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For the same atoms in a three-dimensional volume $V$, the one-atom translational partition function is $q_{\mathrm{trans}}^{(3)}=V/{\Lambda }^3$. Comparing the same $N$ atoms at the same temperature cancels the common $N!$ in their canonical partition functions. Therefore the molar adsorption confinement free energy, with the sign defined as adsorbed minus free, isThe logarithm is dimensionless: $V/\sigma$ and $\Lambda$ are both lengths. For a macroscopic gas depth $V/\sigma \gg \Lambda$, confinement raises the translational Helmholtz free energy. Equivalently, the translational state count is reduced by a factor $\sigma \Lambda /V$ per atom. The question supplies neither a volume nor an adsorption binding energy, so this answer is symbolic and describes the confinement cost, not a numerical net adsorption free energy.

For example, a binding energy $D>0$ per atom would supply a separate $-N_{\mathrm{A}}D$ to $\Delta F_m$, together with any normal-motion internal contribution retained by a more detailed model. Adsorption can then be favourable despite the loss of free translational phase space. Merely calling the change a loss must not reverse the sign of the explicitly defined $F_{\mathrm{ads}}-F_{\mathrm{gas}}$.

Treat each species as a dilute classical ideal gas: molecules of the same species are indistinguishable, while different species are distinguishable. Let $N_i$ be the number of molecules of species $i$, $q_i(T,V)$ its molecular partition function with energies measured from a chosen internal reference, and ${\epsilon }^{0,i}$ that reference energy per molecule relative to a common chemical energy convention. Then the mixture's canonical partition function factorizes:The $N_i!$ corrects the overcounting of indistinguishable particles. The exponential restores the species-dependent reference energies that matter for reactions. Thus the Helmholtz free energy isUsing Stirling's approximation, $\ln (N_i!)\simeq N_i\ln N_i-N_i$, and differentiating at fixed $T,V,N_{j\ne i}$ yieldsThe ideal-gas $q_i$ does not depend on the number of molecules, so no additional $\partial q_i/\partial N_i$ term appears. These are chemical potentials per molecule; multiply by the Avogadro constant for a molar chemical potential. The result assumes the thermodynamic, nondegenerate-gas limit.

For identical, non-interacting molecules, the translational canonical partition function is $Q_N=q_{\mathrm{trans}}^N/N!$. Stirling's approximation gives the Helmholtz free energyAt fixed $V$, $q_{\mathrm{trans}}\propto T^{3/2}$, so differentiation gives $S=Nk[\log (q_{\mathrm{trans}}/N)+5/2]$. For one mole at the standard-state molar volume,Here the quoted molar translational quantity is per mole, and its ratio to $N_A$ is dimensionless. The derivative defining entropy is taken at fixed volume before inserting the standard-state volume; differentiating along a constant-pressure path would give the wrong additive constant.

The surface pressure follows by differentiating the logarithm of the canonical partition function with respect to container area:Its formal divergence is therefore $\varphi =1/2$. This lies outside the low-density regime used to derive the expression. It is a model pole, not an actual geometrical close-packing fraction.

Insert the stipulated near-close-packing relation for $R$ into the hard-disc cage approximation:Writing $y=\sqrt{{\varphi }_{\mathrm{cp}}a/(N{a}_d)}$, we have ${\partial }_{a}y=y/(2a)$ and therefore ${\partial }_a\ln \mathcal{A}=y/[a(y-1)]$. From the supplied crystalline canonical partition function,Its close-packing scale is set by the triangular lattice packing of discs, with ${\varphi }_{\mathrm{cp}}=\pi /(2\sqrt{3})$ in the ideal geometry.

Freeze neighbouring particles and approximate the mobile particle's accessible area by $a_c$. Its one-particle translational contribution to the canonical partition function is $a_c/{\Lambda }^2$ in two dimensions. Neglecting correlations between different cages,There is no additional uncancelled $N!$: assigning particles to cages removes permutations, while restoring all $N!$ assignments cancels the indistinguishability factor. The result is a free-volume approximation; collective cage motion is excluded by construction.

