Exercises — Bose-Einstein condensation — concept

(b). For bosons $\bar{n}(\epsilon )=\frac{1}{e^{(\epsilon -\mu )/k_{B}T}-1}$ must be non-negative, which forces $\mu \le 0$. Why is the capacity finite? Look at the integral. The most atoms the excited states can hold (at $\mu =0$) is the area under this curve: $N_{\text{exc}}^{\max }={\int }_0^{\infty }\underset{\propto \sqrt{\epsilon }}{\underbrace{g(\epsilon )}}\underset{\text{occupation}}{\underbrace{\frac{1}{e^{\epsilon /k_{B}T}-1}}}d\epsilon .$ Substituting $x=\epsilon /k_{B}T$ turns this into a pure number times a temperature factor: ${\int }_0^{\infty }\frac{\sqrt{x}}{e^x-1}dx=\Gamma (3/2)\zeta (3/2)$, a finite value. The figure below shows why it is finite: near $x=0$ the occupation $\frac{1}{e^x-1}\approx \frac{1}{x}$ blows up, but the density-of-states factor $\sqrt{x}$ pushes the product down to $x^{-1/2}$, whose area near zero is finite (the shaded region has no infinite spike). With a finite ceiling, surplus atoms have nowhere to go but the $\epsilon =0$ ground state. No attraction, no gravity — pure statistics.

False. If $\mu >0$ then for the ground state (${\epsilon }_0=0$) we would get $e^{-\mu /k_{B}T}<1$, making the denominator $e^{-\mu /k_{B}T}-1<0$ and hence $\bar{n}({\epsilon }_0)<0$ — a negative number of atoms, which is impossible. So $\mu$ pins itself at $0^{-}$, and the divergence of $N_0=\frac{1}{e^{-\mu /k_{B}T}-1}$ as $\mu \to 0^{-}$ is the "escape valve" that swallows the surplus.

$n{\lambda }_T^3=\zeta (3/2)\approx \mathrm{2.612.}$ $n{\lambda }_T^3$ is the number of atoms sitting inside one "quantum volume" ${\lambda }_T^3$. When this reaches $\sim 2.6$, the matter-wave clouds of neighbouring atoms overlap — the atoms can no longer be told apart as separate wavepackets — and condensation begins.

Use $k_B{T}_c=\frac{2\pi {\hslash }^2}{m}{\left(\frac{n}{g\zeta (3/2)}\right)}^{2/3}$. Inner ratio: $\frac{n}{\zeta (3/2)}=\frac{1.0\times 10^{19}}{2.612}=3.83\times 10^{18}{\text{m}}^{-3}$. Its $2/3$ power: $(3.83\times 10^{18}{)}^{2/3}=2.44\times 10^{12}{\text{m}}^{-2}$. Prefactor: $\frac{2\pi {\hslash }^2}{m}=\frac{2\pi (1.0546\times 10^{-34}{)}^2}{1.44\times 10^{-25}}=4.85\times 10^{-43}{\text{J⋅m}}^2$. So $k_B{T}_c=4.85\times 10^{-43}\times 2.44\times 10^{12}=1.18\times 10^{-30}\text{J}$. Divide by $k_B$: $T_c=\frac{1.18\times 10^{-30}}{1.381\times 10^{-23}}\approx 8.6\times 10^{-8}\text{K}\approx 86\text{nK}$. Order of magnitude ~$10^{-7}$ K — sub-microkelvin, exactly as the parent note said.

$\frac{N_0}{N}=1-{\left(\frac{T}{T_c}\right)}^{3/2}=1-(0.5{)}^{3/2}=1-0.3536=\mathrm{0.6464.}$ ≈ 64.6% of atoms are already condensed at half the critical temperature. The steep early rise reflects the infinite initial slope of the fraction curve just below $T_c$.

