Worked examples — Quantum statistics — distinguishable vs indistinguishable particles

Before anything, recall the one master formula the parent derived:

Let me name one convenient shorthand once so I never sneak it in later: $x\equiv \beta (\epsilon -\mu )=\frac{\epsilon -\mu }{k_{B}T}.$ It is a pure number (energy over energy). Positive $x$ means "this level costs more than the going rate $\mu$"; negative $x$ means "this level is a bargain, below the price of a particle."

The picture below shows all three master curves at once — keep glancing back at it as we work each cell.

Reading the figure: plotted against $x$, the blue Fermi–Dirac curve stays trapped between $0$ and $1$ (exclusion), the pink Bose–Einstein curve rockets up as $x\to 0^{+}$ (bunching), and the pale-yellow Maxwell–Boltzmann curve $e^{-x}$ threads between them. For large $x$ (right side) all three merge — that is the classical corner.

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The scenario matrix

Every question this topic throws at you lands in one of these cells. Each example is tagged with the cell it covers.

Cell	Regime being tested	Which example
A	Sign of $x$: level above $\mu$ ($x>0$)	Ex 1
B	Sign of $x$: level exactly at $\mu$ ($x=0$)	Ex 2
C	Sign of $x$: level below $\mu$ ($x<0$)	Ex 3
D	Limit $T\to 0$ (cold): Fermi step and Bose freeze-out	Ex 4
E	Limit $T\to \infty$ / dilute: all three collapse to classical	Ex 5
F	Degenerate edge: boson at $\epsilon \to \mu$ (divergence / BEC trigger)	Ex 6
G	Pure counting from scratch (microstate census)	Ex 7
H	Real-world word problem (photon gas, $\mu =0$)	Ex 8
I	Exam twist: Gibbs $1/N!$ and entropy of mixing	Ex 9

Prerequisite links if a cell feels shaky: Grand Canonical Ensemble, Fermi Gas & Fermi Energy, Bose–Einstein Condensation, Pauli Exclusion Principle, Blackbody Radiation (Photon Gas), Gibbs Paradox & Entropy of Mixing, Maxwell–Boltzmann Distribution — and always the parent topic note.

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Example 1 — Cell A: a level above the chemical potential

Forecast: the level sits above $\mu$ by $0.2\text{eV}$, which is twice the thermal energy. Guess: much less than half-full — a small number.

1. Compute $x$. $x=\frac{\epsilon -\mu }{k_{B}T}=\frac{5.2-5.0}{0.10}=2$. Why this step? Every distribution depends only on the dimensionless ratio $x$. Getting $x$ turns a physics problem into arithmetic.
2. Plug into Fermi–Dirac ($a=+1$): $⟨n⟩=\frac{1}{e^2+1}$. Why this step? Electrons are fermions (half-integer spin, Pauli Exclusion Principle), so we use $a=+1$.
3. Evaluate. $e^2\approx 7.389$, so $⟨n⟩=\frac{1}{8.389}\approx 0.119$. Why this step? This is the actual number a measurement of that level's average filling would give.

Verify: $x>0$ must give $⟨n⟩<\frac{1}{2}$ for fermions (levels above $\mu$ are more empty than full). $0.119<0.5$ ✓. Also $0<⟨n⟩<1$, respecting exclusion. ✓

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Example 2 — Cell B: exactly at the chemical potential

Forecast: the "price" $\mu$ is often called the Fermi level here. Guess: exactly half-full, independent of $T$.

1. Compute $x$. $x=\frac{\mu -\mu }{k_{B}T}=0$. Why this step? Notice $T$ cancels — this is why the answer will be temperature-independent.
2. Plug in: $⟨n⟩=\frac{1}{e^0+1}=\frac{1}{1+1}=\frac{1}{2}$. Why this step? $e^0=1$ exactly, so the Fermi function passes through $\frac{1}{2}$ at $\epsilon =\mu$ for all $T>0$.

