Introduction to Second Quantization

Until now, we have represented the state of a quantum system by a function and observables by operators. This formalism is called first quantization. This is rather intuitive, because we can directly interpret the wavefunction as the probability amplitude. However, it has several problems:

1. It carrys too much information that makes calculations with many particles unnecessarily complicated.
2. The number of particles must be fixed.
3. Not compatible with relativity.

In this section, we will introduce a new formalism called second quantization. It is more abstract, but was made to solve the problems listed above.

Note: The term second quantization is a bit misleading, because it is not a quantization of a classical theory. It is just a different way to represent quantum states and operators. It is equivalent to the first quantization formalism.

Note: We will only derive the formalism for fermions, but it can be generalized to bosons by replacing the antisyymetric operations with symmetric ones.

The Fock Space

We already know that a first quantization state with $N$ particles lives in the Hilbert space \(\hat{A} \mathcal{H}^{\otimes N}\), where \(\hat{A}\) is the antisymmetrization operator and \(\mathcal{H}\) is a single-particle Hilbert space. The notation \(\mathcal{H}^{\otimes N}\) means that we take the tensor product of \(\mathcal{H}\) with itself \(N\) times, i.e. $$ \mathcal{H}^{\otimes N} = \underbrace{\mathcal{H} \otimes \cdots \otimes \mathcal{H}}_{n\ \text{times}} $$

Since we want to allow an arbitrary number of particles in second quantization, we must enter a larger space, the so-called Fock space. The Fock space is defined as the direct sum of all \(N\)-particle Hilbert spaces: $$ \mathcal{F} = \bigoplus_{N=0}^\infty \mathcal{H}^{\otimes N} $$ We also include the \(N=0\) case, which only contains the vacuum state \(| \mathrm{vac} \rangle\). A determinant is represented by an occupation-number (ON) vector \(\vec{k}\), $$ | k \rangle = | k_1, k_2, \cdots, k_M \rangle, \quad k_p \in {0, 1} $$ with \(M\) single-particle states. The occupation numbers \(k_p\) can either be 0 (empty) or 1 (occupied) due to the Pauli exclusion principle. The single-particle state is also called a site and can e.g. be an orbital.

Because we do not want to do difficult calculations with state vectors (excepct for the vacuum state), we must represent states by some (combination of) operators, something that generate all possible states in the Fock space from the vacuum state.

Creation and Annihilation Operators

We want to define a creation operator \(a_p^{\dagger}\) that creates a particle at site \(p\). It should also take care of Pauli exclusion, i.e. it should destroy the state if site \(p\) is already occupied. This leads to the following definition: $$ \begin{align} a_p^{\dagger} | k_1, \cdots, 0_p, \cdots, k_M \rangle &= \Gamma_p^{\vec{k}} | k_1, \cdots, 1_p, \cdots, k_M \rangle \\ a_p^{\dagger} | k_1, \cdots, 1_p, \cdots, k_M \rangle &= 0 \end{align} $$ where the phase factor \(\Gamma_p^{\vec{k}}\) is defined as $$ \Gamma_p^{\vec{k}} = \prod_{q=1}^{p-1} (-1)^{k_q} $$ So for every already occupied site \(q\) before site \(p\), we have to multiply the state with \(-1\). This is necessary to ensure antisymmetry.

These two equations can be combined into a single one: $$ a_p^{\dagger} | \vec{k} \rangle = \delta_{k_p, 0} \Gamma_p^{\vec{k}} | k_1, \cdots, 1_p, \cdots, k_M \rangle $$ From the definition, it trivially follows that $$ a_p^{\dagger} a_p^{\dagger} | \vec{k} \rangle = 0 $$

For two different creation operators \(a_p^{\dagger}\) and \(a_q^{\dagger}\), we can show the following: $$ (a_p^{\dagger} a_q^{\dagger} + a_q^{\dagger} a_p^{\dagger}) | \vec{k} \rangle = 0 $$

The action of Hermitian adjoints \(a_p\) on \(|\vec{k} \rangle\) can be understood by inserting the identity $$ a_p | \vec{k} \rangle = \sum_{\vec{m}} |\vec{m}\rangle \langle \vec{m} | a_p | \vec{k} \rangle $$

The matrix element can be evaluated as

$$ \langle \vec{m} | a_p | \vec{k} \rangle = \langle \vec{k} | a_p^\dagger | \vec{m} \rangle ^* = \delta_{m_p 0} \Gamma_p^{\vec{m}} \langle \vec{k} | m_1, \cdots, 1_p, \cdots \rangle^* $$

The overlap \(\langle \vec{k} | m_1, \cdots, 1_p, \cdots \rangle\) is nonzero if and only if both vectors are identical, which is the case when \(]\vec{k} \rangle\) and \(| \vec{m} \rangle\) only differ at position \(p\), where \(k_p = 1\) and \(m_p = 0\). Therefore, the matrix element can also be written as

$$ \langle \vec{m} | a_p | \vec{k} \rangle = \delta_{k_p 1} \Gamma_p^{\vec{k}} \langle \vec{m} | k_1, \cdots, 0_p, \cdots \rangle $$

where we used the fact \(\Gamma_p^{\vec{k}} = \Gamma_p^{\vec{m}}\), which follows directly from its definition. Thus only one term in the sum survives and we can conclude $$ a_p |\vec{k} \rangle = \delta_{k_p 1} \Gamma_p^{\vec{k}} | k_1, \cdots, 0_p, \cdots k_M \rangle $$

The operator \(a_p\) removes the occupation at \(p\) if occupied and returns zero if unoccupied. Therefore, we shall call it annihilation operator. A special case of the equation above is $$ a_p |\mathrm{vac} \rangle = 0 $$ which states that the annihilation operator destroys the vacuum state.

