Are smearing functions in QFT operator independent? For a scalar field $\hat{\phi}(x)$, the smearing is performed by convoluting the operator $\hat{\phi}(x)$ with a smooth function $f(x')$ which has support in the neighborhood of the point $x$.
Is this smearing function dependent on the operator at hand, or depends only on the point $x$ about which it is defined? In case it depends on the operator at hand, is the existence of a smearing function which is able to smear all possible operators in the theory guaranteed?
My specific problem is this: Suppose if I want to look at the smeared field operator and its conjugate momenta at a general point $x$. Can I use the same smearing function for both the operators, or is the smearing function necessarily different for both these operators? 
 A: The idea of not using the scalar fields $\Phi(x)$ itself but the operator-valued distributions they define
$$f \mapsto \Phi(f) = \int f(x) \Phi(x) d x $$
is to make all amplitudes distributions. In this approach you are interested maybe in Correlators
$$\langle \Phi(f) \Phi(g) \rangle = \int \langle \Phi(x) \Phi(y) \rangle f(x) g(y) dx dy $$
which takes to arbitrary text functions $f,g$. You may choose $f = g$, but calculating this will in general give you not enough information to infer $\langle \Phi(x) \Phi(y) \rangle$. This is analogous to how in the calculus of variations, one needs to take into account arbitrary variations in order to obtain equations of motion (cf. the fundamental lemma of the calculus of variations https://en.wikipedia.org/wiki/Fundamental_lemma_of_calculus_of_variations).
One a more physical level, one might want to reconstruct $\langle \Phi(x) \Phi(y) \rangle$ by considering $\langle \Phi(f) \Phi(g) \rangle$ and $f,g$ becoming more and more peaked at $x$ and $y$ respectively. This doesn't work anymore if all we have is one function $f$.
