We introduce the time-independent operators \( a_\alpha^{\dagger} \) and \( a_\alpha \) which create and annihilate, respectively, a particle in the single-particle state \( \varphi_\alpha \). We define the fermion creation operator \( a_\alpha^{\dagger} \)
$$ \begin{equation} a_\alpha^{\dagger}|0\rangle \equiv |\alpha\rangle \label{eq:2-1a}, \end{equation} $$and
$$ \begin{equation} a_\alpha^{\dagger}|\alpha_1\dots \alpha_n\rangle_{\mathrm{AS}} \equiv |\alpha\alpha_1\dots \alpha_n\rangle_{\mathrm{AS}} \label{eq:2-1b} \end{equation} $$In Eq. \eqref{eq:2-1a} 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. \eqref{eq:2-1b} \( a_\alpha^{\dagger} \) 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.
$$ \begin{equation} |\alpha_1\dots \alpha_n\rangle_{\mathrm{AS}} = a_{\alpha_1}^{\dagger} a_{\alpha_2}^{\dagger} \dots a_{\alpha_n}^{\dagger} |0\rangle \label{eq:2-2} \end{equation} $$It is easy to derive the commutation and anticommutation rules for the fermionic creation operators \( a_\alpha^{\dagger} \). Using the antisymmetry of the states \eqref{eq:2-2}
$$ \begin{equation} |\alpha_1\dots \alpha_i\dots \alpha_k\dots \alpha_n\rangle_{\mathrm{AS}} = - |\alpha_1\dots \alpha_k\dots \alpha_i\dots \alpha_n\rangle_{\mathrm{AS}} \label{eq:2-3a} \end{equation} $$we obtain
$$ \begin{equation} a_{\alpha_i}^{\dagger} a_{\alpha_k}^{\dagger} = - a_{\alpha_k}^{\dagger} a_{\alpha_i}^{\dagger} \label{eq:2-3b} \end{equation} $$Using the Pauli principle
$$ \begin{equation} |\alpha_1\dots \alpha_i\dots \alpha_i\dots \alpha_n\rangle_{\mathrm{AS}} = 0 \label{eq:2-4a} \end{equation} $$it follows that
$$ \begin{equation} a_{\alpha_i}^{\dagger} a_{\alpha_i}^{\dagger} = 0. \label{eq:2-4b} \end{equation} $$If we combine Eqs. \eqref{eq:2-3b} and \eqref{eq:2-4b}, we obtain the well-known anti-commutation rule
$$ \begin{equation} a_{\alpha}^{\dagger} a_{\beta}^{\dagger} + a_{\beta}^{\dagger} a_{\alpha}^{\dagger} \equiv \{a_{\alpha}^{\dagger},a_{\beta}^{\dagger}\} = 0 \label{eq:2-5} \end{equation} $$The hermitian conjugate of \( a_\alpha^{\dagger} \) is
$$ \begin{equation} a_{\alpha} = ( a_{\alpha}^{\dagger} )^{\dagger} \label{eq:2-6} \end{equation} $$If we take the hermitian conjugate of Eq. \eqref{eq:2-5}, we arrive at
$$ \begin{equation} \{a_{\alpha},a_{\beta}\} = 0 \label{eq:2-7} \end{equation} $$What is the physical interpretation of the operator \( a_\alpha \) and what is the effect of \( a_\alpha \) on a given state \( |\alpha_1\alpha_2\dots\alpha_n\rangle_{\mathrm{AS}} \)? Consider the following matrix element
$$ \begin{equation} \langle\alpha_1\alpha_2 \dots \alpha_n|a_\alpha|\alpha_1'\alpha_2' \dots \alpha_m'\rangle \label{eq:2-8} \end{equation} $$where both sides are antisymmetric.
We distinguish between two cases. The first (1) is when \( \alpha \in \{\alpha_i\} \). Using the Pauli principle of Eq. \eqref{eq:2-4a} it follows
$$ \begin{equation} \langle\alpha_1\alpha_2 \dots \alpha_n|a_\alpha = 0 \label{eq:2-9a} \end{equation} $$The second (2) case is when \( \alpha \notin \{\alpha_i\} \). It follows that an hermitian conjugation
$$ \begin{equation} \langle \alpha_1\alpha_2 \dots \alpha_n|a_\alpha = \langle\alpha\alpha_1\alpha_2 \dots \alpha_n| \label{eq:2-9b} \end{equation} $$Eq. \eqref{eq:2-9b} holds for case (1) since the lefthand side is zero due to the Pauli principle. We write Eq. \eqref{eq:2-8} as
$$ \begin{equation} \langle\alpha_1\alpha_2 \dots \alpha_n|a_\alpha|\alpha_1'\alpha_2' \dots \alpha_m'\rangle = \langle \alpha_1\alpha_2 \dots \alpha_n|\alpha\alpha_1'\alpha_2' \dots \alpha_m'\rangle \label{eq:2-10} \end{equation} $$Here we must have \( m = n+1 \) if Eq. \eqref{eq:2-10} has to be trivially different from zero.
