5. Fock Space and Second Quantisation

When working with basis vectors using the occuppation number representation, we might consider dropping the overall constraint \(\sum_{j=1}^L n_j = N\). This amounts to working in a larger Hilbert space, which is known as the Fock space, and consists of the direct sum of all physical (symmetrised or antisymmetrised) Hilbert spaces \(\mathbb{H}^{(N)}\) for different particle numbers \(N\), going all the way from \(N=0\):

\[\mathbb{H} = \bigoplus_{N=0}^{+\infty} \mathbb{H}^{(N)}\]

In the case of fermions and with a finite-dimensional single-particle Hilbert space \(\mathbb{H}^{(1)} \cong \mathbb{C}^L\), the upper limit in the direct sum is \(N=L\), i.e. there are no states with \(N > L\) and so the associated Hilbert spaces are zero-dimensional. This direct sum furthermore also contains the case \(N=0\), which we have not discussed before. In the previous subsection we started the construction of \(\mathbb{H}^{(N)}\) from a given single particle Hilbert space \(\mathbb{H}^{(1)}\). When there are no particles in the system, there is only a single state in which it can be, corresponding to having all occupation numbers \(n_j = 0\) for all \(j\). Hence, for \(N=0\) particles, the Hilbert space \(\mathbb{H}^{(0)}\) is spanned by a single state, which we typically denote as \(\ket{\Omega}=\ket{0,0,\dots,0}\) and refer to as the vacuum state. Note that this vacuum state is normalised, and is thus very different from an actual zero vector of the vector space, which has norm zero.

The Fock space becomes a Hilbert space simply by incorporating the inner product from each of its summands. States within the different summands of this direct sum are defined to be orthogonal, i.e. \(\braket{\varphi^{(M)} \vert \psi^{(N)}}=0\) for all \(M\)-particle states \(\ket{\varphi^{(M)}}\) and \(N\)-particle states \(\ket{\psi^{(N)}}\) with \(M \neq N\).

The main benefits of using the formalism of second quantisation are not about losing the overall particle number constraint, but for working with operators, in particular to describe (interacting) Hamiltonians. In first quantisation, we need to specify a Hamiltonian given a particular number of particles, i.e.\ the number of particles is an external parameter of the system. Using the Fock space, we can now define operators in such a way that their action is immediately defined for states with an arbitrary number of particles, including even states which are superpositions over different particle numbers.

Hereto, we first introduce operators that enable us to connect the different particle number sectors, by creating (adding) or annihilating (removing) particles in the system. In particular, we denote with \(\hat{a}_j^+\) the operator that adds a new particle in the mode \(j\) in the system and with \(\hat{a}_j^-\) the operator that removes a particle that is in mode \(j\) from the system. As it turns out that both operators are related via the adjoint, i.e. \(\braket{\Phi| \hat{a}_j^+ \Psi} = \braket{\hat{a}_j^- \Phi | \Psi}\), we use the simpler notation \(\hat{a}_j\) for the annihilation operator and \(\hat{a}_j^\dagger\) for the creation operator. To construct these operators in a mathematically precise and constructive way is actually somewhat tedious (but see Wikipedia). We just summarize their main properties. In particular, we want to have the property that the (anti)symmetrized states satisfy

\[\ket{j_1,j_2, \ldots, j_N} = \hat{a}_{j_1}^\dagger \hat{a}_{j_2}^\dagger \cdots \hat{a}_{j_N}^\dagger\ket{\Omega}.\]

It is immediately clear that, because of the (anti)symmetry, this requires that

\[[\hat{a}_i^\dagger, \hat{a}_j^\dagger] = 0\ (\text{bosons})\quad\text{or}\quad \{\hat{a}_i^\dagger,\hat{a}_j^\dagger\} = 0\ (\text{fermions}).\]

From the normalisation of these states, it also follows that

\[[\hat{a}_i, \hat{a}_j^\dagger] = \delta_{i,j}\ (\text{bosons})\quad\text{or}\quad \{\hat{a}_i,\hat{a}_j^\dagger\} = \delta_{i,j}\ (\text{fermions}).\]

