Editing Normal order (section)

====Examples====
1. We again start with the simplest cases:
:<math> : \hat{f}^\dagger \, \hat{f} : \,= \hat{f}^\dagger \, \hat{f} </math>
This expression is already in normal order so nothing is changed. In the reverse case, we introduce a minus sign because we have to change the order of two operators:

:<math> : \hat{f} \, \hat{f}^\dagger : \,= -\hat{f}^\dagger \, \hat{f} </math>

These can be combined, along with the anticommutation relations, to show
:<math> \hat{f} \, \hat{f}^\dagger \,= 1 - \hat{f}^\dagger \, \hat{f} = 1 + :\hat{f} \,\hat{f}^\dagger :</math>
or
:<math> \hat{f} \, \hat{f}^\dagger -  : \hat{f} \, \hat{f}^\dagger : = 1.</math>

This equation, which is in the same form as the bosonic case above, is used in defining the contractions used in [[Wick's theorem]].

2. The normal order of any more complicated cases gives zero because there will be at least one creation or annihilation operator appearing twice. For example:
:<math> : \hat{f}\,\hat{f}^\dagger \, \hat{f} \hat{f}^\dagger  : \,= -\hat{f}^\dagger \,\hat{f}^\dagger \,\hat{f}\,\hat{f} = 0 </math>