Ideal Bose Gas

The ideal Bose gas is a system of noninteracting identical bosons. Its Hamiltonian is a sum of one-particle energies, but its thermodynamics is not classical because every mode can hold any nonnegative number of bosons.

For a uniform gas in a cubic box of volume $V=L^3$,

$$H = \sum_{\mathbf k} \epsilon_{\mathbf k} a_{\mathbf k}^\dagger a_{\mathbf k}, \qquad \epsilon_{\mathbf k} = \frac{\hbar^2k^2}{2m}.$$

The number operator is

$$N = \sum_{\mathbf k} a_{\mathbf k}^\dagger a_{\mathbf k}.$$

The model is exactly solvable because all number operators commute and the many-body eigenstates are occupation-number states. Its importance is disproportionate to its simplicity: it isolates the thermodynamic consequences of Bose statistics, supplies the reference point for weakly interacting gases, and shows how a macroscopic ground-mode occupation can emerge from a one-particle spectrum.

Here we derive the thermodynamics of one spinless species in a three-dimensional periodic cubic box at positive temperature, with ground energy zero. The thermodynamic limit takes $N,V\to\infty$ at fixed density. In grand-canonical equations $N$ and $N_0$ denote mean populations; an exactly fixed total number requires the canonical ensemble. The two-mode canonical example makes the difference explicit in a finite system.

Helpful background. Bose–Einstein statistics supplies the mode weights, and the thermodynamic limit explains why a finite spectrum can develop a phase transition.

The periodic-box spectrum

Periodic boundary conditions give normalized one-particle eigenfunctions

$$\phi_{\mathbf k}(\mathbf r) = \frac{1}{\sqrt V} e^{i\mathbf k\cdot\mathbf r},$$

with allowed wavevectors

$$\mathbf k = \frac{2\pi}{L} \mathbf n, \qquad \mathbf n \in \mathbb Z^3.$$

The lowest mode is $\mathbf k=\mathbf0$. Its energy is chosen to be zero:

$$\epsilon_{\mathbf0} = 0.$$

An occupation configuration

$$\lvert\{n_{\mathbf k}\}\rangle$$

has

$$n_{\mathbf k} = 0,1,2,\ldots,$$

total particle number

$$N = \sum_{\mathbf k} n_{\mathbf k},$$

and energy

$$E = \sum_{\mathbf k} \epsilon_{\mathbf k}n_{\mathbf k}.$$

The particles do not interact. All nonclassical thermodynamics comes from bosonic state counting and the density of one-particle levels.

Why the Ground Mode Must Be Separated

The excited levels become dense as $L\to\infty$, so their sum can be replaced by an energy integral. The ground state remains one distinguished mode and should be kept explicit:

$$N = N_0 + N_{\mathrm{ex}}.$$

Here

$$N_0 = \langle n_{\mathbf0}\rangle$$

is the ground-mode occupation, while $N_{\mathrm{ex}}$ sums all $\mathbf k\neq\mathbf0$ modes.

The first periodic-box shells have energies $\epsilon/\epsilon_L=0,1,2,3,4,5$ and degeneracies $1,6,12,8,6,24$, where $\epsilon_L=2\pi^2\hbar^2/(mL^2)$. The lower curve is the smooth continuum density of excited states in arbitrary vertical units, not the exact discrete spectrum. A macroscopically occupied ground mode must be kept separately.

Replacing the entire spectrum by one continuum integral can erase the very mode whose macroscopic occupation defines ideal-gas condensation.

Grand Partition Function

Introduce the fugacity

$$z \equiv e^{\beta\mu}, \qquad \beta = \frac{1}{k_{\mathrm B}T}.$$

Because the ground energy is zero, convergence requires

$$0 < z < 1,$$

or equivalently

$$\mu < 0$$

for a finite grand-canonical system.