Use units $k_B=\hslash =c=1$. In critical light-cone gauge in string theory, each chiral sector of the closed bosonic string has 24 transverse string oscillator species. Their generating function isEach factor sums the independent occupation numbers of the string oscillators at level $m$. For $q=e^{-t}$ with $t\downarrow 0$, the modular transformation of the Dedekind eta function givesExtract the coefficient by a contour integral and apply the saddle-point approximation to the exponent $(N-1)t+4{\pi }^2/t$. The saddle is $t_{\ast }=2\pi /\sqrt{N-1}$, with exponent $4\pi \sqrt{N-1}$ and second derivative $(N-1{)}^{3/2}/\pi$. The factor $t^{12}$ gives $(N-1{)}^{-6}$, and the Gaussian width gives $(N-1{)}^{-3/4}/\sqrt{2}$. HenceFor a noncompact closed string, closed-string level matching imposes equal left and right levels, but does not correlate their transverse polarizations. Therefore the large-level degeneracy of a closed bosonic string isThe mass relation is ${\alpha }^{\prime }{M}^2=4(N-1)$. Thus $8\pi \sqrt{N-1}={\beta }_{H}M$ withDistinguish degeneracy per discrete level from a smoothed density per unit rest mass. Since $dN/dM={\alpha }^{\prime }M/2$, the latter isThe exponential growth comes from the many ways of distributing string oscillator energy along a string; it is much faster than a fixed finite collection of particle species. The corresponding high-energy entropy is $S(E)={\beta }_{H}E+O(\log E)$, so its microcanonical temperature tends to $T_H$. Additional energy can go into very long strings rather than raising the temperature substantially.

The Boltzmann suppression competes with this exponential density. For the string oscillator rest-mass sum alone the large-$M$ tail is $\int dM{M}^{-26}{e}^{-(\beta -{\beta }_H)M}$. It converges for $\beta >{\beta }_H$ and fails for $\beta <{\beta }_H$. To include translational states in 25 noncompact spatial dimensions, take the one-string partition function per unit spatial volume. At large $M$,This follows by expanding the energy as $M+p^2/(2M)+\cdots$ and doing a Gaussian momentum integral. Its high-mass contribution is consequently proportional toThis integral is finite at $\beta ={\beta }_H$, although nonanalytic there: its first twelve derivatives in $\beta$ are finite, while the thirteenth diverges. The endpoint expansion contains a term proportional to $(\beta -{\beta }_H{)}^{25/2}$. Thus one must not assert that the canonical partition function necessarily diverges exactly at $T_H$; the polynomial prefactor and the spatial spectrum matter. This is the finite canonical partition function at the bosonic Hagedorn threshold distinction. The multi-string ideal gas has the same leading Hagedorn threshold, since the higher occupation-cycle terms probe $\ell \beta$, $\ell \ge 2$, and are nonsingular at ${\beta }_H$.

There is also a spacetime diagnosis. Euclidean temperature makes time a circle of radius $R_{\beta }=\beta /(2\pi )$. A string oscillator ground state winding once around that circle, with zero momentum number, hasIt becomes massless at ${\beta }_H$ and tachyonic for $T>T_H$. This thermal winding instability of a closed bosonic string signals failure of the perturbative low-temperature gas and motivates a transition to a different high-energy phase, often described in terms of long-string or winding-mode condensation. Free state counting fixes the threshold and its singular behavior, not the interacting phase's endpoint or the order of a transition.

Finally, the ordinary bosonic tachyon is already an infrared instability at zero temperature. The above canonical discussion is a formal analysis of the high-level spectrum with that separate infrared problem regulated. It does not establish a stable thermal vacuum of the unmodified bosonic theory; an actual equilibrium transition requires an appropriate stable setting and control of interactions.

The numerical contour length is $L=N{b}_0=10^{10}\times 10^{-8}\mathrm{m}=100\mathrm{m}$, and $k_{B}T=4.14\times 10^{-21}\mathrm{J}$. Begin with a three-dimensional freely jointed chain in which each chemical monomer is also an independent statistical link of length $b_0$. This neglects excluded volume, hydrodynamic effects on equilibrium statistics, and stretching of the chemical backbone.

An applied force $F$ gives a one-link canonical partition functionThe mean extension is $z=k_{B}T{\partial }_F\log Z_1^N=N{b}_0(\coth \xi -1/\xi )$. Expanding the Langevin function at small force gives $z=N{b}_0^{2}F/(3k_{B}T)+O(F^3)$. The three-dimensional entropic chain stiffness is thusThe unforced root-mean-square end-to-end distance is $\sqrt{N}{b}_0=10^{-3}\mathrm{m}$, not the contour length. This response is entropic straightening of a fluctuating coil. Axial stretching of a straight material rod instead has stiffness $EA/L$, and is a different phenomenon. Aqueous solvent does not by itself prove ideal-chain statistics: a good-solvent self-avoiding walk has a different coil size and susceptibility. The estimate above states explicitly the model under which the supplied link data suffice.

