In quantum field theory a product of quantum fields, or equivalently their creation and annihilation operators, is usually said to be normal ordered (also called Wick order) when all creation operators are to the left of all annihilation operators in the product. The process of putting a product into normal order is called normal ordering (also called Wick ordering). The terms antinormal order and antinormal ordering are analogously defined, where the annihilation operators are placed to the left of the creation operators.
Normal ordering of a product of quantum fields or creation and annihilation operators can also be defined in many other ways. Which definition is most appropriate depends on the expectation values needed for a given calculation. Most of this article uses the most common definition of normal ordering as given above, which is appropriate when taking expectation values using the vacuum state of the creation and annihilation operators.
The process of normal ordering is particularly important for a quantum mechanical Hamiltonian. When quantizing a classical Hamiltonian there is some freedom when choosing the operator order, and these choices lead to differences in the ground state energy. That's why the process can also be used to eliminate the infinite vacuum energy of a quantum field.
Notation
If <math>\hat{O}</math> denotes an arbitrary product of creation and/or annihilation operators (or equivalently, quantum fields), then the normal ordered form of <math>\hat{O}</math> is denoted by <math>\mathopen{:} \hat{O} \mathclose{:}</math>.
An alternative notation is <math> \mathcal{N}(\hat{O})</math>.
Note that normal ordering is a concept that only makes sense for products of operators. Attempting to apply normal ordering to a sum of operators is not useful as normal ordering is not a linear operation.
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 (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
that factorial powers <math>\hat n^{\underline{k</math> are equal to normal-ordered (raw) 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 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