Is quantum field operator $\psi$ same as quantum field $\psi$? So in QFT, quantum field operator $\psi$ is there. $\psi$ seems to take the role of wavefunction in QM, which now acts upon vacuum state. Then, in lagrangian of various quantum field theories, $\psi$ appears, but now it is called quantum field. So is quantum field operator no different from quantum field here? If it is not different, then how can $\psi$ can have scalar $|\psi|$ in any assumption as it only operates upon vacuum state? Assuming that vacuum state is represented by some $n \times 1$ vector (let us forget about infinite-dimension for now), operator should be of the form $n \times n$. 
 A: The quantum field has nothing to do with the wavefunction.
This is a peculiar confusion that seems to arise quite often.
The wavefunction $\psi(x)$ is a way of representing a quantum state $\lvert \psi \rangle$ in a Hilbert space $\mathcal{H}_{\mathrm{QM}}$ that is equipped with a position basis $\{\lvert x \rangle \rvert x \in \mathbb{R}\}$ by setting $\psi(x) = \langle x \vert \psi \rangle $. It encodes the quantum state of a (mostly one-particle) system.
The quantum field $\psi(x)$ is an operator-valued distribution, i.e. to every spacetime-point $x$, it associates an operator $\psi(x)$ acting upon the Hilbert space $\mathcal{H}_{\mathrm{QFT}}$ of the quantum field theory. For free theories, it has a mode expansion into creation and annhiliation operators of one-particle states with definite momenta. The quantum field creates the quantum states, it does not represent them.
As Emilio Pisanty points out, one might restrict to a special case: In a free (scalar) theory, the field $\psi^\dagger(x)$ is really the Fourier transform of the creation/annihilation operators of definite momenta, and thus, when applied to the vacuum state $\lvert \Omega \rangle$, creates a particle of definite position, that is to say, the QM space of states (at time $t = 0$) is the subspace spanned by $\{ \psi^\dagger(0,\vec x)\lvert \Omega \rangle \vert \vec x \in \mathbb{R}^3\}$ and one has the identification $\lvert \vec x \rangle = \psi^\dagger(0,\vec x)\lvert \Omega \rangle$, so the wavefunction of a state $\lvert \phi \rangle$ would be $\phi(\vec x) = \langle \vec x \vert \phi \rangle = \langle \Omega \vert \psi(0,\vec x) \vert \phi \rangle $, i.e. the QM wavefunction is a range of expectation values of the quantum field.
(Note that this is not a Lorentz invariant way of writing anything here.)
