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The action is dimensionless and $[d^{4}x]=-4$, so $[\mathcal{L}]=4$. A canonical scalar kinetic term then requires $2+2[\varphi ]=4$, and the electromagnetic kinetic term requires $2[F_{\mu \nu }]=4$. ThusThe brackets here denote mass dimension. The Lagrangian has no explicit coordinate dependence, and all indices are contracted as Lorentz scalars. Its continuous spacetime symmetry is the Poincaré group. Noether's theorem gives conserved four-momentum from translations and conserved angular momentum, including the boost generators, from Lorentz transformations. The nonzero mass $M$ and dimensionful couplings prevent a dilation symmetry.

Sign qualification: the printed electromagnetic kinetic term has the opposite healthy relative sign to the scalar kinetic terms. With the stated mostly-minus Minkowski metric, a physical Maxwell Lagrangian is $-F_{\mu \nu }{F}^{\mu \nu }/4$. The printed plus sign instead gives photon Hamiltonian density $-({\mathbf{E}}^2+{\mathbf{B}}^2)/2$; a nonzero purely electric configuration already has negative energy, unbounded below as its amplitude grows. The physical decay widths below use the healthy Maxwell sign with the given interaction normalization. The dimensional and spacetime-symmetry conclusions themselves do not depend on this sign correction.

Let $\beta =1/(k_{B}T)$ and use the two spin values $+1,-1$ as row and column labels. Define the spin-chain transfer matrixIt is a real symmetric matrix because $E_0(s,t)=E_0(t,s)$; every entry is strictly positive for finite bond energies at $T>0$. The Boltzmann factor of a spin configuration factorizes into the product $W_{{\sigma }_1{\sigma }_2}\cdots W_{{\sigma }_L{\sigma }_1}$. Summing the intermediate spin labels is exactly matrix multiplication, and the final closure imposes periodic boundary conditions. ThusWriting $W=\left(\begin{array}{cc}a& b\\ b& d\end{array}\right)$, its eigenvalues areA positive-entry matrix need not have two positive eigenvalues; ${\lambda }_{-}$ can be negative. The Perron–Frobenius theorem ensures ${\lambda }_{+}>|{\lambda }_{-}|$ here. Therefore the Helmholtz free energy is $F_L=-k_{B}T\ln Z_L$, and its value per spin in the thermodynamic limit is $-k_{B}T\ln {\lambda }_{+}$.

For an open chain there are $L-1$ bonds, with both endpoint Ising spins independently summed. Retain the same spin-chain transfer matrix, but replace the trace by endpoint vectors. With $1=(1,1{)}^{\mathsf{T}}$,This follows directly by expanding the product of $L-1$ Boltzmann factors. If ${\mathbf{v}}_{\pm }$ are orthonormal eigenvectors of the symmetric matrix $W$, the equivalent spectral decomposition isFixed endpoint Ising spins replace the corresponding $1$ by a basis vector; extra endpoint energies replace its entries by the corresponding Boltzmann factors. The leading bulk Helmholtz free energy per spin agrees with the periodic result, because the boundary changes only subextensive terms. In particular, $L=1$ gives $Z_1^{\mathrm{open}}=1^{\mathsf{T}}1=2$, as required for one free Ising spin.

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.

With no pairing, the Hamiltonian is diagonal in the fermionic occupation numbers. The Pauli exclusion principle allows each momentum mode to be either empty or occupied once. Without a fixed fermion number constraint, occupying a mode lowers the energy precisely when its single-particle energy is negative. Therefore, with $|0⟩$ the Fock vacuum, a ground state isChoose a fixed ordering of the fermionic creation operators; changing it alters only the overall phase. Any zero-energy modes may be occupied or empty without changing the ground-state energy.

To identify the occupied modes, complete the square:Here $\beta$ is the dispersion parameter, not inverse temperature. If ${\beta }^2<4\mu$, every mode has positive energy and the unique ground state is the Fock vacuum, with $E_{\mathrm{GS}}=0$. If ${\beta }^2=4\mu$, there are still no negative-energy modes, but $p=\pm \sqrt{\beta /2}$ have zero energy if they lie on the allowed momentum grid. Each such mode doubles the ground-state degeneracy.

