# Second quantization

## Contents

# Second quantization#

## Fock space#

The way we described quantum mechanical states so-far contains a redundancy: it attributes individual particles (electrons) to a given state (spin-orbital), despite the fact that electrons are undistinguishable. This undistinguishability is included *a posteriori* by antisymmetrizing the state. This approach, called first quantization, while perfectly valid, can be cumbersome.

A more natural approach is to completely avoid identifying particles and instead simply ask “how many particles are present in a given quantum state?”. To this end, we introduce the concept of occupation number vector, in which we list all possible states of the system (spin-orbitals) and their occupation. Note that for fermions, like electrons, the occupation can only be 0 or 1. We then define the Fock space as the abstract vector space of occupation vectors. Clearly there is a unique mapping between occupation number vectors and Slater Determinants as long as a spin-orbital basis is selected. Note however that the Fock space includes states with different total number of particles.

Let us for instance consider a Fock space consisting of three spin-orbitals: \(\psi_1(\mathbf{x})\), \(\psi_2(\mathbf{x})\), and \(\psi_3(\mathbf{x})\). The Slater determinant for two electrons occupying \(\psi_1(\mathbf{x})\) and \(\psi_2(\mathbf{x})\) spin-orbitals \(1/\sqrt{2} |\psi_1(\mathbf{x}) \psi_2(\mathbf{x}) |\) can be represented in Fock space as the occupation number vector \(|\mathbf{k}\rangle = | 1 1 0 \rangle\).

The Fock space is a linear vector space, and consequently has the following properties:

two occupation number vectors \(| \mathbf{k} \rangle\) and \(| \mathbf{m} \rangle\) are orthogonal i.e., \(\langle \mathbf{k} | \mathbf{m} \rangle = \delta_{\mathbf{k}, \mathbf{m}}\)

occupation number vectors constitutes complete basis i.e. arbitrary vector \(|\mathbf{g}\rangle\) can be expressed as linear combination of occupation number vectors

and outer product of occupation number vectors resolve to identity operator $\( 1 = \sum_\mathbf{k} | \mathbf{k} \rangle \langle \mathbf{k} | \)$

The Fock space consisting of \(N\) spin orbitals, \(F(N)\), can be partitioned into sub-spaces as

where \(F(M, N)\) is the set of occupation number vectors with \(M\) occupied spin-orbitals. The Fock subspace \(F(M,N)\) can be viewed as an \(M\)-electron wave function expressed in occupation number vectors, as it contains all Slater determinant generated by distributing \(M\) electrons in \(N\) spin orbitals. Among the Fock subspaces, the \(F(0, N)\) subspace corresponds to a completely unoccupied Slater determinant and is called a vacuum state \(| vac \rangle\). The vacuum state has the following properties

## Creation and annihilation operators#

Second quantization also introduces a very convenient way to manipulate states through creation and annihilation operators. As their name imply, those operators can be seen as creating and destroying electrons in specific spin-orbital. In terms of occupation vector manipulation, these actions are described

creation operator \(\hat{a}_p^\dagger\):

annihilation operator \(\hat{a}_p\):

The \(\delta_{k_p, 0}\) ensures that the creation operator cannot add an electron to a state that is already populated while \(\delta_{k_p, 1}\) prevents the destruction of a non-existing electron. The phase factor \(\prod_{i = 1}^{p} (-1)^{k_p}\) ensures the antisymmetry of the wavefunction. However, the main strength of this formulation is that the phase factor is a direct consequence of the anti-commutator relationship of the creation and annihilation operators:

which are primary tools to manipulate occupation number vectors.

Having defined those operators, it is easy to see that any arbitrary state can be simply formed by applying a string of creation operators on the vacuum state.

Similarly, the standard operators of quantum mechanics can be represented in second quantization using the creation and annihilation operators. Any operator \(\hat{O}\) depending on the coordinates (both spatial and spin) \(\mathbf{x}\) of a single electron can be expressed as a product of the integral of the operator in the chosen basis \(O_{pq}= \int \psi_p^* (\mathbf{x}) \hat{O} (\mathbf{x}) \psi_q (\mathbf{x}) d\mathbf{x}\) and the excitation operator \(\hat{a}_p^\dagger \hat{a}_q\):

A similar formula can be used for two-electron operators. For example, the Hamiltonian of a molecular system can be written as:

where \(h_{pq}\) is the one-electron part of electronic Hamiltonian

and \(g_{pqrs}\) is the two-electron part of electronic Hamiltonian

At the end, second quantization is a completely equivalent way to represent quantum mechanics, but by using a unified description of operator and states and reducing all the complications of the antisymmetry to simple commutation rules between the creation and annihilation operators, it is often a significantly more straightforward and compact way to derive various wave function models than the conventional approach based on the direct application of Slater-Condon rules to individual Slater determinants. For example, derivation of working equations for coupled-cluster singles and doubles wave function in second quantization becomes a single lecture affair, compared to the conventional way, which is several days undertaking. Thus, the second quantization is a valuable mathematical tool for working with correlated wave functions.

## Illustration#

To illustrate the power and simplicity of this formulation, we can derive the energy of a Slater Determinant \(| 0 \rangle\):

For this we only need to evaluate the density matrices \(D_{pq} = \langle 0| \hat{a}_p^\dagger \hat{a}_q| 0\rangle\) and \(d_{pqrs} = \langle 0| \hat{a}_p^\dagger \hat{a}_r^\dagger \hat{a}_s \hat{a}_q | 0\rangle\).

For the one-particle density matrix \(D_{pq}\) we immediately notice that the orbital \(q\) has to be occupied for the element to be non-zero when the annihilation operator is acting on the right on \(| 0\rangle\) (the same is true for \(p\) when \(\hat{a}_p^\dagger\) is acting on the left on \(\langle 0|\)). But then, we can apply the commutation rule to get:

Now, \(\hat{a}_q\) acts as a creation operator on the left on \(\langle 0|\) and since we said \(q\) was occupied, this last term gives zero. The only term left is thus \(\delta_{p,q}\). The one-electron term thus gives:

By using the indices \(i, j, k, l\) to refer to occupied orbitals, we can simply write this term as \(\sum_{i} h_{ii}\)

We can now turn our attention the two-electron terms. Here both \(q\) and \(s\) have to be occupied. We will focus on \(s\) which we will call \(i\) and we will use the same trick of moving \(\hat{a}_i\) to the left to generate a zero:

The last term will now disappear and we are left with only terms depending on 2 creation or annihilation operators, which we have already solved earlier:

Our final energy is thus: