Suppose we have a complete set of states \(\psi_n(\vec{r}) \}\) or \(\{ |n\rangle \}\). Then an \(n\)-particle fermionic state can be written as a Slater determinant. Note that (\(\vec{r}_i\)) here denotes space and spin coordinates: \[\Psi(\vec{r}_1, \ldots, \vec{r}_N) = \frac{1}{\sqrt{N!}} \sum_P (-1)^P \psi_{n_1}(P \vec{r}_1) \psi_{n_2}(P \vec{r}_2) \cdots \psi_{n_N}(P \vec{r}_N),\] with the sum over all \(N!\) permutations \(P\) of \(N\) objects. This wave function is obviously anti-symmetric under the interchange of any two particles. Notationally, it is painful to deal with Slater determinants. So a new formalism was developed called the occupation number representation, where we denote each of the wave functions to be included in the Slater determinant (as a shorthand): \[|1, 0, 0, \dots \rangle = \psi_1(\vec{r}_1),\] \[|0, 1, 1, 0, \dots \rangle = \frac{1}{\sqrt{2}} \big( \psi_2(\vec{r}_1) \psi_3(\vec{r}_2) - \psi_3(\vec{r}_1) \psi_2(\vec{r}_2) \big).\] We introduce abstract operators in this space in the spirit of Dirac: \[\hat{c}_k^\dagger |n_1, n_2, \dots, n_k, \dots \rangle = |n_1, n_2, \dots, n_k+1, \dots \rangle\] where \(c^\dagger_k\) creates the state \(k\), and \[\hat{c}_k |n_1, n_2, \dots, n_k, \dots \rangle = |n_1, n_2, \dots, n_k-1, \dots \rangle\] where \(\hat{c}_k\) destroys the state \(k\). The Pauli exclusion principle says \(n_k = 0\) or \(1\) only, so \[(\hat{c}_k^\dagger)^2 = (\hat{c}_k)^2 = 0.\] Now, we can define the vacuum state as: \[|0\rangle = |0, 0, 0, \dots \rangle.\] Acting on this vacuum state gives us: \[\hat{c}_k^\dagger |0\rangle = \psi_k(\vec{r}_1)\] and: \[\hat{c}_{k'}^\dagger \hat{c}_k^\dagger |0\rangle = \frac{1}{\sqrt{2}} \left( \psi_{k'}(\vec{r}_1) \psi_k(\vec{r}_2) - \psi_k(\vec{r}_1) \psi_{k'}(\vec{r}_2) \right),\] but \[\hat{c}_k^\dagger \hat{c}_{k'}^\dagger |0\rangle = \frac{1}{\sqrt{2}} \left( \psi_k(\vec{r}_1) \psi_{k'}(\vec{r}_2) - \psi_{k'}(\vec{r}_1) \psi_k(\vec{r}_2) \right),\] which shows us that \(\hat{c}_{k'}^\dagger \hat{c}_k^\dagger \neq \hat{c}_k^\dagger \hat{c}_{k'}^\dagger\), meaning the operators don’t commute and we get the anticommutation relation: \[(\hat{c}_{k'}^\dagger ,\hat{c}_k^\dagger)_+ =0\] In general, we get the following anticommutation relations: \[(\hat{c}_k^\dagger, \hat{c}_{k'}^\dagger)_+ = 0, \quad (\hat{c}_k, \hat{c}_{k'})_+ = 0, \quad (\hat{c}_k^\dagger, \hat{c}_{k'})_+ = \delta_{kk'}.\] From the definitions above of the creation and annhiliation operators, it follows further that: \[\begin{cases} \hat{c}_k^\dagger |0_k\rangle = |1_k\rangle\\ \hat{c}_k^\dagger |1_k\rangle = 0\\ \hat{c}_k |0_k\rangle = 0\\ \hat{c}_k |1_k\rangle = |0_k\rangle \end{cases}\] so \(\hat{c}_k^\dagger \hat{c}_k |0_k\rangle = 0\) and \(\hat{c}_k^\dagger \hat{c}_k |1_k\rangle = 1 \cdot |1_k\rangle\). As a result, we can think of \(\hat{c}_k^\dagger \hat{c}_k\) as counting the occupation number at state \(k\). We can thus define the so-called number operator \(\hat{n}_k = \hat{c}_k^\dagger \hat{c}_k\) which has the following commutation relations: \[\begin{cases} [\hat{n}_k, \hat{c}_k^\dagger] = \hat{c}_k^\dagger \hat{c}_k \hat{c}_k^\dagger - \hat{c}_k^\dagger \hat{c}_k^\dagger \hat{c}_k = \hat{c}_k^\dagger (\hat{c}_k, \hat{c}_k^\dagger)_+ = \hat{c}_k^\dagger \\ [\hat{n}_k, \hat{c}_k^\dagger] = \hat{c}_k^\dagger \\ [\hat{n}_k, \hat{c}_k] = \hat{c}_k^\dagger \hat{c}_k \hat{c}_k - \hat{c}_k \hat{c}_k^\dagger \hat{c}_k = -\hat{c}_k (\hat{c}_k^\dagger, \hat{c}_k)_+ = -\hat{c}_k\\ [\hat{n}_k, \hat{c}_k] = -\hat{c}_k \end{cases}\] Notice that these operators are thus very similar to raising and lowering operators for the simple harmonic oscillator.