For ${\beta }^2>4\mu$, defineBoth are positive, and the occupied Fermi sea consists of the two intervals $p_{-}<|p|<p_{+}$. With $p=2\pi n/L$, the requested finite-size ground-state energy is explicitlyThe factor two counts opposite momenta. Endpoints are excluded from this sum because they have zero energy; when an endpoint is allowed, its optional occupation adds ground-state degeneracy. A narrow negative-energy interval can contain no allowed finite-size momenta, in which case the sum is empty. The zero-momentum mode is always empty since $ϵ(0)=\alpha \mu >0$.

Since every quasiparticle energy is positive, the ground state is the vacuum of all the ${\alpha }_p$ annihilation operators, including ${\alpha }_0=c_0$. Its zero-momentum mode is therefore empty. For each positive momentum, the inverse Bogoliubov transformation givesBoth annihilate the normalized pair state $(u_p+i{v}_p{c}_p^{†}{c}_{-p}^{†})|0⟩$. For example, applying ${\alpha }_p$ produces the two cancelling terms $i{u}_p{v}_p{c}_{-p}^{†}|0⟩$. ThusThe factors have even fermion parity and commute for disjoint momentum pairs. Each normalized factor is a superposition of zero and two fermions, whose pair-occupation probability is $v_p^2$. ThereforeThere is no zero-mode contribution. For fixed $L$, the sum converges since $v_p^2=O(p^{-4})$.

A normalized simple excited state is $|k⟩={\alpha }_k^{†}|\Omega ⟩$ for $k>0$. Its excitation energy is $E_k$, and its total momentum is $k$: both terms in ${\alpha }_k^{†}=u_k{c}_k^{†}+i{v}_k{c}_{-k}$ add momentum $k$. Acting on the $k,-k$ pair givesThe affected pair now contains exactly one fermion instead of mean number $2v_k^2$; all other pairs are unchanged. ConsequentlyThis is the fermion number of a Bogoliubov excitation. Its change in mean fermion number lies between zero and one for $\mu >0$, with the value one at $\Delta =0$. The Bogoliubov quasiparticle coherently mixes adding a fermion and removing the opposite-momentum fermion; it is not an eigen-excitation of the original fermion number operator. Indeed,so fermion number is not conserved when $\Delta \ne 0$, while fermion parity is conserved. The generally noninteger change is an expectation value in a coherent paired background, rather than a fractional eigenvalue of the fermion number operator. In the weak-pairing limit at fixed $k$, it tends to one, as expected for an ordinary added fermion.

For one triangle, equal base spins $s=s^{\prime }=\sigma$ give energy $J+2\bar{J}\sigma \bar{s}$. Their minimum occurs at $\bar{s}=-\sigma$ and is $J-2\bar{J}$. Opposite base spins give energy $-J$, independently of the apex spin. The all-equal configurations have energy $J+2\bar{J}$.

When $\bar{J}=J$, the six configurations with precisely one spin different from the other two all have minimum energy $-J$. For a fixed first base spin, there is one minimum choice if the next base spin is the same, and two choices of apex if it is opposite. The ground-configuration counting transfer matrix is thereforeThis ground-state counting of the frustrated sawtooth Ising chain proves the extensive geometric frustration and the residual entropy $k_B{\lim }_{L\to \infty }\log {\Omega }_L/(2L)=(k_B/2)\log 3$.

When $\bar{J}>J$, $J-2\bar{J}<-J$. Every triangle must have equal base spins and the opposite apex spin. Shared base spins propagate that choice around the chain, leaving just two global ground states, of energy $L(J-2\bar{J})$. Their entropy per spin is $k_B\log 2/(2L)\to 0$.

For couplings favoring ferromagnetism, write $J=-K$, $\bar{J}=-\bar{K}$ with $K,\bar{K}>0$. Both orientations of an all-equal triangle minimize every bond simultaneously. The common base spins then make the whole chain uniform. There are two global ground states for either $K=\bar{K}$ or $K\ne \bar{K}$, so no corresponding distinction in residual entropy occurs. This statement assumes all bonds remain nonzero.

