }_0)+O(T^4)$; fixed chemical potential instead has $\delta \mu =0$. Thus ${\widetilde{ϵ}}_a={ϵ}_a+O(T^2)$ in either case, provided the kernels are smooth and no instability intervenes. This is the Sommerfeld expansion; singular spectra require separate treatment.^[ref]

The leading entropy is therefore that of quasiparticles with their zero-temperature effective mass:$D^{\ast }$ here is the total, spin-summed density of states. The plotted low-temperature slopes require the enhanced masses $5.4m$ at molar volume $26.2\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and $2.7m$ at $36.8\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, rather than the bare-mass predictions. At fixed molar particle number, ${\gamma }_{\mathrm{mol}}\propto m^{\ast }{V}_{\mathrm{mol}}^{2/3}$. Agreement is restricted to the low-temperature limit: the measured curves depart markedly from a fixed-mass ideal-gas extrapolation at higher temperature. This supports long-lived quasiparticles with smooth low-energy parameters, but heat capacity alone does not measure their lifetime or prove adiabatic continuity.

For a classical ideal monatomic gas, $C_{V,\mathrm{He}}=3N{k}_B/2$ and $C_{p,\mathrm{He}}=5N{k}_B/2$. Treat copper as having one conduction electron per atom. Its degenerate-electron contribution isConsequently the electron-only estimates at5K areThe constant-pressure gas versus electronic metal heat capacity convention makes the second numerator appropriate to a constant-pressure gas measurement. For the solid, the low-temperature difference between $C_p$ and $C_V$ is usually small, but its phonon contribution is not fixed by $E_F$. In a Debye model,The supplied $E_F$ alone therefore does not determine the ratio to copper's total heat capacity; the electronic-only result is conditional.^[ref] An estimate using additional material data is possible: the specific-heat fit in Material Properties: OFHC Copper by National Institute of Standards and Technology gives $c_{p,\mathrm{Cu}}(5\mathrm{K})\simeq 0.155\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Using copper's rounded molar mass $0.0635\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$ gives $C_{p,\mathrm{Cu}}\simeq 9.85\times 10^{-3}\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, and henceThis supplementary estimate includes copper's lattice contribution and retains the ideal-gas approximation for helium. The NIST fit reports a 10% deviation from its source data below 15 K; neither that fit nor the stated pressure alone supplies helium nonideality corrections.^[ref]

Lattice energy is transported by phonons; long-wavelength acoustic modes usually dominate low-temperature conduction. Their speed is set by elastic wave propagation. In the Debye model their volumetric heat capacity grows as $T^3$ at low temperature and approaches the classical lattice limit at high temperature. With boundary-limited mean free path, this gives a low-temperature $T^3$ conductivity; increasing phonon scattering can instead produce a peak and a declining high-temperature trend.

In metals, electrons near the Fermi surface carry both heat and charge. Their heat capacity is linear in T and their typical transport speed is the Fermi velocity, not a classical gas thermal speed. Degeneracy restricts the active thermal fraction but the large speed and long mean free path can make electronic heat conduction dominant. Insulators have few mobile electrons, so phonon conduction usually dominates. The approximate parallel-channel sum is $\kappa ={\kappa }_e+{\kappa }_{\mathrm{ph}}$; which channel dominates is material- and temperature-dependent. ^[ref]

identical particles have the same intrinsic properties, including mass, charge and spin. Their labels are bookkeeping devices rather than measurable identities. When wave functions overlap, quantum mechanics supplies no observable trajectory that permanently identifies “particle 1” independently of “particle 2”. A many-particle description must therefore treat exchange consistently. The Hamiltonian and physical observables are invariant under relabelling identical particles.

For two particles, the exchange operator obeys ${\hat{P}}_{12}^2=\hat{I}$. The usual three-dimensional symmetrization postulate selects states symmetric under exchange for bosons and antisymmetric for fermions. In relativistic quantum theory, the Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions under its locality and positivity assumptions; the connection is not derived merely from ${\hat{P}}_{12}^2=\hat{I}$ in elementary nonrelativistic mechanics. For orthonormal one-particle states $u$ and $v$, the two-particle spatial combinations areThe sign concerns the complete state, including spin and other internal variables, not necessarily the spatial factor alone.