The mode factor is

$$\xi_{\mathbf k} = \frac{1}{ 1-ze^{-\beta\epsilon_{\mathbf k}} }.$$

Therefore

$$\Xi = \prod_{\mathbf k} \frac{1}{ 1-ze^{-\beta\epsilon_{\mathbf k}} },$$

and

$$\ln\Xi = - \sum_{\mathbf k} \ln \left( 1-ze^{-\beta\epsilon_{\mathbf k}} \right).$$

Separating the ground mode gives

$$\ln\Xi ={} -\ln(1-z) - \sum_{\mathbf k\neq\mathbf0} \ln \left( 1-ze^{-\beta\epsilon_{\mathbf k}} \right).$$

The first term is the exact ground-mode contribution. At finite volume and positive temperature the quadratic spectrum makes the excited-mode sum converge; this is the additional condition required beyond convergence of each geometric factor. The excited contribution becomes extensive in the thermodynamic limit.

Mode Occupations and Number Equation

Every momentum mode has mean occupation

$$\overline n_{\mathbf k} = \frac{1}{ z^{-1}e^{\beta\epsilon_{\mathbf k}}-1 }.$$

For the ground mode,

$$N_0 = \frac{z}{1-z}.$$

The exact finite-volume number equation is

$$N = \frac{z}{1-z} + \sum_{\mathbf k\neq\mathbf0} \frac{1}{ z^{-1}e^{\beta\epsilon_{\mathbf k}}-1 }.$$

At specified mean $N$, volume $V$, and temperature $T$, this grand-canonical equation determines $z$ and therefore $\mu$. It does not impose an exact particle count in every configuration.

The equation also exposes the two distinct ways to store particles: distribute them among excited modes, or accumulate them in the single ground mode as $z\to1^-$.

Energy-zero covariance

The general convergence condition is $\mu<\epsilon_0$, where $\epsilon_0$ is the lowest one-particle energy. Changing that energy zero by a constant $C$ requires shifting the chemical potential along with the spectrum:

$$\epsilon_{\mathbf k}'=\epsilon_{\mathbf k}+C, \qquad \mu'=\mu+C.$$

Then $H'=H+CN$, and the grand-canonical generator is unchanged:

$$H'-\mu'N=H-\mu N.$$

Every occupation and grand-canonical probability therefore stays the same. Shifting the spectrum alone at fixed $\mu$ changes the state. At exactly fixed $N$, the added energy $CN$ is instead common to all configurations, so it cancels from their normalized canonical probabilities.

Counting One-Particle States

In a large periodic box, one momentum state occupies volume

$$\left( \frac{2\pi}{L} \right)^3$$

in $\mathbf k$-space. The leading continuum approximation to the number of spinless modes with magnitude below $k$ is

$$\mathcal N(k) = \frac{V}{(2\pi)^3} \frac{4\pi k^3}{3} = \frac{Vk^3}{6\pi^2}.$$

Using

$$k = \frac{\sqrt{2m\epsilon}}{\hbar},$$

the integrated density of states is

$$\mathcal N(\epsilon) = \frac{V}{6\pi^2} \left( \frac{2m\epsilon}{\hbar^2} \right)^{3/2}.$$

Differentiation gives the three-dimensional density of states

$$D(\epsilon) = \frac{V}{4\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt\epsilon.$$

An internal degeneracy $g$ multiplies both $\mathcal N$ and $D$ when the $g$ components are independent and share the same dispersion and chemical potential.

When the Continuum Approximation Applies

The first nonzero momentum has magnitude $2\pi/L$, so the first excitation scale is

$$\epsilon_L = \frac{2\pi^2\hbar^2}{mL^2}.$$

The excited-state sum is well approximated by an integral when thermally relevant functions vary slowly across the level spacing. A useful condition is

$$k_{\mathrm B}T \gg \epsilon_L.$$

Then

$$\sum_{\mathbf k\neq\mathbf0} F(\epsilon_{\mathbf k}) \longrightarrow \int_0^\infty d\epsilon\, D(\epsilon)F(\epsilon).$$

Finite-size corrections become important near the lowest excitations, in small traps or boxes, and when a precise transition crossover is required.

