Now give Fock space operators that change the occupations. For each single-particle state $i$, the creation operator $\hat{a}^{\dagger}_{i}$ adds a boson to it and the annihilation operator $\hat{a}_{i}$ removes one: $\hat{a}^{\dagger}_{i}$ turns $n_{i}$ into $n_{i}+1$ with the amplitude $\sqrt{n_{i}+1}$, and $\hat{a}_{i}$ turns it into $n_{i}-1$ with $\sqrt{n_{i}}$. These are the ladder operators of the oscillator, one for each state, with the same square roots for the same reason, and the same commutator: $[\hat{a}_{i},\hat{a}^{\dagger}_{j}]=\delta_{ij}$, while operators of different states commute. Every state of Fock space is built from the vacuum by creation operators. And the symmetry of bosons comes out by itself: $\hat{a}^{\dagger}_{a}\hat{a}^{\dagger}_{b}\ket{0}=\hat{a}^{\dagger}_{b}\hat{a}^{\dagger}_{a}\ket{0}$ because the operators commute — the order in which the particles are added does not matter, which is what a symmetric state is. The square roots have a meaning: adding a boson to a state that already holds $n$ has amplitude $\sqrt{n+1}$, so bosons prefer company.

A queue attracts a queue: the longer the line at a stall, the likelier a passer-by is to join it. For bosons, the probability of joining a state that already holds $n$ is $n+1$ times that of entering an empty one.

The occupations $n=0,1,2,3,4$ of one bosonic state, from the bottom. Each step up is taken by the creation operator, with the amplitude $\sqrt{n+1}$ written beside it: .

For bosons in single-particle states $i=1,2,\dots$, the operators $\hat{a}_{i}$ and $$ act on occupation-number states as

• $\hat{n}_{i}=\hat{a}^{\dagger}_{i}\hat{a}_{i}$ counts the bosons in state , and counts them all.

Move the annihilation operator to the right through the $n$ creation operators. Each passage leaves the commutator 1 behind, so $\hat{a}(\hat{a}^{\dagger})^{n}\ket{0}=n(\hat{a}^{\dagger})^{n-1}\ket{0}$, since kills the vacuum. Applied times inside the squared norm, this gives .

Move $\hat{a}$ past one $\hat{a}^{\dagger}$ at a time; each passage leaves $[\widehat{a},{\widehat{a}}^{†}]=$.

$\hat{a}\ket{0}=0$.

The squared norm, with the last line used on the right-most $\hat{a}$.

Repeat down to $\braket{0|0}=1$; so the normalised state carries $1/\sqrt{n!}$.

The creation operator adds one boson to a state, with the amplitude $\sqrt{n+1}$ when $n$ are already there.

Creation operators of bosons commute, so the order in which bosons are added never matters: the states they build are symmetric.

An annihilation operator acting on the vacuum gives the zero vector — not the vacuum.

$n$ bosons made in one state by $n$ creation operators have squared norm $n$!, so the normalised state carries $1/\sqrt{n!}$.

Great work! You now know how to:

• act with bosonic creation and annihilation operators on occupation-number states
• build any state from the vacuum and normalise it
• explain why bosonic states come out symmetric