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Fock space · Quantum Mechanics | Math Infinitum

Intuition

Collect the states of every particle number into one space: the vacuum with no particles, the one-particle states, the symmetric or antisymmetric two-particle states, and on up. Their direct sum is Fock space. Its basis is the occupation-number states $\ket{n_{1},n_{2},\dots}$ , and they are orthonormal: two lists give orthogonal states unless they agree entry by entry. A state of Fock space may be a superposition of different particle numbers — the coherent state of a laser mode, met with the oscillator, is one, with an uncertain number of photons. The total number operator $\hat{N}=\sum\hat{n}_{i}$ measures how many particles there are. A Hamiltonian that commutes with it keeps every sector of fixed $N$ apart; processes that make or destroy particles move between sectors. For fermions the space is finite whenever the single-particle states are: each of d states is empty or filled, $2^{d}$ states in all.

A library shelves pamphlets, novels and encyclopaedias in one catalogue while keeping each book what it is. Fock space shelves the states of every particle number side by side, each sector kept apart from the others.

The $2^{4}=16$ states of Fock space for fermions with four single-particle states, one dot each, in columns by the number of particles $N$ , written under each column: $1,4,6,4,1$ states, the binomial coefficients .

Fock space

The Fock space of identical particles is the direct sum of the spaces of $N$ particles, symmetric for bosons and antisymmetric for fermions, beginning with the vacuum $\ket{0}$ :

\begin{aligned}\mathcal{F}&=\mathcal{H}^{(0)}\oplus\mathcal{H}^{(1)}\oplus\mathcal{H}^{(2)}\oplus\cdots\\ \braket{n'_{1},n'_{2},\dots|n_{1},n_{2},\dots}&=\delta_{n'_{1}n_{1}}\,\delta_{n'_{2}n_{2}}\cdots\end{aligned}

Properties

The vacuum $\ket{0}$ holds no particles; it is a state of norm 1, not the zero vector.
The occupation-number states form an orthonormal basis of $\mathcal{F}$ .
$\hat{N}=\sum_{i}\hat{n}_{i}$ has the eigenvalue on , so states with different numbers of particles are orthogonal.

Each fermion state doubles the Fock space

\dim\mathcal{F}_{\text{fermions}}=\sum_{N=0}^{d}\binom{d}{N}=2^{d}

Add the dimensions of the sectors: $N$ fermions in $d$ states have $\binom{d}{N}$ states, and the binomial theorem sums them. The same count comes directly from the occupation numbers: each of the $d$ states is empty or filled, two choices each.

Proof steps

\dim\mathcal{F}=\sum_{N=0}^{d}\dim\mathcal{H}^{(N)}_{\text{anti}}

The sectors are orthogonal, so their dimensions add.

\dim\mathcal{H}^{(N)}_{\text{anti}}=\binom{d}{N}

The count of the last lesson.

\sum_{N=0}^{d}\binom{d}{N}=(1+1)^{d}=2^{d}

The binomial theorem with both numbers equal to 1.

\begin{aligned}&\ket{n_{1},\dots,n_{d}},\ \ n_{i}\in\{0,1\}:\\ &2\times2\times\cdots\times2=2^{d}\end{aligned}

The same count made directly: two choices for each state.

Applications

Practice

Empty or Filled

Each fermionic single-particle state is empty or filled, so $d$ of them give $2^{d}$ occupation-number states in all, every particle number included.

\dim\mathcal{F}=2^{d}

Try it

How many states does the Fock space of fermions with 5 single-particle states have?