Thermal de Broglie Wavelength

Define the thermal de Broglie wavelength

$$\lambda_T \equiv \sqrt{ \frac{2\pi\hbar^2}{mk_{\mathrm B}T} } = \frac{h}{ \sqrt{2\pi mk_{\mathrm B}T} }.$$

It is the natural length scale obtained by comparing thermal kinetic energy with quantum wavelength. The dimensionless phase-space density is

$$\mathcal D \equiv n\lambda_T^3.$$

When

$$\mathcal D \ll 1,$$

the relevant modes have small mean occupation and Maxwell–Boltzmann thermodynamics is accurate. This criterion concerns phase-space occupation, not the width of an arbitrarily chosen wave packet. Quantum degeneracy becomes important when $\mathcal D$ is of order one.

The convention for $\lambda_T$ matters. Factors of $2\pi$ move if a different thermal-wavelength definition is used, so formulas should not mix conventions.

Bose Integrals and Polylogarithms

For $0<z<1$, define

$$\operatorname{Li}_s(z) = \sum_{\ell=1}^{\infty} \frac{z^\ell}{\ell^s}.$$

For real $s>0$ and $0<z<1$, an equivalent integral representation is

$$\operatorname{Li}_s(z) = \frac{1}{\Gamma(s)} \int_0^\infty dx\, \frac{x^{s-1}}{z^{-1}e^x-1}.$$

These functions are also called Bose functions in statistical mechanics. At $z=1$ and $s>1$,

$$\operatorname{Li}_s(1) = \zeta(s).$$

The limiting values needed most often are

$$\zeta(3/2) \simeq 2.612, \qquad \zeta(5/2) \simeq 1.341.$$

Excited-State Number

Using the density of states,

$$N_{\mathrm{ex}} = \int_0^\infty d\epsilon\, \frac{D(\epsilon)}{ z^{-1}e^{\beta\epsilon}-1 }.$$

Set

$$x = \beta\epsilon.$$

The integral becomes

$$N_{\mathrm{ex}} = \frac{V}{\lambda_T^3} \operatorname{Li}_{3/2}(z).$$

Thus the thermodynamic-limit number equation is

$$n = \frac{N_0}{V} + \frac{1}{\lambda_T^3} \operatorname{Li}_{3/2}(z).$$

In the normal phase, $N_0/V$ vanishes in the thermodynamic limit and

$$n\lambda_T^3 = \operatorname{Li}_{3/2}(z).$$

This equation determines $z$ from density and temperature. The MB-B006 ideal-gas benchmark supplies numerical checks of the Bose functions, a normal-state fugacity root, and the condensate fraction. Bracket a normal-state root within $0<z<1$ and compare its number residual with the numerical integration or series error. If the specified density exceeds the excited-state capacity, separate the ground mode instead of searching for an unphysical $z>1$.

Maximum Excited-State Density

Because $z<1$ and $\operatorname{Li}_{3/2}(z)$ increases with $z$,

$$n_{\mathrm{ex}} \leq \frac{\zeta(3/2)}{ \lambda_T^3 }.$$

The excited modes therefore have a finite maximum density at fixed temperature:

$$n_{\mathrm{ex,max}} = \frac{\zeta(3/2)}{ \lambda_T^3 }.$$

If the total density exceeds this value, the excess cannot be accommodated by changing $z$ while keeping $z<1$. In the thermodynamic limit, $z\to1$ and the excess occupies the ground mode macroscopically.

The corresponding ideal-gas temperature scale is

$$T_c = \frac{2\pi\hbar^2}{mk_{\mathrm B}} \left[ \frac{n}{\zeta(3/2)} \right]^{2/3}.$$

For the uniform three-dimensional ideal gas, the thermodynamic-limit condensate fraction below this scale is

$$\frac{N_0}{N} = 1- \left( \frac{T}{T_c} \right)^{3/2}.$$

For the uniform gas, the one-body density matrix is diagonal in momentum and its eigenvalues are these mode occupations. The Penrose–Onsager criterion states condensation more generally through an extensive eigenvalue of that density matrix, without relying on a plane-wave basis. The condensation page develops its relation to coherence, experiments, and interactions.

