Normal order

In quantum field theory , the normal order (also Wick order or normal product ) describes the state in which all creation operators are to the left of the annihilation operators . The anti-normal order is defined analogously if the annihilation operators are to the left of the creation operators.

notation

The notation denotes the normal order of , where is an arbitrarily arranged product of creation and annihilation operators (or quantum fields). Alternatively, the notation is also used. ${\displaystyle {\mathopen {:}}{\hat {O}}{\mathclose {:}}}$ ${\displaystyle {\hat {O}}}$ ${\displaystyle {\hat {O}}}$ ${\displaystyle {\mathcal {N}}({\hat {O}})}$

Bosons

When using the bosonic creation and annihilation operators, the following notation is used:

• ${\displaystyle {\hat {b}}^{\dagger }}$: Creation operator.
• ${\displaystyle {\hat {b}}}$: Annihilation operator.

These fulfill the typical commutator relations for bosons.

Examples

1. The simplest example is : ${\displaystyle {\mathopen {:}}{\hat {b}}^{\dagger }{\hat {b}}{\mathopen {:}}}$

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

Here is not changed, because the expression is already present in the normal order. The creation operator is already to the left of the annihilation operator . ${\displaystyle {\hat {b}}^{\dagger }{\hat {b}}}$ ${\displaystyle {\hat {b}}^{\dagger }}$ ${\displaystyle {\hat {b}}}$

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

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

Here, the normal order operation rearranges the expression so that it appears to the left of . ${\displaystyle {\hat {b}}^{\dagger }}$ ${\displaystyle {\hat {b}}}$

These two results can be used together with the commutator relations mentioned above

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

or

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

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

3. If several operators are involved, the result is:

${\displaystyle {:}{\hat {b}}^{\dagger }{\hat {b}}{\hat {b}}{\hat {b}}^{\dagger }{\hat {b}}{\hat {b}}^{\dagger }{\hat {b}}{:}={\hat {b}}^{\dagger }{\hat {b}}^{\dagger }{\hat {b}}^{\dagger }{\hat {b}}{\hat {b}}{\hat {b}}{\hat {b}}=({\hat {b}}^{\dagger })^{3}{\hat {b}}^{4}.}$

4. A simple example shows that normal order is not a linear operation :

${\displaystyle {\hat {b}}^{\dagger }{\hat {b}}={:}{\hat {b}}{\hat {b}}^{\dagger }{:}={:}1+{\hat {b}}^{\dagger }{\hat {b}}{:}\neq {:}1{:}+{:}{\hat {b}}^{\dagger }{\hat {b}}{:}=1+{\hat {b}}^{\dagger }{\hat {b}}}$

5. If several bosons are involved, we get:

1. ${\displaystyle {:}{\hat {b}}_{1}^{\dagger }{\hat {b}}_{2}{:}={\hat {b}}_{1}^{\dagger }{\hat {b}}_{2}}$
2. ${\displaystyle {:}{\hat {b}}_{2}{\hat {b}}_{1}^{\dagger }{:}={\hat {b}}_{1}^{\dagger }{\hat {b}}_{2}}$
3. ${\displaystyle {:}{\hat {b}}_{1}^{\dagger }{\hat {b}}_{2}{\hat {b}}_{3}{:}={\hat {b}}_{1}^{\dagger }{\hat {b}}_{2}{\hat {b}}_{3}}$
4. ${\displaystyle {:}{\hat {b}}_{2}{\hat {b}}_{1}^{\dagger }{\hat {b}}_{3}{:}={\hat {b}}_{1}^{\dagger }{\hat {b}}_{2}{\hat {b}}_{3}}$
5. ${\displaystyle {:}{\hat {b}}_{3}{\hat {b}}_{2}{\hat {b}}_{1}^{\dagger }{:}={\hat {b}}_{1}^{\dagger }{\hat {b}}_{2}{\hat {b}}_{3}}$

Fermions

Single fermions

When using fermions , the operators become

• ${\displaystyle {\hat {f}}^{\dagger }}$: Fermionic creation operator,
• ${\displaystyle {\hat {f}}}$: Fermionic annihilation operator

used. These fulfill the typical anti-commutator relations for fermions:

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

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

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

where defines the anti- commutator. These can be rewritten to ${\displaystyle \left[A,B\right]_{+}\equiv AB+BA}$

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

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

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

In order to define the normal order for fermions, the number of exchanges must be taken into account, since a minus sign appears for each exchange.