Furthermore, we can define the operator \(\hat{N} = \sum_k \hat{n}_k\) which is called the total number operator. It acts as: \[\hat{N} |n_1, n_2, \dots\rangle = \sum_{k=1}^\infty n_k |n_1, n_2, \dots\rangle.\] So \(\hat{N}\) counts the number of occupied states.
Consider the one electron operators \(\hat{O}_1\) given by: \[\langle l \sigma | \hat{O} | l' \sigma' \rangle = \int dr_1\, \psi_l(\vec{r}_1) \chi_\sigma^\dagger \hat{O}(\vec{r})_{\sigma\sigma'} \chi_{\sigma'} \psi_{l'}(\vec{r}_1)\] where \(l\) labels spatial wave functions, \(\sigma\) labels spin wave functions. We thus see that: \[\hat{O} \iff \sum_{l \sigma l' \sigma'} \langle l \sigma | \hat{O} | l' \sigma' \rangle \hat{c}_{l \sigma}^\dagger \hat{c}_{l' \sigma'}\] Let us consider an example.
Example: Consider the momentum operator \(\hat{\vec{p}} = -i \hbar \vec{\nabla}\) and suppose that the states are plane waves \(\frac{1}{\sqrt{V}} e^{i \vec{k} \cdot \vec{r}} \chi_\sigma = \phi_{k \sigma}\). Then we have: \[\langle \phi_{k \sigma} | \hat{\vec{p}} | \phi_{k' \sigma'} \rangle = \frac{1}{V} \int dr\, e^{-i \vec{k} \cdot \vec{r}} \chi_\sigma^\dagger \left(-i \hbar \vec{\nabla} \right) \chi_{\sigma'} e^{i \vec{k}' \cdot \vec{r}}= \delta_{\sigma \sigma'} \hbar \vec{k} \delta(\vec{k} - \vec{k}')\] As a result, we have: \[\hat{\vec{p}} \iff \sum_{k \sigma k' \sigma'} \delta_{\sigma \sigma'} \delta_{kk'}\hbar \vec{k} \hat{c}_{k \sigma}^\dagger \hat{c}_{k' \sigma'} = \sum_{k \sigma} \hbar \vec{k} \hat{c}_{k \sigma}^\dagger \hat{c}_{k' \sigma'} = \sum_{k \sigma} \hbar \vec{k} \hat{n}_{k \sigma}\]
Recall our familiar spin operators in matrix form. We can equivalently represent them as in our second quantization notation. As a example, take \(\hat{S}_z\), \(\hat{S}_+\), and \(\hat{S}_-\): \[\hat{S}_z = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \iff \frac{1}{2} \sum_\ell \left( \hat{c}_{\ell \uparrow}^\dagger \hat{c}_{\ell \uparrow} - \hat{c}_{\ell \downarrow}^\dagger \hat{c}_{\ell \downarrow} \right) = \frac{1}{2} \sum_\ell \left( \hat{n}_{\ell \uparrow} - \hat{n}_{\ell \downarrow} \right)\] \[\hat{S}_+ = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \iff \sum_\ell \hat{c}_{\ell \uparrow}^\dagger \hat{c}_{\ell \downarrow}\] \[\hat{S}_- = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \iff \sum_\ell \hat{c}_{\ell \downarrow}^\dagger \hat{c}_{\ell \uparrow}\]
Above we consider one-particle operators, but what about two particle
operators of the form: \[\hat{O}_2 =
\frac{e^2}{r_{ij}} = \frac{e^2}{|\vec{r}_i - \vec{r}_j|}.\] We
can express them as: \[\sum_{l_1 l_2 l_3 l_4}
\sum_{\sigma_1 \sigma_2 \sigma_3 \sigma_4}
\langle l_1 \sigma_1, l_2 \sigma_2 | \hat{O}_2 | l_3 \sigma_3, l_4
\sigma_4\rangle
\hat{c}_{l_1 \sigma_1}^\dagger \hat{c}_{l_2 \sigma_2}^\dagger
\hat{c}_{l_3 \sigma_3} \hat{c}_{l_4 \sigma_4}\] where \(1\) and \(4\) correspond to \(\vec{r}_1\), \(2\) and \(3\) to \(\vec{r}_2\). Furthermore: \[\langle l_1 \sigma_1, l_2 \sigma_2 | \hat{O}_2 |
l_3 \sigma_3, l_4 \sigma_4 \rangle =
\int d\vec{r}_1 \int d\vec{r}_2 \,
\phi_{l_1 \sigma_1}^*(\vec{r}_1) \phi_{l_2 \sigma_2}^*(\vec{r}_2)
\hat{O}_2(\vec{r}_1, \vec{r}_2)
\phi_{l_3 \sigma_3}(\vec{r}_2) \phi_{l_4 \sigma_4}(\vec{r}_1)\]
Again, let us consider an example for clarity.
