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Normal ordering
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If A is a formal linear combination of products of numbers, creation
operators and annihilation operators, the normal ordering :A: of A is
the operator defined by moving in each product term the creation
operators to the left of the annihilation operators.
In particular,
:AB: = :BA: = 0 for any A,B,
since AB and BA contain the same creation and annihilation operators
with the same multiplicity, only in different order. Therefore, we
also have :[A,B]: = :AB:-:BA: = 0.
This may look a bit strange since :[a,a^*]:=0 and :1:=1 although
[a,a^*]=1. But nothing is wrong here, since [a,a^*] and 1, although
equal as operators, are different as formal expressions.
Conclusion:
Normal ordering (or time-ordering) is applied only to _expressions_
involving operators, not to operators themselves. The normal ordering
of an operator is not well-defined unless you specify a particular
form of writing it.
But one can do all the algebra that doesn't depend on specific
equations relating the symbols. For example, if one writes phi and pi
as linear combinations of c/a operators, expands an expression
:H(phi,pi):, and normally orders the results, one gets correct results
as long as one uses neither commutation relations nor field equations.