Examples

1. At the beginning again the simplest case:

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

A normal order is already in place. Conversely, however, a minus sign is introduced due to the exchange of both operators:

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

This can be used together with the anti-commutator relations to

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

or

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

to show. This equation is also used in Wick's theorem to introduce the contraction.

2. The normal order of any complicated case is zero, since at least one creation or annihilation operator occurs twice. For example:

${\displaystyle {:}{\hat {f}}{\hat {f}}^{\dagger }{\hat {f}}{\hat {f}}^{\dagger }{:}=-{\hat {f}}^{\dagger }{\hat {f}}^{\dagger }{\hat {f}}{\hat {f}}=0}$

Multiple fermions

When using different fermions there are operators: ${\displaystyle N}$ ${\displaystyle 2N}$

• ${\displaystyle {\hat {f}}_{i}^{\dagger }}$: the creation operator of the -th fermion. ${\displaystyle i}$
• ${\displaystyle {\hat {f}}_{i}}$: the annihilation operator of the -th fermion. ${\displaystyle i}$

Where . ${\displaystyle i=1,\ldots ,N}$

These fulfill the commutator relations:

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

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

${\displaystyle \left[{\hat {f}}_{i},{\hat {f}}_{j}^{\dagger }\right]_{+}=\delta _{ij}}$

where and denotes the Kronecker delta . ${\displaystyle i,j=1,\ldots ,N}$ ${\displaystyle \delta _{ij}}$

This can be rewritten to:

${\displaystyle {\hat {f}}_{i}^{\dagger }{\hat {f}}_{j}^{\dagger }=-{\hat {f}}_{j}^{\dagger }{\hat {f}}_{i}^{\dagger }}$

${\displaystyle {\hat {f}}_{i}{\hat {f}}_{j}=-{\hat {f}}_{j}{\hat {f}}_{i}}$

${\displaystyle {\hat {f}}_{i}{\hat {f}}_{j}^{\dagger }=\delta _{ij}-{\hat {f}}_{j}^{\dagger }{\hat {f}}_{i}.}$

Examples

1. For two different fermions ( ) results ${\displaystyle N=2}$

${\displaystyle {:}{\hat {f}}_{1}^{\dagger }{\hat {f}}_{2}{:}={\hat {f}}_{1}^{\dagger }{\hat {f}}_{2}}$

Since the expression is already in normal order, nothing changes.

${\displaystyle {:}{\hat {f}}_{2}{\hat {f}}_{1}^{\dagger }{:}=-{\hat {f}}_{1}^{\dagger }{\hat {f}}_{2}}$

A minus sign must be inserted here because the order of two operators has been reversed.

${\displaystyle {:}{\hat {f}}_{2}{\hat {f}}_{1}^{\dagger }{\hat {f}}_{2}^{\dagger }{:}={\hat {f}}_{1}^{\dagger }{\hat {f}}_{2}^{\dagger }{\hat {f}}_{2}=-{\hat {f}}_{2}^{\dagger }{\hat {f}}_{1}^{\dagger }{\hat {f}}_{2}}$

In contrast to the boson case, the order in which the operators are written plays a role here.

2. With three different fermions ( ) we get: ${\displaystyle N=3}$

${\displaystyle {:}{\hat {f}}_{1}^{\dagger }{\hat {f}}_{2}{\hat {f}}_{3}{:}={\hat {f}}_{1}^{\dagger }{\hat {f}}_{2}{\hat {f}}_{3}=-{\hat {f}}_{1}^{\dagger }{\hat {f}}_{3}{\hat {f}}_{2}}$

${\displaystyle {:}{\hat {f}}_{2}{\hat {f}}_{1}^{\dagger }{\hat {f}}_{3}{:}=-{\hat {f}}_{1}^{\dagger }{\hat {f}}_{2}{\hat {f}}_{3}={\hat {f}}_{1}^{\dagger }{\hat {f}}_{3}{\hat {f}}_{2}}$

${\displaystyle {:}{\hat {f}}_{3}{\hat {f}}_{2}{\hat {f}}_{1}^{\dagger }{:}={\hat {f}}_{1}^{\dagger }{\hat {f}}_{3}{\hat {f}}_{2}=-{\hat {f}}_{1}^{\dagger }{\hat {f}}_{2}{\hat {f}}_{3}}$

literature

• F. Mandl, G. Shaw, Quantum Field Theory, John Wiley & Sons, 1984.
• Wolfgang Nolting, Basic Course Theoretical Physics 7, Springer Berlin Heidelberg, 2009