Example: Consider the two particle isotropic interaction given
by the operator \(\hat{O}_2 = V(|\vec{r}_1 -
\vec{r}_2|)\). We can evaluate in a plane-wave basis by: \[\begin{aligned}
\langle \phi_{k_1 \sigma_1} \phi_{k_2 \sigma_2} | V(|\vec{r}_1 -
\vec{r}_2|) | \phi_{k_3 \sigma_3} \phi_{k_4 \sigma_4} \rangle &=
\frac{1}{V^2} \langle \sigma_1 \sigma_4 | \sigma_2 \sigma_3 \rangle
\int e^{-i (\vec{k}_1 \cdot \vec{r}_1 + \vec{k}_2 \cdot \vec{r}_2)}
V(|\vec{r}_1 - \vec{r}_2|) e^{i (\vec{k}_3 \cdot \vec{r}_2 + \vec{k}_4
\cdot \vec{r}_1)} d^3r_1 d^3r_2 \\
&= \frac{1}{V^2} \delta_{\sigma_1 \sigma_4} \delta_{\sigma_2
\sigma_3}
\int e^{-i (\vec{k}_1 - \vec{k}_4) \cdot \vec{r}_1} e^{-i (\vec{k}_2 -
\vec{k}_3) \cdot \vec{r}_2}
V(|\vec{r}_1 - \vec{r}_2|) d^3r_1 d^3r_2
\end{aligned}\] Now let \(\vec{R} =
\frac{\vec{r}_1 + \vec{r}_2}{2}\) and \(\vec{r} = \vec{r}_1 - \vec{r}_2\). We get:
\[\begin{aligned}
&= \frac{1}{V^2} \delta_{\sigma_1 \sigma_4} \delta_{\sigma_2
\sigma_3}
\int e^{-i(\vec{k}_1 + \vec{k}_2 - \vec{k}_3 - \vec{k}_4) \cdot \vec{R}}
e^{-i(\vec{k}_1 - \vec{k}_3) \cdot \vec{r}/2} e^{i(\vec{k}_2 -
\vec{k}_4) \cdot \vec{r}/2}
U(r) \, dR \, dr \\
&= \frac{1}{V} \delta_{\sigma_1 \sigma_4} \delta_{\sigma_2 \sigma_3}
\int e^{-i(\vec{k}_1 - \vec{k}_4) \cdot \vec{r}} U(r) \, dr \,
\delta(\vec{k}_1 + \vec{k}_2 - \vec{k}_3 - \vec{k}_4) \\
&= \frac{1}{V} \delta_{\sigma_1 \sigma_4} \delta_{\sigma_2 \sigma_3}
\delta(\vec{k}_1 + \vec{k}_2 - \vec{k}_3 - \vec{k}_4) V_{FT}(\vec{k}_1 -
\vec{k}_4)
\end{aligned}\] where we set \(U(r) =
\frac{e^2}{r}\) and \(V_{FT}(\vec{k}_1
- \vec{k}_4) = \frac{4\pi e^2}{|\vec{k}_1 - \vec{k}_4|^2}
\frac{1}{V}\). As a result, we find the Coulomb
operator to be: \[\frac{1}{2}
\sum_{k k'} \sum_{\sigma \sigma'} \frac{4 \pi e^2}{q^2 V}
\hat{c}_{k+q \sigma}^\dagger \hat{c}_{k' -q\sigma'}^\dagger
\hat{c}_{k'\sigma'} \hat{c}_{k \sigma}\] where the \(\frac{1}{2}\) avoids double counting.
This method of dealing with creation and annihilation operators is called second quantization. As before, we will find working with this operator formalism will make life easier than working with wave functions.