Bosons and fermions

Intuition

Of the two eigenvalues, nature uses each for a different kind of particle, and never mixes them: the states of identical bosons are all symmetric, the states of identical fermions all antisymmetric, whatever the particles do. Which kind a particle is follows from its spin. Particles of whole-number spin — photons, the helium-4 atom, the pion — are bosons; particles of half-odd spin — electrons, protons, neutrons, quarks, the helium-3 atom — are fermions. This spin–statistics connection is a theorem of relativistic quantum field theory and is taken here as given. A composite of an even number of fermions is a boson, of an odd number a fermion. The two kinds count states differently: two bosons in $d$ single-particle states have $d(d+1)/2$ states, two fermions only $d(d-1)/2$ .

Bosons are like identical coins in a purse — two of them can sit in the same pocket; fermions are like guests with numbered seats, one to a seat, and the seating chart does not care which guest is which.

Every state of $N$ identical particles has the same symmetry under the exchange of any two, $\hat{P}_{ij}$ , fixed by the kind of particle:

\text{bosons:}\ \ \hat{P}_{ij}\Psi=+\Psi,\qquad \text{fermions:}\ \ \hat{P}_{ij}\Psi=-\Psi

Properties

Spin and statistics: whole-number spin gives bosons, half-odd spin fermions — a theorem of relativistic quantum field theory, stated here.
A composite particle made of an even number of fermions is a boson, of an odd number a fermion: helium-4 is a boson, helium-3 a fermion.
Two particles in $d$ single-particle states have $d^{2}$ product states: $d(d+1)/2$ symmetric and antisymmetric.

Counting two-particle states

\dim\mathcal{H}_{\text{sym}}=\frac{d(d+1)}{2},\qquad \dim\mathcal{H}_{\text{anti}}=\frac{d(d-1)}{2}

Group the $d^{2}$ product states in pairs that the exchange swaps, and singles that it leaves alone. Each pair gives one symmetric and one antisymmetric combination; each single is symmetric.

Proof steps

\ket{i}\ket{j},\qquad i,j=1,\dots,d:\ \ d^{2}\ \text{states}

The product basis of two particles.

i<j:\ \ \tfrac{1}{\sqrt2}\bigl(\ket{i}\ket{j}\pm\ket{j}\ket{i}\bigr)

$d(d-1)/2$ pairs, each giving one symmetric and one antisymmetric state.

i=j:\ \ \ket{i}\ket{i}

$d$ more states, all symmetric; their antisymmetric combination would be zero.

\begin{aligned}&\frac{d(d-1)}{2}+d=\frac{d(d+1)}{2}\\ &\frac{d(d+1)}{2}+\frac{d(d-1)}{2}=d^{2}\end{aligned}

The symmetric count, and the check that nothing was lost.

Applications

Practice

Two Kinds

States of identical bosons are symmetric under every exchange, states of identical fermions antisymmetric. Whole-number spin means boson, half-odd spin fermion.

Try it

Electrons have spin $\tfrac12$ . Which kind of particle are they?