The antisymmetric state vanishes if both fermions occupy the same one-particle state: this is the Pauli exclusion principle. For $N$ fermions in orthonormal spin orbitals ${\chi }_j$, the normalized Slater determinant $\Psi =(N!{)}^{-1/2}\det [{\chi }_j(i)]$ supplies the required exchange sign. Bosonic occupation has no corresponding upper limit, and symmetrized products describe many bosons. Fermionic occupation is zero or one per complete one-particle state; two electrons can still occupy the same spatial atomic orbital with opposite spin.

Spin and spatial symmetry are linked. A two-electron spin singlet state is antisymmetric, so its spatial factor must be symmetric. A symmetric triplet state instead requires an antisymmetric spatial factor. Exchange then changes pair correlations and the energies available when interactions depend on particle separation, even without introducing a new classical “exchange force”. It contributes to atomic shell structure, bonding and magnetic behaviour.

In an occupation-number description, bosonic creation and annihilation operators satisfy $[{\hat{a}}_i,{\hat{a}}_j^{†}]={\delta }_{ij}$, whereas fermionic operators satisfy $\{{\hat{c}}_i,{\hat{c}}_j^{†}\}={\delta }_{ij}$ and anticommute among themselves. The latter imply $({\hat{c}}_i^{†}{)}^2=0$. This formalism implements the same exchange constraints without assigning permanent particle labels.

For noninteracting particles in thermal equilibrium, the mean occupation of a state of energy $ϵ$ isFermi-Dirac statistics produces a filled Fermi sea at zero temperature and degeneracy pressure; Bose-Einstein statistics permits macroscopic Ground state occupation when the density of available states and thermodynamic conditions allow condensation. The different statistics therefore have measurable collective consequences, not just different notation.^[ref]

When phase-space occupation is dilute, both distributions reduce to the Maxwell–Boltzmann statistics limit. Likewise, well-separated localized wave packets have negligible overlap and small exchange corrections, so effectively labelled classical trajectories can become a useful approximation. In two spatial dimensions more general exchange statistics are possible; the standard boson/fermion account here concerns ordinary three-dimensional particles and their total states.

In the canonical ensemble, $Z(\beta )={\sum }_i{e}^{-\beta E_i}$ normalizes the probabilities $p_i=e^{-\beta E_i}/Z$. For temperature-independent levels,The derivative holds external parameters, including magnetic field, fixed.

For one independent symmetric three-level system put $ϵ=\mu B$ and $x=\beta ϵ$. Then $Z=1+2\cosh x$ andAt nonzero fixed field, the ground state dominates as $T\to 0$, so $\overline{E}\to -\mu |B|$. At high temperature all three levels become nearly equally populated and their mean tends to zero from below. At fixed positive temperature the energy is even in $B$, is quadratic near zero field, and approaches $-\mu |B|$ at large $|B|$ as the lowest level becomes occupied. The heat capacity vanishes both when excitations are frozen out and when populations are almost equal, with a positive Schottky anomaly between those limits.

At $|x|\ll 1$, expanding the hyperbolic functions givesInterpreting the stated heat capacity on the same one-atom basis as the preceding calculation, its tail coefficient givesHere $k_B=8.61733\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$ is used consistently with $a$ in electronvolt kelvin. For a sample of $N$ independent atoms, the total coefficient would be $N$ times the one-atom coefficient; it must first be divided by $N$.

