Why does the subclassification of fields under parity require the quantum theory? The fields of relativistic field theory (scalars, vectors, tensors, and spinors) are all defined via their transformation properties under the restricted Lorentz group (which excludes discrete spacetime transformations, namely, parity and time-reversal). This classification is done completely classically.
However, when we make further classification of fields into sub-categories, for example, a scalar field into true scalars and pseudoscalars, by asking how they transform under parity, I see that the QFT books start by pretending that we have already quantized the theory, there exists a vacuum state $|0\rangle$, and the action the creation and annihilation operators $|0\rangle$ is $a_{{\vec k}}|0\rangle\sim |{\vec k}\rangle$ are known. For a reference, see Peskin and Schroeder's QFT, page $65-67$.

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*Why does this sub-classification require the quantum theory? Can we do it classically? Any help or hint will be highly appreciated.

 A: Common misconception. You cannot derive how a field transforms, under any transformation. All you can do is make definitions.
One often hears that, under a (connected) Lorentz transformation $\Lambda\in Spin(1,d-1)$, a field must transform as
$$
q(x)\mapsto D(\Lambda)q(x)\tag1
$$
where $D$ is a finite-dimensional irreducible representation. This is not true. Generic fields can transform in any way you want. Say, for example, you have two fields transforming as in $(1)$, e.g., a scalar $\phi$ (with $D=1$) and a vector $A$ (with $D=\Lambda$). Then the field $q(x)=\phi(x)+A_0(x)+A_1(x^2)A_2(x+1)$ does not transform as $(1)$, but rather it transforms in a weird non-linear way. General fields do not transform as $(1)$, instead they typically transform in very complicated ways. So what we do is: if a field transform as in $(1)$, we give it a special name: we call it a Lorentz field (e.g., a Lorentz scalar, a Lorentz spinor, a Lorentz vector, etc.). Most fields are not Lorentz fields. #NotAllFields.
Other symmetries are exactly analogous. For example, one often hears that, under a flavour symmetry $U\in SU(N)$, fields transform as
$$
q_i(x)\mapsto U_{ij}q_j(x)\tag2
$$
Again, this is not true. If we are given a set of fields that do transform this way, then the combination $q_1(x)+q_2(x)q_3(-x)-q_4(x^2)$ no longer transforms this way, and this is still a valid field. The correct statement is that, if a field $q_i(x)$ transforms as in $(2)$, we give it a special name (e.g., we say that $q$ transforms in the vector of $SU(N)$ or something like that).
Finally, we have parity. By definition, such transformation involves $x\mapsto \mathcal Px$, with $\mathcal P=\mathrm{diag}(1,-1,-1,-1)$. But the action of parity on fields is in principle arbitrary. There are some ways a field can transform that are simple, and occur often enough, that merit being given a name. For example, if a Lorentz scalar transform as
$$
\phi(x)\mapsto \pm \phi(\mathcal Px)\tag3
$$
then we call it a true scalar if $+$, and a pseudo-scalar if $-$. Of course, other transformations are perfectly possible, e.g. we could have a theory that is invariant under $x\mapsto \mathcal Px$, but only if the fields transform as
$$
q_i(x)\mapsto A_{ij}q_j(\mathcal Px)\tag4
$$
for some matrix $A$. The choices $A=\pm 1$, corresponding to (true/pseudo)scalars are particularly common, but by no means unique. If is perfectly consistent to have a theory that is not invariant for $A=\pm1$, but it is for some other choice of $A$. In that case, we would need to introduce a new name for the field $q$, it is neither a true scalar nor a pseudo-scalar. So, to reiterate the first sentence, you cannot derive how a field transforms under any given transformation, including parity. The best you can do is to study different possibilities, and give a special name to those that you liked the most, say because they are particularly simple or because they appear in phenomenologically relevant theories.
When we say a theory is invariant under some symmetry, it is not enough to specify the symmetry group. One must also assign explicit transformation properties to the fields, as these transformations can in principle take any form you want. So if I want to study, say, a $\phi^4$ theory, it is not enough to claim that the theory is invariant under parity; I must specify how exactly parity acts on $\phi$, for otherwise the claim is empty. The theory could be invariant under $\phi\mapsto A\phi(\mathcal Px)$ for some $A$, but not for some other $A'$. It could even be invariant under some non-linear transformation, such as $\phi\mapsto \phi^2+\phi+1$ or something like that. The transformation properties of fields under a given symmetry are never fixed given the symmetry group alone.
Let me wrap this up with three final comments.

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*All I said above is true both classically and quantum-mechanically. One can have classical spinors, and quantum mechanical spinors, and all other types of fields that transform in any conceivable way whatsoever. There is no fundamental difference between the two worlds, classical vs. quantum. The only difference is that in the first case the fields are $c$-numbers while in the second case they are operators. Their transformation properties under symmetries doesn't care about the nature of these operators.


*It is sometimes the case that, if you construct a quantum theory by quantizing a classical theory, a given symmetry can be lost. For example, you can have a theory that is invariant under parity (for a given assignment of $A$), but the quantum theory has some anomaly that breaks the symmetry. Even more interestingly, one may sometimes recover the symmetry by changing the transformation properties of the fields, say, by choosing a different $A$. This mostly requires a case-by-case analysis. The general idea is that, if you have a classical symmetry for a given assignment of transformation properties, the symmetry may be completely broken, partially broken, or preserved, in the quantum theory, and you may sometimes have to modify how fields transform.


*We abuse the word "scalar" in physics. It typically means "just a number", but this definition is obviously wrong: for example, all the individual components of the position vector are "just numbers", but they are definitely not scalars. The correct definition of scalar is that it transforms trivially under some transformation. But this obviously requires specifying what transformation we have in mind. A given object may be a scalar under some transformation, but a non-scalar under some other transformation, e.g., a set $\phi_i(x)$ may be scalars under Lorentz transformations, but transform as a vector of $SU(N)$, such as $\phi_i\mapsto U_{ij}\phi_j$, with $U\in SU(N)$. So when we say that something is a scalar, we must specify a scalar under what. A Lorentz scalar is a scalar under Lorentz, $Spin(1,d-1)$. A "true-scalar" typically means a scalar under the orientation cover of Lorentz, $Pin(1,d-1)$.
