Everything so far fixed the number of particles: one electron in hydrogen, two in helium, $N$ in a Slater determinant. Three facts push beyond that. First, particles are made and destroyed: an excited atom emits a photon that did not exist before, a photon of enough energy near a nucleus becomes an electron and a positron, a neutron decays into a proton, an electron and an antineutrino. A wavefunction with a fixed list of arguments cannot say that. Second, for identical particles the labels are a burden. A symmetrised function of $N$ positions carries the same information $N$! times over, since swapping labels changes nothing; all that matters is how many particles occupy each single-particle state — the occupation numbers. Third, light came in quanta from the start, with Planck and Einstein, yet no Schrödinger equation for one photon ever appeared: photons come from quantising the electromagnetic field. The way forward is to describe states by occupation numbers and to let operators change them. The name, second quantisation, is historical; what it does is bookkeeping.

A cloakroom that takes identical umbrellas needs no tickets: all it records is how many umbrellas hang on each hook. Occupation numbers are that record for identical particles.

The lowest state of four identical particles in the same five single-particle levels. Four bosons, left, all occupy the lowest level: occupations $\ket{4,0,0,0,0}$. Four fermions, right, can hold at most one to a level and fill the four lowest: $\ket{1,1,1,1,0}$.

For identical particles with single-particle states $\ket{\varphi_{1}},\ket{\varphi_{2}},\dots$, a symmetric or antisymmetric state is fixed by how many particles occupy each:

• The occupation numbers say everything a symmetric or antisymmetric state says: which label each particle carries is not a physical question.
• $N$ bosons in $d$ single-particle states have $\binom{N+d-1}{N}$ states; fermions have .

A symmetric state is fixed by its occupation numbers, so count the lists $(n_{1},\dots,n_{d})$ of whole numbers adding up to $N$. Write stars in a row and cut them into groups with bars: every arrangement of the stars and bars is one list, and every list is one arrangement. For fermions each is 0 or 1, so a state is a choice of occupied states out of .

A symmetric state is fixed by its occupation numbers, and each list gives one.

$N$ stars cut into $d$ groups by $d-1$ bars; here $N=6$ and $d=4$.

Choose which of the $N+d-1$ places hold the bars.

Choose the $N$ occupied states out of $d$.

The counts of symmetric and antisymmetric two-particle states found in the chapter on identical particles.

A state of identical bosons is fixed by how many occupy each single-particle state. $N$ bosons in $d$ states can be arranged in $C(N+d-1,N)$ ways.

An atom emits a photon that did not exist before; a photon of enough energy near a nucleus turns into an electron and a positron. The number of particles is not fixed.

For identical particles, which particle is in which state is not a physical question. Only how many particles occupy each state is.

Each occupation number of fermions is 0 or 1, so $N$ fermions in $d$ states have $C(d,N)$ states: a choice of which states are filled.

A list of how many particles occupy each single-particle state, in a fixed order of the states, describes a state of identical particles completely.

Great work! You now know how to:

• say why a fixed number of particles is not enough
• describe identical particles by occupation numbers
• count the states of bosons and of fermions