quantum field theory - Normal Order of Normal Order

in first volume of Polchinski page 39 we can read a compact formula to perform normal-order for bosonic fields $$:\cal F:=\underbrace{\exp\left\{\frac{α'}{4}∫\mathrm{d}^2z\mathrm{d}^2w\log|z-w|^2\frac{δ}{δφ(z,\bar z)}\frac{δ}{δφ(w,\bar zw)} \right\}}_{:=\mathcal{O}}\cal F, \tag{1}$$

What I do not understand it is that I would like to have (bearing in mind the definition involving $a$ and $a^†$ $$::\cal F::=:\cal F:\tag{2}$$ but with this formula $$\cal O^2\cal F≠\cal O \cal F.\tag{3}$$

EXAMPLE

$$:φ(z)φ(w):=φ(z)φ(w)-\frac{α'}{2}\log|z-w|^2\tag{4}$$ but $$::φ(z)φ(w)::=:φ(z)φ(w):-\frac{α'}{2}\log|z-w|^2=φ(z)φ(w)-α'\log|z-w|^2\tag{5}$$

Answer

1. 
   

   Short explanation: Polchinski's eq. (1) is not a formula that transforms no normal order into normal order: The expression ${\cal F}$ 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.

   
   


2. 
   
   

   Longer explanation: When dealing with non-commutative operators, say $\hat{X}$ and $\hat{P}$, the "function of operators" $f(\hat{X},\hat{P})$ 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 $$\begin{array}{c} \text{Symbols/Functions}\cr\cr \updownarrow\cr\cr\text{Operators}\end{array}\tag{A}$$ (E.g. the correspondence map from Weyl symbols to operators is explained in this Phys.SE post.) To define an operator $\hat{\cal O}$ on operators, one often give the corresponding operator ${\cal O}$ on symbols/functions, i.e., $$\begin{array}{ccc} \text{Normal-Ordered Symbols/Functions}&\stackrel{\cal O}{\longrightarrow} & \text{Radial-Ordered Symbols/Functions} \cr\cr \updownarrow &&\updownarrow\cr\cr \text{Normal-Ordered Operators}&\stackrel{\hat{\cal O}}{\longrightarrow} & \text{Radial-Ordered Operators}\end{array}\tag{B}$$ E.g. Polchinski's differential operator ${\cal O}$ does strictly speaking only make sense if it acts on symbols/functions. The identification (A) of symbols and operators is implicitly implied in Polchinski.

   
   


3. 
   

   Concerning idempotency of normal ordering, see also e.g. this related Phys.SE post.