Normal order
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia article "Normal_order".

For other uses, see Normal order (disambiguation).

In quantum field theory a product of creation and annihilation operators is in normal order (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.

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.

content

1 Notation
2 Bosons
- 2.1 Single bosons
  - 2.1.1 Examples
- 2.2 Multiple bosons
  - 2.2.1 Examples
3 Fermions
- 3.1 Single fermions
  - 3.1.1 Examples
- 3.2 Multiple fermions
  - 3.2.1 Examples
4 Uses in quantum field theory
- 4.1 Free fields
- 4.2 Wick's theorem
5 References

Notation

If $\hat{O}$ denotes an arbitrary operator, then the normal ordered form of $\hat{O}$ is denoted by $\mathcal{N}(\hat{O})$ .

An alternative notation involves placing the operator inside two colons denoted by $: \hat{O} :$

Bosons

Bosons satisfy Bose-Einstein statistics.

Single bosons

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

$\hat{b}^\dagger$ : the boson's creation operator.
$\hat{b}$ : the boson's annihilation operator.

These satisfy the commutator relationship

$\left[\hat{b}^\dagger, \hat{b}^\dagger \right]_- = 0$

$\left[\hat{b}, \hat{b} \right]_- = 0$

$\left[\hat{b}, \hat{b}^\dagger \right]_- = 1$

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

Examples

1. We'll consider the simplest case first. This is the normal ordering of $\hat{b}^\dagger \hat{b}$ :

$: \hat{b}^\dagger \, \hat{b} : \,= \hat{b}^\dagger \, \hat{b}.$

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

2. A more interesting example is the normal ordering of $\hat{b} \, \hat{b}^\dagger$ :

$: \hat{b} \, \hat{b}^\dagger : \,= \hat{b}^\dagger \, \hat{b}.$

Here the normal ordering operation has reordered the terms by placing $\hat{b}^\dagger$ to the left of $\hat{b}$ .

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

$\hat{b} \, \hat{b}^\dagger = \hat{b}^\dagger \, \hat{b} + 1 =\, : \hat{b} \, \hat{b}^\dagger : + 1.$

$\hat{b} \, \hat{b}^\dagger - : \hat{b} \, \hat{b}^\dagger : \,= 1.$

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

3. An example with multiple operators is:

$: \hat{b}^\dagger \, \hat{b} \, \hat{b} \, \hat{b}^\dagger \, \hat{b} \, \hat{b}^\dagger \, \hat{b}:\, = \hat{b}^\dagger \, \hat{b}^\dagger \, \hat{b}^\dagger \, \hat{b} \, \hat{b} \, \hat{b} \, \hat{b} \, \hat{b} = (\hat{b}^\dagger)^3 \, \hat{b}^5.$

4. A more complicated example shows how we can normal order functions of operators by expanding them out in a series and normal ordering each term:

$: \exp (\lambda \hat{a}^\dagger \hat{a}) : \,= \sum^\infty_{n=0} \frac{\lambda^n}{n!} \hat{a}^{\dagger n} \hat{a}^n$

Multiple bosons

If we now consider $N$ different bosons there are $2 N$ operators:

$\hat{b}_i^\dagger$ : the $i t h$ boson's creation operator.
$\hat{b}_i$ : the $i t h$ boson's annihilation operator.

Here $i = 1,\ldots,N$ .

These satisfy the commutation relations:

$\left[\hat{b}_i^\dagger, \hat{b}_j^\dagger \right]_- = 0$

$\left[\hat{b}_i, \hat{b}_j \right]_- = 0$

$\left[\hat{b}_i, \hat{b}_j^\dagger \right]_- = \delta_{ij}$

where $i,j = 1,\ldots,N$ and $δ i j$ denotes the Kronecker delta.