For the last case, the minus and plus signs apply when the sequence \( \alpha ,\alpha_1, \alpha_2, \dots, \alpha_n \) and \( \alpha_1', \alpha_2', \dots, \alpha_{n+1}' \) are related to each other via even and odd permutations. If we assume that \( \alpha \notin \{\alpha_i\} \) we obtain
$$ \begin{equation} \langle\alpha_1\alpha_2 \dots \alpha_n|a_\alpha|\alpha_1'\alpha_2' \dots \alpha_{n+1}'\rangle = 0 \label{eq:2-12} \end{equation} $$when \( \alpha \in \{\alpha_i'\} \). If \( \alpha \notin \{\alpha_i'\} \), we obtain
$$ \begin{equation} a_\alpha\underbrace{|\alpha_1'\alpha_2' \dots \alpha_{n+1}'}\rangle_{\neq \alpha} = 0 \label{eq:2-13a} \end{equation} $$and in particular
$$ \begin{equation} a_\alpha |0\rangle = 0 \label{eq:2-13b} \end{equation} $$If \( \{\alpha\alpha_i\} = \{\alpha_i'\} \), performing the right permutations, the sequence \( \alpha ,\alpha_1, \alpha_2, \dots, \alpha_n \) is identical with the sequence \( \alpha_1', \alpha_2', \dots, \alpha_{n+1}' \). This results in
$$ \begin{equation} \langle\alpha_1\alpha_2 \dots \alpha_n|a_\alpha|\alpha\alpha_1\alpha_2 \dots \alpha_{n}\rangle = 1 \label{eq:2-14} \end{equation} $$and thus
$$ \begin{equation} a_\alpha |\alpha\alpha_1\alpha_2 \dots \alpha_{n}\rangle = |\alpha_1\alpha_2 \dots \alpha_{n}\rangle \label{eq:2-15} \end{equation} $$The action of the operator \( a_\alpha \) from the left on a state vector is to to remove one particle in the state \( \alpha \). If the state vector does not contain the single-particle state \( \alpha \), the outcome of the operation is zero. The operator \( a_\alpha \) is normally called for a destruction or annihilation operator.
The next step is to establish the commutator algebra of \( a_\alpha^{\dagger} \) and \( a_\beta \).
The action of the anti-commutator \( \{a_\alpha^{\dagger} \),$a_\alpha\}$ on a given \( n \)-particle state is
$$ \begin{align} a_\alpha^{\dagger} a_\alpha \underbrace{|\alpha_1\alpha_2 \dots \alpha_{n}\rangle}_{\neq \alpha} &= 0 \nonumber \\ a_\alpha a_\alpha^{\dagger} \underbrace{|\alpha_1\alpha_2 \dots \alpha_{n}\rangle}_{\neq \alpha} &= a_\alpha \underbrace{|\alpha \alpha_1\alpha_2 \dots \alpha_{n}\rangle}_{\neq \alpha} = \underbrace{|\alpha_1\alpha_2 \dots \alpha_{n}\rangle}_{\neq \alpha} \label{eq:2-16a} \end{align} $$if the single-particle state \( \alpha \) is not contained in the state.
If it is present we arrive at
$$ \begin{align} a_\alpha^{\dagger} a_\alpha |\alpha_1\alpha_2 \dots \alpha_{k}\alpha \alpha_{k+1} \dots \alpha_{n-1}\rangle &= a_\alpha^{\dagger} a_\alpha (-1)^k |\alpha \alpha_1\alpha_2 \dots \alpha_{n-1}\rangle \nonumber \\ = (-1)^k |\alpha \alpha_1\alpha_2 \dots \alpha_{n-1}\rangle &= |\alpha_1\alpha_2 \dots \alpha_{k}\alpha \alpha_{k+1} \dots \alpha_{n-1}\rangle \nonumber \\ a_\alpha a_\alpha^{\dagger}|\alpha_1\alpha_2 \dots \alpha_{k}\alpha \alpha_{k+1} \dots \alpha_{n-1}\rangle &= 0 \label{eq:2-16b} \end{align} $$From Eqs. \eqref{eq:2-16a} and \eqref{eq:2-16b} we arrive at
$$ \begin{equation} \{a_\alpha^{\dagger} , a_\alpha \} = a_\alpha^{\dagger} a_\alpha + a_\alpha a_\alpha^{\dagger} = 1 \label{eq:2-17} \end{equation} $$The action of \( \left\{a_\alpha^{\dagger}, a_\beta\right\} \), with \( \alpha \ne \beta \) on a given state yields three possibilities. The first case is a state vector which contains both \( \alpha \) and \( \beta \), then either \( \alpha \) or \( \beta \) and finally none of them.
