Exercises — Quantum statistics — distinguishable vs indistinguishable particles

The rule (from the Spin–Statistics Theorem): integer spin → boson (any number per state); half-integer spin → fermion (at most one).

• (a) electron, spin $\frac{1}{2}$ → fermion, at most one. (Governs the Pauli Exclusion Principle.)
• (b) photon, spin $1$ → boson, any number (why lasers work).
• (c) ${}^4$He: an even number (6) of fermions → total spin integer → boson (why it undergoes Bose–Einstein Condensation).
• (d) proton, spin $\frac{1}{2}$ → fermion, at most one. Key move: count the fermions inside a composite. Even count ⇒ boson; odd count ⇒ fermion.

$a=+1$ (Fermi–Dirac). Since $e^{\beta (\epsilon -\mu )}>0$ always, the denominator $e^{\beta (\epsilon -\mu )}+1>1$, so $⟨n⟩<1$. This is Pauli exclusion showing up in the arithmetic: a level holds at most one fermion, so its average filling is between 0 and 1. Contrast $a=-1$ (Bose): the denominator can shrink toward $0$, so $⟨n⟩$ can blow up — unlimited occupation.

The figure below draws out the 2-in-3 case explicitly, box by box, so you can see why the three counts differ; the 2-in-4 answers then follow from the same formulas.

Figure s01 — what it shows: three side-by-side panels, each showing rows of "3 boxes" with the two particles placed inside. Left panel (orange, distinguishable): particles carry labels A and B, so "A in box 1, B in box 2" and "B in box 1, A in box 2" are drawn as separate rows — 9 rows in total (a representative sample is drawn). Middle panel (blue, bosons): identical dots with no labels, and rows where both dots sit in one box are allowed — 6 rows. Right panel (green, fermions): identical dots but no box ever holds two — only the 3 rows with one dot in each of two different boxes survive. Reading left→right you literally watch arrangements disappear as the rules tighten: $9\to 6\to 3$.

2-in-3 (the figure): distinguishable $3^2=9$, bosons $\left(\genfrac{}{}{0ex}{}{3+2-1}{2}\right)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$, fermions $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$.

2-in-4 (the posed problem):

• (a) Distinguishable: each particle independently picks one of 4 boxes → $4^2=\boxed{16}$.
• (b) Bosons (unordered, repeats allowed) → $\left(\genfrac{}{}{0ex}{}{4+2-1}{2}\right)=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=\boxed{10}$.
• (c) Fermions (unordered, no repeats) → $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=\boxed{6}$. Notice the ordering $6<10<16$: fermions are the most "spread out," distinguishable the most numerous — the same pattern the figure shows for 2-in-3.

With $\epsilon =\mu$, $\beta (\epsilon -\mu )=0$, so $e^0=1$: $⟨n{⟩}_{FD}=\frac{1}{1+1}=\boxed{\frac{1}{2}}.$ It is exactly $\frac{1}{2}$ at every temperature — the level at the Fermi energy is always half-filled. This is the pivot point of the Fermi function: below $\mu$ levels are mostly full, above $\mu$ mostly empty, and right at $\mu$ it is a coin flip.

$e^{\ln 2}=2$, so $⟨n{⟩}_{BE}=\frac{1}{2-1}=\boxed{1}.$ One particle on average. Compare Fermi at the same argument: $1/(2+1)=\frac{1}{3}$. Same energy, but bosons pile up more — the $-1$ instead of $+1$ increases the count.

Recall the classical form from the definition callout above: $⟨n{⟩}_{MB}=e^{-\beta (\epsilon -\mu )}$ (the master formula with $a=0$). Here $\beta (\epsilon -\mu )=3$, so $e^3\approx 20.0855$.

• Classical: $⟨n{⟩}_{MB}=e^{-3}\approx \boxed{0.0498}$.

• Fermi: $\frac{1}{20.0855+1}\approx \boxed{0.0474}$.

• Bose: $\frac{1}{20.0855-1}\approx \boxed{0.0524}$. All three agree to within $\sim 5\%$. Why? When $e^{\beta (\epsilon -\mu )}\gg 1$, the $\pm 1$ is a tiny correction — this is exactly the classical regime: dilute, few double occupations, so distinguishable-vs-not stops mattering. Fermi sits below classical (exclusion suppresses), Bose above (bunching enhances).