Verify: This is the defining property of the chemical potential for fermions — the level that is exactly half-occupied. On the master figure at the top, the blue curve crosses $\frac{1}{2}$ exactly at $x=0$. ✓

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Example 3 — Cell C: a level below the chemical potential

Forecast: below $\mu$ means a "bargain" level; fermions should fill it almost completely. Guess: close to 1.

1. Compute $x$. $x=\frac{4.8-5.0}{0.10}=-2$. Why this step? Negative $x$ is the mirror image of Ex 1 — good to see the symmetry.
2. Plug in: $⟨n⟩=\frac{1}{e^{-2}+1}=\frac{1}{0.1353+1}=\frac{1}{1.1353}\approx 0.881$. Why this step? Fills the "below $\mu$" cell so you have seen both signs.
3. Notice the symmetry. $⟨n(x)⟩+⟨n(-x)⟩=1$: here $0.119+0.881=1.000$. Why this step? The Fermi function is antisymmetric about $(\mu ,\frac{1}{2})$; this is a fast exam sanity check.

Verify: $0.881>\frac{1}{2}$ (level below $\mu$ is more than half full) ✓ and it pairs with Ex 1 to sum to exactly 1 ✓.

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Example 4 — Cell D: the cold limit $T\to 0$ — Fermi step and Bose freeze-out

Forecast: fermions pack into the lowest levels — full below $\mu$, empty above, a cliff at $\mu$. Bosons, being able to pile up, should all abandon excited levels and crash into the lowest one — every excited level empties out.

Part (a) — Fermi step.

1. Below $\mu$: $\epsilon <\mu \Rightarrow x<0\Rightarrow x\to -\infty$ as $T\to 0$. Then $e^x\to 0$, so $⟨n⟩=\frac{1}{0+1}=1$. Why this step? Cold means $\beta$ huge; it magnifies the sign of $(\epsilon -\mu )$ into $\pm \infty$.
2. Above $\mu$: $\epsilon >\mu \Rightarrow x\to +\infty$, so $e^x\to \infty$ and $⟨n⟩=\frac{1}{\infty +1}=0$. Why this step? Same logic, opposite sign — this produces the step.
3. The step. The Fermi function becomes a unit step: 1 up to $\mu$, then 0. The value at $\mu$ stays $\frac{1}{2}$ (Ex 2). At $T=0$, $\mu$ is called the Fermi energy ${\epsilon }_F$. Why this step? This step is the entire basis of white-dwarf and metal-electron physics.

Part (b) — Bose freeze-out.

4. Any excited level with $\epsilon >\mu$: $x>0$ and $x\to +\infty$ as $T\to 0$, so $e^x\to \infty$ and $⟨n{⟩}_{BE}=\frac{1}{e^x-1}\to \frac{1}{\infty -1}=0$. Why this step? Every excited boson level empties as $T\to 0$ — this is the "freeze-out." The particles do not vanish; they all fall into the lowest level (the ground state).
5. The lowest level: to hold all $N$ particles it needs $⟨n⟩$ macroscopic, which (from Ex 6) requires $x\to 0^{+}$, i.e. $\mu \to {\epsilon }_{\min }$ from below. This macroscopic ground-state pile-up is Bose–Einstein Condensation. Why this step? It closes the cell: fermions form a step (spread out), bosons condense (collapse into one state) — the two cold limits are opposites.

Verify: In the figure below, the blue cold Fermi curve is nearly a vertical cliff at $\epsilon =\mu$ passing through $\frac{1}{2}$; the pink cold Bose curve for excited levels is pinned near $0$ except where $\epsilon \to \mu$, where it spikes — exactly the two behaviours just derived. ✓

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Example 5 — Cell E: the classical corner (all three agree)

Forecast: big $x$ means the $\pm 1$ is a rounding error next to $e^5\approx 148$. Guess: all three answers within about $1\%$.