Choose the unitary discrete Fourier transformThe finite-sum orthogonality identity gives $[b_k,b_q^{†}]={\delta }_{kq}$. It also gives ${\sum }_j{a}_{j+1}^{†}{a}_j={\sum }_k{e}^{-ik}{b}_k^{†}{b}_k$. Substituting $A,B$ from solution therefore gives the dispersion relationUsing the opposite sign in the Fourier transform reverses $k$ and the sine term together, without changing the spectrum.

For $u\gg 1$ every $ϵ(k)>0$, so the ground state of this quadratic Hamiltonian is the bosonic vacuum, $b_k|\mathrm{GS}⟩=0$ for every $k$. In the original spin variables it is the fully $+z$ polarized product state. Its energy is zero after the constant subtraction. Repeating the construction about the fully $-z$ polarized state describes the other easy-axis sector of the original Hamiltonian.

A one-magnon excitation $b_k^{†}|\mathrm{GS}⟩$ has energy $ϵ(k)$ above the reference ground state. A normalized many-magnon Fock state with occupations $n_k$ has excitation energy ${\sum }_k{n}_kϵ(k)$, within the dilute regime of the linear spin-wave approximation. The lowest excitation in this sector has the minimum allowed $ϵ(k)$.

Let $R=\sqrt{J^2+D^2}$ and $k_0=\arctan (D/J)$. Since $J\cos k+D\sin k=R\cos (k-k_0)$, the continuum minimum of the dispersion relation isFor $D>0$, the energy gap closes at a positive $u_c$, at wave vector $k_0$. Below it, the quadratic expansion about the polarized state has a negative mode and is no longer a stable ground state approximation; higher-order terms and a different ordered reference state are needed.

Two qualifications are required by the actual assumptions. First, $D=0$ is permitted, in which case $u_c=0$; it is a concrete counterexample to the printed strict inequality $u_c>0$. Second, a finite ring only has discrete momenta. Its finite-ring spin-wave threshold with antisymmetric exchange isThis approaches $u_c$ in the thermodynamic limit and equals it when $k_0$ is allowed. For example, $L=4$ and $0<D<J$ give $u_c(L)=0$, since the maximum of $1,D/J,-1,-D/J$ is one. Thus the continuum formula and its positive threshold require the large-ring interpretation. The two symmetry-related polarized ground states already have equal exact energy; the calculated energy gap refers to excitations about one chosen orientation.

Assume three-dimensional isotropic gapless excitations, zero chemical potential and one independent branch. Momentum-state counting givesSubstituting $x=ϵ/(k_{B}T)$ into the Bose energy integral shows $U/V\propto T^{1+3/n}$ and $C/V\propto T^{3/n}$. The observed exponent therefore gives $n=2$.

For the quadratic-boson heat-capacity coefficient, the required Bose integral isThe equality follows by expanding $(e^x-1{)}^{-1}={\sum }_{r\ge 1}{e}^{-rx}$ and integrating term by term. HenceThe supplied units make its $C$ a heat capacity per volume. Equating its coefficient to $220\mathrm{J}{\mathrm{K}}^{-5/2}{\mathrm{m}}^{-3}$ givesEquivalently $ϵ=p^2/(2m_{\ast })$ has $m_{\ast }=1.09\times 10^{-28}\mathrm{kg}$. For $q_b$ identical branches, the measured coefficient fixes $q_b/{\alpha }^{3/2}$, so instead $\alpha =(4.58\times 10^{27}\mathrm{k}{\mathrm{g}}^{-1})q_b^{2/3}$. The one-branch convention is necessary for the quoted numerical value. The supplied integral with $x^4$ is not the moment arising for the exponent determined here.

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.

For two localized energy levels $0,\delta$ with degeneracies $d_0,d_1$, let $x=\delta /(k_{B}T)$. Canonical equilibrium givesHere $p_1$ is the total upper-level occupation probability and $u,s,c$ are per unit. For nondegenerate levels the entropy rises from zero to $k_B\ln 2$, while the heat capacity vanishes at both low and high temperatures and has a Schottky maximum near $k_{B}T=0.417\delta$. A spin one-half in a field, or a localized defect with two configurations, supplies an example. Higher levels must be thermally inaccessible; unequal degeneracies change the populations and peak. Distinguishable localized units have total partition function $z^N$, unlike a gas of indistinguishable particles sharing just two orbitals.

