Editing Normal order (section)

===Single bosons===
If we start with only one type of boson there are two operators of interest:

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

These satisfy the [[commutator]] relationship

:<math>\left[\hat{b}^\dagger, \hat{b}^\dagger \right]_- = 0</math>
:<math>\left[\hat{b}, \hat{b} \right]_- = 0</math>
:<math>\left[\hat{b}, \hat{b}^\dagger \right]_- = 1</math>

where <math>\left[ A, B \right]_- \equiv AB - BA</math> denotes the [[commutator]]. We may rewrite the last one as: <math>\hat{b}\, \hat{b}^\dagger = \hat{b}^\dagger\, \hat{b} + 1.</math>

====Examples====
1. We'll consider the simplest case first. This is the normal ordering of <math>\hat{b}^\dagger \hat{b}</math>:

:<math> {:\,}\hat{b}^\dagger \, \hat{b}{\,:} = \hat{b}^\dagger \, \hat{b}. </math>

The expression <math>\hat{b}^\dagger \, \hat{b}</math> has not been changed because it is ''already'' in normal order - the creation operator <math>(\hat{b}^\dagger)</math> is already to the left of the annihilation operator <math>(\hat{b})</math>.

2. A more interesting example is the normal ordering of <math>\hat{b} \, \hat{b}^\dagger </math>: 
:<math> {:\,}\hat{b} \, \hat{b}^\dagger{\,:} = \hat{b}^\dagger \, \hat{b}. </math>
Here the normal ordering operation has ''reordered'' the terms by placing <math>\hat{b}^\dagger</math> to the left of <math>\hat{b}</math>.

These two results can be combined with the commutation relation obeyed by <math>\hat{b}</math> and <math>\hat{b}^\dagger</math> to get

:<math> \hat{b} \, \hat{b}^\dagger = \hat{b}^\dagger \, \hat{b} + 1 = {:\,}\hat{b} \, \hat{b}^\dagger{\,:} \; + 1.</math>
or
:<math> \hat{b} \, \hat{b}^\dagger -  {:\,}\hat{b} \, \hat{b}^\dagger{\,:} = 1.</math>

This equation is used in defining the contractions used in [[Wick's theorem]].

3. An example with multiple operators is:

:<math> {:\,}\hat{b}^\dagger \, \hat{b} \, \hat{b} \, \hat{b}^\dagger \, \hat{b} \, \hat{b}^\dagger \, \hat{b}{\,:} = \hat{b}^\dagger \, \hat{b}^\dagger \, \hat{b}^\dagger \, \hat{b} \, \hat{b} \, \hat{b} \, \hat{b} = (\hat{b}^\dagger)^3 \, \hat{b}^4.</math>

4. A simple example shows that normal ordering cannot be extended by linearity from the monomials to all operators in a self-consistent way. Assume that we can apply the commutation relations to obtain:

:<math> {:\,}\hat{b} \hat{b}^\dagger{\,:} = {:\,}1 + \hat{b}^\dagger \hat{b}{\,:}.</math>

Then, by linearity,

: <math>{:\,}1 + \hat{b}^\dagger \hat{b}{\,:} = {:\,}1{\,:} + {:\,}\hat{b}^\dagger \hat{b}{\,:} = 
1 + \hat{b}^\dagger \hat{b} \ne \hat{b}^\dagger \hat{b}={:\,}\hat{b} \hat{b}^\dagger{\,:},</math>

a contradiction.

The implication is that normal ordering is not a linear function on operators, but on the [[free algebra]] generated by the operators, i.e. the operators do not satisfy the [[canonical commutation relations]] while inside the normal ordering (or any other ordering operator like [[time-ordering]], etc).