# Second Quantization: Do fermion operators on different sites HAVE to anticommute?

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.

• 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. – Jahan Claes Jun 17 '16 at 21:02

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.