Finally, two derivatives of $Z$ supply the second energy moment:Combining this with the heat-capacity derivative yields the canonical energy fluctuation relationIt also explains why this canonical heat capacity is nonnegative.^[ref]

The classical equipartition theorem assigns mean energy $k_{B}T/2$ to each independent quadratic term in the Hamiltonian, provided the canonical distribution and its relevant boundary integrals exist. For a quadratic term $a{q}^2$, the Boltzmann distribution is proportional to $e^{-\beta a{q}^2}$, with $\beta =1/(k_{B}T)$. Integrating a total derivative givesAfter dividing by the normalization,The same argument applies to quadratic momentum terms. Other independent canonical coordinates factor out of the ratio. It is a classical statement: widely spaced quantum energy levels can freeze out a contribution at low temperature.

The confining quartic potential energy $U(x)=Q{x}^4$, with $Q>0$, is not quadratic, so the $k_{B}T/2$ rule cannot be assigned to it. This does not invalidate the canonical Boltzmann distribution or its generalized integration-by-parts identity. Indeed,Hence equipartition for a homogeneous potential givesEquivalently, $Z_x=\Gamma (1/4)(\beta Q{)}^{-1/4}/2$ and $-\partial \ln Z_x/\partial \beta =1/(4\beta )$. This is the potential contribution only; an additional standard kinetic term $p^2/(2m)$ would contribute a further $k_{B}T/2$ to energy and $k_B/2$ to heat capacity.

For an intact diatomic molecule, the classical degrees of freedom comprise three translational quadratic terms, two rotational quadratic terms, and two terms for its one bond-stretching vibration: one kinetic and one potential. Thus complete classical equipartition predictsThe heat capacity of a diatomic gas has lower-temperature regimes because rotational and vibrational levels are quantized. In a dilute molecular gas, the expected progression isThe rotational energy scale is normally much smaller than the vibrational scale, so these steps are separated. The high-temperature value treats the vibration as harmonic and excludes dissociation and electronic excitation; the lowest-temperature gas regime is meaningful only while the sample remains dilute and gaseous.^[ref]

The plotted example uses a quantum rigid rotor and harmonic vibration, with an illustrative ratio of characteristic temperatures. It is a model calculation, not measured data for a particular molecular species. For the vibrational part,which tends to zero on freezing out and to one when many vibrational levels are accessible. The classical equipartition theorem is recovered when the relevant level spacings are small compared with $k_{B}T$.

An equation of state relates thermodynamic variables within a material model; a phase diagram records which equilibrium phase or phases are stable under specified controls. For a pure substance at fixed temperature and pressure, stability means minimizing Gibbs free energy. Two phases coexist when their molar chemical potentials agree. The equation of state and the free-energy stability condition must therefore be used together: an algebraic relation between $p,V,T$ alone need not identify the stable branch.

A pressure-temperature phase diagram contains single-phase regions separated by phase coexistence curves. For a one-component system the Gibbs phase rule gives $F=3-P$: a single phase has two independent controls, two-phase coexistence lies on a curve, and three phases coexist at a triple point. The liquid-gas coexistence curve ends at a liquid-gas critical point; beyond it a continuous path connects liquidlike and gaslike states without crossing a first-order coexistence boundary. A fixed $p,T$ point on a coexistence curve does not specify the proportions of the two phases: an additional extensive constraint, such as total volume, fixes the phase fraction.

At a first-order phase transition, first derivatives of Gibbs free energy, such as molar entropy or volume, are discontinuous. The conversion has latent heat $L=T\Delta s$. At a continuous transition the first derivatives match, while response functions associated with higher derivatives can be singular or discontinuous. At an ordinary liquid-gas critical endpoint, the two coexisting densities approach one another and the latent heat tends to zero; the material does not retain a sharp liquid-gas boundary above it.

The coexistence slope follows directly from equality of the two molar chemical potentials. Since $d\mu =-sdT+vdp$, differentiating ${\mu }_{\alpha }={\mu }_{\beta }$ along the boundary givesThis is the exact Clapeyron equation, also commonly called the Clausius-Clapeyron relation. For vaporization, both $L$ and $\Delta v$ are positive, so the saturation pressure rises with temperature. For ordinary ice melting into liquid water, $L>0$ but the molar volume decreases, giving the familiar negative melting-curve slope. The sign follows from the volume change, not from a different law.