Count oscillator polarizations at fixed center-of-mass momentum. In a critical closed bosonic string, each chiral sector has 24 transverse oscillator species. A species at frequency $m$ can be occupied $0,1,2,\dots$ times, so its contribution is $(1-q^m{)}^{-1}$. The full chiral generating function is thereforeThe two sectors are independent, while closed-string level matching restricts their oscillator numbers to the same $N$. Thus the number of states at the closed-string mass level is $g_N=d_N^2$, not $d_{2N}$.

Put $q=e^{-\beta }$, with $\beta >0$. The Dedekind eta function obeysThe transformed argument $2\pi i/\beta$ has large imaginary part, so its product approaches one and $\eta (2\pi i/\beta )\sim e^{-{\pi }^2/(6\beta )}$. HenceThis establishes the singular exponential as well as its power prefactor. The corrections to the transformed eta product are exponentially small in $1/\beta$.

The Cauchy coefficient formula, followed by $q=e^{-\beta }$, gives a steepest-descent integral with leading formLet $n=N-1$. The saddle of $S(\beta )=n\beta +4{\pi }^2/\beta$ isAlong the vertical steepest-descent direction $\beta ={\beta }_{\ast }+it$, the quadratic part is $-S^{\prime \prime }({\beta }_{\ast })t^2/2$. Thus the saddle-point approximation gives the Gaussian prefactorThe point $q=1$ supplies the largest exponential contribution. At another fixed root of unity of order $k>1$, the corresponding modular singularity gives a smaller exponential $e^{4\pi \sqrt{n}/k}$, so it does not change this leading asymptotic. Squaring the chiral result gives the large-level degeneracy of a closed bosonic stringReplacing $N-1$ by $N$ gives the equivalent usual leading asymptotic. The power $-27/2$ matters: the two-sector degeneracy is not just an unspecified exponential.

Since $M=2\sqrt{(N-1)/{\alpha }^{\prime }}$, the exponential is $e^{{\beta }_{H}M}$ withHere Boltzmann's constant is one. Coarse-graining the levels into a density per unit rest mass introduces the Jacobian $dN/dM={\alpha }^{\prime }M/2$. ConsequentlyThis is distinct from the degeneracy per level, whose power in $M$ is $-27$. The internal-state contribution to a canonical partition function behaves at large mass asIt converges for $\beta >{\beta }_H$ and diverges for $\beta <{\beta }_H$. At $\beta ={\beta }_H$, this particular internal rest-mass integral converges because of the power $M^{-26}$; one should not discard the prefactor and claim divergence there without specifying the full thermodynamic ensemble. For example, including continuous momenta in 25 spatial dimensions multiplies the high-mass integrand by a factor proportional to $M^{25/2}$, still leaving convergence of this ideal one-string integral at the endpoint, although sufficiently high derivatives are singular. Multi-string effects, volume and interactions affect the detailed limiting thermodynamics.

The leading microcanonical entropy grows as $S(E)\sim {\beta }_{H}E$, with logarithmic corrections, so increasingly large energies are stored in increasingly excited, long strings rather than in an arbitrarily hot ordinary gas. The Hagedorn temperature is the boundary of canonical convergence from above in inverse temperature; extrapolating the free-string canonical description to $T>T_H$ fails. This counting does not settle the interacting phase above that scale. Also, the bosonic ground-state tachyon is a separate vacuum instability; the high-level asymptotic is meaningful as a formal spectrum calculation without asserting a stable bosonic thermal vacuum.

Heat capacity at fixed volume is $C_V=(\partial E/\partial T{)}_{V,N}=T(\partial S/\partial T{)}_{V,N}$. From the Helmholtz free energy differential $dF=-SdT-PdV$ at fixed particle number, equality of mixed derivatives gives $(\partial S/\partial V{)}_T=(\partial P/\partial T{)}_V$. Differentiating once more givesFor the canonical partition function, $F=-k_{B}T\log Z$, $E=k_B{T}^2{\partial }_T\log Z$, and $S=k_B(\log Z+T{\partial }_T\log Z)$. Thus, for $V>aN$,andNo Stirling approximation is needed. The heat capacity is independent of volume, while the pressure is linear in temperature, so both sides of the Maxwell-derived identity vanish.

