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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)?

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  • $\begingroup$ Do you have a reason to think the equality doesn't hold? my understanding is that operator equations hold inside of expectation values(upto contact terms) and this is precisely what makes normal ordering linear. $\endgroup$
    – Anonjohn
    Oct 9, 2020 at 14:51
  • $\begingroup$ @Anonjohn The problem about linearity is that in this case $:\hat{a}\hat{a}^\dagger-\hat{a}^\dagger\hat{a}:=:1:\Rightarrow:\hat{a}\hat{a}^\dagger:-:\hat{a}^\dagger\hat{a}:=:1:$, but $:1:=1$ and so you would have $0=1$. $\endgroup$
    – Rob Tan
    Oct 9, 2020 at 14:57
  • $\begingroup$ Forgive me for obtuse, but how did you conclude that $:\hat{a}\hat{a}^{\dagger}:=:\hat{a}^{\dagger}\hat{a}:$? The very equations you write indicate that they differ by the identity operator in a free theory. And indeed they do? $\endgroup$
    – Anonjohn
    Oct 9, 2020 at 15:01
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    $\begingroup$ @Anonjohn No no I am obtuse ahah For what I read the fact that $:\hat{a}\hat{a}^\dagger:=:\hat{a}^\dagger\hat{a}:=\hat{a}^\dagger\hat{a}$ is just a definition $\endgroup$
    – Rob Tan
    Oct 9, 2020 at 15:03

2 Answers 2

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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.

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  • $\begingroup$ Thank you for your answer. Yes I read the post you cite, but I came out with more doubts. I don't understand also the second point of your answer here, sorry. At the same time, if this is just an operation that takes symbols to operators, why cannot I define $\hat{b}^\dagger=\hat{a}$ ending with the paradoxical result at the end of my question? $\endgroup$
    – Rob Tan
    Oct 9, 2020 at 15:42
  • $\begingroup$ I don't know if I understood, I will think about it; meanwhile, thanks! $\endgroup$
    – Rob Tan
    Oct 9, 2020 at 19:35
  • $\begingroup$ So on your third point you are saying that the dag in itself is not important, because what the normal ordering operation is actually looking at is at the creation and annihilation operators, right? I continue to don't understand the first two points. If normal ordering is a linear mapping how can this be solved $:\hat{a}\hat{a}^\dagger:=:\hat{a}^\dagger\hat{a}:+1\Rightarrow 0=1$? Sorry for the dumbness $\endgroup$
    – Rob Tan
    Oct 10, 2020 at 8:28
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Oct 10, 2020 at 9:02
  • $\begingroup$ Ok I just understand that the whole thing is much much more complicated than just acting on symbols. I just cannot go over the really basic incoherences that I encounter: such as the prescription of linearity, but the apparent paradoxical results. I will think about it, thanks $\endgroup$
    – Rob Tan
    Oct 10, 2020 at 9:34
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@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.

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  • $\begingroup$ Thanks a lot! I need to think about this too $\endgroup$
    – Rob Tan
    Oct 10, 2020 at 20:51

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