Question bank — Bose-Einstein condensation — concept

False — ideal-gas BEC has zero interaction; the crowding comes from Bose-Einstein statistics permitting unlimited shared occupation, not from a potential.

False — $\mu$ is pinned at $0^{-}$; a positive $\mu$ would make $\bar{n}({\epsilon }_0)=\frac{1}{e^{-\mu /k_{B}T}-1}$ negative, which is forbidden.

False — the Density of states $g(\epsilon )\propto \sqrt{\epsilon }$ vanishes at $\epsilon =0$, so the integral silently discards the ground state; below $T_c$ you must add $N_0$ back by hand.

False — in 2D the excited-state integral diverges at small energy, so excited states never saturate and there is no BEC for a uniform 2D ideal gas.

False — at $T_c$ the condensate fraction is just reaching zero; $N_0/N=1-(T/T_c{)}^{3/2}=0$ there and grows only for $T<T_c$.

True — $N_0/N=1-(T/T_c{)}^{3/2}\to 1$ as $T\to 0$, so all atoms occupy the single lowest state.

Misleading — it is atoms in the lowest momentum/quantum state, a single delocalised wavefunction spanning the trap, not a spatial pile.

True — it is a purely statistical phase transition; the phase-space density $n{\lambda }_T^3$ crossing $g\zeta (3/2)\approx 2.612$ (for $g=1$) is the whole trigger.

False — fermions obey Fermi-Dirac statistics with $\bar{n}=\frac{1}{e^{(\epsilon -\mu )/k_{B}T}+1}\le 1$ per state (Pauli exclusion), so no macroscopic pile-up in one state is possible.

True — but "overlap" is made precise: condensation begins exactly when $n{\lambda }_T^3=g\zeta (3/2)\approx 2.612$ (for degeneracy $g=1$), i.e. the interparticle spacing is comparable to ${\lambda }_T$; the "$\sim 1$" slogan just means this order-unity number.

Error: $\mu$ is bounded by $\mu \le 0$; once excited states saturate at $\mu \to 0^{-}$, $\mu$ can rise no further and the surplus has no choice but the ground state.

Error: the integrand near $\epsilon =0$ behaves like $\sqrt{\epsilon }/\epsilon =1/\sqrt{\epsilon }$, which is integrable in 3D — the excited integral stays finite, so it saturates. Only the discrete $N_0$ term diverges.

Error: the Boltzmann form places no upper bound on excited occupation ($\mu$ unrestricted), so there is no overflow and no condensation.

Error: $T_c\propto n^{2/3}$, so lower density gives lower $T_c$ — that's why real alkali BECs (tiny $n$) demand sub-microkelvin cooling.

Error: with a constant density of states the excited integral ${\int }_0^{\infty }\frac{dx}{e^x-1}$ diverges logarithmically at small $x$, so excited states absorb any number — no saturation, no condensate.

Error: the Riemann zeta function $\zeta (3/2)$ comes from the energy integral ${\int }_0^{\infty }\frac{\sqrt{x}dx}{e^x-1}=\Gamma (3/2)\zeta (3/2)$ after the substitution $x=\epsilon /k_{B}T$, not from spatial volume.

Error: they are linked but distinct — Superfluidity involves interactions and flow without dissipation; ideal-gas BEC is non-interacting and need not be superfluid.

Because $g(\epsilon )\to 0$ at $\epsilon =0$ gives the ground state zero weight in the continuum integral, yet below $T_c$ it holds a macroscopic $N_0$ — so it must be added as a discrete term.

Because the saturation condition $n=g\zeta (3/2)/{\lambda }_T^3$ ties density and temperature together through ${\lambda }_T$; only their combination measures when matter waves overlap.

Because as $\mu \to 0^{-}$ the denominator $\to 0^{+}$, so $N_0\to \infty$ — a single state provides an "escape valve" of unbounded capacity.

To pull all temperature dependence out front as $(k_{B}T{)}^{3/2}$ and reduce the remaining integral to a pure dimensionless number ($\Gamma (3/2)\zeta (3/2)$); see the line-by-line box at the top.

Because $N_0/N=1-(T/T_c{)}^{3/2}$; near $T_c$ the deviation $1-(T/T_c{)}^{3/2}\approx \frac{3}{2}(1-T/T_c)$ grows from zero as a fractional power in the underlying integral, a non-analytic onset that flags a genuine phase transition rather than smooth crossover.

Because it emerges solely from the symmetrization of identical-boson wavefunctions (allowing shared states), with no potential energy or classical mechanism involved.

Higher $T$ shrinks ${\lambda }_T$ (since ${\lambda }_T\propto T^{-1/2}$), lowering $n{\lambda }_T^3$ below $g\zeta (3/2)$; the excited states regain enough room to hold everyone.

Approaching from above, $\mu$ climbs toward $0$; at and below $T_c$ it locks at $0^{-}$ and stays there, which is the hallmark of the condensed phase.

Zero — there is no condensate above $T_c$; excited states still have spare room, so $N_0$ is microscopic (order 1, not macroscopic).

That $N_0/N$ stays finite (nonzero) as $N\to \infty$ — a single quantum state holding a fixed fraction of an unbounded total.

The excited-state integral diverges even more strongly than in 2D, so a uniform 1D ideal gas has no saturation and no condensation.

Yes — trapping alters the effective density of states, allowing BEC in a 2D harmonic trap where the uniform 2D gas fails; geometry, not just spatial dimension, sets the power of $\epsilon$.

It reduces to the Boltzmann form $e^{-(\epsilon -\mu )/k_{B}T}$, since the $-1$ becomes negligible — high-energy tails always look classical.

Distinguishability removes the statistical enhancement; each species condenses independently by its own boson statistics, but they don't merge into one shared BEC state.