Use the arc length $s$ and the unit tangent $\mathbf{t}={\partial }_s\mathbf{r}$ of an inextensible filament. In the high-force state choose its positive longitudinal branch. The constraint givesThe extension is $z={\int }_0^L{t}_{z}ds=L-\frac{1}{2}{\int }_0^L|{\mathbf{t}}_{\perp }{|}^{2}ds+\cdots$. The curvature of a space curve obeys ${\kappa }^2=|{\partial }_s\mathbf{t}{|}^2$. Since ${\partial }_s{t}_z=-{\mathbf{t}}_{\perp }\cdot {\partial }_s{\mathbf{t}}_{\perp }+\cdots$ is second order in small fluctuations, its squared contribution is fourth order. Thus the worm-like chain energy through quadratic order isPositive force suppresses transverse orientation fluctuations, while the filament bending modulus penalizes their spatial variation.

To apply the equipartition theorem without missing either component or a mode-counting factor, expand each component in a real orthonormal basis of eigenfunctions of $-{\partial }_s^2$, $t_{\alpha }(s)={\sum }_q{a}_{\alpha q}{w}_q(s)$, with $\int w_q{w}_{q^{\prime }}ds={\delta }_{q{q}^{\prime }}$ and $\alpha =x,y$. The quadratic energy is $\frac{1}{2}{\sum }_{\alpha ,q}(A{q}^2+f)a_{\alpha q}^2$. Each real quadratic degree of freedom contributes $k_{B}T/2$, soFor a chain long compared with the force-dependent correlation length, the interior is translation invariant and the mode sum per length becomes $\int dq/(2\pi )$. Equivalently, for the Fourier transform ${\tilde{t}}_{\alpha }(q)=\int e^{-iqs}{t}_{\alpha }(s)ds$,The negative and positive wavenumbers here represent the real-field mode count consistently, rather than two independent complex modes. Summing both transverse components gives the high-force worm-like chain elasticitywhere the integral follows by setting $q=\sqrt{f/A}u$ and using ${\int }_{-\infty }^{\infty }(1+u^2{)}^{-1}du=\pi$. ThereforeHere $L_p$ is the three-dimensional zero-force persistence length. The result requires $f{L}_p\gg k_{B}T$, negligible backbone stretching, and a long-chain bulk approximation. It does not describe the eventual bond-stretching regime. For a continuum polymer description, the force-dependent length should also exceed the microscopic cutoff.

For comparison, a three-dimensional freely jointed chain has contour length $L=Nb$ and independent link orientations. At force $f$, a link of polar angle $\vartheta$ has Boltzmann factor $e^{x\cos \vartheta }$ with $x=fb/(k_{B}T)$. Its angular canonical partition function isThe total canonical partition function is $Z_1^N$. Differentiating with respect to force gives the force-extension of a three-dimensional freely jointed chain:ConsequentlyThe extension deficits scale as $f^{-1}$ and $f^{-1/2}$ respectively. Inverting them gives force divergences proportional to the first and second inverse powers of the deficit. Freely jointed link rotations cost no bending energy; the worm-like chain instead has a continuum of coupled bending modes whose contributing wavelengths change with force. Setting a long-wavelength coarse-grained link length $b=2L_p$ can match weak-force statistics, but does not make these high-force laws identical.

For the transverse tangent correlation of a stretched worm-like chain, use a separation $r$ consistently: the source writes $C(y)$ but integrates a separation denoted $r$. In the translation-invariant bulk the spatial average is redundant andTo evaluate the integral without an assumed decay law, let $G(r)$ be the inverse Fourier transform of $(A{q}^2+f{)}^{-1}$. Then $(-A{\partial }_r^2+f)G=\delta (r)$. Away from zero, decay at both infinities requires $G=B{e}^{-|r|/\xi }$ with $\xi =\sqrt{A/f}$. Integrating across zero yields $-A[G^{\prime }(0^{+})-G^{\prime }(0^{-})]=1$, so $B=1/(2\sqrt{Af})$. ThusAt $r=0$ this agrees with the two-component equipartition theorem calculation. The correlation length decreases with force and is distinct from the zero-force persistence length. The full tangent correlation has a nonzero aligned background; the exponentially decaying quantity here is specifically the transverse correlation.