With respect to the normalized basis vectors, using the occupation representation, we have

\[\ket{n_1, n_2, \ldots, n_L} = \frac{1}{\sqrt{n_1! n_2! \cdots n_L!}} (\hat{a}_1^\dagger)^{n_1} (\hat{a}_2^\dagger)^{n_2} \cdots (\hat{a}_L^\dagger)^{n_L} \ket{\Omega}\]

which can be summarized using

\[\begin{split}\hat{a}_j \ket{n_1, n_2, \ldots, n_j, \ldots, n_L} &= (\pm 1)^{n_1 + n_2 + \ldots + n_{j-1}} \sqrt{n_j} \ket{n_1, n_2, \ldots, n_j - 1, \ldots, n_L},\\ \hat{a}_j^\dagger \ket{n_1, n_2, \ldots, n_j, \ldots, n_L} &= (\pm 1)^{n_1 + n_2 + \ldots + n_{j-1}} \sqrt{n_j+1} \ket{n_1, n_2, \ldots, n_j + 1, \ldots, n_L}.\end{split}\]

It then follows easily that the operator \(\hat{n}_j = \hat{a}_j^\dagger \hat{a}_j\) satisfies

\[\hat{n}_j \ket{n_1, n_2, \ldots, n_j, \ldots, n_L} = n_j \ket{n_1, n_2, \ldots, n_j, \ldots, n_L}\]

and thus measures the number of particles in mode \(j\). The operators \(\hat{n}_j\) are referred to as number operators. The total number of particles can then be measured using

\[\hat{N} = \sum_{j=1}^{L} \hat{n}_j\]

but the Fock space does of course contain states which are superpositions over different particle numbers (and which are thus not eigenstates of \(\hat{N}\)).

Furthermore, by studying how single particle states \(\ket{j} \equiv \hat{a}_j^\dagger \ket{\Omega}\) change under a change of single particle basis, or thus, a transformation to a new set of modes, we can deduce how the associated creation and annihilation operators transform. Suppose we have a different single-particle basis, which for clarity we label with greek letters \(\kappa = 1,\ldots,L\). We then find

\[\ket{\kappa} = \hat{a}_\kappa^\dagger \ket{\Omega} = \sum_{j} \ket{j} \braket{j\vert\kappa} = \sum_{j} \braket{j\vert \kappa} \hat{a}_j^\dagger \ket{\Omega}\]

from which we obtain

\[\hat{a}_\kappa^\dagger = \sum_{j} \braket{j\vert \kappa} \hat{a}_j^\dagger, \qquad\hat{a}_\kappa = \sum_{j} \braket{\kappa\vert j} \hat{a}_j.\]

Note that the transformation matrix \(\braket{j \vert \kappa}\) between two orthonormal bases correspond to a unitary matrix. These transformation rules will be employed often, for example, to switch between a position and momentum space representation.

Note

The bosonic creation and annihilation operators are of course reminiscent from the operators introduced for diagonalising the single particle harmonic oscillator model. Indeed, out of the bosonic creation and annihilation operator associated to every mode \(j\) we can build two Hermitian operators

\[\hat{X}_j = \frac{1}{\sqrt{2}}(\hat{a}_j + \hat{a}_j^\dagger),\quad \hat{P}_j = \frac{-\mathrm{i}}{\sqrt{2}}(\hat{a}_j - \hat{a}_j^\dagger)\]

which than satisfy the well known commutation relations \(\left[\hat{X}_j, \hat{P}_k\right] = \mathrm{i} \delta_{j,k}\). In second quantisation, the Fock space of bosons built from a single particle system with \(L\) modes can equivalently be thought of as a regular tensor product space of \(L\) distinguishable quantum particles moving on the real line, or technically, as \(\left(L^2(\mathbb{R})\right)^{\otimes L}\).