Since the annihilation operator is the Hermitian adjoint of the creation operator, we can follow $$ (a_p a_q + a_q a_p) | \vec{k} \rangle = 0 $$

We can then examine the action of the action of one creation and one annihilation operator on a state vector \(| \vec{k} \rangle\). After several algebraic transformations, we can show $$ (a_p^\dagger a_q + a_q a_p^\dagger) | \vec{k} \rangle = \delta_{pq} $$

We have now worked out the action of two ladder operators (creation or annihilation operator) on an arbitrary state vector. Since it works for all possible states, we can as well write the equations solely in terms of operators: $$ \begin{align} \{a_p^\dagger, a_q^\dagger\} &= 0 \\ \{a_p, a_q\} &= 0 \\ \{a_p^\dagger, a_q\} &= \delta_{pq} \end{align} $$ where \({A, B} = AB + BA\) is the anticommutator. These relations are known as the anticommutation relations of the ladder operators and can be viewed as the defining property of the ladder operators.

Now we know how to represent states using operators in second quantization. But how do observables look like?

Operators in Second Quantization

We shall at first take a look at one-particle operators \(\hat{O}_1 = \sum_{i} \hat{o}(i)\), where \(\hat{o}(i)\) is the one-particle operator acting on the \(i\)-th particle. By inserting the identity to the left and right of the operator, we obtain $$ \hat{O}_1 = \sum_{i} \hat{o}(i) = \sum_{i} \sum_{\vec{k}, \vec{k}'} | \vec{k} \rangle \langle \vec{k} | \hat{o}(i) | \vec{k}' \rangle \langle \vec{k}' | $$

Suppose the \(i\)-th particle is on the \(p\)-th site in the state \(| \vec{k} \rangle\) and \(q\)-th site in the state \(| \vec{k}' \rangle\). Then we can replace the sum over \(i\) by a sum over \(p\) and \(q\). Because the operator \(\hat{o}(i)\) acts only on the \(i\)-th particle, the matrix element \(\langle \vec{k} | \hat{o}(i) | \vec{k}' \rangle\) evaluates to $$ \langle \vec{k} | \hat{o}(i) | \vec{k}' \rangle = \langle \vec{k} \backslash p | \langle{p} | \hat{o}(i) | q \rangle | \vec{k}' \backslash q \rangle = o_{pq} \langle \vec{k} \backslash p | \vec{k}' \backslash q \rangle $$ where \(| \vec{k} \backslash p \rangle\) (sloppily) denotes the state vector \(| \vec{k} \rangle\) with the \(p\)-th site removed. But We already know how to (formally) remove a site from a state vector, namely by applying the annihilation operator \(a_p\). Therefore, we can write the overlap as $$ \langle \vec{k} \backslash p | \vec{k}' \backslash q \rangle = \langle a_p \vec{k} | a_q \vec{k}' \rangle = \langle \vec{k} | a_p^\dagger a_q | \vec{k}' \rangle $$

Putting everything together, we obtain $$ \hat{O}_1 = \sum_{pq} o_{pq} a_p^\dagger a_q $$

We can see that a one-particle operator is represented by linear combination of creation-annihilation pairs weighted by the matrix elements of the one-site operator \(\hat{o}(i)\).

We can perform the same calculating for two-particle operators by inserting a total of four identities and arrive at $$ \hat{O}_2 = \frac{1}{2} \sum_{pqrs} o_{pqrs} a_p^\dagger a_r^\dagger a_s a_q $$ with $$ o_{pqrs} = (pq | \hat{o}(i, j) | rs) $$

So, a molecular electronic Hamiltonian in the language of second quantization is given by $$ \begin{align} \hat{H} &= \hat{h} + \hat{g} \\ &= \sum_{pq} h_{pq} a_p^\dagger a_q + \frac{1}{2} \sum_{pqrs} g_{pqrs} a_p^\dagger a_r^\dagger a_s a_q \end{align} $$ where $$ \begin{align} h_{pq} &= \int \phi_p^{*}(x) \left( -\frac{1}{2} \nabla^2 - \sum_{I=1}^{N} \frac{Z_I}{|r - R_I|} \right) \phi_q(x)\ \mathrm{d}x \\ g_{pqrs} &= \int \phi_p^{*}(x_1) \phi_r^{*}(x_2) \frac{1}{|r_1 - r_2|} \phi_q(x_1) \phi_s(x_2)\ \mathrm{d}x_1 \mathrm{d}x_2 \end{align} $$

Now with the representation of the observables and states in second quantization, we can evaluate matrix elements. Since all states are represented by a product of creation operators acting on the vacuum, a matrix element of, say, a one-electron operator would look like this: $$ \langle \mathrm{vac} | a_{r_1} \cdots a_{r_k} \ (o_{pq} a_p^\dagger a_q)\ a_{s_1}^\dagger \cdots a_{s_l}^\dagger | \mathrm{vac} \rangle $$ Ignoring the number \(o_{pq}\), we have a product of ladder operators in the middle, also known as a string. For two-electron operators, the situation is similar, just with two additional operators.