What does it mean "to smear an operator" in QFT? If possible, try to keep it physical (not too mathy). If I'm right, this smearing is necessary to determine the position of a particle in QFT. Why is that necessary? And please, spare my poor soul from a lot a math ;-)
 A: Fundamental particles in QFT are mathematical points with no size. This leads to so-called "ultraviolet" (or short-distance) divergences (infinities) as the distance between two particles becomes arbitrarily small. The "smoothing" procedure is one way to "regulate" (or make finite) the divergences by averaging the position of the particle over some small but finite volume. This effectively treats the particle as if it had some size, which mathematically tames the divergences. Of course we are ultimately interested in taking the limit where the size of the particle goes to zero, so smoothing is usually just the first step in a process to understand how to take this limit consistently.
You may be familiar with the potential of a point charge in electromagnetism
\begin{equation}
V(r) = k \frac{q}{r}
\end{equation}
where $k$ is Coulomb's constant, $q$ is the charge of the particle, and $r$ is the distance from the particle. As $r\rightarrow 0$, $V\rightarrow \infty$. This is an example of an ultraviolet divergence as you get arbitrarily close to a point particle. One manifestation of this divergence is that the self-energy of the point charge $U = \frac{1}{8\pi k}\int d^3 x E^2$ (where $E$ is the electric field) is infinite.
An example of a "smoothing" procedure would be to replace the charged particle with a ball of radius $a$ with the same charge. Then the maximum value of the potential is $V(a) = k q / a$; inside the ball, the potential drops smoothly to zero, rather than racing off to infinity. The self energy is finite.
A: Based on the comments, my understanding is that you're asking about so-called operator-valued distributions, of the form
$$\Phi^\dagger(f):= \int dx \ f(x) \hat \phi_x^\dagger$$
In the canonical approach to QFT, one can loosely understand the quantum field operators $\hat \phi^\dagger_x$ and  $\hat \phi_x$ as respectively creating and destroying a particle at the point $x$.  In other words, if we take $|0\rangle$ to be the vacuum state which has no particles, then we might think of $\phi^\dagger_x |0\rangle$ as the state which contains a single particle at the point $x$ and $\phi^\dagger_x \phi^\dagger_y|0\rangle$ as the state which contains two particles - one at $x$ and one at $y$.
This is ultimately problematic, however. Just as in non-relativistic QFT, it is not possible to have a particle which is exactly localized to a point - it must be "smeared out" over some non-pointlike region of space.  As a result, $\phi_x^\dagger |0\rangle$ is not actually a physically realizable state, and $\phi_x^\dagger$ is not a well-behaved operator.
An example of a true operator is
$$\int \mathrm dx \ \frac{1}{\sqrt{\pi}}e^{-x^2} \hat \phi^\dagger_x$$
which acts on $|0\rangle$ to produce a single-particle state with wavefunction $\frac{e^{-x^2}}{\sqrt{\pi}}$. This is a perfectly well-defined state, which we obtained by integrating the "quantum field operator" $\phi^\dagger_x$ against a smooth, square-integrable function. This is what is generally meant by "smearing" a quantum field operator.
With this in mind, we define an operator-valued distribution $\Phi^\dagger$ to be an object which eats some well-behaved function $f$ and spits out a well-behaved operator
$$\Phi^\dagger(f):=\int \mathrm dx \ f(x) \hat\phi_x^\dagger$$
These are the well-defined objects which are necessary in a more mathematically rigorous formulation of QFT - though it should be said, they are rarely referenced in a first course (in my experience).
