How does the following definition of contruction/destruction operators for the Fock state of Fermions account for the anti-symmetrization?

Consider the following definitions of the creation and destruction operators for a Fermionic Fock state: $$\hat a^{\dagger}_n \left|N_0 N_1...\right.\rangle = (-1)^{\sum_{k

$$\hat a_n \left|N_0 N_1...\right.\rangle = (-1)^{\sum_{k

I understand how the prefactors of $$(1-N_n)$$ and $$N_n$$ come because of the fact that the $$N_i$$s can either be just $$0$$ or $$1$$ in case of Fermions. but I do not know why the $$(-1)^{\sum_{k factors also come and if they somehow account for the minus that comes from anti-symmetrization.

For instance if I look at the action of the creation operator on the following Fock state as: $$\hat a^{\dagger}_3 \left|1 0 0...\right.\rangle = - \left|1 0 1...\right.\rangle$$

What is the signifcance of the minus sign that comes on the right hand side of the above equation?

• Have you tried a simple example? Have you checked a (nother) book? Have you checked some internet sources, e.g. Wikipedia? Commented Jan 22, 2023 at 7:00

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The creation operator $$\:\hat a^\dagger_i\:$$ increases the occupation number of the level $$\:i\:$$ by $$\:1\:$$. If the state $$\:i\:$$ was empty then [see equations \eqref{ft-01}, \eqref{ft-02} in the Footnote] \begin{align} &\hat a^\dagger_i\vra{n_1,n_2,\cdots,0_i,\cdots} \e \nonumber\\ &\e \!\!\!\overbrace{\hat a^\dagger_i(\hat a^\dagger_1)^{n_1}}^{\plr{\m 1}^{n_1}(\hat a^\dagger_1)^{n_1}\hat a^\dagger_i}\!\!\!(\hat a^\dagger_2)^{n_2}\cdots(\hat a^\dagger_i)^0\cdots\vra 0 \nonumber\\ &\e \plr{\m 1}^{n_1}(\hat a^\dagger_1)^{n_1}\!\!\!\overbrace{\hat a^\dagger_i(\hat a^\dagger_2)^{n_2}}^{\plr{\m 1}^{n_2}(\hat a^\dagger_2)^{n_2}\hat a^\dagger_i}\!\!\!(\hat a^\dagger_3)^{n_3}\cdots(\hat a^\dagger_i)^0\cdots\vra 0 \nonumber\\ & \e \plr{\m 1}^{n_1\p n_2}(\hat a^\dagger_1)^{n_1}(\hat a^\dagger_2)^{n_2}\!\!\overbrace{\hat a^\dagger_i(\hat a^\dagger_3)^{n_3}}^{\plr{\m 1}^{n_3}(\hat a^\dagger_3)^{n_3}\hat a^\dagger_i}\!\!(\hat a^\dagger_4)^{n_4}\cdots(\hat a^\dagger_i)^0\cdots\vra 0 \nonumber\\ & \e \plr{\m 1}^{n_1\p n_2\p n_3}(\hat a^\dagger_1)^{n_1}(\hat a^\dagger_2)^{n_2}(\hat a^\dagger_3)^{n_3}\overbrace{\hat a^\dagger_i(\hat a^\dagger_4)^{n_4}}^{\plr{\m 1}^{n_4}\cdots}\cdots(\hat a^\dagger_i)^0\cdots\vra 0 \nonumber\\ &\e \plr{\m 1}^{n_1\p n_2\p n_3\p n_4}\cdots\cdots \tl{01} \end{align} that is \begin{align} \hat a^\dagger_i\vra{n_1,n_2,\cdots,0_i,\cdots}&\e\plr{\m 1}^{\sum\limits_{k\e 1}^{k\e i\m 1}\!\!n_k}(\hat a^\dagger_1)^{n_1}(\hat a^\dagger_2)^{n_2}\cdots(\hat a^\dagger_i)\cdots\vra 0 \nonumber\\ & \e \plr{\m 1}^{\sum\limits_{k\e 1}^{k\e i\m 1}\!\!n_k}\vra{n_1,n_2,\cdots,1_i,\cdots} \tl{02} \end{align} If, on the other hand, the level $$\:i\:$$ was already occupied then the application of $$\:\hat a^\dagger_i\:$$ destroys the state vector: $$$$\hat a^\dagger_i\vra{n_1,n_2,\cdots,1_i,\cdots} \e \plr{\m 1}^{\sum\limits_{k\e 1}^{k\e i\m 1}\!\!n_k}(\hat a^\dagger_1)^{n_1}(\hat a^\dagger_2)^{n_2}\cdots(\hat a^\dagger_i)^2\cdots\vra 0\e 0. \tl{03}$$$$
Note that \eqref{02} contains a phase factor $$\:(\m 1)^{\sum\limits_{k\e 1}^{k\e i\m 1}\!\!n_k}\:$$ which has no physical significance as it depends on the labelling of the states $$\:i\:$$ which is arbitrary.
Since $$\:\hat a^\dagger_i\hat a^\dagger_j\e \m\hat a^\dagger_j\hat a^\dagger_i\:$$ we have \begin{align} &\hat a^\dagger_i(\hat a^\dagger_j)^{n_j} \e \plr{\hat a^\dagger_i\hat a^\dagger_j}(\hat a^\dagger_j)^{n_j\m 1}\e \plr{\m\hat a^\dagger_j\hat a^\dagger_i}(\hat a^\dagger_j)^{n_j\m 1}\e \nonumber\\ &\plr{\m 1}\hat a^\dagger_j\!\!\overbrace{\hat a^\dagger_i(\hat a^\dagger_j)^{n_j\m 1}}^{\plr{\m 1}\hat a^\dagger_j\hat a^\dagger_i(\hat a^\dagger_j)^{n_j\m 2}}\!\!\!\e \plr{\m 1}^2(\hat a^\dagger_j)^2\hat a^\dagger_i(\hat a^\dagger_j)^{n_j\m 2}\e \cdots\e \nonumber\\ & \plr{\m 1}^m(\hat a^\dagger_j)^m\hat a^\dagger_i(\hat a^\dagger_j)^{n_j\m m}\e\cdots\e \plr{\m 1}^{n_j}(\hat a^\dagger_j)^{n_j}\hat a^\dagger_i \tl{ft-01} \end{align} that is $$$$\boxed{\:\:\hat a^\dagger_i(\hat a^\dagger_j)^{n_j}\e\plr{\m 1}^{n_j}(\hat a^\dagger_j)^{n_j}\hat a^\dagger_i\:\:\vp} \tl{ft-02}$$$$ i.e. the operators $$\:\hat a^\dagger_i\:$$ and $$\:(\hat a^\dagger_j)^{n_j}\:$$ commute (anticommute) if $$\:n_j\:$$ is even (odd).