What do smeared field operators act on? If we have a fock space and a vacuum state $|0\rangle$, it is obvious what a field operator does. It acts on this as though a position eigenstate and creates a particle, which is made more rigorous by expanding $\phi(\vec{x}, t)$ into a superposition of creation and annihilation operators.
Now I'm confused about how we can translate this into the smeared case. When we take $\int dx f(x) \phi(x)$, how does
$$\phi(f)|0\rangle$$
look? Does it allow the same expansion? I don't know much about operator valued distributions in QFT, only reading up to second quantization. Any help is appreciated.
 A: 
It acts on this as though a position eigenstate and creates a particle [...]

You must be careful here - as you should recall from nonrelativistic quantum mechanics, position eigenstates don't actually exist as physical states.  Their "wavefunctions" are delta functions, which are not square-integrable and therefore do not correspond to allowed states. The same is true for quantum field operators, where $\hat\phi(x)$ is a singular object for the same reason that $|x\rangle$ is.
Sweeping such subtleties under the rug, $\hat\phi^\dagger(x)$ should be thought of as a position-space creation operator which acts on the vacuum $|0\rangle$ to produce $|x\rangle$.  In that sense, it would be somewhat more consistent to call this operator $\hat \phi^\dagger_x$, but conventions vary.
As such, the operator-valued distribution $\Phi^\dagger$ eats a (square-integrable) function $f$ and produces a single-particle state with $f$ as its wavefunction:
$$\Phi^\dagger(f)|0\rangle:=\int dx\ f(x) \hat\phi^\dagger(x) |0\rangle = \int dx \ f(x) |x\rangle\equiv |f\rangle$$
More generally, it acts on an $n$-particle state to produce an $(n+1)$-particle state by adding $|f\rangle$ in the appropriately symmetrized or antisymmetrized way.  Explicitly, if we are working with a bosonic field, the action of $\Phi^\dagger(f)$ on the one-particle state $|\psi\rangle = \int dx \ \psi(x)|x\rangle$ would be
$$\Phi^\dagger(f)|\psi\rangle = \frac{1}{2}\left(|f\rangle\otimes |\psi\rangle + |\psi\rangle\otimes |f\rangle\right) = \int dx \int dy \ \frac{f(x)\psi(y)+\psi(x)f(y)}{2}|x\rangle\otimes |y\rangle$$
The adjoint operator $\Phi(f)$ would annihilate the vacuum; its action on one and two-particle states would be
$$\Phi(f)|\psi\rangle = \langle f|\psi\rangle$$
$$\Phi(f)\left[\frac{|\psi\rangle\otimes|\rho\rangle +|\rho\rangle\otimes |\psi\rangle}{2}\right]= \frac{|\rho\rangle \cdot \langle f|\psi\rangle+ |\psi\rangle \cdot \langle f|\rho\rangle }{2}$$
One can show more generally that $[\Phi(f),\Phi^\dagger(g)] = i\hbar \int dx\ f^*(x) g(x)$; if we plug in delta functions, we find
$$[\hat \phi(x),\hat \phi^\dagger(y)] = [\Phi(\delta_x),\Phi^\dagger(\delta_y)]=i\hbar \int dz\ \delta(x-z) \delta (y-z) = i\hbar \delta(x-y)$$
in accordance with the field-operator commutation relations we already know.