Set $\frac{N_0}{N}=\frac{1}{2}$: $\frac{1}{2}=1-{\left(\frac{T}{T_c}\right)}^{3/2}\Rightarrow {\left(\frac{T}{T_c}\right)}^{3/2}=\frac{1}{2}.$ Raise both sides to the $2/3$ power: $\frac{T}{T_c}={\left(\frac{1}{2}\right)}^{2/3}=\mathrm{0.630.}$ So at $T\approx 0.63T_c$ the gas is exactly half condensed.

$T_c\propto n^{2/3}$ (everything else fixed). So $\frac{T_{c,B}}{T_{c,A}}={\left(\frac{n_B}{n_A}\right)}^{2/3}=8^{2/3}=(2^3{)}^{2/3}=2^2=4.$ $T_c$ is 4× larger. Denser gas → smaller interparticle spacing → the matter waves overlap at a higher temperature (larger ${\lambda }_T$ not needed as much), so condensation happens sooner (warmer).

In 2D, counting states in a $k$-disc gives $N(k)\propto k^2\propto \epsilon$, so $g(\epsilon )=\frac{dN}{d\epsilon }=$ constant (no $\sqrt{\epsilon }$ factor). The maximum excited population at $\mu =0$ is then $N_{\text{exc}}^{\max }\propto {\int }_0^{\infty }\frac{g(\epsilon )d\epsilon }{e^{\epsilon /k_{B}T}-1}\propto {\int }_0^{\infty }\frac{dx}{e^x-1}.$ Near $x=0$, $e^x-1\approx x$, so the integrand behaves like $1/x$, and ${\int }_0\frac{dx}{x}=\ln x{|}_0$ diverges. A divergent maximum means the excited states can absorb unbounded numbers of atoms — there is no saturation, hence no overflow, no condensate. Contrast with 3D, where the extra $\sqrt{x}$ from $g(\epsilon )\propto \sqrt{\epsilon }$ tames the small-$x$ behaviour ($\sqrt{x}/x=x^{-1/2}$ is integrable), giving a finite number $\Gamma (3/2)\zeta (3/2)$. Here $\Gamma$ is the Gamma function, simply the smooth continuation of the factorial to non-integer arguments — it is defined by the very integral we are doing, $\Gamma (s)={\int }_0^{\infty }{x}^{s-1}{e}^{-x}dx$, and it satisfies $\Gamma (n)=(n-1)!$ for whole numbers; the specific value we need is $\Gamma (3/2)=\frac{\sqrt{\pi }}{2}\approx 0.886$. So the "finite number" $\Gamma (3/2)\zeta (3/2)$ is just (a fixed shape factor from the integral) × (the zeta value $2.612$). Dimensionality decides everything.

The intuition first (what are we even doing?). The two boxed formulas look different — one is about wavelengths overlapping, the other about temperature — but they must be the same physics written twice. Our job is to start from "matter waves overlap" and turn the crank until "temperature" pops out. Each algebra step below is just isolating $T_c$; watch how ${\lambda }_T$ (which hides $T$ inside a square root) gets unwrapped layer by layer, like peeling the temperature out of its packaging. Step 1 — state the overlap condition. $n{\lambda }_{T_c}^3=\zeta (3/2)$, i.e. ${\lambda }_{T_c}^3=\frac{\zeta (3/2)}{n}$. Why: this is the physical starting point — the moment the clouds touch. Step 2 — expose the hidden $T$. Insert ${\lambda }_{T_c}=\frac{h}{\sqrt{2\pi m{k}_B{T}_c}}$, because $T$ lives inside ${\lambda }_T$ and we want it out in the open: ${\lambda }_{T_c}^3=\frac{h^3}{(2\pi m{k}_B{T}_c{)}^{3/2}}=\frac{\zeta (3/2)}{n}.$ Step 3 — undo the $3/2$ power. Raise both sides to the $2/3$ power to strip the exponent off the temperature bracket (this is why the final formula carries a $2/3$): $2\pi m{k}_B{T}_c={\left(\frac{n{h}^3}{\zeta (3/2)}\right)}^{2/3}={\left(\frac{n}{\zeta (3/2)}\right)}^{2/3}{h}^2.$ Step 4 — isolate $k_B{T}_c$ and convert $h\to \hslash$. Divide by $2\pi m$; then use $h=2\pi \hslash$ so $h^2=4{\pi }^2{\hslash }^2$ and $\frac{h^2}{2\pi m}=\frac{2\pi {\hslash }^2}{m}$: $k_B{T}_c=\frac{h^2}{2\pi m}{\left(\frac{n}{\zeta (3/2)}\right)}^{2/3}=\frac{2\pi {\hslash }^2}{m}{\left(\frac{n}{\zeta (3/2)}\right)}^{2/3}✓$ The two boxed formulas are the same statement — one phrased in the geometric language of overlapping waves, the other in the thermal language of a critical temperature.

