Exercises — Bose-Einstein statistics — bosons

A boson has integer spin ($0,1,2,\dots$). Check each:

• Photon, spin $1$ → boson. ✓
• Electron, spin $\frac{1}{2}$ → half-integer → fermion (see Fermi-Dirac statistics — fermions).
• Helium-4, net spin $0$ → boson. ✓
• Proton, spin $\frac{1}{2}$ → fermion.
• Phonon, spin $0$ → boson. ✓

Bosons: photon, He-4, phonon. These are exactly the "pile-up" particles that obey our formula.

The signature is the ==$-1$== in the denominator. It comes from summing the boson geometric series ${\sum }_{n=0}^{\infty }{x}^n=1/(1-x)$, which gives $\bar{n}=x/(1-x)$. Switch the sign to $+1$ and you get Fermi-Dirac, whose sum $1+x$ (only $n=0,1$) gives $\bar{n}=x/(1+x)$.

Mnemonic: Boson → Bunch → Below by one → $-1$.

What we do: plug into $\bar{n}=1/(e^{\beta (\epsilon -\mu )}-1)$ with $\mu =0$ and $\beta \epsilon =2$. Why: for photons number is not conserved, so $\mu =0$ — a fact worth memorising. $\bar{n}=\frac{1}{e^2-1}=\frac{1}{7.389-1}=\frac{1}{6.389}\approx \mathrm{0.157.}$ What it means: a mode at twice the thermal energy holds far less than one photon on average — the high-frequency starvation that makes blackbody intensity drop (see Planck radiation law).

Bose: ${\bar{n}}_{BE}=\frac{1}{e^{0.05}-1}=\frac{1}{1.05127-1}=\frac{1}{0.05127}\approx \mathrm{19.50.}$ Classical: ${\bar{n}}_{MB}=e^{-0.05}=\mathrm{0.95123.}$ Ratio: ${\bar{n}}_{BE}/{\bar{n}}_{MB}\approx 19.50/0.9512\approx \mathrm{20.5.}$ What it means: very close to $\mu$ the bosons bunch ~20× harder than the classical guess — this "extra crowding" is the physical seed of lasers and Bose-Einstein condensation.

Step — what & why: write ${\bar{n}}_{BE}=1/(e^y-1)$ with $y=\beta (\epsilon -\mu )$. For large $y$, $e^y\gg 1$, so $e^y-1\approx e^y$. This is legitimate because we are dropping a $1$ next to a huge number. ${\bar{n}}_{BE}=\frac{1}{e^y-1}=\frac{e^{-y}}{1-e^{-y}}\approx e^{-y}\left(1+e^{-y}+\dots \right)\approx e^{-y}={\bar{n}}_{MB}.$ Fractional error at $y=4$: exact ${\bar{n}}_{BE}=1/(e^4-1)=1/53.598=0.018659$; classical $e^{-4}=0.018316$. $\frac{{\bar{n}}_{BE}-{\bar{n}}_{MB}}{{\bar{n}}_{BE}}=\frac{0.018659-0.018316}{0.018659}\approx 0.0184=1.84\%.$ What it means: already at $4k_{B}T$ above $\mu$ the quantum bunching correction is under $2\%$ — the "dilute limit" where all three statistics merge (see Maxwell-Boltzmann statistics).

What & why: at the lowest level $\epsilon ={\epsilon }_{\min }$, as $\mu \to {\epsilon }_{\min }^{-}$ we have ${\epsilon }_{\min }-\mu \to 0^{+}$, so $x=e^{-\beta ({\epsilon }_{\min }-\mu )}\to 1^{-}$. Then $\bar{n}=x/(1-x)$ has denominator $\to 0$, so $\bar{n}\to +\infty$. Numeric check at $\beta ({\epsilon }_{\min }-\mu )=0.001$: ${\bar{n}}_0=\frac{1}{e^{0.001}-1}=\frac{1}{1.0010005-1}=\frac{1}{0.0010005}\approx \mathrm{999.5.}$ What it means: a nearly-infinite crowd rushes into the single lowest state — the mathematical divergence of ${\mathcal{Z}}_0=1/(1-x)$ is the onset of Bose-Einstein condensation. This is why $\mu$ can approach but never touch ${\epsilon }_{\min }$. Look at the left edge of the figure below.

Step 1 — what & why. Probabilities are $P(n)=x^n/{\mathcal{Z}}_i$ with ${\mathcal{Z}}_i=1/(1-x)$. We already know $\bar{n}=x{\partial }_x\ln {\mathcal{Z}}_i=x/(1-x)$ (the generating-function trick from the parent note). We now apply $x{\partial }_x$ again to reach the second moment. Step 2 — the machinery. A clean identity: $\mathrm{Var}(n)=x\frac{\partial \bar{n}}{\partial x}$. Why this works: $\bar{n}=x{\partial }_x\ln \mathcal{Z}$, and differentiating the log-partition function a second time yields the fluctuation. Apply it: $\bar{n}=\frac{x}{1-x}⟹\frac{\partial \bar{n}}{\partial x}=\frac{(1-x)+x}{(1-x{)}^2}=\frac{1}{(1-x{)}^2}.$ $\mathrm{Var}(n)=x\cdot \frac{1}{(1-x{)}^2}=\frac{x}{(1-x{)}^2}.$ Step 3 — factor it into $\bar{n}$ form. $\frac{x}{(1-x{)}^2}=\frac{x}{1-x}\cdot \frac{1}{1-x}=\bar{n}\cdot (1+\bar{n}),$ using $\frac{1}{1-x}=1+\frac{x}{1-x}=1+\bar{n}$. $\boxed{\mathrm{Var}(n)=\bar{n}(1+\bar{n}).}$ Interpretation: a classical (Poisson) count has $\mathrm{Var}=\bar{n}$. Bosons carry an extra ${\bar{n}}^2$: they fluctuate more. That excess is photon bunching — bosons arrive in clumps, the statistical fingerprint measured in the Hanbury Brown–Twiss experiment.