These may be rewritten as:

$\hat{b}_i^\dagger \, \hat{b}_j^\dagger = \hat{b}_j^\dagger \, \hat{b}_i^\dagger$

$\hat{b}_i \, \hat{b}_j = \hat{b}_j \, \hat{b}_i$

$\hat{b}_i \,\hat{b}_j^\dagger = \hat{b}_j^\dagger \,\hat{b}_i + \delta_{ij}.$

Examples

1. For two different bosons ( $N = 2$ ) we have

$: \hat{b}_1^\dagger \,\hat{b}_2 : \,= \hat{b}_1^\dagger \,\hat{b}_2$

$: \hat{b}_2 \, \hat{b}_1^\dagger : \,= \hat{b}_1^\dagger \,\hat{b}_2$

2. For three different bosons ( $N = 3$ ) we have

$: \hat{b}_1^\dagger \,\hat{b}_2 \,\hat{b}_3 : \,= \hat{b}_1^\dagger \,\hat{b}_2 \,\hat{b}_3$

Notice that since (by the commutation relations) $\hat{b}_2 \,\hat{b}_3 = \hat{b}_3 \,\hat{b}_2$ the order in which we write the annihilation operators does not matter.

$: \hat{b}_2 \, \hat{b}_1^\dagger \, \hat{b}_3 : \,= \hat{b}_1^\dagger \,\hat{b}_2 \, \hat{b}_3$

$: \hat{b}_3 \hat{b}_2 \, \hat{b}_1^\dagger : \,= \hat{b}_1^\dagger \,\hat{b}_2 \, \hat{b}_3$

Fermions

Fermions satisfy Fermi-Dirac statistics.

Single fermions

For a single fermion there are two operators of interest:

$\hat{f}^\dagger$ : the fermion's creation operator.
$\hat{f}$ : the fermion's annihilation operator.

These satisfy the anticommutator relationships

$\left[\hat{f}^\dagger, \hat{f}^\dagger \right]_+ = 0$

$\left[\hat{f}, \hat{f} \right]_+ = 0$

$\left[\hat{f}, \hat{f}^\dagger \right]_+ = 1$

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

$\hat{f}^\dagger\, \hat{f}^\dagger = 0$

$\hat{f} \,\hat{f} = 0$

$\hat{f} \,\hat{f}^\dagger = 1 - \hat{f}^\dagger \,\hat{f} .$

To define the normal ordering of an product of fermionic creation and annihilation operators we must take into account the number of interchanges between neighbouring operators. We get a minus sign for each such interchange.

Examples

1. We again start with the simplest cases:

$: \hat{f}^\dagger \, \hat{f} : \,= \hat{f}^\dagger \, \hat{f}$

This expression is already in normal order so nothing is changed.

$: \hat{f} \, \hat{f}^\dagger : \,= -\hat{f}^\dagger \, \hat{f}$

Here we introduce a minus sign because we have changed the order of two operators.

These can be combined, along with the anticommutation relations, to show

$\hat{f} \, \hat{f}^\dagger \,= 1 - \hat{f}^\dagger \, \hat{f} = 1 + :\hat{f} \,\hat{f}^\dagger :$

$\hat{f} \, \hat{f}^\dagger - : \hat{f} \, \hat{f}^\dagger : = 1.$

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:

$: \hat{f}\,\hat{f}^\dagger \, \hat{f} \hat{f}^\dagger : \,= \hat{f}^\dagger \,\hat{f}^\dagger \,\hat{f}\,\hat{f} = 0$

Multiple fermions

For $N$ different fermions there are $2 N$ operators:

$\hat{f}_i^\dagger$ : the $i t h$ fermion's creation operator.
$\hat{f}_i$ : the $i t h$ fermion's annihilation operator.

Here $i = 1,\ldots,N$ .

These satisfy the commutation relations:

$\left[\hat{f}_i^\dagger, \hat{f}_j^\dagger \right]_+ = 0$

$\left[\hat{f}_i, \hat{f}_j \right]_+ = 0$

$\left[\hat{f}_i, \hat{f}_j^\dagger \right]_+ = \delta_{ij}$

where $i,j = 1,\ldots,N$ and $δ i j$ denotes the Kronecker delta.

These may be rewritten as:

$\hat{f}_i^\dagger \, \hat{f}_j^\dagger = -\hat{f}_j^\dagger \, \hat{f}_i^\dagger$

$\hat{f}_i \, \hat{f}_j = -\hat{f}_j \, \hat{f}_i$

$\hat{f}_i \,\hat{f}_j^\dagger = \delta_{ij} - \hat{f}_j^\dagger \,\hat{f}_i .$

When calculating the normal order of products of fermion operators we must take into account the number of interchanges 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 ( $N = 2$ ) we have

$: \hat{f}_1^\dagger \,\hat{f}_2 : \,= \hat{f}_1^\dagger \,\hat{f}_2$

Here the expression is already normal ordered so nothing changes.