For the magnetic atom, choose level labels with $E_m=-mg{\mu }_{B}B$, $m=-J,\dots ,J$. Put $x=T_B/T=g{\mu }_{B}B/(k_{B}T)$. Its partition function and exact entropy of a finite magnetic multiplet areThe ratio form has removable limit $Z_J(0)=2J+1$. Since $s_J$ is a function of $x$ only, it depends on field and temperature solely through $B/T$. For fixed $J$ at small $x$, symmetry removes the odd powers and the supplied sum of $m^2$ givesFor large $J$, the expansion requires $J|x|\ll 1$, not merely $|x|\ll 1$.

For demagnetization with a linear-heat-capacity load, integrate its specified $C_L/T$:Isothermal magnetization at $T_i$ removes spin entropy; the bath receives the corresponding heat. During the following reversible, thermally isolated demagnetization, it is the combined entropy of spins and load that stays constant. At the minimum-field endpoint $B_f=0$, the ideal independent spins reach $s_{\max }=k_B\ln (2J+1)$. ThereforeThe bracket is the initial magnetic entropy deficit. The PDF's intermediate equality uses the initial entropy itself instead; with its preceding definition of $s$, that equality omits the maximum-entropy term. The corrected balance recovers its final temperature formula:Without the small-field expansion, the exact independent-spin prediction is $T_i-(N\theta /M)[\ln (2J+1)-s_J(T_i,B_i)/k_B]$. As a check, zero initial field gives zero entropy deficit and no cooling. The incorrectly printed intermediate balance would fail this check.

The result requires the load law throughout the temperature interval, hence $T_i\ll \theta$; a small initial Zeeman splitting, quantitatively $J{T}_{B,i}/T_i\ll 1$; positive final temperature; negligible additional heat capacities and heat leaks; thermal equilibration between subsystems; and a reversible sweep slow enough to maintain equilibrium. A negative extrapolated temperature is a breakdown of this approximation, not a cooling prediction. Spins alone would keep $B/T$ fixed, but the attached load changes that adiabat by transferring entropy to them as it cools.

Interactions invalidate the product of independent one-site partition functions. The entropy must be computed from the many-body energy spectrum or a justified interacting approximation. Correlations and possible magnetic ordering introduce an interaction scale and can remove the independent-spin entropy available for cooling near zero field. Internal fields and ordering commonly limit practical demagnetization. The detailed correction and any residual entropy depend on the interaction Hamiltonian, dimensionality and frustration; neither a universal ordering temperature nor a numerical final-temperature correction is specified here.

In the classical equilibrium circuit, the capacitor stores $C{V}^2/2$. Equipartition assigns this quadratic term mean energy $k_{B}T/2$, soThe mean voltage is zero and its variance is $k_{B}T/C$. Resistance determines the equilibration time and noise spectrum, not this full-band equilibrium variance. A finite-bandwidth or appreciably loading readout can measure less than the ideal result.

Let $\Delta =\hslash \Omega$ and choose the nondegenerate energy levels as 0 and $\Delta$. For N independent identical units, $Z=(1+e^{-\beta \Delta }{)}^N$; set N=1 for a single two-level system. Differentiating its logarithm givesAt low temperature, $U\sim N\Delta e^{-x}$ and $C_V\sim N{k}_B{x}^2{e}^{-x}$: the upper level is frozen out. At high temperature, the two levels approach equal populations, giving $U\to N\Delta /2$ while $C_V\sim N{k}_B{x}^2/4\to 0$. The energy therefore rises monotonically to a plateau, while the heat capacity has a Schottky peak:Unequal degeneracies would change the populations and peak; the usual equal-degeneracy model is stated here. Shifting both energy levels changes the energy baseline but not heat capacity.

Equipartition gives mean energy $k_{B}T/2$ for each independent quadratic term in a classical canonical Hamiltonian. For example, $p^2/(2m)$ and $k{x}^2/2$ each contribute this amount in a harmonic oscillator. It is a thermal average, not an instantaneous equality; constraints, nonquadratic terms or quantum level spacings large compared with $k_{B}T$ require a different calculation.