1. Fermi–Dirac: $\frac{1}{e^5+1}=\frac{1}{148.413+1}=\frac{1}{149.413}\approx 0.006693$.
2. Bose–Einstein: $\frac{1}{e^5-1}=\frac{1}{147.413}\approx 0.006784$.
3. Maxwell–Boltzmann: $e^{-5}=\frac{1}{148.413}\approx 0.006738$. Why these steps? Same $x$, only $a$ changes ($+1,-1,0$). The whole point is to see the answers converge.
4. Relative spread. $\frac{0.006784-0.006693}{0.006738}\approx 0.0135$, about $1.35\%$. Why this step? Quantifies "they agree." As $x$ grows the spread shrinks like $e^{-x}$.

Verify: Ordering must be $⟨n{⟩}_{BE}>⟨n{⟩}_{MB}>⟨n{⟩}_{FD}$ (bosons bunch, fermions avoid), and indeed $0.006784>0.006738>0.006693$ ✓. This is the classical regime the parent flagged, i.e. the far-right of the master figure.

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Example 6 — Cell F: the boson edge $\epsilon \to \mu$ (divergence)

Forecast: the parent warned $⟨n{⟩}_{BE}$ diverges as $\epsilon \to \mu$. Guess: the numbers blow up like $1/x$.

1. $x=1.0$: $⟨n⟩=\frac{1}{e^1-1}=\frac{1}{1.71828}\approx 0.5820$.
2. $x=0.1$: $⟨n⟩=\frac{1}{e^{0.1}-1}=\frac{1}{0.10517}\approx 9.508$.
3. $x=0.01$: $⟨n⟩=\frac{1}{e^{0.01}-1}=\frac{1}{0.010050}\approx 99.50$. Why these steps? Watch each tenfold shrink in $x$ multiply $⟨n⟩$ by roughly ten — that is the $1/x$ divergence.
4. Small-$x$ law. For tiny $x$, $e^x-1\approx x$, so $⟨n⟩\approx 1/x\to \infty$. Why this step? Explains the runaway: an unbounded pile-up in the lowest level is exactly Bose–Einstein Condensation.
5. Why $x<0$ is forbidden. If $x<0$ then $e^x-1<0$ and $⟨n⟩<0$ — an impossible negative occupation. So $\mu <{\epsilon }_{\min }$ always for bosons. Why this step? Fixes the parent's mistake-callout: boson $\mu$ is bounded above by the ground-state energy.

Verify: $⟨n⟩$ is positive and increasing as $x\to 0^{+}$: $0.582<9.508<99.50$ ✓, and each equals $\approx 1/x$ for small $x$ ($1/0.1=10\approx 9.5$, $1/0.01=100\approx 99.5$) ✓.

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Example 7 — Cell G: microstate census from scratch

Forecast: distinguishable should be the biggest ($g^N=4^3$), bosons next, fermions smallest. Guess $64,20,4$.

1. Distinguishable. Each of the $N=3$ labeled particles independently picks 1 of $g=4$ boxes: $g^N=4^3=64$. Why this step? "Order matters, repetition allowed" — the classical count from the parent.
2. Bosons. Unordered, repetition allowed (stars-and-bars): $\left(\genfrac{}{}{0ex}{}{N+g-1}{N}\right)=\left(\genfrac{}{}{0ex}{}{3+4-1}{3}\right)=\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$. Why this step? Indistinguishable particles → count unordered fillings; bosons may share a box.
3. Fermions. Unordered, no repetition: $\left(\genfrac{}{}{0ex}{}{g}{N}\right)=\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$. Why this step? Pauli Exclusion Principle: at most one per box → just choose which $N=3$ of the $g=4$ boxes are occupied.

Verify: Monotone ordering $64>20>4$ ✓ (each restriction removes microstates). Sanity check the fermion count another way: choosing which single box is empty also gives $\left(\genfrac{}{}{0ex}{}{4}{1}\right)=4$ ✓.

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Example 8 — Cell H: real-world word problem (photon gas, $\mu =0$)

Forecast: photons love to bunch (bosons), but this mode costs twice the thermal energy, so not many. Guess: a bit under 1.