(a) $\frac{N_0}{N}=1-(0.30{)}^{3/2}=1-0.1643=0.8357$, so ≈ 83.6%. (b) $N_0=0.8357\times 2.0\times 10^4=1.67\times 10^4$ atoms, i.e. ≈ 16 700 atoms share the single lowest state. (c) $N_{\text{exc}}=N-N_0=2.0\times 10^4-1.67\times 10^4=3.29\times 10^3\approx 3300$ atoms (equivalently $N(T/T_c{)}^{3/2}=2.0\times 10^4\times 0.1643$).

At fixed $n$ and $g$, $T_c\propto \frac{1}{m}$ (from $k_B{T}_c=\frac{2\pi {\hslash }^2}{m}(\cdots )$). $\frac{T_{c,\text{Na}}}{T_{c,\text{Rb}}}=\frac{m_{\text{Rb}}}{m_{\text{Na}}}=\frac{1.44\times 10^{-25}}{3.82\times 10^{-26}}=\mathrm{3.77.}$ Sodium has the higher $T_c$, about 3.8× that of rubidium. Lighter atoms have a larger ${\lambda }_T$ at the same temperature, so their matter waves overlap sooner (warmer). This is why lighter species are easier to condense.

At onset $n{\lambda }_{T_c}^3=\zeta (3/2)$, so ${\lambda }_{T_c}={\left(\frac{\zeta (3/2)}{n}\right)}^{1/3}={\left(\frac{2.612}{1.0\times 10^{19}}\right)}^{1/3}=(2.612\times 10^{-19}{)}^{1/3}\text{m}.$ $(2.612\times 10^{-19}{)}^{1/3}=6.39\times 10^{-7}\text{m}\approx 0.64\mu \text{m}$. Interparticle spacing: $d=n^{-1/3}=(1.0\times 10^{19}{)}^{-1/3}=4.64\times 10^{-7}\text{m}\approx 0.46\mu \text{m}$. Ratio $\frac{{\lambda }_{T_c}}{d}=\zeta (3/2{)}^{1/3}=2.612^{1/3}=1.38$. The wavelength is ~1.4× the spacing — i.e. the matter waves have just grown large enough to overlap. This is the geometric meaning of "phase-space density of order one."

Substituting $x=\epsilon /k_{B}T$, the shape integral is ${\int }_0^{\infty }\frac{x^{s-1}}{e^x-1}dx=\Gamma (s)\zeta (s)$, where $\Gamma (s)={\int }_0^{\infty }{x}^{s-1}{e}^{-x}dx$ is the Gamma function introduced in L3.2 (the factorial extended to real $s$).