The first case results in
$$ \begin{align} a_\alpha^{\dagger} a_\beta |\alpha\beta\alpha_1\alpha_2 \dots \alpha_{n-2}\rangle = 0 \nonumber \\ a_\beta a_\alpha^{\dagger} |\alpha\beta\alpha_1\alpha_2 \dots \alpha_{n-2}\rangle = 0 \label{eq:2-18a} \end{align} $$The second case gives
$$ \begin{align} a_\alpha^{\dagger} a_\beta |\beta \underbrace{\alpha_1\alpha_2 \dots \alpha_{n-1}}_{\neq \alpha}\rangle =& |\alpha \underbrace{\alpha_1\alpha_2 \dots \alpha_{n-1}}_{\neq \alpha}\rangle \nonumber \\ a_\beta a_\alpha^{\dagger} |\beta \underbrace{\alpha_1\alpha_2 \dots \alpha_{n-1}}_{\neq \alpha}\rangle =& a_\beta |\alpha\beta\underbrace{\beta \alpha_1\alpha_2 \dots \alpha_{n-1}}_{\neq \alpha}\rangle \nonumber \\ =& - |\alpha\underbrace{\alpha_1\alpha_2 \dots \alpha_{n-1}}_{\neq \alpha}\rangle \label{eq:2-18b} \end{align} $$Finally if the state vector does not contain \( \alpha \) and \( \beta \)
$$ \begin{align} a_\alpha^{\dagger} a_\beta |\underbrace{\alpha_1\alpha_2 \dots \alpha_{n}}_{\neq \alpha,\beta}\rangle &=& 0 \nonumber \\ a_\beta a_\alpha^{\dagger} |\underbrace{\alpha_1\alpha_2 \dots \alpha_{n}}_{\neq \alpha,\beta}\rangle &=& a_\beta |\alpha \underbrace{\alpha_1\alpha_2 \dots \alpha_{n}}_{\neq \alpha,\beta}\rangle = 0 \label{eq:2-18c} \end{align} $$For all three cases we have
$$ \begin{equation} \{a_\alpha^{\dagger},a_\beta \} = a_\alpha^{\dagger} a_\beta + a_\beta a_\alpha^{\dagger} = 0, \quad \alpha \neq \beta \label{eq:2-19} \end{equation} $$We can summarize our findings in Eqs. \eqref{eq:2-17} and \eqref{eq:2-19} as
$$ \begin{equation} \{a_\alpha^{\dagger},a_\beta \} = \delta_{\alpha\beta} \label{eq:2-20} \end{equation} $$with \( \delta_{\alpha\beta} \) is the Kroenecker \( \delta \)-symbol.
The properties of the creation and annihilation operators can be summarized as (for fermions)
$$ a_\alpha^{\dagger}|0\rangle \equiv |\alpha\rangle, $$and
$$ a_\alpha^{\dagger}|\alpha_1\dots \alpha_n\rangle_{\mathrm{AS}} \equiv |\alpha\alpha_1\dots \alpha_n\rangle_{\mathrm{AS}}. $$from which follows
$$ |\alpha_1\dots \alpha_n\rangle_{\mathrm{AS}} = a_{\alpha_1}^{\dagger} a_{\alpha_2}^{\dagger} \dots a_{\alpha_n}^{\dagger} |0\rangle. $$The hermitian conjugate has the folowing properties
$$ a_{\alpha} = ( a_{\alpha}^{\dagger} )^{\dagger}. $$Finally we found
$$ a_\alpha\underbrace{|\alpha_1'\alpha_2' \dots \alpha_{n+1}'}\rangle_{\neq \alpha} = 0, \quad \textrm{in particular } a_\alpha |0\rangle = 0, $$and
$$ a_\alpha |\alpha\alpha_1\alpha_2 \dots \alpha_{n}\rangle = |\alpha_1\alpha_2 \dots \alpha_{n}\rangle, $$and the corresponding commutator algebra
$$ \{a_{\alpha}^{\dagger},a_{\beta}^{\dagger}\} = \{a_{\alpha},a_{\beta}\} = 0 \hspace{0.5cm} \{a_\alpha^{\dagger},a_\beta \} = \delta_{\alpha\beta}. $$