Grand Potential and Pressure

For the excited modes,

$$\ln\Xi_{\mathrm{ex}} = \frac{V}{\lambda_T^3} \operatorname{Li}_{5/2}(z).$$

The ground-mode term

$$\ln\Xi_0 = -\ln(1-z)$$

is not extensive for fixed $z<1$. Along a condensed sequence with $N_0$ proportional to $V$, $1-z=1/(N_0+1)$, so this term grows only as $\ln V$ and again vanishes after division by $V$. The grand potential density in the homogeneous thermodynamic limit is therefore

$$\frac{\Omega_{\mathrm G}}{V} = - \frac{k_{\mathrm B}T}{\lambda_T^3} \operatorname{Li}_{5/2}(z).$$

Using $\Omega_{\mathrm G}=-PV$ gives

$$P = \frac{k_{\mathrm B}T}{\lambda_T^3} \operatorname{Li}_{5/2}(z).$$

At finite volume mechanical pressure is $P=-(\partial\Omega_{\mathrm G}/\partial V)_{T,\mu}$, not automatically $-\Omega_{\mathrm G}/V$. The periodic-box ground energy is independent of volume, so its grand-potential term has zero such derivative. The zero-momentum particles carry no kinetic energy and no mechanical pressure in this ideal model.

Internal Energy

The ground state has zero one-particle energy, so

$$U = \int_0^\infty d\epsilon\, D(\epsilon) \frac{\epsilon}{ z^{-1}e^{\beta\epsilon}-1 }.$$

Evaluating the integral gives

$$U = \frac{3}{2} \frac{Vk_{\mathrm B}T}{\lambda_T^3} \operatorname{Li}_{5/2}(z).$$

Therefore

$$U = \frac{3}{2}PV.$$

This relation follows from the quadratic nonrelativistic dispersion in three dimensions. It is not universal for lattice bands, relativistic particles, harmonic traps, or other dispersions.

Entropy

The grand-potential identity gives

$$S = \frac{U+PV-\mu N}{T}.$$

In the normal phase, where the ground-mode density is negligible,

$$\frac{S}{V} ={} \frac{k_{\mathrm B}}{\lambda_T^3} \Biggl[ \frac{5}{2} \operatorname{Li}_{5/2}(z) - \ln z\, \operatorname{Li}_{3/2}(z) \Biggr].$$

In the condensed thermodynamic-limit ideal gas, $z\to1$ and the ground mode contributes no extensive entropy. A finite-volume grand-canonical ground-mode mixture can have subextensive entropy, while the thermal cloud carries the entropy density.

Equation of State in the Normal Phase

When $N_0/V$ is negligible,

$$n = \frac{1}{\lambda_T^3} \operatorname{Li}_{3/2}(z),$$

and

$$P = \frac{k_{\mathrm B}T}{\lambda_T^3} \operatorname{Li}_{5/2}(z).$$

Hence

$$\frac{P}{nk_{\mathrm B}T} = \frac{ \operatorname{Li}_{5/2}(z) }{ \operatorname{Li}_{3/2}(z) }.$$

For $0<z<1$, this ratio is less than one. At the same density and temperature, Bose statistics lowers the ideal-gas pressure relative to the Maxwell–Boltzmann result because it favors multiple occupation of already populated states.

Dilute pressure correction

Expanding the Bose integrals and eliminating fugacity gives

$$\frac{P}{nk_{\mathrm B}T}=1-\frac{n\lambda_T^3}{2^{5/2}} +O[(n\lambda_T^3)^2].$$

The negative correction is statistical, not an attractive interparticle potential. The classical-limit derivation gives the series inversion, internal-degeneracy factor, and interaction qualifications.

