# Action of Fermionic Creation and Annihilation Operators

I'm currently trying to understand the action of fermionic creation and annihilation operators on states in the occupation number representation.

Well, I understand why the bosonic creation operator for example acts by

$$a_i^+ |..., n_i, ...\rangle = |..., n_i+1, ...\rangle$$

on a state given in occupation number representation.

But why is the action of the corresponding fermionic creation operator given by

$$a_i^+ |..., n_i, ...\rangle = (1-n_i) (-1)^{\sum_{j<i} n_j}|..., n_i+1, ...\rangle~?$$

How can one explain the factor $(-1)^{\sum_{j<i} n_j}$?

• It's because $\{a, a^\dagger \} = 1$. – DanielSank Apr 22 '17 at 17:42
• You will have to move it through $\sum_{j<i}n_i$ fermionic creation operatorsm and you pick up a sign from the anticommutator each time you pass it through one fermionic operator. – ZeroTheHero Apr 22 '17 at 18:01