For a finite open chain, the source's integral samples $s+r$ beyond $L$ and supplies no endpoint tangent conditions. An exact finite-length correlation is therefore not uniquely specified. The result above is the intended bulk form when $L\gg \xi$. As one explicit finite convention, periodically extending the tangent fluctuations gives the periodic finite-length tangent correlation of a stretched worm-like chainwhere $d_L(r)$ is the distance from zero around the periodic interval. Then $C_L(0)=(k_{B}T/\sqrt{Af})\coth [L/(2\xi )]$, and the extension formula acquires this same finite-length factor. The bulk law is recovered as $L/\xi \to \infty$. This makes the endpoint assumption explicit rather than claiming an unspecified finite-chain correlation is exactly translationally invariant.

Introduce a constant filament bending modulus $B$, which sets the mechanical force scale. The filament has intrinsic curvature of an elastic filament ${\kappa }_0(s)$, and its actual signed curvature in a small-slope Monge representation is $\kappa \simeq {\zeta }_{xx}$. To quadratic order one may identify material arc length $s$ with $x$ in ${\kappa }_0$, whileAt fixed positive extension force, the mechanical potential is bending energy minus $FL$. Replacing $L$ by $\mathcal{L}-(\mathcal{L}-L)$ leaves, up to a force-dependent constant,Its first variation isThe higher-order Euler-Lagrange equation is thereforeFor a Fourier transform relation, take a periodic bulk segment, or a long-filament bulk approximation with negligible endpoint contributions; these are necessary boundary conditions because the question specifies no filament end conditions. Using $f(x)=\int \hat{f}(q)e^{iqx}dq/(2\pi )$, every nonzero wavenumber obeysA zero displacement mode is an arbitrary translation. In a periodic representation a constant intrinsic-curvature mode contributes a constant bending cost but no bulk slope, so it does not contribute to the following deficit. The Parseval identity givesFor clarity about finite normalization, on a periodic interval of length $L$ define ${\kappa }_n=L^{-1}{\int }_0^L{\kappa }_0(x)e^{-i{q}_{n}x}dx$, $q_n=2\pi n/L$. Then the equivalent exact quadratic-order formula isThis is force-extension of an intrinsically curved filament, determined by its particular fixed shape rather than by a thermal orientation entropy.

For an ensemble with quenched intrinsic curvature, first minimize each realization and then average the last expression, replacing $|{\kappa }_n{|}^2$ by its ensemble mean. For stationary bulk disorder, let $S_{\kappa }(q)$ be the power spectrum, normalized so that its correlation is $⟨{\kappa }_0(x){\kappa }_0(x+r)⟩=\int S_{\kappa }(q)e^{iqr}dq/(2\pi )$. The measurable mean deficit per unit projected length isIf the curvature is smooth enough that the second spectral moment is finite, one can take the large-force limit under this integral, obtaining the smooth-curvature high-force extension lawFor a stationary ensemble the integral is $L⟨({\kappa }_{0,x}{)}^2⟩$. Equivalently, the bulk slope is ${\zeta }_x\simeq -(B/F){\kappa }_{0,x}$; a plot of the mean deficit against $F^{-2}$ measures the mean squared intrinsic-curvature gradient, if $B$ is known. The $F^{-2}$ law is a smooth-curvature bulk result, not a universal law for every random filament. For example, ideal white intrinsic-curvature disorder with $S_{\kappa }(q)=S_0$ has an infinite second spectral moment; retaining the full integral instead gives $⟨\mathcal{L}-L⟩/L=S_0\sqrt{B}/(8\sqrt{F})$ in that ideal continuum model.

Endpoint conditions also matter. If each end is moment-free and has no transverse applied force, the natural conditions from the variation areat each end. These can produce a free-end curvature boundary layer of width $\ell =\sqrt{B/F}$. For a constant intrinsic filament curvature $\kappa$, integration of the higher-order Euler-Lagrange equation with zero transverse force gives $B({\zeta }_x{)}_{xx}-F{\zeta }_x=0$ and henceThe last formula follows by integrating ${\sinh }^2[(x-L/2)/\ell ]$, and tends to ${\kappa }^2{\ell }^3/2\propto F^{-3/2}$, despite the vanishing bulk curvature gradient. Thus a finite free-ended measurement must include or suppress endpoint bending before interpreting a bulk $F^{-2}$ fit.

