Normal order

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For the evaluation strategy, see Normal order evaluation.

When a quantum mechanical Hamiltonian is written in normal order, it means that all creation operators are placed to the left of all annihilation operators in the expression. The process of putting a Hamiltonian into normal order is called normal ordering. The terms antinormal order and antinormal ordering are analogously defined, where the annihilation operators are placed to the left of the creation operators.

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.

1 Notation
2 Bosons
3 fermions
4 Multimode normal ordering
- 4.1 Bosons
- 4.2 Fermions
5 Free fields
6 Free fields
7 Wick's theorem

[edit] 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

$: \hat{O} :$

as illustrated above. Sometimes, the former expression is preferred because the addition of a period makes the latter difficult to read, viz.

$: \hat{O} :.$

[edit] Bosons

Normal ordering in bosons depends on the commutator relationship

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

so that

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

[edit] fermions

Normal ordering in fermions depends on the anticommutator relationship

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

so that

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

[edit] Multimode normal ordering

For the multimode case, the relationship

[edit] Bosons

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

so that

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

[edit] Fermions

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

so that

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

[edit] Free fields

Normal order only applies to free field theories. The normal order of the operators is the choice that leads to zero ground state energy. It puts all annihilation operators to the right, and all creation operators to the left, leading to a ground state expectation value of 0:

$\langle 0 | H | 0 \rangle = 0$

if H is in normal order.

The operator for putting an expression in normal order is N. It has the effect of moving (by commutation relations) all creation operators to the left and all annihilation operators to the right with all commutators (for bosons) and anticommutators (for fermions) temporarily set to 0.

No matter what order some expression K is in,

$0 = \langle 0 | N(K) | 0 \rangle.$

[edit] 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):.