= Solution
For $i<j$, interchange the entries in slots $i,j$ while preserving the order of all intermediate <fermionic creation operators>. Move $a_j^\dagger$ left through $j-i$ factors, then move $a_i^\dagger$ right through the remaining $j-i-1$ intermediate factors. The <canonical anticommutation relations> give the overall <fermionic sign>
$$
(-1)^{(j-i)+(j-i-1)}=(-1)^{2(j-i)-1}=-1.
$$
Hence
$$
\boxed{|1,\ldots,j,\ldots,i,\ldots,n\rangle=-|1,\ldots,i,\ldots,j,\ldots,n\rangle.}
$$
More generally any <permutation> $\pi$ gives
$$
\boxed{|\pi(1),\ldots,\pi(n)\rangle=\operatorname{sgn}(\pi)|1,\ldots,n\rangle.}
$$
The <sign of a permutation> is independent of its decomposition into adjacent interchanges. Thus the $n$-<particle> sector is an <exterior power> of the one-<particle> state space, the defining antisymmetry of <fermionic Fock space>. Repeated complete mode labels make the state zero, so they satisfy the same identity trivially.
Back to article page
