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The extension of the one-dimensional freely jointed chain isAs the integer $n_{+}$ runs from $0$ to $n$, the possible extensions are thereforeFor fixed $n_{+}$ and $n_{-}=n-n_{+}$, choosing which $n_{+}$ of the $n$ links point in the positive direction determines the microstate. The state degeneracy is consequently the binomial coefficient

Several microstates can have the same extension. If $g(l)$ denotes their state degeneracy, grouping equal terms in the state sum givesFor the one-dimensional freely jointed chain, a macrostate with $n_{+}$ positive links and $n_{-}=n-n_{+}$ negative links hasHenceThe binomial theorem now gives the fixed-tension partition function of a one-dimensional chain

Separation with Dirichlet walls givesFor $a<b<c$, the ground state $(1,1,1)$ has a nondegenerate energy. The cheapest excitation changes the quantum number in the longest direction, so the first excited state $(1,1,2)$ also has a nondegenerate energy.

If $a<b=c$, the ground-state energy remains nondegenerate, but $(1,2,1)$ and $(1,1,2)$ have equal first-excited energy, giving degeneracy two.

Measure single-particle energy from the lowest state and write ${\epsilon }_k=c{k}^{\alpha }$, with $c>0$. In a homogeneous system, the number of wavevector states in a thin spherical shell is proportional to $k^{d-1}dk$. Changing variables from $k$ to $\epsilon$ gives the density of states per unit spatial volume,where $S_{d-1}$ is the area of the unit sphere in $d$ dimensions; an additional internal state degeneracy would multiply this expression by a constant.

The Bose-Einstein distribution has $\mu \le 0$. At saturation, $\mu \to 0^{-}$, the maximum excited-state number density isOnly the lower endpoint can cause divergence. There the integrand is proportional toThe improper integral converges precisely when its power exceeds $-1$. ThusWhen $d=\alpha$, the divergence is logarithmic; when $d<\alpha$, it is a power divergence. In either case the excited states can accommodate any fixed finite density at every positive temperature, so there is no bulk Bose-Einstein condensation at nonzero temperature.

For completeness, putting $s=d/\alpha >1$ and using the Gamma function and Riemann zeta function givesAt fixed density $n$, the right-hand side reaches $n$ at $T_c$; below it the excess particles occupy the lowest state, with condensate fraction $1-(T/T_c{)}^s$. This is Bose condensation with power-law dispersion in the homogeneous thermodynamic limit. A finite vessel or a trapping potential changes the low-energy state count and must not be silently substituted for that setting.

Canonical probabilities. A canonical ensemble describes a system of fixed particle number and fixed external parameters such as volume, weakly coupled to a much larger thermal reservoir. They exchange energy and reach thermal equilibrium, while the reservoir's temperature stays effectively constant. The interaction energy is negligible in the state counting. The combined isolated system is described by equal probabilities for its accessible microstates.

If its total energy is $E_{\mathrm{tot}}$, the probability of a particular system microstate $i$ is proportional to the reservoir multiplicity at $E_{\mathrm{tot}}-E_i$. Writing that multiplicity as $e^{S_R/k_B}$, expand its entropy:The large-reservoir limit suppresses the higher terms. Normalization therefore givesThe sum counts individual microstates; when grouping them by energy, their state degeneracies must be included. This $Z$ is the canonical partition function.

Classical orientational heat capacity. Hold the magnetic field fixed. Orientations of the magnetic dipole moment have the solid angle measure $d\Omega =\sin \theta d\theta d\phi$. With $x=\beta mB$, the angular partition function isChoosing a normalized angular measure would remove $4\pi$, with no effect on the energy derivatives. Differentiating at fixed $B$,The angular average $⟨\cos \theta ⟩=\coth x-1/x$ is the Langevin function. Since $dx/dT=-x/T$,This is the heat capacity of a classical magnetic dipole, in $\mathrm{J}{\mathrm{K}}^{-1}$ for one magnetic moment. The word specific here refers to that single moment; $N$ independent magnetic moments have $N$ times this heat capacity. At $B=0$ the energy is independent of orientation and $C_B=0$, obtained by the continuous $x\to 0$ limit. At fixed nonzero $B$, the classical low-temperature limit instead gives $C_B\to k_B$.

Equipartition and its proof. The equipartition theorem states that an independent quadratic term $a{q}^2/2$ in a classical Hamiltonian, with $a>0$, contributes $k_{B}T/2$ to its mean energy. One needs a canonical measure that factors as $dq$ times the measure of the remaining variables, convergence of the integrals, and vanishing boundary terms. The coordinate must not be restricted by a thermally important hard boundary or a coupled constraint.

