In Eq. (1) the operator \( a_\alpha^{\dagger} \) acts on the vacuum state \( |0\rangle \), which does not contain any particles. Alternatively, we could define a closed-shell nucleus or atom as our new vacuum, but then we need to introduce the particle-hole formalism, see the discussion to come.

In Eq. (2) \( a_\alpha^{\dagger} \) acts on an antisymmetric \( n \)-particle state and creates an antisymmetric \( (n+1) \)-particle state, where the one-body state \( \varphi_\alpha \) is occupied, under the condition that \( \alpha \ne \alpha_1, \alpha_2, \dots, \alpha_n \). It follows that we can express an antisymmetric state as the product of the creation operators acting on the vacuum state. $$ \begin{equation} |\alpha_1\dots \alpha_n\rangle_{\mathrm{AS}} = a_{\alpha_1}^{\dagger} a_{\alpha_2}^{\dagger} \dots a_{\alpha_n}^{\dagger} |0\rangle \tag{3} \end{equation} $$