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For low temperature, factor out the largest Boltzmann factor from the dominant eigenvalue:Thus the Helmholtz free energy per site has the expansionDifferentiate at fixed $J$, remembering that $K=J/(k_{B}T)$. The entropy per site isThe nonzero residual entropy comes from an exponentially large ground-state degeneracy, generated by geometrical frustration. A triangle with antiferromagnetic energy $J[st+(s+t)\tau ]$ has minimum energy $-J$: if $s=t$, the apex must have the opposite sign; if $s=-t$, either apex sign is permitted. The two all-equal configurations instead have energy $3J$. Thus exactly one of a minimum-energy triangle's bonds is frustrated. The bridge can simultaneously minimize its energy at $-J$ by requiring $t=-s^{\prime }$, where $s^{\prime }$ is the next cell's first base spin.

For fixed $s$ and $s^{\prime }$, the number of allowed apex choices is therefore two when $s=s^{\prime }$ and one when $s=-s^{\prime }$. The zero-temperature counting transfer matrix on successive first base spins isThe eigenvalues of $G$ are $3$ and $1$. Dividing $k_B\ln g_M$ by the $3M$ sites and taking the thermodynamic limit gives $k_B\ln 3/3$, independently confirming the derivative calculation. This does not assert a unique ordered ground state: the extensive choice of frustrated bonds is precisely what leaves entropy at zero temperature. At high temperature, all Ising spins become independent and $s\to k_B\ln 2$ per site.

The Boltzmann factor factorizes over bonds. Define the spin-chain transfer matrix byHere ${\delta }_{ab}$ is the Kronecker delta. Summing the product of these matrix elements over all configurations, and closing the final bond by periodic boundary conditions, gives the canonical partition functionHence, writing $I$ for the identity matrix and $P$ for the all-ones matrix,Every element of $P^2$ is a sum of $Q$ ones, while every diagonal element of $P$ is one. ThusMore explicitly, for $u=(1,\dots ,1{)}^T$, $Pu=Qu$. If ${\sum }_a{v}_a=0$, then $Pv=0$. These complementary eigenspaces have dimensions $1$ and $Q-1$. Equivalently, $P/Q$ is the orthogonal projector onto the linear span of $u$. The eigenvalues and their eigenvalue multiplicities are thereforeSince the two matrices share these eigenvectors, the exact finite-ring canonical partition function isBoth terms matter in the finite-ring low-temperature limit: discarding the second would lose the correct ground state degeneracy factor and introduce a spurious single-wall contribution.

Put $x=e^{-\beta J}$. Factoring out the ground state Boltzmann factor from the exact canonical partition function givesAt fixed finite $L$, the binomial theorem givesTaking the logarithm yields the finite-ring low-temperature expansion of the Helmholtz free energy,The constant term is the ground state energy. The term $-{\beta }^{-1}\log Q$ is the contribution of the ground state degeneracy factor. The first thermal correction follows directly by retaining the first excited state manifold:Each pair of domain walls costs $2J$, explaining the absence of an $e^{-\beta J}$ term. For $Q=2$, every configuration has an even number of domain walls, so the displayed remainder improves to $O_L(e^{-4\beta J}/\beta )$.

This is an expansion for fixed $L$, not a uniform thermodynamic limit. In particular, the small correction requires $\left(\genfrac{}{}{0ex}{}{L}{2}\right)(Q-1)x^2\ll 1$, with higher terms also controlled as $x\to 0$ at fixed $L,Q$. Taking $L\to \infty$ first instead selects ${\lambda }_{+}$ and givesIts leading correction is of order $x$. The finite-ring pair constraint and the extensive population of domain walls in an arbitrarily long chain explain the different ordering of these two limits.

After the shift $x(\tau )=x_0(\tau )+y(\tau )$, the endpoint values of $y$ are both zero. Expanding the Euclidean action givesBy integration by parts, the cross term isThe first term vanishes by the homogeneous Dirichlet boundary conditions, and the second by the Euler-Lagrange equation. The translation has unit Jacobian determinant in the time-sliced configuration-space path integral, so $\mathcal{D}x=\mathcal{D}y$. Consequently classical-path factorization of a quadratic path integral is exact here:All dependence on the common endpoint $x$ lies inThe endpoint domain and the integrand defining $Z$ are the same for every $x$. This $Z$ is a zero-endpoint Gaussian path integral; the canonical partition function instead integrates the entire diagonal kernel over $x$. The final sentence of the printed part omits $/\hslash$ from its exponential: the physical Boltzmann factor has the dimensionless exponent $-S/\hslash$.

