Editing Normal order (section)

==Fermions==

[[Fermions]] are particles which satisfy [[Fermi–Dirac statistics]]. We will now examine the normal ordering of fermionic creation and annihilation operator products.

===Single fermions===
For a single fermion there are two operators of interest:

* <math>\hat{f}^\dagger</math>: the fermion's creation operator.
* <math>\hat{f}</math>: the fermion's annihilation operator.

These satisfy the [[anticommutator]] relationships
:<math>\left[\hat{f}^\dagger, \hat{f}^\dagger \right]_+ = 0</math>
:<math>\left[\hat{f}, \hat{f} \right]_+ = 0</math>
:<math>\left[\hat{f}, \hat{f}^\dagger \right]_+ = 1</math>

where <math>\left[A, B \right]_+ \equiv AB + BA</math> denotes the [[anticommutator]]. These may be rewritten as

:<math>\hat{f}^\dagger\, \hat{f}^\dagger =  0 </math>
:<math>\hat{f} \,\hat{f}  =  0 </math>
:<math>\hat{f} \,\hat{f}^\dagger =  1 - \hat{f}^\dagger \,\hat{f} .</math>

To define the normal ordering of a product of fermionic creation and annihilation operators we must take into account the number of [[Transposition (mathematics)|interchange]]s between neighbouring operators. We get a minus sign for each such interchange.

====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>

===Multiple fermions===
For <math>N</math> different fermions there are <math>2 N</math> operators:
* <math>\hat{f}_i^\dagger</math>: the <math>i^{th}</math> fermion's creation operator.
* <math>\hat{f}_i</math>: the <math>i^{th}</math> fermion's annihilation operator.
Here <math>i = 1,\ldots,N</math>.

These satisfy the anti-commutation relations:
:<math>\left[\hat{f}_i^\dagger, \hat{f}_j^\dagger \right]_+ = 0 </math>
:<math>\left[\hat{f}_i, \hat{f}_j \right]_+ = 0 </math>
:<math>\left[\hat{f}_i, \hat{f}_j^\dagger \right]_+ = \delta_{ij} </math>
where <math>i,j = 1,\ldots,N</math> and <math>\delta_{ij}</math> denotes the [[Kronecker delta]].

These may be rewritten as:
:<math>\hat{f}_i^\dagger \, \hat{f}_j^\dagger = -\hat{f}_j^\dagger \, \hat{f}_i^\dagger </math>
:<math>\hat{f}_i \, \hat{f}_j = -\hat{f}_j \, \hat{f}_i </math>
:<math>\hat{f}_i \,\hat{f}_j^\dagger = \delta_{ij} - \hat{f}_j^\dagger \,\hat{f}_i .</math>

When calculating the normal order of products of fermion operators we must take into account the number of [[Transposition (mathematics)|interchange]]s of neighbouring operators required to rearrange the expression. It is as if we pretend the creation and annihilation operators anticommute and then we reorder the expression to ensure the creation operators are on the left and the annihilation operators are on the right - all the time taking account of the anticommutation relations.

====Examples====
1. For two different fermions (<math>N=2</math>) we have
:<math> : \hat{f}_1^\dagger \,\hat{f}_2 : \,= \hat{f}_1^\dagger \,\hat{f}_2 </math>
Here the expression is already normal ordered so nothing changes.

:<math> : \hat{f}_2 \, \hat{f}_1^\dagger  : \,= -\hat{f}_1^\dagger \,\hat{f}_2 </math>
Here we introduce a minus sign because we have interchanged the order of two operators.

:<math> : \hat{f}_2 \, \hat{f}_1^\dagger \, \hat{f}^\dagger_2  : \,= \hat{f}_1^\dagger \, \hat{f}_2^\dagger \,\hat{f}_2 = -\hat{f}_2^\dagger \, \hat{f}_1^\dagger \,\hat{f}_2  </math>
Note that the order in which we write the operators here, unlike in the bosonic case, ''does matter''.

2. For three different fermions (<math>N=3</math>) we have
:<math> : \hat{f}_1^\dagger \, \hat{f}_2 \, \hat{f}_3 : \,= \hat{f}_1^\dagger \,\hat{f}_2 \,\hat{f}_3 = -\hat{f}_1^\dagger \,\hat{f}_3 \,\hat{f}_2</math>
Notice that since (by the anticommutation relations) <math>\hat{f}_2 \,\hat{f}_3 = -\hat{f}_3 \,\hat{f}_2</math> the order in which we write the operators ''does matter'' in this case.

Similarly we have
:<math> : \hat{f}_2 \, \hat{f}_1^\dagger \, \hat{f}_3  : \,= -\hat{f}_1^\dagger \,\hat{f}_2 \, \hat{f}_3 = \hat{f}_1^\dagger \,\hat{f}_3 \, \hat{f}_2</math>
:<math> : \hat{f}_3 \hat{f}_2 \, \hat{f}_1^\dagger  : \,= \hat{f}_1^\dagger \,\hat{f}_3 \, \hat{f}_2 = -\hat{f}_1^\dagger \,\hat{f}_2 \, \hat{f}_3 </math>