The fluctuation-dissipation theorem connects equilibrium fluctuations with the dissipative part of linear response. If an applied force couples as $-f(t)x$, use response to $e^{-i\omega t}$ and define the two-sided angular-frequency spectrum $S_{xx}(\omega )=\int e^{i\omega t}⟨\delta x(t)\delta x(0)⟩dt$. The classical relation isFor a viscously damped particle, $m\dot{v}=-\zeta v+\xi (t)$, the corresponding white-noise force satisfies $⟨\xi (t)\xi (t^{\prime })⟩=2\zeta k_{B}T\delta (t-t^{\prime })$. Noise replenishes energy lost by friction; their strengths cannot be varied independently at fixed equilibrium temperature.

Electrically, the one-sided voltage-noise spectrum per hertz of an equilibrium resistor is $4k_{B}TR$. A capacitor in parallel filters it toThe same equilibrium voltage variance follows from equipartition. Increasing R makes the spectrum taller and narrower without changing its integrated area. This thermodynamic connection between dissipative conductance and thermal electrical fluctuations is the content of the classical noise calculation.^[ref]

A black body absorbs incident electromagnetic radiation and, in equilibrium, emits a spectrum determined by temperature alone. A cavity with a small aperture approximates this ideal. Photon number is not conserved in thermal equilibrium, so its chemical potential is zero. Combining the Bose occupation with the two-polarization electromagnetic mode density gives the spectral energy densityChanging variable to $x=\hslash \omega /(k_{B}T)$ shows directly why the total energy density is proportional to $T^4$:Three powers of T arise from the number of thermally populated modes and one from their characteristic energy. Isotropic radiation incident on a surface has outward hemispherical flux $cu/4$, givingThe factor 1/4 is the hemispherical angular average; multiplying energy density by c alone would overcount. The spectrum's frequency scale grows with T. Classical equipartition over all modes would instead produce the Rayleigh-Jeans ultraviolet divergence; the Planck occupation suppresses high-frequency modes. Real materials can have wavelength-dependent emissivity, and a body's net radiative exchange also includes absorbed environmental radiation.

Write $z=e^{\beta (\mu -ϵ)}$. For one nondegenerate fermionic quantum state, the allowed occupations 0 and 1 give ${\Xi }_F=1+z$. For a bosonic state, summing all nonnegative occupations gives ${\Xi }_B=(1-z{)}^{-1}$, requiring $z<1$ or $\mu <ϵ$. Differentiating the logarithm,The interacting point defect has two distinct spin states, not one fermionic state. Its four microstates have grand-canonical weights $1,z,z,q{z}^2$, with $q=e^{-\beta U}$. The first two single-electron weights are separate up/down spin possibilities; the double state has one electron of each spin. ThusFor $U=0$, $\Xi =(1+z{)}^2$ and $\bar{n}=2z/(1+z)$. The sketch is a single smooth step from 0 to 2, centred on $\mu =ϵ$, where $\bar{n}=1$. As $T\to 0$ its width tends to zero.

For positive U with $\beta U\gg 1$, the lowest grand energies are successively $0$, $ϵ-\mu$ and $2ϵ+U-2\mu$. Away from the two crossings,Finite temperature smooths each step over an energy range of order $k_{B}T$. Near the lower step double occupation is exponentially suppressed, so $\bar{n}\simeq 2z/(1+2z)$. Near the upper step put $y=e^{\beta (\mu -ϵ-U)}$ and neglect the empty state, giving $\bar{n}\simeq 1+y/(2+y)$. Spin degeneracy shifts the half-step positions to $ϵ-k_{B}T\ln 2$ and $ϵ+U+k_{B}T\ln 2$.

For a grand-canonical probability $p_s\propto e^{-\beta (E_s-\mu n_s)}$, differentiation gives ${\partial }_{\mu }{p}_s=\beta (n_s-\bar{n})p_s$. Multiplying by $n_s$ and summing provesFor this defect, direct state counting givesAt $\mu =ϵ$, $z=1$ and thereforeThe same limit gives $\bar{n}=2/3$: empty, up and down states have equal weights, while the double state is suppressed. Replacing this degeneracy by a two-outcome equal-weight model would incorrectly give variance $1/4$ at this chemical potential.

