Schrodinger equation for free EM field The question come from the fact that I've seen for the first time in my life the quantization of a field, in particular of the free em field. 
I've study how it is possible to write the energy of the em field as a function of canonical coordinates $p_\lambda$ and $q_\lambda$ and how to substitute them with operators $\hat Q_\lambda$ and$\hat P_\lambda$ that obey appropriate commutation rules. Than with some substitution I've studied how to introduce the operators $\hat b$ and $\hat b^\dagger$. With some other substitution I found the operators corresponding to the classical quantity  $\vec A$, $\vec B$, $\vec E$ and so on.
Now the problem is that I feel like I have all the operators of the em field and the commutation rules for them, but i don't know what Is the equation that the wave function of this operators should follow, so I feel like everything I derived is useless.
Is maybe the Schrodinger equation valid for free EM field? Let's say that  $|\Psi\rangle $ is the state of the free EM field and that $\hat H$ is its hamiltonian operator, it it true that $i\hbar\,\partial_t |\Psi\rangle = \hat{H}|\Psi\rangle $?
If not what is the right equation for the wave function of this operators?
 A: Yeah, that equation is correct. The form of the Hamiltonian is
$$
H = \int \frac{d^3 p}{(2\pi)^3} |p| \cdot \sum_{i=1}^{2} a_i^{\dagger}(p) a_i(p),
$$
with $a$ and $a^{\dagger}$ annihilation and creation operators for photons with polarization $i$ respectively.
In fact, the equation that you're written down (usually referred to as the generalized Schrodinger equation, though annav@ seems to disagree) is very general. It is valid for any system which is symmetric under translations in time.
Due to Wigner, in Quantum Mechanics, symmetries can always be represented by either a unitary linear operator, or by an antiunitary antilinear map. Since time translations are a strongly continuous 1-parametric group, they can only be represented by a unitary 1-parametric group acting on the Hilbert space. By Stone's theorem, any such group is generated by a self-adjoint operator $H$, which is your Hamiltonian.
Note that similar correspondence holds in the classical theory: invariance under time translations leads to energy conservation (by Noether's theorem). The energy observable is nothing other than the classical Hamiltonian.
This equation
$$
i \hbar \frac{d}{dt} \left| \psi \right> = H \left| \psi \right>
$$
is an expression of the invariance of the theory under time translations. Any theory with time translation symmetry obeys an equation like that. It is valid for a non-relativistic particle, for the relativistic particle, for the relativistic field, etc.
In fact, it is only for General Relativistic systems it seems to not hold. In General Relativity, time is described dynamically rather than being a static background. Time translations become a redundancy rather than a genuine symmetry, and an equation like that doesn't exist.
A: The Schroedinger  equation is nothing but the statement of the Stone theorem when the one-parameter group is the time evolution. It does not matter if the system is relativistic or classic, it only matters that you know the Hamiltonian $H$ of the system. The equation has always the form
$$\frac{d}{dt} \psi = -iH \psi .$$
For free quantum field theories the Schroedinger equation can be stated both for a pure state of one particle or for a pure state of many particles. It is false that equations as Klein Gordon and Dirac are the Schroedinger equations because they are not a statement of Stone's theorem. To obtain that statement you usually have to transform  these equations. For instance to produce the Schroedinger equation of a KG particle we have to correctly recognize its Hamiltonian and its Hilbert space (made of only positive frequency solutions of the KG equation). This procedure produces a Hamiltonian operator which is non-local and it is represented by a pseudo-differential operator 
$$H = \sqrt{-\Delta +m^2}.$$
The field equations as KG or Dirac, in this context of Schroedinger picture for states and in counterposition to the Heisenberg picture for operators, are more complicated equations which permit to write the evolution into an apparently local form. E.g. the KG equation can be written
$$(\frac{d}{dt} -iH)(\frac{d}{dt} + iH)\psi =0\:.$$
A: 
That's the first time I see field quantization so it's possible that
  I'm wrong about the concept... I'm in the habit that if I have
  operators for the observables than I have wave functions on which the
  operators can be applied. And the wave function should respect the
  Schrodinger equation.

I'll make an attempt to untangle this a bit.
First, it's true that in the Schrodinger picture of quantum mechanics, the state vector $|\psi(t)\rangle$ of the system is time dependent and obeys the abstract Schrodinger equation:
$$H|\psi(t)\rangle = i\hbar\frac{d}{dt}|\psi(t)\rangle $$
On the (one-particle) coordinate basis $|\mathbf{x}\rangle$, the state vectors are wave functions, and the Hamiltonian is a differential operator:
$$\left(-\frac{\hbar^2}{2m}\nabla^2+V(\mathbf{x})\right)\psi(\mathbf{x},t) = i\hbar\frac{\partial}{\partial t}\psi(\mathbf{x},t)$$
There is another picture, the Heisenberg picture where the state vector of the system is not time dependent; in this picture the operators are time dependent.
Now, in the Heisenberg picture, instead of Schrodinger equation that governs the time evolution of the state vector, there is an operator equation of motion (since the operators carry the time dependence rather than the state vectors).
However, there is a (time independent) state vector $|\psi_H\rangle$ (the subscript H indicates the Heisenberg picture) that satisfies
$$|\psi_H\rangle = |\psi(0))\rangle$$
That is, the state vectors in the Heisenberg and Schrodinger pictures are equal at time $t=0$.
Finally, for the dynamics, the Heisenberg equation of motion for operators is* 
$$\frac{d}{dt}A(t)=\frac{i}{\hbar}[H,A(t)]$$

With that out of the way, a point that (to me) is too quickly passed over in many introductions to quantum fields is that in passing from a wave equation for a wave function (Schrodinger picture) to a wave equation for a quantum field, we change pictures to the Heisenberg picture (where the operators are time dependent).
That is, the equation of motion that the (operator valued) free photon field must satisfy is not due to a Schrodinger's equation (which, again, governs the time evolution of state vectors).
That isn't to say that there isn't a Schrodinger picture (representation) of the free photon field with time dependent wavefunction(al)s that obey a Schrodinger's equation. I'll quote from chapter 10 (Free Fields in the Schrodinger Representation) of Brian Hatfield's Quantum Field Theory of Point Particles and Strings:

For field theory in the Schrodinger representation, we must only
  substitute the word "functional" for "function". Differential
  representations of the canonical commutators are obtained by replacing
  conjugate momenta with functional derivatives. Coordinate
  representations of state vectors or elements of Fock space are wave
  functionals. The Schrodinger equation is a functional differential
  equation whose solutions, the eigenfunctionals of the Hamiltonian
  functional differential operator, represent possible states of the
  system.

(emphasis mine)

Looking around here for some relevant Q & A (or for anything that might be helpful - I'll update from time to time):
Field operator in Schrodinger Picture
State, Dynamics and Interpretation in QFT
Schrodinger functional