Figure s02 — what it shows: all three occupation curves plotted against the single variable $x=\beta (\epsilon -\mu )$ on the horizontal axis (moving right = colder or higher-energy level), with $⟨n⟩$ on the vertical axis. Blue is Bose $1/(e^x-1)$, which shoots up toward infinity as $x\to 0^{+}$ (the condensation divergence, marked with a red dashed line at $x=0$). Green is Fermi $1/(e^x+1)$, which flattens to the dotted ceiling $⟨n⟩=1$ on the far left and passes through the marked point $⟨n⟩=\frac{1}{2}$ at $x=0$. Orange dashed is classical $e^{-x}$, sitting between them. A gray dashed line at $x=3$ marks the point used in this problem: there the three curves have visually merged — the classical limit. Near $x=0$ they fan apart, which is where quantum differences dominate.

Requirement: occupation can't be negative, $⟨n{⟩}_{BE}=\frac{1}{e^{\beta (\epsilon -\mu )}-1}\ge 0$. This needs the denominator positive: $e^{\beta (\epsilon -\mu )}-1>0\Rightarrow e^{\beta (\epsilon -\mu )}>1\Rightarrow \beta (\epsilon -\mu )>0\Rightarrow \mu <\epsilon .$ Since this must hold for every level including the lowest, $\mu <{\epsilon }_{\min }$. As $\mu \to {\epsilon }_{\min }$: for the ground level $\beta ({\epsilon }_{\min }-\mu )\to 0^{+}$, so $e^{(\cdots )}\to 1^{+}$ and $⟨n⟩\to \infty$ — a macroscopic number of particles crash into the single lowest state. That is Bose–Einstein Condensation. Fermions have no such bound: their denominator has $+1$, always positive, so $\mu$ (the Fermi level) can be any sign. Edge case — how strict is the inequality? In the thermodynamic limit (infinite system) the strict $\mu <{\epsilon }_{\min }$ holds and $\mu$ can only approach ${\epsilon }_{\min }$ from below, never touch it. In a finite real system the lowest level holds a finite maximum number, so $\mu$ may creep microscopically close, within $O(1/N)$ of ${\epsilon }_{\min }$, and the divergence is replaced by a large-but-finite ground-state occupation. So "$\mu <{\epsilon }_{\min }$ strictly" is exact only as $N\to \infty$; for finite $N$ it is "as close as the ground level's capacity allows." This is why real BEC is a sharp-but-smoothed crossover, not a mathematically perfect singularity.

Step A — where the entropy formula comes from. Using the free energy just defined, $S=-\partial F/\partial T$ with $F=-k_{B}T\ln Z_N$ gives $S=k_B(\ln Z_N+\beta ⟨E⟩)$. For an ideal gas the single-particle partition function factorises as $Z_1=V/{\lambda }^3$, where $\lambda =h/\sqrt{2\pi m{k}_{B}T}$ is the thermal wavelength, so $Z_1$ is proportional to $V$. Step B — apply the $N!$ patch. With $Z_N=Z_1^N/N!$, $\ln Z_N=N\ln Z_1-\ln N!\approx N\ln \frac{V}{{\lambda }^3}-(N\ln N-N)=N\ln \frac{V}{N{\lambda }^3}+N,$ using Stirling $\ln N!\approx N\ln N-N$. The energy term adds $\frac{3}{2}N$ (equipartition, $⟨E⟩=\frac{3}{2}N{k}_{B}T$). Collecting constants gives the Sackur–Tetrode entropy: $S=N{k}_B\left[\ln \frac{V}{N}+\frac{3}{2}\ln \frac{2\pi m{k}_{B}T}{h^2}+\frac{5}{2}\right].$ The crucial feature this derivation earns: the $-\ln N!$ turned $\ln V$ into $\ln (V/N)$ — entropy depends on density $V/N$, not raw volume. Step C — mix identical gas.

• Before: two boxes, each $(N,V)$: $S_i=2\cdot N{k}_B\left[\ln \frac{V}{N}+\cdots \right]$.
• After: one box, $(2N,2V)$: $S_f=2N{k}_B\left[\ln \frac{2V}{2N}+\cdots \right]=2N{k}_B\left[\ln \frac{V}{N}+\cdots \right]$.
• $\Delta S=S_f-S_i=\boxed{0}$. Correct — nothing physical changed when you mix a gas with itself. Without $N!$ the entropy would keep $\ln V$ (not $\ln (V/N)$), giving the fake mixing term $\Delta S_{\text{wrong}}=2N{k}_B\ln 2>0$ — the Gibbs paradox. The $N!$ is indistinguishability leaking into classical bookkeeping.

