Quantum statistics — distinguishable vs indistinguishable particles

WHAT are we even asking?

WHY does indistinguishability arise? Because the probability density $|\psi|^2$ must be unchanged when we swap two identical particles. That forces the wavefunction to be either symmetric ($+$, bosons) or antisymmetric ($-$, fermions) under exchange.

HOW the wavefunction encodes it (derive from scratch)

Take two particles in single-particle states $a$ and $b$. Let $\hat{P}$ swap particle 1 and 2.

Step 1 — physical requirement. Swapping identical particles can't change any measurable thing: $|\psi(1,2)|^2 = |\psi(2,1)|^2.$ Why this step? Probability density is the only observable about position; identical particles must give identical predictions.

Step 2 — solve it. That means $\psi(2,1) = e^{i\phi}\psi(1,2)$. Swap twice → back to start, so $e^{2i\phi}=1 \Rightarrow e^{i\phi}=\pm1$. Why this step? Two swaps is the identity operation, $\hat P^2 = \mathbb{1}$.

Step 3 — build the states. $\psi_\pm = \frac{1}{\sqrt2}\big[\psi_a(1)\psi_b(2) \pm \psi_a(2)\psi_b(1)\big].$ $+$ → symmetric (bosons), $-$ → antisymmetric (fermions). Why this step? These are the only combinations of the product states obeying $\psi(2,1)=\pm\psi(1,2)$.

Step 4 — Pauli exclusion falls out for free. Set $a=b$ in the antisymmetric case: $\psi_- = \frac{1}{\sqrt2}[\psi_a(1)\psi_a(2) - \psi_a(2)\psi_a(1)] = 0.$ Why this step? A state that is zero everywhere has zero probability — two fermions cannot occupy the same single-particle state.

The counting that defines the statistics

Imagine 2 particles going into 3 single-particle states (boxes). Count the allowed microstates.

The occupation-number distributions (derive the key results)

Use the grand canonical ensemble: each single-particle level of energy $\varepsilon$ exchanges particles with a reservoir at temperature $T$, chemical potential $\mu$. Let $\beta = 1/k_BT$. The grand partition function for one level is a sum over its occupation $n$: $\mathcal Z = \sum_n e^{-\beta(\varepsilon-\mu)n}.$ Why this step? Levels are independent, so we treat one level and sum/multiply later. The Boltzmann factor weights each occupation by its energy and particle number.

The mean occupation is $\langle n\rangle = \frac{\sum_n n\,e^{-\beta(\varepsilon-\mu)n}}{\sum_n e^{-\beta(\varepsilon-\mu)n}} = -\frac{1}{\beta}\frac{\partial}{\partial\mu}\ln\mathcal Z \ \text{(equivalently)}.$

Fermions: $n\in\{0,1\}$. $\mathcal Z = 1 + e^{-\beta(\varepsilon-\mu)},\qquad \langle n\rangle = \frac{e^{-\beta(\varepsilon-\mu)}}{1+e^{-\beta(\varepsilon-\mu)}}.$

Bosons: $n\in\{0,1,2,\dots\}$, a geometric series (needs $\varepsilon>\mu$): $\mathcal Z = \sum_{n=0}^\infty e^{-\beta(\varepsilon-\mu)n} = \frac{1}{1-e^{-\beta(\varepsilon-\mu)}}.$ $\langle n\rangle = \frac{e^{-\beta(\varepsilon-\mu)}}{1-e^{-\beta(\varepsilon-\mu)}}.$

One-line unifier: $\displaystyle \langle n\rangle = \frac{1}{e^{\beta(\varepsilon-\mu)}+a}$, with $a=+1$ (Fermi), $a=-1$ (Bose), $a=0$ (classical).

The Gibbs $1/N!$ patch (why classical counting was wrong)

Common mistakes (Steel-man + fix)

Flashcards

Connections

• Bose–Einstein Condensation
• Fermi Gas & Fermi Energy
• Grand Canonical Ensemble
• Gibbs Paradox & Entropy of Mixing
• Pauli Exclusion Principle
• Blackbody Radiation (Photon Gas)
• Spin–Statistics Theorem
• Maxwell–Boltzmann Distribution

Concept Map

Hinglish (regional understanding)