Do the operators $\hat{P}^i=-i\partial^i, ~\hat{J}^{ij}=i(x^i\partial^j-x^j\partial^i),$ ever arise in QFT as momentum and angular momentum operators? The generators of the translation and the rotation group, acting on a classical field (say, a scalar field), are given by the differential operators $$\hat{P}^i=-i\partial^i, ~~\hat{J}^{ij}=i(x^i\partial^j-x^j\partial^i),$$ respectively. These do not have any physical meaning in a classical context. In particular, it has nothing to do with the momentum density $P^i$ and the angular momentum density $J^{ij}$ of the classical field, which are given by the Noether charges (numbers, not operators)
$$P^i=\int d^3x T^{0i},~~J^{ij}=\int d^3x\left(x^iT^{0j}-x^jT^{0i}\right)$$ respectively. In quantum field theory, I believe, the momentum and angular momentum operators for the quantum field is obtained from the second expressions as $$ \hat{P}^i=\int d^3x \hat{T}^{0i},~~\hat{J}^{ij}=\int d^3x\left(x^i\hat{T}^{0j}-x^j\hat{T}^{0i}\right).$$ So in the QFT context, the first equation does not arise as the momentum or the angular momentum operators of the field although these are the same as the momentum and angular momentum operators of ordinary quantum mechanics.
If this understanding is flawed, I want to know why. I want to know whether the set of operators described in the first equation ever appears in quantum field theory as momentum and angular momentum operators.
 A: The point is that symmetry transformations in QFT, say Poincare transformations, have (infinite-dimenstional unitary irreducible) representations acting on the quantum Hilbertspace and as "classical" transformations on fields. E.g. for translations
$$
\hat P^i  = \int d^3x~\hat T^{0i} 
$$
is an operator on the Hilbert-space, i.e. it can be written as an integral over annihilation and creation operators. The non-relativistic QM analogues are for example the Hamilton operator or the angular momentum operator, which you are surely familiar with.
As mentioned above, transformations also have a representation acting on fields, say for translations
$$
\mathcal P^i = -i\partial^i
$$
E.g. a finite translation of a classical field can be written as
$$
\phi(x) \rightarrow e^{i a\cdot \mathcal P} \phi(x) = e^{a \cdot \partial} \phi(x) = \sum_j \frac{1}{j!} (a \cdot \partial)^j \phi(x) = \phi(x+a).
$$
The important point is that these transformations on classical fields, denote $\Lambda$, must "match" the corresponding Hilbert-space operator, that is
$$
\hat U(\Lambda)^{\dagger} \hat \phi \hat U(\Lambda) \equiv \widehat{\Lambda \phi}
$$
(Here $\hat U$ maps the abstract group element $\Lambda$ to the corresponding Hilbert-space operator.)
Or equivalently, in terms of a generator $\mathcal G$, given by $\Lambda = e^{i\alpha \mathcal G}$ or $\hat U(\Lambda) = e^{i \alpha \hat G}$ we must have
$$
[\hat G, \hat \phi] = \widehat{\mathcal G \phi}.
$$
For example for translations we have
$$
e^{-i\hat P \cdot a}\hat \phi(x) e^{i \hat P \cdot a} = \hat \phi(x+a) = \widehat{e^{ia\cdot \mathcal P} \phi(x)}.
$$
The first equality is a well-known relation that you probably know. It can be shown explicitly in the canonical formalism (also in the path integral formalism). Note that it is equivalent to the relation
$$
[\hat P^i, \hat \phi] = -i \partial^i \hat \phi.
$$
Can you show this, say for free KG theory?
Hint: show first that $\hat P^i = \int d^3\vec x~\hat \pi(\vec x) \partial^i\hat \phi(\vec x)$, where $\hat \pi$ is the canonical conjugate field to $\hat \phi$. Then use the (postulated) canonical equal time commutation relations.
As it turns out the quantum operator realizations of the classical transformations are the corresponding conserved classical Noether charges turned operators (in some sensible way) by the quantization $\phi \rightarrow \hat \phi$.