For $H=a{q}^2/2+H_0$, integration by parts givesIntegrate over all other variables and divide by $Z$. It follows thatThe same reasoning applies to quadratic momenta among the canonical variables. It is the number of quadratic energy contributions that is counted, not an unrestricted count of coordinate labels.

Two soft orientational directions. When $x\gg 1$, the moment is concentrated near alignment. Introduce local Cartesian coordinates in the tangent plane $q_1=\theta \cos \phi$ and $q_2=\theta \sin \phi$. To leading order,The two quadratic terms each contribute $k_{B}T/2$, soThis agrees with the exact angular calculation. Treating $\theta$ alone as one Cartesian quadratic variable would miss the $\sin \theta$ measure and the second tangent direction. There are no extra rotational kinetic terms in the stipulated Hamiltonian; adding such terms would be a different model.

Quantum spin one. The spin projections are $m_s=-1,0,1$, with $S_z=\hslash m_s$. Put $\Delta =\hslash |\gamma |B>0$, where $\gamma$ is the gyromagnetic ratio. The three energies are $-\Delta ,0,+\Delta$, irrespective of the sign of $\gamma$. Relative to the ground state they are $0,\Delta ,2\Delta$. Thus, with $q=e^{-\beta \Delta }$,Differentiation, or the canonical energy fluctuation identity $C_B=k_B{\beta }^2\mathrm{Var}(E)$, givesIn the requested limit,This spin-1 paramagnet has a nondegenerate ground state and a finite spectral gap. Excitations are exponentially rare, so its heat capacity tends to zero. Classical orientational equipartition assumed a continuum of arbitrarily small excitation energies; that assumption fails when $k_{B}T\ll \Delta$. This is a change in the accessible state spectrum, not a violation of the equipartition theorem within its classical domain.

A rigid rotor restricted to one angular coordinate $\varphi$. Single-valued spatial states $e^{im\varphi }$ have $m\in \mathbb{Z}$ and energy $E_m={\hslash }^2{m}^2/(2I)$, with moment of inertia $I$. The zero-angular-momentum level has state degeneracy one; every nonzero $|m|$ has two directions. For labelled molecular ends its high-temperature rotational partition function is $\sqrt{\pi T/{\theta }_{\mathrm{rot}}}$, where ${\theta }_{\mathrm{rot}}={\hslash }^2/(2I{k}_{\mathrm{B}})$. Equivalent orientations and nuclear spin statistics require a further molecular counting convention. One quadratic angular momentum gives rotational heat capacity $k_{\mathrm{B}}/2$ per molecule, rather than $k_{\mathrm{B}}$ for an unrestricted linear rigid rotor.

When a single-particle energy is independent of its spin projection and all projections are available, the state degeneracy due to spin is $g_s=2S+1$. A spin-half gas has $g_s=2$ if both projections are available, while a fully polarized gas can have one populated spin species. A magnetic field that resolves Zeeman splitting requires counting the separate energies instead of multiplying one common occupancy by $g_s$.

A tight binding chain of one species of fermions with nearest-neighbour hopping and density interaction. At half filling and $V/|t|\to +\infty$, minimizing the interaction requires alternating occupied and empty sites. An even periodic chain has two such configurations at $t=0$; finite hopping admixes defects through virtual hopping. A finite ring can select superpositions rather than one ordered product state. Two uncoupled copies have four limiting ordered choices when each is half filled, without a general claim of exact finite-hopping fourfold state degeneracy.

For $s_1=s_2=3/2$, angular momentum addition gives one copy of $j=0,1,2,3$. The identity $2{\mathbf{S}}_1\cdot {\mathbf{S}}_2={\mathbf{J}}^2-{\mathbf{S}}_1^2-{\mathbf{S}}_2^2$ gives the four eigenvalues in the title, with state degeneracy $1,3,5,7$. These exhaust the $4\times 4$ spin space. If the particles are identical fermions, an appropriate spatial exchange parity accommodates each sector; only a specified spatial restriction can remove sectors.

Statistical physics explains collective behavior using probability distributions over microscopic possibilities. Its central surprise is that incomplete microscopic knowledge can still produce precise predictions for internal energy, entropy and heat capacity. The bridge is statistical mechanics: specify the available microstates, assign their probabilities, then compute the expected values of measured quantities.

A two-level system already displays the idea. Warming it makes its higher energy more likely; its heat capacity first rises, then falls as the two probabilities approach equality. With two distinguishable units, merely counting microstates explains why the middle energy has twice the state degeneracy. These small examples expose the machinery before the much larger sums needed for a photon gas or an Ising model.

The appeal is economical: the same microscopic energies and probabilities explain both an average reading and its variance. A measurement's spread therefore becomes useful physical information rather than an inconvenience to discard.