An optional normalization check uses the Dirichlet oscillator fluctuation determinant. Its eigenvalues are $m[(n\pi /\mathcal{T}{)}^2+{\omega }^2]$, $n\ge 1$. Relative to the free-particle Gaussian path integral, the hyperbolic-sine infinite product givesThus $Z=\sqrt{m\omega /[2\pi \hslash \sinh (\beta \hslash \omega )]}$. Integrating the resulting diagonal thermal kernel of a quantum harmonic oscillator by a Gaussian integral yields $\mathrm{Tr}{e}^{-\beta H}=[2\sinh (\beta \hslash \omega /2){]}^{-1}$, including the zero-point energy.

Because $S[x_0]=0$ at $x=0$, the factor $Z$ cancels from the ratio. Write $z=\beta \hslash \omega$ and $q=\sqrt{m\omega /\hslash }x$. The exact answer isAt high temperature, $z\ll 1$. Expanding the hyperbolic tangent givesThe displayed leading expression is the ratio of classical Boltzmann factors for potential energy $V(x)=m{\omega }^2{x}^2/2$ and $V(0)=0$. Indeed the classical canonical ensemble givesThe momentum Gaussian integral is independent of $x$ and cancels in the ratio. This is the classical phase-space limit of the thermal kernel. Here $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant; the associated classical position variance is $1/(\beta m{\omega }^2)$. At fixed $x$, $R$ ultimately tends to one as temperature tends to infinity, while its width grows.

At low temperature, $z\gg 1$. Now $\tanh (z/2)=1-2e^{-z}+O(e^{-2z})$, so at fixed $q$Unlike the classical high-temperature expression, this limiting width remains nonzero at zero temperature, reflecting the ground state of the quantum harmonic oscillator.

Insert the complete orthonormal basis of energy eigenstates on either side of the Boltzmann factor. The spectral theorem givesAt low temperature, the gap $\hslash \omega$ suppresses excited energy eigenstates. The harmonic oscillator eigenfunction of the ground state isThereforeThe common zero-point energy factor cancels. Dividing the kernel by its trace would give the position probability density of the canonical ensemble; at zero temperature this tends to $|{\phi }_0(x){|}^2$, whose variance is $\hslash /(2m\omega )$. Thus zero temperature projects onto the ground state and retains its quantum spread, rather than concentrating at the minimum of the classical potential energy.

There is also a check on the first correction. With $q=\sqrt{m\omega /\hslash }x$, the first excited harmonic oscillator eigenfunction satisfies $|{\phi }_1(x){|}^2=2q^2|{\phi }_0(x){|}^2$ and ${\phi }_1(0)=0$. Hence the spectral theorem gives $R=e^{-q^2}[1+2q^2{e}^{-z}+O(e^{-2z})]$, in agreement with part (c).

Write $K=\beta J$, where $\beta =1/(k_{B}T)$ is the inverse temperature, $T$ is temperature and $k_B$ is the Boltzmann constant. In the canonical ensemble, each spin configuration has probability $e^{-\beta H}/Z_N$. The nearest-neighbor Boltzmann factors multiply, so the spin-chain transfer matrix, in the order $+1,-1$, isMultiplying its entries around the ring and summing over every Ising spin yields the partition function:The trace implements the periodic boundary conditions; replacing it by an endpoint sum would describe a different partition function. The normalized eigenvectors and corresponding eigenvalues areThe negative eigenvalue is essential for the antiferromagnetic Ising model. Introduce their ratioThe exact finite-$N$ partition function and Helmholtz free energy areFor finite $K>0$, $|t|<1$, so $Z_N>0$ for both even and odd $N$. The last term retains the finite-ring effect of geometric frustration.

To insert an Ising spin into this trace, use the diagonal matrix $\mathsf{S}=\mathrm{diag}(1,-1)$. In the eigenbasis of $\mathsf{T}$, it exchanges $|+⟩$ and $|-⟩$, and hence has vanishing diagonal entries. The one-point correlation function is thereforeThe same conclusion follows directly from spin inversion symmetry: reversing every Ising spin preserves the energy and the Boltzmann factor while reversing $S_1$.

