Some questions about Normal Ordering in QFT I have some questions about normal ordering in quantum field theory: I already read this very good question with very very good answers and this other question with other very good answers (I read also this one and many others, but without understanding much).
For what I understood, normal ordering is more a simbolic operation than an operator, so if for example I have
\begin{equation*}
\left[\hat{a},\hat{a}^\dagger\right]
=
1
\end{equation*}
Then I'm not authorized to say that $:\hat{a}\hat{a}^\dagger:=:\hat{a}^\dagger\hat{a}:+:1:$ (where I think that $:1:=1$ demonstrated through unitary operators). What I don't understand here is

*

*The main fact is that this operation is non-linear? So (even if here the answer from Sebastiano Peotta seems to say the opposite)
\begin{equation*}
:\hat{a}^\dagger\hat{a}:+:1:
\neq
:\hat{a}^\dagger\hat{a}+1:
\,?
\end{equation*}


*Or the main fact is that this operation doesn't care about operatorial equalities? In that case I would just have
\begin{equation*}
:\hat{a}\hat{a}^\dagger:
\neq
:\hat{a}^\dagger\hat{a}+1:
\,?
\end{equation*}
At the same time I wasn't able to find this kind of question, that is the main doubt that I have:

*

*what if I rename the operator with the following substitution $\hat{b}^\dagger=\hat{a}$? In that case $:\hat{a}\hat{a}^\dagger:=\hat{a}^\dagger\hat{a}$, but $:\hat{b}^\dagger\hat{b}:=\hat{b}^\dagger\hat{b}=\hat{a}\hat{a}^\dagger$!

Is this crazy or am I doing something wrong (very likely)?
 A: The main point is that the normal ordering procedure $:~:$ does not take operators to operators, but symbols/functions to operators, cf. e.g. this & this Phys.SE posts.

*

*With this understanding, the normal ordering procedure $:~:$ becomes a linear map, cf. OP's first bullet point.


*A commutator of symbols/functions is manifestly zero, which resolves OP's second bullet point.


*The third bullet point seems to be mostly a notation/semantic issue: There is no reason why creation (annihilation) operators should have (not have) a dagger, respectively. One could in principle chose an opposite non-standard notation.
It should perhaps be stressed that the choice of normal order prescription is tied to the choice of vacuum, cf. e.g. my Phys.SE answer here.
A: @Qmechanic 's impeccable answer above may be illustrated by the basic rule of normal ordering, which is not a functor. That is, normal orderings of commutators vanish; the objects inside : : are commutative entities, "(Weyl) symbols", and not operators, so are reprieved  through commutative rules, as Sebastiano Peotta's answer reviews. So linearity  of  :  : applies to its arguments, which are commutative symbols, and not operators.
Consequently, inside : : , you indeed "don't care about operator equalities", as you say, such as $\left[\hat{a},\hat{a}^\dagger\right]=1$,
$$
:\left[\hat{a},\hat{a}^\dagger\right]:~~ = ~~:0:~~=0 \neq 1= ~~:1:~.
$$
Thus your second bullet is correct,
$$
:\hat{a}\hat{a}^\dagger: -:\hat{a}^\dagger\hat{a}:~~\neq ~~ :1: ~,
$$
because inside : : you only care about (trivial) symbol commutators, instead of operator commutator equalities.
I don't want to confuse you, but you might take off the operator hats inside  : :, even if only in your mind, to remind you they represent symbols, playing by their own rules; that is, think of the symbols as semiclassical objects, not operators, obeying a trivial commutation relation, instead of the usual quantum one. A gonzo quantization rule, yielding inconsistent alternate answers.
