Why are quantized fields somehow unique? A classical field 
$$\phi(x)  =\int\frac{d^3p}{\sqrt{E_p}} (a_pe^{-ipx} + a^*_pe^{ipx})$$
is not a unique thing. We could "choose" $a_p$ and $a^*_p$ conveniently so that it represent field of our specific problem.
But when we quantize it we get
$$\phi(x)  =\int\frac{d^3p}{\sqrt{E_p}} (a_pe^{-ipx} + a^{\dagger}_pe^{ipx})$$
then $a_p$ and $a^{\dagger}_p$ are unique thing (ladder operators of a SHO). So who this field represent our specific quantum system?! We are in Heisenberg picture so we expect dynamics in operators but rigidity of $a_p$ and $a^{\dagger}_p$ don't let any dynamic even for interacting fields. What is the overall picture of QFT in this manner?
 A: After the quantization the fields are not describing the state of the system anymore, but they are linear operators that act on it. Your question is about something we already observe in standard quantum mechanics, take for example the harmonic oscillator. The classical state variables $x,p$ are associated with Operators $X$ and $P$ composed in a very similar fashion to the fields you mentioned, see Wikipedia. The don't wiggle around as they do in a classical harmonical oscillator, instead they are a fixed objects and so are the fields in QFT. Instead the dynamic happens in the Hilbert space of our system. For a free QFT the Hilbert space is given by a Fock space and instead of saying my (classical) system is in the state $\phi(x)$ you say my (quantum) system is in the state $|\psi\rangle\in\mathcal{H}$. The be precise here, the classical states are given by solutions of your equations of motion, thus fields, while quantum states are given by a density operator $\rho$ on the Hilbert space of your system (or an element of your Hilbert space, depending on how you precise you want to be).
I hope this can answer your question.
Cheers!