Compressibility and Density Fluctuations

In the normal phase,

$$\left( \frac{\partial n}{\partial\mu} \right)_T = \frac{\beta}{\lambda_T^3} \operatorname{Li}_{1/2}(z).$$

As $z\to1^-$,

$$\operatorname{Li}_{1/2}(z) \longrightarrow \infty.$$

The ideal gas therefore becomes extremely compressible near the condensation boundary. Through the grand-canonical fluctuation relation, this is accompanied by enhanced number fluctuations.

A stable weak repulsive interaction gives a finite bulk inverse compressibility away from critical singularities and qualitatively changes the low-energy response. The infinite compressibility of the condensed uniform ideal gas is one reason the noninteracting model is not a complete description of a physical superfluid.

Below the Ideal Condensation Scale

In the thermodynamic limit below $T_c$,

$$z \to 1, \qquad \mu \to 0^-.$$

The excited population is

$$N_{\mathrm{ex}} = \frac{V}{\lambda_T^3} \zeta(3/2),$$

and the pressure is

$$P = \frac{k_{\mathrm B}T}{\lambda_T^3} \zeta(5/2).$$

Because $\lambda_T^{-3}\propto T^{3/2}$,

$$U \propto T^{5/2}, \qquad C_V \propto T^{3/2}$$

within the uniform ideal model below $T_c$.

The condensate changes the density without changing the ideal-gas pressure at fixed temperature. For a stable dilute repulsive gas, the interaction energy supplies a density-dependent pressure and removes this ideal-gas softness away from critical singularities.

Why Dimensionality Matters

For a quadratic dispersion in $d$ spatial dimensions, the low-energy density of states scales as

$$D_d(\epsilon) \propto \epsilon^{d/2-1}.$$

At $z=1$, the low-energy excited-number integral behaves as

$$N_{\mathrm{ex}} \sim \int_0 d\epsilon\, \epsilon^{d/2-2}.$$

This integral converges at the lower limit only when

$$d > 2.$$

Consequently, a uniform ideal Bose gas with quadratic dispersion has no ordinary finite-temperature Bose–Einstein condensation in the one- or two-dimensional thermodynamic limit.

This statement has precise boundaries. Finite systems have crossovers, trapping changes the density of states, and interacting two-dimensional systems can exhibit Berezinskii–Kosterlitz–Thouless physics rather than ideal-gas condensation. Low-Dimensional Quantum Gases develops those boundaries and the associated one- and two-dimensional state counting.

Internal Degeneracy

Suppose there are $g$ independent internal components with equal masses, dispersions, and chemical potentials. In the normal phase,

$$n = \frac{g}{\lambda_T^3} \operatorname{Li}_{3/2}(z),$$

and

$$P = \frac{gk_{\mathrm B}T}{\lambda_T^3} \operatorname{Li}_{5/2}(z).$$

The condensation structure depends on whether the internal states interconvert, whether their populations are separately conserved, and whether the ground state is degenerate. A degeneracy factor should not be inserted mechanically without specifying those constraints.

Finite Size and Ensemble Choice

At finite $L$:

• the spectrum is discrete;
• a normalizable positive-temperature grand-canonical state requires $\mu$ strictly below the ground energy;
• $N_0$ is finite;
• all thermodynamic functions are smooth;
• there is no exact nonanalytic phase transition.

The grand-canonical ground-mode variance is

$$\operatorname{Var}(N_0) = N_0(1+N_0).$$

If $N_0$ is macroscopic, this predicts relative fluctuations of order one:

$$\frac{ \sqrt{\operatorname{Var}(N_0)} }{ N_0 } \longrightarrow 1.$$

This is the grand-canonical condensate-fluctuation problem, sometimes called the grand-canonical catastrophe. It reflects the combination of an ideal zero-energy mode and unrestricted particle exchange. A fixed-$N$ ensemble suppresses total-number fluctuations, and interactions alter condensate fluctuations further.

Ensemble equivalence must therefore be stated observable by observable. Bulk pressure and energy can agree while the condensate-number variance does not.

