Does normal ordering (not) conflict with canonical commutators?

Question

Normal ordering is pretty useful to stop expressions from diverging in quantum field theory and works out perfectly fine regarding this, but there is this little problem: Consider for example an annihilation operator $\widehat{a}$ and the corresponding creation operator $\widehat{a}^\dagger$ with their canonical commutator: $$[\widehat{a},\widehat{a}^\dagger] =i\hbar.$$ Applying normal ordering (and reordering $\widehat{a}\widehat{a}^\dagger$ into $\widehat{a}^\dagger\widehat{a}$) makes the left side vanish, while the right side doesn't change, resulting in a contradiction. In general, having a longer expression of operators, swapping an annihilation with a creation operator using this commutator, knowing it won't change their normal ordering at all, would just create a term with a $\delta$ distribution out of nowhere.

It therefore seems like normal ordering can map same inputs to different outputs (which is horror for any mathematician) and therefore the operators already have to be in a very specific order before normal ordering can be applied, if there should be no arbitariness in the result.

Since post like here (commutators are forbidden inside normal ordering) and here (proper definition of normal ordering) already give an answer to this specific problem and I could only use more and maybe different clarification concerning it, the focus is on the following questions, on which I haven't yet found anything:

Is normal ordering indeed arbitary? Is the arbitrariness arising here maybe exactly why it is able to prevent diverging expressions in the first place? What are the possible changes of its definition to prevent this and more importantly how can they be justified? Are there motivations other than to just avoid contradictions? Since the exact same also holds for its introduction to avoid divergence in the first place: Is there a suitable physical interpretation for using it?

Answer 1

The only thing normal ordering (in QFT) is in conflict with is quite possibly unwarranted expectations. It is perfectly formally and physically consistent, and self-evident in deformation quantization, but, since this is not familiar to many students, I'll discuss it in the parallel equivalent track of Hilbert space. Fortuitously, you have kept the parameter ℏ in your commutation relation, which makes this discussion simpler.

1. Given an arbitrary QFT operator $\hat F(a,a^\dagger)$ you may first merely rearrange it by use of the commutation relation into a unique, consistent equal expression $c(\hbar)+\sum_{m,n}b_{m,n}(\hbar)a^{\dagger ~m} a^n$, where not both m and n are null, and c is a constant. It is then evident that $\langle 0|\hat F|0\rangle =c$. This is a mere arrangement prescription, as it is used in deformation quantization, which you might not care about.

2. In QFT, you may then set $\hbar\to 0$, that is, discard all terms that arose out of a commutation in your rearrangement, and call the result $:\hat F:$, so that many-many-many different $\hat F$s may yield the same $:\hat F:$ (surjection). This is the dross that is isolated and jettisoned after application of Wick's theorem, since $\langle 0|:\hat F:|0\rangle =c(0)$. In this second step, you have essentially collapsed all commutators to zero, the crucial property of normal ordering. In your commutation expression, you thus find 0=0, quite consistent.