Note

For fermions, we can also construct Hermitian operators out of the creation and annihilation operators, which we denote as

\[\hat{\eta}^{(1)}_j = \frac{1}{\sqrt{2}}(\hat{a}_j + \hat{a}_j^\dagger),\quad \hat{\eta}^{(2)}_j = \frac{-\mathrm{i}}{\sqrt{2}}(\hat{a}_j - \hat{a}_j^\dagger).\]

In this case, we find that they satisfy the anticommutation relation

\[\{ \hat{\eta}^{(\alpha)}_j, \hat{\eta}^{(\beta)}_k \} = \delta_{\alpha,\beta} \delta_{j,k}\]

so that the \(\hat{\eta}^{(1)}\) type operators and \(\hat{\eta}^{(2)}\) type operators behave similarly. In that case, one often uses a different notation by setting

\[\hat{\chi}_{2j-1} = \hat{\eta}^{(1)}_j = \frac{1}{\sqrt{2}}(\hat{a}_j + \hat{a}_j^\dagger),\quad \hat{\chi}_{2j} = \hat{\eta}^{(2)}_j = \frac{-\mathrm{i}}{\sqrt{2}}(\hat{a}_j - \hat{a}_j^\dagger)\]

and thus \(\{\hat{\chi}_k, \hat{\chi}_l\} = \delta_{k,l}\) for all \(k, l = 1,\ldots, 2L\). These Hermitian fermionic operators are referred to as Majorana operators.

Note furthermore that the Fock space of fermions built from a single particle system with \(L\) modes looks remarkably like a system of \(L\) qubits, i.e. the tensor product \((\mathbb{C}^2)^{\otimes L}\). While this is true for how the occupation number basis vectors are labelled, one important fact is that the operators \(\hat{a}_j\) and \(\hat{a}_j^\dagger\) should not be thought of as local operators that act nontrivially on the single site \(j\), and as the identity elsewhere, since they do not mutually commute, but rather anticommute. It is possible to map these fermionic creation and annihilation operators to ‘nonlocal’ qubit operators using the Jordan-Wigner transformation.

With these creation and annihilation operators, we can now represent general operators in a way that does not depend on the precise number of particles in the system. The simplest case are ‘single-particle’ operators, i.e. operators that were defined with respect to the single-particle Hilbert space \(\mathbb{H}^{(1)}\). The easiest case are operators which are are diagonal with respect to the chosen single-particle basis. In that case, every particle one of the eigenmodes of the single-particle operator will give a contribution that equals the associated eigenvalue. Hence, the many-body representation of such an operator is given by

\[\hat{O}^{(1)} = \sum_{j} \lambda_j \ket{j}\bra{j} \quad \rightarrow \quad \hat{O} = \sum_{j} \lambda_j \hat{a}_j^\dagger \hat{a}_j.\]

However, we can easily transform away from the basis of eigenmodes to a general set of modes, and then find

\[\hat{O}^{(1)} = \sum_{j,k} O_{j,k} \ket{j}\bra{k} \quad \rightarrow \quad \hat{O} = \sum_{j,k} O_{j,k} \hat{a}_j^\dagger \hat{a}_k.\]

Vice versa, if you are given an operator that only contains terms where every term contains exactly one creation and one annilation operator, then it is especially easy to diagonalise this operator, since one only needs to diagonalise the corresponding single-particle version of the operator. When the Hamiltonian of the many-body system is of this form, the system is said to be free or noninteracting.

Note

There is a larger class of operators that can easily be diagonalised, namely operators where every term is quadratic in the creation and annihilation operators. This means that every term contains either a creation and an annilation operator, or two creation operators, or two annihilation operators. Such Hamiltonians are said to be quadratic or Gaussian, and can be diagonalised using a Bogoliubov transformation.

Similarly, there exist two-particle operators, in particular, typical interaction terms in the Hamiltonian such as the Coulomb interaction between electrons. Such operators take the form

\[\hat{O}^{(2)} = \sum_{j,k,l,m} O_{j,k; l,m} \ket{j,k} \bra{l,m}\]

and can be translated to act on the full Fock space as

\[\hat{O} = \sum_{j\leq k; l\leq m} O_{(j,k); (l,m)} \hat{a}_{j}^\dagger \hat{a}_k^\dagger \hat{a}_m \hat{a}_l = \frac{1}{4} \sum_{j, k; l, m} O_{(j,k); (l,m)} \hat{a}_{j}^\dagger \hat{a}_k^\dagger \hat{a}_m \hat{a}_l.\]

As soon as such type of operators are present in the Hamiltonian (which thus contain more than two creation of annihilation operators), it becomes impossible to diagonalise the Hamiltonian based on a simple calculation in the single-particle Hilbert space, and the exponentially large many-body Hilbert space need to be considered.