# 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

$| \mathbf{g} \rangle = \sum_\mathbf{k} c_\mathbf{k} | \mathbf{k} \rangle$

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

$F(N) = F(0,N) \oplus F(1,N) \oplus ... \oplus F(N,N)$

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

$| vac \rangle = | 0_1, 0_2, ..., 0_N \rangle \ \mathrm{and} \ \langle vac | vac \rangle = 1$

## 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 $$\hat{a}_p^\dagger$$:

$\hat{a}_p^\dagger | \mathbf{k} \rangle = \delta_{k_p, 0} \prod_{i = 1}^{p} (-1)^{k_p} | k_1 k_2...1_p ...k_N \rangle$
• annihilation operator $$\hat{a}_p$$:

$\hat{a}_p | \mathbf{k} \rangle = \delta_{k_p, 1} \prod_{i = 1}^{p} (-1)^{k_p} | k_1 k_2...0_p ...k_N \rangle$

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:

$\begin{eqnarray*} \hat{a}_p^\dagger \hat{a}_q + \hat{a}_q \hat{a}_p^\dagger &=& \delta_{p,q} \ , \\ \hat{a}_p^\dagger \hat{a}_q^\dagger + \hat{a}_q^\dagger \hat{a}_p^\dagger &=& 0 \ , \\ \hat{a}_p \hat{a}_q + \hat{a}_q \hat{a}_p &=& 0 \ , \end{eqnarray*}$

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 $$\hat{a}_p^\dagger \hat{a}_q$$:

$\hat{O} = \sum_{pq} O_{pq} \hat{a}_p^\dagger \hat{a}_q$

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

$\hat H = \sum_{pq} h_{pq} \hat{a}_p^\dagger \hat{a}_q + \frac{1}{2} \sum_{pqrs} g_{pqrs} \hat{a}_p^\dagger \hat{a}_r^\dagger \hat{a}_s \hat{a}_q \ ,$

where $$h_{pq}$$ is the one-electron part of electronic Hamiltonian

$h_{pq} = \int \psi_p(\mathbf{x})^* \bigg \{ - \frac{1}{2} \hat{\nabla}^2 - \sum_K \frac{Z_k}{|\mathbf{r} - \mathbf{R}_K |} \bigg \} \psi_q(\mathbf{x}) d \mathbf{x} \ ,$

and $$g_{pqrs}$$ is the two-electron part of electronic Hamiltonian

$g_{pqrs} = \int \int \psi_p(\mathbf{x})^* \psi_r(\mathbf{x}')^* \frac{1}{|\mathbf{r} - \mathbf{r}'|} \psi_q(\mathbf{x}) \psi_s(\mathbf{x}') d \mathbf{x} d \mathbf{x}' \ .$

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 $$| 0 \rangle$$:

$\langle 0| \hat H | 0\rangle = \sum_{pq} h_{pq} \langle 0| \hat{a}_p^\dagger \hat{a}_q| 0\rangle + \frac{1}{2} \sum_{pqrs} g_{pqrs} \langle 0| \hat{a}_p^\dagger \hat{a}_r^\dagger \hat{a}_s \hat{a}_q | 0\rangle$

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:

$\langle 0| \hat{a}_p^\dagger \hat{a}_q | 0 \rangle = \delta_{p,q} \langle 0| 0 \rangle - \langle 0| \hat{a}_q \hat{a}_p^\dagger | 0 \rangle$

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:

$\sum_{pq} h_{pq} \langle 0| \hat{a}_p^\dagger \hat{a}_q| 0\rangle = \sum_{pq \in \mathrm{occupied}} h_{pq} \delta_{p,q} = \sum_{p \in \mathrm{occupied}} h_{pp}$

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:

\begin{align*} \hat{a}_p^\dagger \hat{a}_r^\dagger \hat{a}_i \hat{a}_q &= \delta_{r,i} \hat{a}_p^\dagger \hat{a}_q - \hat{a}_p^\dagger \hat{a}_i \hat{a}_r^\dagger \hat{a}_q\\ &= \delta_{r,i} \hat{a}_p^\dagger \hat{a}_q - \delta_{p,i} \hat{a}_r^\dagger \hat{a}_q + \hat{a}_i \hat{a}_p^\dagger \hat{a}_r^\dagger \hat{a}_q \end{align*}

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:

\begin{align*} \frac{1}{2} \sum_{pqrs} g_{pqrs} \langle 0| \hat{a}_p^\dagger \hat{a}_r^\dagger \hat{a}_s \hat{a}_q | 0\rangle &= \frac{1}{2} \sum_{ipq} g_{pqii} \langle 0|\hat{a}_p^\dagger \hat{a}_q | 0\rangle - \sum_{iqr} g_{iqri} \langle 0| \hat{a}_r^\dagger \hat{a}_q | 0\rangle \\ & = \frac{1}{2} \sum_{ij} g_{jjii} - g_{ijji} \end{align*}

Our final energy is thus:

$\langle 0| \hat H | 0\rangle = \sum_i h_{ii} + \frac{1}{2} \sum_{ij} g_{jjii} - g_{ijji}$