For the freely jointed chain, rigid links have independent orientations, so the restoring force is entropic. Let $T$ now denote absolute temperature, and $\beta =1/(k_{B}T)$ with Boltzmann constant $k_B$. In the usual three-dimensional chain, a segment at polar angle $\theta$ has force energy $-Fb\cos \theta$. Its angular canonical partition function and the full chain partition function areDifferentiating $\log Z_N$ with respect to $\beta F$ yields the exact mean projected displacementHere $\mathcal{L}$ is the Langevin function; this is force-extension of a three-dimensional freely jointed chain, in the fixed-force canonical ensemble. At high extension, $\coth \xi =1+2e^{-2\xi }+\cdots$, soThus the smooth quenched-curvature bulk deficit falls as $F^{-2}$, whereas the three-dimensional freely jointed chain deficit falls as $F^{-1}$ and depends explicitly on temperature. The distinction is mechanical straightening of a fixed intrinsic shape versus orientational entropy; thermal bending fluctuations of a stiff filament have not been included in the first calculation.

If the comparison chain is also restricted to two dimensions, it is a planar freely jointed chain rather than the three-dimensional model. In this case,The last identity follows directly by differentiating the integral defining $I_0$; $I_0$ and $I_1$ are Modified Bessel functions of the first kind. At large $\xi$ the angular weight is proportional to $e^{-\xi {\theta }^2/2}$, giving $⟨\cos \theta ⟩\simeq 1-⟨{\theta }^2⟩/2=1-1/(2\xi )$. Consequently $Nb-⟨L⟩\sim N{k}_{B}T/(2F)$: the same inverse-force exponent, with half the three-dimensional coefficient. Both versions assume inextensible links and forces below the regime where the links themselves stretch.

For the Hamiltonian $p^2/(2m)+V(q)$, the real-time configuration-space path integral iswith its measure defined by time slicing. Wick rotation gives the Euclidean path integral and the thermal trace.

For the infinite square well on $(0,L)$, the Dirichlet boundary conditions select the normalized energy eigenstatesThere is no $n=0$ state: the corresponding sine is identically zero. Taking the trace in this orthonormal basis gives the canonical partition function

With $\beta =1/(k_{B}T)$, the canonical partition function and equilibrium probabilities areDifferentiating $T\log Z$, with fixed energy levels, gives $S=k_B\log Z+⟨E⟩/T$. Since $\log p(n)=-\beta E_n-\log Z$, normalization yieldsThe sums include degeneracy by labeling states rather than distinct energy values, and a zero probability contributes zero by continuity.

For the canonical partition function $Z={\sum }_{\{s\}}{e}^{-\beta E}$, differentiation at fixed temperature gives ${\partial }_B\log Z=\beta ⟨{\sum }_i{s}_i⟩$. Thus the average spin per particle, or magnetization in these units, is $\boxed{m=(N\beta {)}^{-1}{\partial }_B\log Z}$.

Set $k_B=1$. After the Gaussian integral over momenta, the canonical partition function factors aswhereas the ideal gas has $Z_{\mathrm{ideal}}=Z_{\mathrm{kin}}{V}^N/N!$. HenceUsing the Helmholtz free energy $F=-T\log Z$ therefore gives

For the Gibbs entropyfirst maximize subject only to ${\sum }_n{p}_n=1$ over $\Omega$ accessible microstates. A Lagrange multiplier givesso every $p_n$ is equal. Normalization yields the microcanonical ensemble

For the canonical ensemble, impose both normalization and the mean-energy constraint ${\sum }_n{p}_n{E}_n=E$. Stationarity ofgives $p_n=C{e}^{-\beta E_n}$. The canonical partition function fixes $C$, soThermodynamic consistency identifies $\beta =(k_{B}T{)}^{-1}$. This is the maximum-entropy derivation of equilibrium ensembles.

Put $x=\beta ϵ$. The one-particle canonical partition function isand independence gives $Z=z^N$. Therefore the independent symmetric three-level system hasand, using part (c),As $T\to \infty$, $x\to 0$, all three levels become equally likely andAs $T\to 0^{+}$, every particle occupies the nondegenerate ground level $-ϵ$, and