For the two-point correlation function, the two insertions divide the ring into paths of $n$ and $N-n$ bonds. ThusIn particular, the exact finite-ring antiferromagnetic Ising correlation isBecause the one-point correlation functions vanish, this is also the connected correlation function. It gives $C_0=1$ and $C_n=C_{N-n}$, as required by $S_1^2=1$ and exchanging the two paths around the ring. At $K=0$, distinct Ising spins are independent and $C_n=0$ for $1\le n\le N-1$; $C_0=1$ is understood separately if a power $0^0$ occurs in the displayed expression.

Full dissociation gives bulk concentrations in $\mathrm{mM}$ of $155$ for sodium, $5$ for potassium, $2$ for calcium, $114$ for chloride, and $25$ for hydrogen phosphate. The positive and negative charge concentrations are both $164\mathrm{mM}$, so the bulk is electrically neutral. At $T=298.15\mathrm{K}$,The Boltzmann distribution then givesFor completeness the coion concentrations are $114e^{-\nu }=6.65196\mathrm{mM}$ and $25e^{-2\nu }=0.0851195\mathrm{mM}$. They are depleted, not zero. Using $1\mathrm{mM}=1\mathrm{mol}{\mathrm{m}}^{-3}$ and multiplying molar concentrations by the Avogadro constant in the Grahame equation givesThis calculation retains every species. Since $|{\varphi }_0|/V_T=2.84$, a linear Debye–Hückel approximation is not justified. The large accumulation, especially of divalent counterions, follows from the valence in the Boltzmann factor. However, multi-molar local concentrations also warn that point-ion mean-field theory neglects finite ion size, hydration, specific adsorption, correlations, and changes in local permittivity. These are predictions of the stated idealized model, not precise measured interfacial concentrations.

Each Cartesian component of one link has zero mean and variance $b^2/3$, with zero cross-component covariance. By the multivariate central limit theorem, for large $N$ on the coil scale the vector probability density isIt is normalized with respect to $d^{3}r$; the radial probability density is $4\pi r^{2}p(\mathbf{r})$. The Gaussian limit of the end-to-end distribution of a freely jointed chain is an approximation, not the exact finite-$N$ law near $r=Nb$.

At prescribed vector separation the configurational entropy is $S(\mathbf{r})=k_B\ln p(\mathbf{r})+\text{constant}$. The ideal chain has no extension-dependent internal energy, so its Helmholtz free energy isThe externally required force is ${\nabla }_{\mathbf{r}}A$, while the chain's restoring force has the opposite sign. ThusThe last equation also follows by multiplying $p(\mathbf{r})$ by the Boltzmann factor $e^{fz/(k_{B}T)}$ and completing the square. This entropic spring resists extension because fewer link configurations realize a stretched chain. At the same small applied force, increasing temperature reduces extension; at the same extension it increases the required force. Do not differentiate $-k_{B}T\ln [4\pi r^{2}p(\mathbf{r})]$ to obtain the fixed-vector Hooke law, since that radial measure includes an additional geometric multiplicity. The low-force condition is $fb/(k_{B}T)\ll 1$, equivalently $⟨z⟩\ll Nb$.

The rotational state populations start at $N_0/N_0=1$. Their $2J+1$ statistical weight factor initially favors higher $J$, while the Boltzmann factor eventually suppresses them, producing one maximum followed by a tail tending to zero. The physically allowed values occur at integer $J$; a continuous curve only guides the eye.

For $k_{\mathrm{B}}T\gg \delta$, the Boltzmann factors of the orientations become almost equal, giving $3^N$ nearly equiprobable microstates. The Boltzmann entropy of that distribution approachesThe exact limiting value is reached only as $T\to \infty$ for a nonzero splitting. The useful crossover scale is $T\sim \delta /k_{\mathrm{B}}$; one needs $T\gg \delta /k_{\mathrm{B}}$ for a good saturation approximation. The binding offset $\epsilon$ is not an additional population scale here because all sites remain occupied and it shifts every orientation equally.

Put $x={\mu }_{\mathrm{B}}B/(k_{\mathrm{B}}T)$. Summing the two Boltzmann factors gives the occupied-site internal partition functionIf the single atom can occupy any of the $M$ identical sites, its complete surface partition function isOne also commonly calls the one-site factor $z$ itself $q_{\mathrm{ads}}$. Both conventions are valid, but the combinatorial factor in part (c) must use the same convention. We use the complete one-atom surface function $q_{\mathrm{ads}}=Mz$ below and give the equivalent one-site expression.