• Small-$x$ convergence: near $x=0$, $e^x-1\approx x$, so integrand $\approx x^{s-1}/x=x^{s-2}$. The integral ${\int }_0{x}^{s-2}dx$ converges iff $s-2>-1$, i.e. $\boxed{s>1}$.
• Large-$x$: the $e^x$ kills the tail always. So the only condition is $s>1$ (also $\zeta (s)$ is finite only for $s>1$ — the same threshold). (i) uniform 3D: $g\propto \sqrt{\epsilon }={\epsilon }^{1/2}$ so $s-1=1/2\Rightarrow s=3/2>1$ → BEC exists ✓. (ii) uniform 2D: $g=$ const $={\epsilon }^0$ so $s-1=0\Rightarrow s=1$, not $>1$ → borderline divergent, no BEC ✓ (matches L3.2). (iii) 3D harmonic trap: $g\propto {\epsilon }^2$ so $s-1=2\Rightarrow s=3>1$ → BEC exists (and indeed harmonic-trap BEC is what experiments actually see). The trap helps by raising $s$.

For general $s$, $N_{\text{exc}}(T)\propto {\int }_0^{\infty }\frac{{\epsilon }^{s-1}}{e^{\epsilon /k_{B}T}-1}d\epsilon \propto T^s$ (pull out $(k_{B}T{)}^s$ in the substitution $x=\epsilon /k_{B}T$; the leftover integral is the pure number $\Gamma (s)\zeta (s)$). So $\frac{N_{\text{exc}}}{N}={\left(\frac{T}{T_c}\right)}^s,\frac{N_0}{N}=1-{\left(\frac{T}{T_c}\right)}^s.$ Thus $\alpha =s$. Uniform 3D: $s=3/2$ (recovers the parent-note exponent ✓). Harmonic trap: $\alpha =s=3$. Evaluate at $T=0.5T_c$ with $\alpha =3$: $\frac{N_0}{N}=1-(0.5{)}^3=1-0.125=\mathrm{0.875.}$ ≈ 87.5% — compared with the uniform gas's 64.6%, the trap condenses a substantially larger fraction at the same reduced temperature. The reason: its steeper $g(\epsilon )\propto {\epsilon }^2$ leaves fewer low-energy excited states to hide in, so atoms are forced into the ground state sooner.

(a) Order parameter: the condensate fraction $N_0/N$. It is exactly $0$ for $T\ge T_c$ and grows continuously (from zero) for $T<T_c$ — the hallmark of a continuous (second-order-like) transition (see Phase transitions). (b) Non-analyticity: the function $N_0/N=1-(T/T_c{)}^{3/2}$ has an infinite slope at $T_c^{-}$: $\frac{d}{dT}\left[1-{\left(\frac{T}{T_c}\right)}^{3/2}\right]=-\frac{3}{2T_c}{\left(\frac{T}{T_c}\right)}^{1/2}\to -\frac{3}{2T_c}\text{as}T\to T_c^{-},$ while for $T>T_c$ it is identically $0$. The derivative jumps — a kink at $T_c$ that no analytic function can have. (More precisely, the heat capacity has a cusp at $T_c$.) This non-analyticity is the definition of a phase transition. (c) Classical invisibility: Boltzmann statistics gives $\bar{n}=e^{-(\epsilon -\mu )/k_{B}T}$, which is finite and positive for any $\mu$ — no cap on $\mu$, hence no saturation, hence no overflow. The special ingredient is the $-1$ in $\frac{1}{e^{(\epsilon -\mu )/k_{B}T}-1}$, which forces $\mu \le 0$ and creates the finite ceiling. Zero interactions needed; the transition lives entirely in the quantum statistics. (Contrast Fermi-Dirac statistics, where the $+1$ forbids piling up and gives no condensate — instead a filled Fermi sea. The condensate's coherence underlies Superfluidity.)

Q: In one breath, what sets $T_c$, and how does the condensate grow below it (general $s$)? A: $T_c$ is fixed by $n{\lambda }_{T_c}^3=g\zeta (3/2)$ (matter waves overlap); below it $N_0/N=1-(T/T_c{)}^s$ where $s=3/2$ (uniform 3D) or $s=3$ (harmonic trap), and there is no BEC unless $s>1$.