Second quantization

Let us consider an operator proportional with \( a_\alpha^{\dagger} a_\beta \) and \( \alpha=\beta \). It acts on an \( n \)-particle state resulting in $$ \begin{equation} a_\alpha^{\dagger} a_\alpha |\alpha_1\alpha_2 \dots \alpha_{n}\rangle = \begin{cases} 0 &\alpha \notin \{\alpha_i\} \\ \\ |\alpha_1\alpha_2 \dots \alpha_{n}\rangle & \alpha \in \{\alpha_i\} \end{cases} \tag{28} \end{equation} $$ Summing over all possible one-particle states we arrive at $$ \begin{equation} \left( \sum_\alpha a_\alpha^{\dagger} a_\alpha \right) |\alpha_1\alpha_2 \dots \alpha_{n}\rangle = n |\alpha_1\alpha_2 \dots \alpha_{n}\rangle \tag{29} \end{equation} $$