From the parent derivation, ${\psi }_{-}=\frac{1}{\sqrt{2}}[{\psi }_a(1){\psi }_b(2)-{\psi }_a(2){\psi }_b(1)].$ Set $b=a$: ${\psi }_{-}=\frac{1}{\sqrt{2}}[{\psi }_a(1){\psi }_a(2)-{\psi }_a(2){\psi }_a(1)]=\frac{1}{\sqrt{2}}\cdot 0=0.$ A wavefunction that is identically zero has probability $|\psi {|}^2=0$ everywhere — the configuration does not exist. That is the Pauli Exclusion Principle, and note it is not a force: no potential term appears, only the antisymmetry of counting. This is why the fermion count in Q2.1 used $\left(\genfrac{}{}{0ex}{}{N}{k}\right)$ (no repeats).

(a) Why $\mu =0$ — via the free energy. In the Grand Canonical Ensemble a level's occupation is weighted by $e^{-\beta (\epsilon -\mu )n}$; here $\mu$ is the Lagrange multiplier that enforces a chosen average particle number $⟨N⟩$. But for photons the number is not a conserved constraint — the cavity walls create and destroy photons freely, so there is no fixed $⟨N⟩$ to enforce. Equilibrium is instead set by letting the system pick its own $N$ to minimise the Helmholtz free energy $F$ (defined above): the minimum condition is ${\left(\partial F/\partial N\right)}_{T,V}=0$, and since $\mu =\partial F/\partial N$, this reads $\mu =0$ identically for photons. (b) Occupation. Put $\mu =0$ into Bose–Einstein: $⟨n⟩=\frac{1}{e^{\beta \epsilon }-1}=\frac{1}{e^{\epsilon /k_{B}T}-1}.$ (c) With $\epsilon =\hslash \omega$ and multiplying by the photon energy and mode density, this is exactly the Planck blackbody spectrum. No $\mu$ appears, and because $\epsilon >0=\mu$ always, there is no condensation of photons.

Set-up. Let $u=e^x$. Since $x>0$ we have $u>1$. Writing the three occupations (using $⟨n{⟩}_{MB}=e^{-x}=1/u$, the $a=0$ master formula from the top of the page): $⟨n{⟩}_{BE}=\frac{1}{u-1},⟨n{⟩}_{MB}=\frac{1}{u},⟨n{⟩}_{FD}=\frac{1}{u+1}.$ Why reciprocals reverse the order. Compare the three denominators: $u-1<u<u+1$ (each is the previous plus one). Now the function $f(y)=1/y$ is strictly decreasing for $y>0$: its derivative is $f^{\prime }(y)=-1/y^2<0$, so a larger input gives a smaller output. All three denominators are positive (since $u>1$), so applying $f$ flips the chain of inequalities: $u-1<u<u+1\stackrel{1/y\text{ decreasing}}{\Rightarrow }\frac{1}{u-1}>\frac{1}{u}>\frac{1}{u+1},$ which is exactly $⟨n{⟩}_{BE}>⟨n{⟩}_{MB}>⟨n{⟩}_{FD}$. Interpretation: bosons always over-occupy, fermions always under-occupy, relative to the classical middle — the whole content of the $\pm 1$. Evaluate at $x=\ln 3$. Then $u=e^{\ln 3}=3$:

• Bose: $\frac{1}{3-1}=\frac{1}{2}=\boxed{0.5}$.
• Classical: $\frac{1}{3}\approx \boxed{0.3333}$.
• Fermi: $\frac{1}{3+1}=\frac{1}{4}=\boxed{0.25}$. Ordering $\frac{1}{2}>\frac{1}{3}>\frac{1}{4}$ confirmed, consistent with the general proof.

Two distinct ideas, kept separate:

• Counting microstates (how many labelled/unlabelled arrangements): distinguishable $N^k$, bosons $\left(\genfrac{}{}{0ex}{}{N+k-1}{k}\right)$, fermions $\left(\genfrac{}{}{0ex}{}{N}{k}\right)$.
• Mean occupation of a level (how full one level sits on average): the single formula $⟨n⟩=1/(e^{\beta (\epsilon -\mu )}+a)$ with $a=+1,-1,0$, bounded by $\mu <{\epsilon }_{\min }$ for bosons and $\mu =0$ for photons. The first answers "how many ways?"; the second answers "how full?" — different questions, same underlying indistinguishability.