Fixed-number counting

For a finite set of modes labeled by $\alpha$, the canonical partition function sums only configurations with the declared total number:

$$Z_N(\beta) =\sum_{\{n_\alpha\}:\,\sum_\alpha n_\alpha=N} e^{-\beta\sum_\alpha\epsilon_\alpha n_\alpha}.$$

The same information is encoded in the grand partition function as a generating series:

$$\Xi(z,\beta)=\sum_{M=0}^{\infty}z^M Z_M(\beta), \qquad Z_N(\beta)=[z^N]\prod_\alpha \frac{1}{1-ze^{-\beta\epsilon_\alpha}}.$$

Here $[z^N]$ means the coefficient of $z^N$. Extracting that coefficient does not introduce a particle reservoir into the fixed-$N$ model. It enforces the number constraint that prevents the canonical trace from factorizing into independent mode sums. The construction also applies to the box spectrum when the traces converge.

The normalizable grand-canonical thermal state of this quadratic Hamiltonian is Gaussian: different momentum modes are statistically independent, and its correlation functions obey Wick factorization. Projecting onto an exact total number generally gives a non-Gaussian state with correlated occupations. Canonical averages must therefore use the constrained weights rather than assume independent Bose–Einstein occupations or apply Gaussian Wick contractions without justification.

Two bosons in two modes

Restrict the one-particle space to a nondegenerate ground mode of energy $0$ and one excited mode of energy $\Delta>0$, and fix $N=2$. The allowed Fock states are

$$|2,0\rangle,\qquad |1,1\rangle,\qquad |0,2\rangle,$$

with energies $0,\Delta,2\Delta$. With $q=e^{-\beta\Delta}$, their weights give

$$Z_2=1+q+q^2, \qquad \langle n_1\rangle=\frac{q+2q^2}{1+q+q^2}.$$

As $T\to0$, $q\to0$ and both particles occupy the ground mode. As $T\to\infty$ in this fixed two-mode model, $q\to1$, the three configurations become equally likely, and $\langle n_1\rangle\to1$. There is no extra multiplicity for “which particle” is excited: the Fock states already count identical bosons. This high-temperature limit refers to the truncated model, not to a full gas in which additional levels become accessible. No phase transition occurs in this finite example.

At $\beta\Delta=\ln2$, the respective probabilities are $4/7$, $2/7$, and $1/7$. Direct averaging gives

$$\langle n_1\rangle=\frac47, \qquad \langle n_1^2\rangle=\frac67,$$

and hence

$$\begin{aligned} \operatorname{Var}(n_0)=\operatorname{Var}(n_1)&=\frac{26}{49},\\ \operatorname{Cov}(n_0,n_1)&=-\frac{26}{49}. \end{aligned}$$

The negative covariance follows from $n_0=2-n_1$ and ensures $\operatorname{Var}(n_0+n_1)=0$. Independent grand-canonical modes instead have zero cross-mode covariance and a fluctuating total number, even if their chemical potential is chosen to reproduce the same mean total population. The Hamiltonian alone does not determine these fluctuation predictions; the ensemble matters.

Traps, lattice bands, and interactions

The uniform continuum formulas depend on the spectrum. In a three-dimensional harmonic trap the semiclassical density of states scales as $\epsilon^2$ above the ground level, giving different condensation powers. Quantum gases in traps derives the corresponding number integrals. On a lattice one instead integrates over the actual band; van Hove singularities and band edges can alter the result.

Trapping can also change the dimensional capacity criterion. The harmonic-trap analysis permits finite excited-state capacity in two dimensions under the appropriate trap thermodynamic limit, unlike the uniform quadratic gas. The trap limit and dimensional crossover must be specified before transferring such conclusions between geometries.