$: \hat{f}_2 \, \hat{f}_1^\dagger : \,= -\hat{f}_1^\dagger \,\hat{f}_2$

Here we introduce a minus sign because we have interchanged the order of two operators.

$: \hat{f}_2 \, \hat{f}_1^\dagger \, \hat{f}^\dagger_2 : \,= \hat{f}_1^\dagger \, \hat{f}_2^\dagger \,\hat{f}_2 = -\hat{f}_2^\dagger \, \hat{f}_1^\dagger \,\hat{f}_2$

Note that the order in which we write the operators here, unlike in the bosonic case, does matter.

2. For three different fermions ( $N = 3$ ) we have

$: \hat{f}_1^\dagger \, \hat{f}_2 \, \hat{f}_3 : \,= \hat{f}_1^\dagger \,\hat{f}_2 \,\hat{f}_3 = -\hat{f}_1^\dagger \,\hat{f}_3 \,\hat{f}_2$

Notice that since (by the anticommutation relations) $\hat{f}_2 \,\hat{f}_3 = -\hat{f}_3 \,\hat{f}_2$ the order in which we write the operators does matter in this case.

Similarly we have

$: \hat{f}_2 \, \hat{f}_1^\dagger \, \hat{f}_3 : \,= -\hat{f}_1^\dagger \,\hat{f}_2 \, \hat{f}_3 = \hat{f}_1^\dagger \,\hat{f}_3 \, \hat{f}_2$

$: \hat{f}_3 \hat{f}_2 \, \hat{f}_1^\dagger : \,= \hat{f}_1^\dagger \,\hat{f}_3 \, \hat{f}_2 = -\hat{f}_1^\dagger \,\hat{f}_2 \, \hat{f}_3$

Uses in quantum field theory

The vacuum expectation value of a normal ordered product of creation and annihilation operators is zero. This is because, denoting the vacuum state by $|0\rangle$ , the creation and annihilation operators satisfy

$\langle 0 | \hat{a}^\dagger = 0 \qquad \textrm{and} \qquad \hat{a} |0\rangle = 0$

(here $\hat{a}^\dagger$ and $\hat{a}$ are creation and annihilation operators (either bosonic or fermionic)).

Any normal ordered operator therefore has a vacuum expectation value of zero. Although an operator $\hat{O}$ may satisfy

$\langle 0 | \hat{O} | 0 \rangle \neq 0$

we always have

$\langle 0 | :\hat{O}: | 0 \rangle = 0$

This is particularly useful when defining a quantum mechanical Hamiltonian. If the Hamiltonian of a theory is in normal order then the ground state energy will be zero: $\langle 0 |\hat{H}|0\rangle = 0$ .

Free fields

With two free fields φ and χ,

$:\phi(x)\chi(y):=\phi(x)\chi(y)-\langle\Omega|\phi(x)\chi(y)|\Omega\rangle$

where | $Ω$ > is the vacuum state. Each of the two terms on the right hand side typically blows up in the limit as y approaches x but the difference between them has a well-defined limit. This allows us to define :φ(x)χ(x):.

Wick's theorem

Main article: Wick's theorem

Wick's theorem states that:

$\phi_{i_1}(x_1)\cdots \phi_{i_N}(x_N)=\sum_\textrm{all\ possible\ pairs\ of\ contractions}:\phi_{i_1}(x_1)\cdots \phi_{i_N}(x_N):$

(with contractions).

This theorem provides a simple method for computing vacuum expectation values of operators. This theorem was the reason normal ordering was defined in the first place.

References

F. Mandl, G. Shaw, Quantum Field Theory, John Wiley & Sons, 1984.
S. Weinberg, The Quantum Theory of Fields (Volume I) Cambridge University Press (1995)

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Contents

Notation

Bosons

Single bosons

Examples

Multiple bosons

Examples

Fermions

Single fermions

Examples

Multiple fermions

Examples

Uses in quantum field theory

Free fields

Wick's theorem

References