1. Set $\mu =0$. Then $x=\beta \epsilon =\epsilon /k_{B}T=2$. Why this step? Photons have $\mu =0$ (they can be created/destroyed freely), which is why we never see photon BEC in a cavity.
2. Bose–Einstein occupation ($a=-1$): $⟨n⟩=\frac{1}{e^2-1}$. Why this step? This is the Planck occupation number — the seed of Planck's radiation law.
3. Evaluate. $e^2-1=6.389$, so $⟨n⟩=\frac{1}{6.389}\approx 0.1565$. Why this step? This is the average photon count in that specific mode at that temperature.

Verify: Compare with the fermion answer at the same $x=2$ from Ex 1 ($0.119$): the boson value $0.1565$ is larger, as bunching demands $⟨n{⟩}_{BE}>⟨n{⟩}_{FD}$ ✓. Positive and finite (since $x>0$, no divergence) ✓.

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Example 9 — Cell I: exam twist — Gibbs $1/N!$ and mixing entropy

Forecast: if the gases are truly identical, mixing them changes nothing physical, so $\Delta S=0$. Guess: the naive un-corrected count gives a spurious positive $\Delta S$; the $1/N!$ cancels it.

1. Naive (labeled) count. Treating particles as distinguishable, removing the partition gives each particle twice the volume, so naive $\Delta S_{\text{naive}}=2N{k}_B\ln 2$. Why this step? This is the wrong, non-extensive answer that creates the Gibbs paradox.
2. Apply Gibbs correction. Dividing $Z_N$ by $N!$ subtracts the relabeling over-count. The number of ways to split $2N$ identical particles into the two labeled halves is $\left(\genfrac{}{}{0ex}{}{2N}{N}\right)$, so the correction removes $k_B\ln \left(\genfrac{}{}{0ex}{}{2N}{N}\right)$. Why this step? The over-count of relabelings is exactly this binomial factor.
3. Cancel via Stirling. For large $N$, $\ln \left(\genfrac{}{}{0ex}{}{2N}{N}\right)\approx 2N\ln 2$, so $\Delta S=2N{k}_B\ln 2-k_B\ln \left(\genfrac{}{}{0ex}{}{2N}{N}\right)\approx 2N{k}_B\ln 2-2N{k}_B\ln 2=0$. Why this step? The spurious mixing entropy vanishes for identical gases — extensivity restored.
4. Numbers, $N=100$. $\ln \left(\genfrac{}{}{0ex}{}{200}{100}\right)=135.80$; and $2N\ln 2=200\times 0.6931=138.63$. Ratio $135.80/138.63=0.980$. Why this step? We compute the actual finite-$N$ binomial and compare it to the Stirling estimate $2N\ln 2$ so the reader sees the cancellation is not exact for finite $N$ — it only becomes exact as $N\to \infty$, where the residual (a $\frac{1}{2}\ln (\pi N)$ Stirling term) becomes negligible next to the $2N\ln 2$ that grows with $N$.

Verify: For distinguishable gases (or two different gases) the correction does not apply and $\Delta S=2N{k}_B\ln 2>0$ — a real, positive mixing entropy — while identical gases give $0$. The numeric check $\ln \left(\genfrac{}{}{0ex}{}{200}{100}\right)\approx 135.80$ vs $138.63$ confirms the ratio $0.980$, approaching $1$ as $N$ grows. ✓

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Recall Rapid self-test

Level above $\mu$, fermion, $x=2$: occupation ::: $1/(e^2+1)\approx 0.119$ Level exactly at $\mu$, fermion: occupation ::: exactly $1/2$, for all $T$ Boson with $x\to 0^{+}$: occupation ::: diverges like $1/x$ → BEC Bose excited levels as $T\to 0$: occupation ::: all go to $0$ (freeze-out) — particles condense into ground state Photon mode ($\mu =0$) at $\epsilon =2k_{B}T$: occupation ::: $1/(e^2-1)\approx 0.157$ 3 particles, 4 states, fermions: microstates ::: $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ 3 particles, 4 states, bosons: microstates ::: $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ Why does mixing entropy vanish for identical gases ::: the $1/N!$ Gibbs factor removes the relabeling over-count