An ideal gas has no collisions to establish equilibrium by itself. Its Gibbs state is a statistical model of an equilibrated or prepared population, not a mechanism for equilibration. It also omits interaction-induced depletion, sound, and vortex-core structure. Condensation alone is not a proof of superfluid behavior: the uniform ideal gas has quadratic excitations and zero Landau critical velocity. The weakly interacting gas explains how stable repulsion supplies a finite inverse compressibility and a linear phonon spectrum. Superfluid response and metastable flow require their own criteria, beyond a condensate count.

Equilibrium photons and phonons usually have $\mu=0$ because their number can change through exchange with the material system. They obey a Bose occupation law, but are not the conserved-particle gas whose number constraint produces the condensation calculation above. Effective conservation or driven preparation requires a separate model.

Exercises

Evaluate the excited-state number

Use the series

$$\frac{1}{z^{-1}e^x-1} = \sum_{\ell=1}^{\infty} z^\ell e^{-\ell x}$$

to show that

$$N_{\mathrm{ex}} = \frac{V}{\lambda_T^3} \operatorname{Li}_{3/2}(z).$$

Solution

After substituting $x=\beta\epsilon$, the excited-state integral is proportional to

$$\int_0^\infty dx\, \frac{x^{1/2}}{z^{-1}e^x-1}.$$

Expand the denominator:

$$I(z) \equiv \int_0^\infty dx\, \frac{x^{1/2}}{z^{-1}e^x-1}.$$

Then

$$\begin{aligned} I(z) &= \sum_{\ell=1}^{\infty} z^\ell \int_0^\infty dx\, x^{1/2}e^{-\ell x} \\ &= \Gamma(3/2) \sum_{\ell=1}^{\infty} \frac{z^\ell}{\ell^{3/2}} \\ &= \Gamma(3/2) \operatorname{Li}_{3/2}(z). \end{aligned}$$

Combining the prefactor from $D(\epsilon)$ with $\beta^{-3/2}$ gives $V/\lambda_T^3$. Hence

$$N_{\mathrm{ex}} = \frac{V}{\lambda_T^3} \operatorname{Li}_{3/2}(z).$$

Diagnose condensate fluctuations

Show that the grand-canonical relative ground-mode fluctuation approaches one when $N_0\gg1$, and explain why this does not imply that every physical condensate has order-one relative fluctuations.

Solution

For one ideal thermal bosonic mode,

$$\operatorname{Var}(N_0) = N_0(1+N_0).$$

Therefore

$$\begin{aligned} \frac{ \sqrt{\operatorname{Var}(N_0)} }{N_0} &= \sqrt{ 1+ \frac{1}{N_0} } \\ &\longrightarrow 1. \end{aligned}$$

The result assumes an ideal zero-energy mode exchanging particles with a reservoir. A canonical ensemble fixes total $N$ and correlates the ground and excited populations. Interactions also impose an energetic cost for density fluctuations. The ideal grand-canonical result is therefore ensemble- and model-specific, not a universal prediction for laboratory condensates.

A large occupation with subextensive entropy

In a grand-canonical ground mode set $N_0=cV$ with fixed positive density $c$. Find $z$, $\mu$, the growth of $\ln\Xi_0$, and the leading ground-mode entropy. Why does $-\Omega_0/V$ not give its finite-volume mechanical pressure?

Solution

From $N_0=z/(1-z)$, $z=N_0/(N_0+1)$ and $\mu=-k_{\mathrm B}T\ln(1+1/N_0)\sim-k_{\mathrm B}T/N_0$. Thus $\ln\Xi_0=\ln(N_0+1)$. The geometric-mode entropy is

$$S_0=k_{\mathrm B}[(N_0+1)\ln(N_0+1)-N_0\ln N_0] =k_{\mathrm B}[\ln N_0+1+O(N_0^{-1})].$$

Both are subextensive. Mechanical pressure differentiates at fixed $T,\mu$, where $z$ is fixed and $\Omega_0=k_{\mathrm B}T\ln(1-z)$ has no explicit volume dependence. Hence $P_0=0$. Differentiating along the imposed path $N_0=cV$ would change $\mu$ and answer a different question.

