Editing Normal order (section)

==Bosons==

[[Bosons]] are particles which satisfy [[Bose–Einstein statistics]]. We will now examine the normal ordering of bosonic creation and annihilation operator products.

===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).

===Multiple bosons===
If we now consider <math>N</math> different bosons there are <math>2 N</math> operators:
* <math>\hat{b}_i^\dagger</math>: the <math>i^{th}</math> boson's creation operator.
* <math>\hat{b}_i</math>: the <math>i^{th}</math> boson's annihilation operator.
Here <math>i = 1,\ldots,N</math>.

These satisfy the commutation relations:
:<math>\left[\hat{b}_i^\dagger, \hat{b}_j^\dagger \right]_- = 0 </math>
:<math>\left[\hat{b}_i, \hat{b}_j \right]_- = 0 </math>
:<math>\left[\hat{b}_i, \hat{b}_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{b}_i^\dagger \, \hat{b}_j^\dagger = \hat{b}_j^\dagger \, \hat{b}_i^\dagger </math>
:<math>\hat{b}_i \, \hat{b}_j = \hat{b}_j \, \hat{b}_i </math>
:<math>\hat{b}_i \,\hat{b}_j^\dagger = \hat{b}_j^\dagger \,\hat{b}_i + \delta_{ij}.</math>

====Examples====
1. For two different bosons (<math>N=2</math>) we have
:<math> : \hat{b}_1^\dagger \,\hat{b}_2 : \,= \hat{b}_1^\dagger \,\hat{b}_2 </math>
:<math> : \hat{b}_2 \, \hat{b}_1^\dagger  : \,= \hat{b}_1^\dagger \,\hat{b}_2 </math>

2. For three different bosons (<math>N=3</math>) we have
:<math> : \hat{b}_1^\dagger \,\hat{b}_2 \,\hat{b}_3 : \,= \hat{b}_1^\dagger \,\hat{b}_2 \,\hat{b}_3</math>
Notice that since (by the commutation relations) <math>\hat{b}_2 \,\hat{b}_3 = \hat{b}_3 \,\hat{b}_2</math> the order in which we write the annihilation operators does not matter.

:<math> : \hat{b}_2 \, \hat{b}_1^\dagger \, \hat{b}_3  : \,= \hat{b}_1^\dagger \,\hat{b}_2 \, \hat{b}_3 </math>
:<math> : \hat{b}_3 \hat{b}_2 \, \hat{b}_1^\dagger  : \,= \hat{b}_1^\dagger \,\hat{b}_2 \, \hat{b}_3 </math>

===Bosonic operator functions===
Normal ordering of bosonic operator functions <math>f(\hat n)</math>, with occupation number operator <math>\hat n=\hat b\vphantom{\hat n}^\dagger \hat b</math>, can be accomplished using [[Factorial power|(falling) factorial powers]] <math>\hat n^{\underline{k}}=\hat n(\hat n-1)\cdots(\hat n-k+1)</math> and [[Newton series]] instead of [[Taylor series]]: 
It is easy to show 
<ref name="Hucht">{{cite journal | last1=König | first1=Jürgen | last2=Hucht | first2=Alfred | title=Newton series expansion of bosonic operator functions | journal=SciPost Physics | publisher=Stichting SciPost | volume=10 | issue=1 | date=2021-01-13 | issn=2542-4653 | doi=10.21468/scipostphys.10.1.007 | page=007| arxiv=2008.11139 | bibcode=2021ScPP...10....7K | s2cid=221293056 | doi-access=free }}</ref> 
that factorial powers <math>\hat n^{\underline{k}}</math> are equal to normal-ordered (raw) [[Exponentiation|powers]] <math>\hat n^{k}</math> and are therefore normal ordered by construction,

: <math>
\hat{n}^{\underline{k}}
 = \hat b\vphantom{\hat n}^{\dagger k} \hat b\vphantom{\hat n}^k
 = {:\,}\hat n^k{\,:},
</math>

such that the Newton series expansion 

: <math>
\tilde f(\hat n) = \sum_{k=0}^\infty \Delta_n^k \tilde f(0) \, \frac{\hat n^{\underline{k}}}{k!}
</math>

of an operator function <math>\tilde f(\hat n)</math>, with <math>k</math>-th [[forward difference]] <math>\Delta_n^k \tilde f(0)</math> at <math>n=0</math>, is always normal ordered. Here, the [[Second_quantization#Action_on_Fock_states|eigenvalue equation]] <math>\hat n |n\rangle = n |n\rangle</math> relates <math>\hat n</math> and <math>n</math>.

As a consequence, the normal-ordered Taylor series of an arbitrary function <math>f(\hat n)</math> is equal to the Newton series of an associated function <math>\tilde f(\hat n)</math>, fulfilling

: <math> 
\tilde f(\hat n) = {:\,} f(\hat n) {\,:},
</math>

if the series coefficients of the Taylor series of <math>f(x)</math>, with continuous <math>x</math>, match the coefficients of the Newton series of <math>\tilde f(n)</math>, with integer <math>n</math>,

: <math>
\begin{align}
       f(x) &= \sum_{k=0}^\infty F_k \, \frac{x^k              }{k!}, \\
\tilde f(n) &= \sum_{k=0}^\infty F_k \, \frac{n^{\underline{k}}}{k!}, \\
F_k &= \partial_x^k f(0) = \Delta_n^k \tilde f(0),
\end{align}
</math>

with <math>k</math>-th [[partial derivative]] <math>\partial_x^k f(0)</math> at <math>x=0</math>.
The functions <math>f</math> and <math>\tilde f</math> are related through the so-called '''[[normal-order transform]]''' <math>\mathcal N[f]</math> according to

: <math>
\begin{align}
\tilde f(n) &= \mathcal N_x[f(x)](n) \\
 &= \frac{1}{\Gamma(-n)} \int_{-\infty}^0 \mathrm d x \, e^x \, f(x) \, (-x)^{-(n+1)} \\
 &= \frac{1}{\Gamma(-n)}\mathcal M_{-x}[e^{x} f(x)](-n),
\end{align}
</math>

which can be expressed in terms of the [[Mellin transform]] <math>\mathcal M</math>, see <ref name="Hucht"/> for details.