What & why: compute $y_j=\beta ({\epsilon }_j-\mu )$ for each level, then ${\bar{n}}_j=1/(e^{y_j}-1)$.

• Level 1: $y_1=\beta ({\epsilon }_1-\mu )=0.2$. ${\bar{n}}_1=1/(e^{0.2}-1)=1/0.22140=\mathrm{4.5167.}$
• Level 2: $y_2=\beta ({\epsilon }_2-\mu )=1.2$. ${\bar{n}}_2=1/(e^{1.2}-1)=1/2.32012=\mathrm{0.43101.}$ Ratio: $\frac{{\bar{n}}_1}{{\bar{n}}_2}=\frac{4.5167}{0.43101}\approx \mathrm{10.48.}$ What it means: level 1 (only $0.2k_{B}T$ above $\mu$) is in the crowded near-$\mu$ regime; level 2 ($1.2k_{B}T$ above) is heading toward the dilute classical tail. A single $k_{B}T$ of energy separation drops occupancy by an order of magnitude — bosons pile disproportionately into the cheapest states.

Step 1 — assemble the pieces. Energy density = (energy per photon) $\times$ (number of modes at that energy) $\times$ (photons per mode): $u(\epsilon )\propto \underset{\text{energy}}{\underbrace{\epsilon }}\cdot \underset{g(\epsilon )}{\underbrace{{\epsilon }^2}}\cdot \underset{\bar{n},\mu =0}{\underbrace{\frac{1}{e^{\epsilon /k_{B}T}-1}}}=\frac{{\epsilon }^3}{e^{\epsilon /k_{B}T}-1}.$ That is Planck's law (see Planck radiation law). Step 2 — why we maximise. The peak of $u$ tells us the colour of a hot object. Let $z=\epsilon /k_{B}T$ (dimensionless) so $u\propto z^3/(e^z-1)$. Set $\frac{d}{dz}\left[\frac{z^3}{e^z-1}\right]=0$. Step 3 — the transcendental condition. Differentiate (quotient rule): $\frac{3z^2(e^z-1)-z^3{e}^z}{(e^z-1{)}^2}=0⟹3(e^z-1)=z{e}^z⟹3-3e^{-z}=z.$ Rearranged: $\boxed{(3-z)e^z=3}$, i.e. $z=3(1-e^{-z})$. Step 4 — solve numerically (why: no closed form). Iterate $z_{k+1}=3(1-e^{-z_k})$ starting $z_0=3$: $3(1-e^{-3})=2.851;3(1-e^{-2.851})=2.827;\to \mathrm{2.821.}$ Converges to $z^{\star }={\epsilon }^{\star }/k_{B}T\approx \mathrm{2.821.}$ What it means: the blackbody energy density peaks at ${\epsilon }^{\star }\approx 2.821k_{B}T$ — hotter body ⇒ higher peak energy ⇒ bluer glow. This single number is Wien's displacement law, emerging purely from the boson $-1$ and the ${\epsilon }^2$ density of states.

The convergence condition: the geometric sum $\mathcal{Z}=\sum x^n=1/(1-x)$ demands $x=e^{-\beta (\epsilon -\mu )}<1$, i.e. $\mu <\epsilon$ for every level. BEC is triggered when $\mu$ is pushed up to touch the ground-state energy ${\epsilon }_{\min }$, forcing macroscopic ground-state occupancy.

For massive bosons: particle number $N$ is conserved. To hold $N$ fixed as $T$ drops, $\mu$ must rise toward ${\epsilon }_{\min }$ — and at a finite critical temperature it reaches it. Condensation happens.

For photons: number is not conserved, so $\mu$ is pinned at $0$ by equilibrium (it is never a free knob to raise). It cannot climb toward ${\epsilon }_{\min }$; instead, as $T$ drops, photons simply disappear into the walls. With $\mu$ frozen at $0$, the ground-state occupancy $1/(e^{\beta {\epsilon }_{\min }}-1)$ stays finite (for ${\epsilon }_{\min }>0$). No divergence ⇒ no condensate.

One-line synthesis: BEC needs a conserved $N$ so that $\mu$ can be driven up to ${\epsilon }_{\min }$; photons lack that conservation, so they thin out rather than condense.

1.1 photon, He-4, phonon · 1.2 the $-1$ · 2.1 $0.157$ · 2.2 $19.5$ (ratio $20.5$) · 3.1 $1.84\%$ error · 3.2 $\approx 999.5$, diverges · 4.1 $\mathrm{Var}=\bar{n}(1+\bar{n})$ · 4.2 ratio $\approx 10.5$ · 5.1 ${\epsilon }^{\star }\approx 2.821k_{B}T$ · 5.2 photons: $\mu =0$ fixed ⇒ no BEC.