Shift the one-particle energy zero

Show that $\overline n_\alpha$ is unchanged under $\epsilon_\alpha\mapsto\epsilon_\alpha+C$ and $\mu\mapsto\mu+C$. What changes if only $\epsilon_\alpha$ is shifted?

Solution

The occupation depends only on the difference:

$$\overline n_\alpha =\frac{1}{e^{\beta(\epsilon_\alpha-\mu)}-1}.$$

Shifting both quantities leaves $\epsilon_\alpha'-\mu'=\epsilon_\alpha-\mu$, so every occupation and grand-canonical weight is unchanged. Shifting the spectrum at fixed $\mu$ changes this difference by $C$ and therefore describes a different grand-canonical equilibrium state. For comparison, at fixed $N$ one has $Z_N'=e^{-\beta CN}Z_N$, and the common factor cancels from all normalized probabilities.

Test a general infrared dispersion

Let $\epsilon\propto k^p$, with $p>0$, near an isolated minimum in $d$ uniform spatial dimensions. Determine when the saturated excited-state number is infrared finite.

Solution

The low-energy density of states scales as $D(\epsilon)\propto\epsilon^{d/p-1}$. At saturation, $(e^{\beta\epsilon}-1)^{-1}\sim(\beta\epsilon)^{-1}$, so the number integral near its lower endpoint is proportional to

$$\int_0 d\epsilon\,\epsilon^{d/p-2}.$$

It converges when $d/p-2>-1$, or $d>p$. Equality gives a logarithmic divergence. For quadratic dispersion, $p=2$, this recovers $d>2$. This local test diagnoses infrared capacity; a full critical temperature additionally requires the density of states over the thermally occupied spectrum and a stated thermodynamic limit.

Find the two-mode canonical energy

For the two-boson example, compute the mean energy from $Z_2$ and find its low- and high-temperature limits.

Solution

Since $q=e^{-\beta\Delta}$,

$$\langle E\rangle =-\frac{\partial\ln Z_2}{\partial\beta} =\Delta\frac{q+2q^2}{1+q+q^2} =\Delta\langle n_1\rangle.$$

As $T\to0$, $q\to0$ and $\langle E\rangle\to0$. As $T\to\infty$ within this two-mode model, $q\to1$ and $\langle E\rangle\to\Delta$, the average of the three equally weighted energies $0,\Delta,2\Delta$.

Bracket the normal-state fugacity

Show that the normal-state benchmark equation $\operatorname{Li}_{3/2}(z)=1$ has exactly one root in $0<z<1$. Why is bracketing preferable to an unconstrained Newton iteration?

Solution

For $0<z<1$,

$$\frac{d}{dz}\operatorname{Li}_{3/2}(z) =\frac{\operatorname{Li}_{1/2}(z)}{z}>0.$$

The function is continuous and strictly increasing from $0$ to $\zeta(3/2)>1$, so it crosses the target exactly once. A bracket with opposite residual signs keeps the iteration in the physical interval. An unconstrained Newton step can leave that interval and encounter an unphysical branch, including $z>1$. The root residual must also be checked against the error in evaluating the Bose function.

References

• F. Dalfovo, S. Giorgini, L. P. Pitaevskii, and S. Stringari, “Theory of Bose–Einstein condensation in trapped gases”, Reviews of Modern Physics 71, 463–512 (1999).
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover (2003).
• K. Huang, Statistical Mechanics, 2nd ed., Wiley (1987).
• M. Kardar, Statistical Physics of Particles, Cambridge University Press (2007).
• L. D. Landau and E. M. Lifshitz, Statistical Physics, Part 1, 3rd ed., Butterworth–Heinemann (1980).
• R. K. Pathria and P. D. Beale, Statistical Mechanics, 3rd ed., Elsevier (2011).
• C. J. Pethick and H. Smith, Bose–Einstein Condensation in Dilute Gases, 2nd ed., Cambridge University Press (2008).
• L. Pitaevskii and S. Stringari, Bose–Einstein Condensation and Superfluidity, Oxford University Press (2016).