For a dilute ideal vapour with $v_g\gg v_l$, $\Delta v\simeq RT/p$. The ideal-vapor Clausius-Clapeyron approximation becomesIf molar latent heat is approximately constant over the interval,These approximations deteriorate close to the critical point, where liquid and gas volumes become comparable, or over a range in which latent heat changes appreciably.^[ref]

The ideal gas law $pV=nRT$ is an equation of state for a dilute gas, but it has no liquid branch and cannot predict condensation. A simple model including repulsion and attraction is the Van der Waals molar equation,where $b>0$ represents excluded molar volume and $a>0$ represents attraction. At its critical point, $\partial p/\partial v={\partial }^{2}p/\partial v^2=0$, givingBelow $T_c$, the homogeneous isotherm has an interval with $\partial p/\partial v>0$. That would have negative isothermal compressibility and is mechanically unstable. Equilibrium liquid-gas coexistence replaces the loop by a constant-pressure segment. Equality of chemical potentials gives the Maxwell construction:The equal-area condition enforces the common tangent to the appropriate free energy; it is a stability construction rather than permission to follow the unstable loop physically. The average molar volume between $v_l$ and $v_g$ varies by changing the phase fraction.

The model captures a liquid-gas critical point and qualitative coexistence, but its simple constants are not a complete quantitative description of real fluids and it has no solid phase. A full phase diagram needs additional solid free energies and can contain several crystal phases. Conversely, an experimentally determined coexistence curve and latent heat constrain equations of state through the Clapeyron equation. This is the useful connection: local thermodynamic response, global free-energy stability and the geometry of phase boundaries describe the same material at different levels.

Write $y=g{\mu }_{B}B/(k_{B}T)$. Summing the Boltzmann factors gives the one-atom partition functionSince $⟨m⟩=-{\partial }_y\log Z_1$, the Brillouin magnetization formula isHere $M$ is the total magnetic moment used by the question; the magnetization per unit volume is $M/V$. For independent distinguishable atoms, $Z_N=Z_1^N$, with no indistinguishability factor.

Putting $J=1/2$ yieldsThe apparent singularities of the hyperbolic cotangents at $B=0$ cancel; the finite sum makes the zero-field limit explicit.^[ref]

At fixed particle number, $dU=TdS-PdV$. The stated Maxwell relation givesFor the intended classical monatomic Van der Waals gas, equipartition supplies $f(T)=3N{k}_{B}T/2$: the attractive mean-field term is temperature independent and there are three translational quadratic degrees of freedom. Consequently,The caloric closure of the van der Waals equation is essential. Caloric assumption: the pressure equation alone determines the volume dependence, but not the function $f(T)$. The monatomic kinetic-energy model is the additional assumption behind the claimed formula.^[ref]

Bose-Einstein condensation is macroscopic occupation of a single lowest-energy one-particle state. For a homogeneous three-dimensional ideal gas of conserved massive bosons, set that energy to zero. The excited-state population isHere $g_{3/2}(z)={\sum }_{r\ge 1}{z}^r/r^{3/2}$. As $\mu \to 0^{-}$, the excited states can hold only $V\zeta (3/2)/{\lambda }_T^3$ particles. Additional particles occupy the ground state. ThusThe isolated ground state must be counted separately from the continuum density of states; otherwise one misses the condensate. A dilute atomic gas is a physical realization of the quantum-degeneracy regime, but trapping geometry and interactions modify the homogeneous ideal-gas formulas.^[ref]

In the homogeneous ideal gas, one and two dimensions do not have finite-temperature condensation in the thermodynamic limit because the low-energy population integral diverges. Finite size or a trap changes that conclusion's assumptions. Condensation and superfluidity are distinct: the ideal quadratic gas has zero Landau critical velocity, while interactions can support a sound mode and superfluid flow. Neither zero viscosity nor phase-rigid flow follows from the occupation formula alone.