For fixed degeneracies and ionization energy,Therefore $K_c$ increases with temperature in this model. Ionization costs positive energy, so it is suppressed at low temperature by its Boltzmann factor. At higher temperature that penalty becomes less important.

Ionization also creates an additional independently translating particle and more ways to distribute the particles and their momenta. This favours the ionized mixture through its translational and mixing entropy, consistent with the $T^{3/2}$ prefactor and increasing ionization at high temperature.

Choose the ground-state energy as zero and write $x=\epsilon /(k_{B}T)$. The molecular partition function must sum individual states, so the triply degenerate excited level contributes three Boltzmann factors:The probabilities of occupying the two levels areEach of the three individual excited states has probability $e^{-x}/q$. At high temperature the four states become equiprobable, so the two levels have probabilities $1/4$ and $3/4$, not $1/2$ each. This is the distinction between a level and its state degeneracy.

Differentiating each Boltzmann factor yieldsThe derivative is positive: warming increases occupation of both excited energy levels. At $100\mathrm{K}$,

Let $x=g{\mu }_{B}B/(k_{B}T)$. The single-ion partition function and mean projected magnetic moment areFor $J>0$, the geometric sum also givesIn the high-temperature, weak-polarization regime $|x|J\ll 1$, expand the Boltzmann factors. The numerator's zero-order term vanishes because $\sum m_J=0$, and the first-order term uses $\sum m_J^2=J(J+1)(2J+1)/3$. The denominator is $2J+1+O(x^2)$. ThereforeThis fixed-field per-ion Curie coefficient describes mean moment per ion at fixed field, so it includes $B$. It is not the field-independent volume-susceptibility Curie constant. For number density $n$ and $B\simeq {\mu }_{0}H$, the latter is $C_{\chi }=n{\mu }_0(g{\mu }_B{)}^{2}J(J+1)/(3k_B)$, giving ${\chi }_H=C_{\chi }/T$. At $J=0$, the specified moment is zero and both expressions vanish.

Assume thermal equilibrium, a three-dimensional nondegenerate semiconductor, one parabolic conduction extremum and one parabolic valence extremum, with two spin states in each. Use the supplied effective masses as scalar density-of-states parameters. With conduction energy $E=E_c+\epsilon$, state counting gives the per-volume density of statesSince $E_c-\mu \gg k_{B}T$, replace the Fermi-Dirac distribution by its Boltzmann factor. Using $x=\epsilon /(k_{B}T)$ and the supplied integral,The electron hole occupation is $1-f(E)$ in the valence band. Applying the same integration below $E_v$ givesTheir product cancels the chemical potential, giving the semiconductor mass-action lawThus, for masses expressed in ordinary mass units,The labels $B,C$ in this last box are exponents, unrelated to the magnetic field or Curie constant above. Valley degeneracy or multiple valence bands must be included in the effective density of states, or absorbed into suitable density-of-states effective masses, not counted twice. Real silicon has anisotropic multiple valleys, so its conductivity mass is not automatically its density-of-states mass; the numerical answers below implement the paper's supplied scalar-mass model.^[ref]

The formula is not an identity for arbitrarily degenerate carriers. For a concrete counterexample, take $\mu -E_c=10k_{B}T$ while holes remain dilute. Fermi occupation gives $n/N_c\simeq 24.1$, whereas the Boltzmann replacement would give $e^{10}\simeq 22026$; the predicted product then differs by about three orders of magnitude. Separate electron and hole quasi-Fermi levels out of equilibrium would also prevent the cancellation used above.

Set $K=\beta J$. The transfer matrix for the one-dimensional Ising model assigns a Boltzmann factor to each bond. In the Ising spin order $(+1,-1)$, it isMultiplying these bond factors and summing every Ising spin implements the canonical ensemble. The periodic boundary conditions close the product into a trace:The eigenvectors $|+⟩=(1,1{)}^T/\sqrt{2}$ and $|-⟩=(1,-1{)}^T/\sqrt{2}$ form an orthonormal basis, with eigenvaluesThe negative second eigenvalue encodes alternating correlations. Defining $q={\lambda }_{-}/{\lambda }_{+}=-\tanh K$, the exact partition function and total Helmholtz free energy areThe parity-dependent term must be retained for finite $N$; in particular it nearly cancels the leading term of $Z_N$ at low temperature when $N$ is odd.

