Normal order

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

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

These satisfy the commutator relationship

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

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

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

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

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

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

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

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

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

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

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

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

or

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

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

3. An example with multiple operators 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 ordering cannot be extended by linearity from the monomials to all operators in a self-consistent way:

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

The implication is that normal ordering is not a linear function on operators.

If we now consider ${\displaystyle N}$ different bosons there are ${\displaystyle 2N}$ operators:

• ${\displaystyle {\hat {b}}_{i}^{\dagger }}$: the ${\displaystyle i^{th}}$ boson's creation operator.
• ${\displaystyle {\hat {b}}_{i}}$: the ${\displaystyle i^{th}}$ boson's annihilation operator.

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

These satisfy the commutation relations:

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

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

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

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

These may be rewritten as:

${\displaystyle {\hat {b}}_{i}^{\dagger }\,{\hat {b}}_{j}^{\dagger }={\hat {b}}_{j}^{\dagger }\,{\hat {b}}_{i}^{\dagger }}$

${\displaystyle {\hat {b}}_{i}\,{\hat {b}}_{j}={\hat {b}}_{j}\,{\hat {b}}_{i}}$

${\displaystyle {\hat {b}}_{i}\,{\hat {b}}_{j}^{\dagger }={\hat {b}}_{j}^{\dagger }\,{\hat {b}}_{i}+\delta _{ij}.}$

1. For two different bosons (${\displaystyle N=2}$) we have

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

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

2. For three different bosons (${\displaystyle N=3}$) we have

${\displaystyle$ :{\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) ${\displaystyle {\hat {b}}_{2}\,{\hat {b}}_{3}={\hat {b}}_{3}\,{\hat {b}}_{2}}$ the order in which we write the annihilation operators does not matter.

${\displaystyle$ :{\hat {b}}_{2}\,{\hat {b}}_{1}^{\dagger }\,{\hat {b}}_{3}:\,={\hat {b}}_{1}^{\dagger }\,{\hat {b}}_{2}\,{\hat {b}}_{3}}

${\displaystyle$ :{\hat {b}}_{3}{\hat {b}}_{2}\,{\hat {b}}_{1}^{\dagger }:\,={\hat {b}}_{1}^{\dagger }\,{\hat {b}}_{2}\,{\hat {b}}_{3}}

For a single fermion there are two operators of interest:

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

These satisfy the anticommutator relationships

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

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

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

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

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

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

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

To define the normal ordering of a 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.

1. We again start with the simplest cases:

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

This expression is already in normal order so nothing is changed. In the reverse case, we introduce a minus sign because we have to change the order of two operators:

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

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

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

or

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

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:

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

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

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

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

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

These may be rewritten as:

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

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

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

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.

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

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

Here the expression is already normal ordered so nothing changes.

${\displaystyle$ :{\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.

${\displaystyle$ :{\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}}

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

2. For three different fermions (${\displaystyle N=3}$) we have

${\displaystyle$ :{\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}}

Notice that since (by the anticommutation relations) ${\displaystyle {\hat {f}}_{2}\,{\hat {f}}_{3}=-{\hat {f}}_{3}\,{\hat {f}}_{2}}$ the order in which we write the operators does matter in this case.

Similarly we have

${\displaystyle$ :{\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}}

With two free fields φ and χ,

${\displaystyle$ :\phi (x)\chi (y):=\phi (x)\chi (y)-\langle 0|\phi (x)\chi (y)|0\rangle }

where ${\displaystyle |0\rangle }$ is again 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 states the existence of a relationship between the time ordered product of ${\displaystyle n}$ fields and a sum of normal ordered products. This may be expressed for ${\displaystyle n}$ even as

${\displaystyle {\begin{aligned}T\left[\phi (x_{1})\cdots \phi (x_{n})\right]=&:\phi (x_{1})\cdots \phi (x_{n}):+\sum _{\textrm {perm}}\langle 0|T\left[\phi (x_{1})\phi (x_{2})\right]|0\rangle$ :\phi (x_{3})\cdots \phi (x_{n}):\\&+\sum _{\textrm {perm}}\langle 0|T\left[\phi (x_{1})\phi (x_{2})\right]|0\rangle \langle 0|T\left[\phi (x_{3})\phi (x_{4})\right]|0\rangle :\phi (x_{5})\cdots \phi (x_{n}):\\\vdots \\&+\sum _{\textrm {perm}}\langle 0|T\left[\phi (x_{1})\phi (x_{2})\right]|0\rangle \cdots \langle 0|T\left[\phi (x_{n-1})\phi (x_{n})\right]|0\rangle \end{aligned}}}

where the summation is over all the distinct ways in which one may pair up fields. The result for ${\displaystyle n}$ odd looks the same except for the last line which reads

${\displaystyle \sum _{\text{perm}}\langle 0|T\left[\phi (x_{1})\phi (x_{2})\right]|0\rangle \cdots \langle 0|T\left[\phi (x_{n-2})\phi (x_{n-1})\right]|0\rangle \phi (x_{n}).}$

This theorem provides a simple method for computing vacuum expectation values of time ordered products of operators and was the motivation behind the introduction of normal ordering.