I want to give a brief overview as to where I am coming from with this question.
When we go from classical mechanics to QM, one the crucial things we consider is that $\vec x$ and $\vec p$ are replaced with operators $\hat{\vec x}$ and $\hat{\vec p}$ and that the poisson brackets transform in the following way:
$$\{x_i,p_j\}=i\delta_{ij}\rightarrow [\hat x_i,\hat p_j]=i\hbar\delta_{ij}. (Eq.1)$$
When we go from QM to QFT the canonical commutation relations are:
$$[\phi(x),\pi(y)]_{x_0=y_0}=i\delta^{3}(\vec x - \vec y) (Eq.2)$$
$$[\phi(x),\phi(y)]_{x_0=y_0}=0 (Eq.3)$$
$$[\pi(x),\pi(y)]_{x_0=y_0}=0 (Eq.4)$$
From (2) one can say $\phi(x)$ plays the role of $x_i$ (or $\vec x$) and $\pi(x)$ that of $p_j$ (or $\vec p$).
But then we can consider the Heisenberg equations in QFT, such as:
$$\partial_\mu \phi(x)=[iP^\mu,\phi(x)] (Eq.5)$$
$$k^\mu a^\dagger(\vec k)=[P^\mu,a^\dagger(\vec k)] (Eq.6)$$
$$-k^\mu a^\dagger(\vec k)=[P^\mu,a(\vec k)] (Eq.7)$$
I have two questions:
Is there any relation between the two sets of commutation relations? And can we make any claim about the framework/picture we are in, i.e Schroedinger picture or Heisenberg one?
If $P^\mu$ the four momentum operator is present in QFT, why don't we have a 4 position operator as well $X^\mu$ ?