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

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

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 $(-1)^{\sum_{k<n}N_k}$ 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?

  • 1
    $\begingroup$ Have you tried a simple example? Have you checked a (nother) book? Have you checked some internet sources, e.g. Wikipedia? $\endgroup$ Commented Jan 22, 2023 at 7:00

1 Answer 1


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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: \begin{equation} \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} \end{equation}

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 \begin{equation} \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} \end{equation} i.e. the operators $\:\hat a^\dagger_i\:$ and $\:(\hat a^\dagger_j)^{n_j}\:$ commute (anticommute) if $\:n_j\:$ is even (odd).


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