Represent insertion of one Ising spin by $\Sigma =\mathrm{diag}(1,-1)$. In the eigenbasis of $T$, this matrix swaps $|+⟩$ and $|-⟩$. Its diagonal entries in that eigenbasis vanish, giving the exact one-point correlation functionEquivalently, configurations pair with their opposites under spin inversion symmetry, with the same Boltzmann factor and opposite $S_1$.

Two insertions separated by $n$ bonds give the two-point correlation functionThus the exact finite antiferromagnetic Ising-ring correlation isThe two numerator terms represent the two arcs between the sites. The formula has $C_N(0)=1$ and $C_N(n)=C_N(N-n)$, as required by the periodic boundary conditions. Since the one-point correlation function is zero, this is also the connected correlation function. At $K=0$, every Ising spin is independent, so $C_N(n)=0$ for distinct sites and $C_N(0)=1$; powers with exponent zero are understood as one.

The Boltzmann distribution for two atomic levels at thermodynamic equilibrium iswhere $\omega$ is the statistical weight of an atomic level. For a resolved level of total angular momentum $J$, $\omega =2J+1$ when magnetic substates are unresolved.

To extend this counting to the continuum, assume an ideal, nondegenerate plasma with classical free electrons and neglect the small difference between the translational masses of the two heavy ionic species. Let $z_e=e^{{\mu }_e/kT}$ be the electron fugacity in the Saha equation, with the electron chemical potential measured relative to its rest energy. The two electron spin states and the phase-space cell volume $h^3$ giveChemical equilibrium for capture equates the bound species' chemical potential to the sum for its parent ion and free electron. Thus applying Boltzmann factors to a bound level at energy $-I_n$ gives $N_n/N_{+}=({\omega }_n/{\omega }_{+})z_e{e}^{I_n/kT}$. Eliminating the electron fugacity proves the level-specific Saha ionization equation:The factor two is the electron spin multiplicity; the factor with power $3/2$ is the inverse free-electron translational state density. A total-ionization Saha equation sums the bound states into an internal partition function, whereas this expression refers to one specified level.

For dielectronic recombination, write the entrance ion as $X^{(z+1)+}$ in ground state $i$, and the intermediate state of $X^{z+}$ as $d=(j,n\ell )$. Capture excites the core from $i$ to $j$ while binding the incident electron. The resonance energy relative to the entrance continuum is $\bar{E}=E_{ij}-I_{n\ell }>0$. The doubly excited state can eject the electron by autoionization, or undergo radiative stabilization to a bound state.

Let $A_{di}^a$ be the partial autoionization rate back to the entrance ion, $A_a$ the total autoionization rate, $A_r$ the total radiative decay rate, and $A_s$ the rate of radiative decays that stabilize into bound levels. For an isolated narrow resonance, the Saha-Boltzmann equilibrium ratio has the bound-state energy replaced by the positive resonance energy:At equilibrium, inverse dielectronic capture and autoionization fluxes balance: $N_e{N}_i{C}_d=N_d^{\mathrm{eq}}{A}_{di}^a$. This establishes the capture coefficient even when the actual plasma is not in local thermodynamic equilibrium. In the low-density, weak-radiation recombining plasma, the intermediate level instead has the stationary balanceMultiplying its population by the successful stabilization rate gives the actual recombination coefficientHere the equilibrium population was used only to determine the capture rate by detailed balance; it was not assumed to equal the intermediate population when photons escape. This is precisely why the competing decay probabilities appear.

Use the supplied core-transition relation between the Einstein coefficients and atomic oscillator strength,Here $\alpha$ is the fine-structure constant and $a_0$ the Bohr radius, not a recombination coefficient. Factor out $A_{ji}$ and defineSubstitution yieldsThis definition accommodates several entrance and decay channels. In the common spectator-electron approximation with one autoionization channel and stabilization by the core transition, $A_s=A_r=A_{ji}$ and $A_{di}^a=A_a$. If all coupled substates of the captured $n\ell$ electron are summed, $g_d=2(2\ell +1)g_j$, givingFor an intermediate state resolved by fine structure, retain its actual $g_d$ instead of that summed weight. The stabilization probability itself is $A_s/(A_a+A_r)$; the requested ${\beta }_{j,n\ell }$ is not just this branching probability, because the prefactor has already factored out the core radiative rate and statistical weights of atomic levels. Radiative transitions to another still-autoionizing state require that state's eventual survival probability, rather than automatically counting every emitted photon as stabilization.