in first volume of Polchinski page 39 we can read a compact formula to perform normal-order for bosonic fields :F:=exp{α′4∫d2zd2wlog|z−w|2δδφ(z,ˉz)δδφ(w,ˉzw)}⏟:=OF,
What I do not understand it is that I would like to have (bearing in mind the definition involving a and a† ::F::=:F:
EXAMPLE
:φ(z)φ(w):=φ(z)φ(w)−α′2log|z−w|2
Answer
Short explanation: Polchinski's eq. (1) is not a formula that transforms no normal order into normal order: The expression F on the right-hand side of eq. (1) is implicitly assumed to be radially ordered. In fact, eq. (1) is a Wick theorem for changing radial order into normal order, cf. e.g. this Phys.SE post.
Longer explanation: When dealing with non-commutative operators, say ˆX and ˆP, the "function of operators" f(ˆX,ˆP) does not make sense unless one specifies an operator ordering prescription (such as, e.g., radial ordering, time-ordering, Wick/normal ordering, Weyl/symmetric ordering, etc.). A more rigorous way is to introduce a correspondence map Symbols/Functions↕Operators
(E.g. the correspondence map from Weyl symbols to operators is explained in this Phys.SE post.) To define an operator ˆO on operators, one often give the corresponding operator O on symbols/functions, i.e., Normal-Ordered Symbols/FunctionsO⟶Radial-Ordered Symbols/Functions↕↕Normal-Ordered OperatorsˆO⟶Radial-Ordered OperatorsE.g. Polchinski's differential operator O does strictly speaking only make sense if it acts on symbols/functions. The identification (A) of symbols and operators is implicitly implied in Polchinski.Concerning idempotency of normal ordering, see also e.g. this related Phys.SE post.
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