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In second quantization, we assume we have fermion operators $a_i$ which satisfy $\{a_i,a_j\}=0$, $\{a_i,a_j^\dagger\}=\delta_{ij}$, $\{a_i^\dagger,a_j^\dagger\}=0$. Another way to say this is that

$$ a_i^\dagger|n_1,...,n_i,...,n_N\rangle = \left\{ \begin{array}{lr} (-1)^{\sum_{j<i} n_j}|n_1,...,n_i+1,...,n_N\rangle & n_i=0\\ 0 &n_i=1 \end{array}\right| $$

$$ a_i|n_1,...,n_i,...,n_N\rangle = \left\{ \begin{array}{lr} (-1)^{\sum_{j<i} n_j}|n_1,...,n_i-1,...,n_N\rangle & n_i=1\\ 0 &n_i=0 \end{array}\right| $$ from which you can derive the relations above.

I understand why the operators on the same sites have to obey the anticommutation relations, since otherwise Pauli exclusion would be violated. I'm not sure I understand why the operators on different sites have to anticommute, however.

Why can't we have an algebra of fermionic operators obeying anticommutation relations for $i=j$, and otherwise obeying the relations $[a_i^{(\dagger)},a_j^{(\dagger)}]=0$? We could define the operators by

$$ a_i^\dagger|n_1,...,n_i,...,n_N\rangle = \left\{ \begin{array}{lr} |n_1,...,n_i+1,...,n_N\rangle & n_i=0\\ 0 &n_i=1 \end{array}\right| $$

$$ a_i|n_1,...,n_i,...,n_N\rangle = \left\{ \begin{array}{lr} |n_1,...,n_i-1,...,n_N\rangle & n_i=1\\ 0 &n_i=0 \end{array}\right| $$ without the sign in front of the ket, from which you can derive the new commutation/anticommutation relations. Is this somehow illegal? Are the operators I've defined not actually well-defined? Is there some way to use the definition I gave to get a contradiction? Or do we just assume the fermion operators anticommute for notational convenience?

So far all the books/pdfs I've looked at prove the anticommutation relations hold for fermion operators on the same site, and then assume anticommutation relations hold on different sites.

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    $\begingroup$ I think operationally, this looks like a Jordan-Wigner transformation operator, just without the "string." So I guess this could be related to the question: what goes wrong if we forget the string in a Jordan-Wigner transformation. $\endgroup$ – Jahan Claes Jun 17 '16 at 21:02
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On the mere level of "second quantization" there is nothing wrong with fermionic operators commuting with other fermionic operators. They don't "know" that they are operators for "the same fermion" on different sites, so they could as well commute.

But the deeper reason that fermionic operators on different sites anticommute is that they are just modes of the same fermionic field in the underlying QFT, and the modes of a spinor field anticommute because the fields themselves anticommute, and this relation is inherited by their modes.

The essentially same argument in another phrasing says that fermionic states must be antisymmetric under exchange of identical fermions. This is a postulate of QM/"second quantization" and becomes a derived statement only in QFT as the spin-statistics theorem. So you must have that swapping $i\leftrightarrow j$ incurs a minus on the state that has one fermionic exictation at $i$ and another at $j$ - and this precisely corresponds to $a^\dagger_i$ and $a^\dagger_j$ anticommuting.

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It is equivalent to ask the operators on different sites to commute or anticommute. Namely, there is always a so-called Klein transformation changing the commutation between different sites. If they anticommute one says they have natural commutation relations.

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