Across the full chemical-potential range the variance has two well-separated narrow peaks and tends to zero on the 0, 1 and 2 plateaux. The limiting shapes are $2z/(1+2z{)}^2$ near the lower step and $2y/(2+y{)}^2$ near the upper one. Each peak has height $1/4$, at the shifted half-step position; the variance at the unshifted crossings is instead $2/9$.

At $T\ll T_F$, Pauli exclusion restricts thermally changed occupations to an energy interval of order $k_{B}T$ about the Fermi energy. The affected fraction is of order $T/T_F$; their typical excitation energy is of order $k_{B}T$, so the thermal energy correction is proportional to $T^2$ and the heat capacity is linear in $T$.

More precisely, with $g(E_F)$ the total density of one-particle states per unit energy, including spin,For a three-dimensional ideal gas with quadratic dispersion, $g(E_F)=3N/(2E_F)$, henceThe fixed-particle-number Sommerfeld expansion gives $U=3N{E}_F/5+({\pi }^2/4)N(k_{B}T{)}^2/E_F+\cdots$. The temperature dependence of the chemical potential must be included when obtaining this expression; holding it artificially fixed changes the intermediate energy correction.

Let $E$ denote the fluctuating microscopic energy and $U=⟨E⟩$ its mean, at fixed $N,V$. In a canonical ensemble with $\beta =1/(k_{B}T)$,Since $dT/d\beta =-k_B{T}^2$, this givesHere $\Delta U$ denotes an energy fluctuation about the mean, as in the question. The relation is not restricted to ideal gases when the Hamiltonian operator has no explicit temperature dependence. Alternatively, the stated equilibrium free-energy curvature gives a Gaussian energy weight proportional to $\exp [-(\Delta U{)}^2/(2k_B{T}^2{C}_V)]$, with the same variance.

For the relative energy fluctuation of a monatomic ideal gas, $U=3N{k}_{B}T/2$ and $C_V=3N{k}_B/2$, henceTaking room conditions as approximately $T=300$ K and $p=10^5$ Pa, with $V=10^{-6}$ m${}^3$, givesThe mean energy is about $0.15$ J and its standard deviation about $2.49\times 10^{-11}$ J. The tiny relative fluctuation explains why macroscopic equilibrium internal energy behaves as a sharply defined thermodynamic quantity. The estimate uses a thermal reservoir, not a particle reservoir; allowing $N$ itself to fluctuate changes the relevant fluctuation problem.

Bose-Einstein condensation is macroscopic occupation of the lowest one-particle state. It is a change in occupation statistics, not ordinary liquid formation and not a requirement that all atoms occupy one spatial point. For ideal bosons,Set the lowest one-particle energy to zero. Then $\mu \le 0$, and the population of excited states in a uniform three-dimensional box isHere $g_s(z)={\sum }_{j\ge 1}{z}^j/j^s$. As $\mu$ approaches zero, the excited-state population saturates at $V\zeta (3/2)/{\lambda }^3$. Any excess particles occupy the ground state. In the thermodynamic limit,Thus the onset occurs when the thermal de Broglie wavelength is comparable to the mean interparticle spacing. At finite $N$, the chemical potential remains slightly below the ground-state energy; setting it equal to zero is a thermodynamic-limit statement.

The 1995 rubidium experiment used magnetic confinement and evaporative cooling of atoms. Absorption images after trap manipulation revealed a narrow, anisotropically expanding component superposed on the broader thermal distribution; the new component grew as cooling proceeded. The thermal component supplied an internal comparison through the same imaging optics. Trap decompression and release geometry mattered, and finite imaging resolution and interactions limited direct comparison with an ideal-gas ground state.^[ref]

A narrow cloud alone is not sufficient evidence. One should check its emergence across a controlled cooling sequence, atom-number and temperature calibration, the thermal background and the predicted expansion anisotropy. These are tests of an occupation transition rather than a simple focusing artifact. The box formula above is not the exact critical-temperature or condensate-fraction formula for the trapped experiment: the trap changes the density of states, and interactions also matter. Condensation and superfluid response are related but distinct properties; the former definition is ground-state occupation.