I must confess that I'm still confused about the question of time evolution in relativistic quantum field theory (RQFT). From symmetry arguments, from the representation of the Poincare group through unitary operators on a Hilbert space, one knows that the generator of time translations is the Hamiltonian operator ($H$) of the field, and the unitary operator corresponding to a finite time translation from $t_0$ to $t_1$ is $U = \exp[-iH(t_1 – t_0)]$ (with $\hbar = 1$). So apparently, from symmetry arguments, the Schrodinger picture is live and well (i.e., it's valid and must work) in RQFT.
However, I've never seen in the literature on RQFT a state vector describing the state of the quantum field at a finite time $t$, that is, I've never seen a $|\psi (t)\rangle$. Then, there is the paper by Dirac in which he takes issue with the Schrodinger time-evolution operator $U = \exp[-iH(t_1 – t_0)]$, practically showing that it doesn't make any sensible sense ("Quantum Electrodynamics without dead wood" published in Phys. Rev., http://journals.aps.org/pr/abstract/10.1103/PhysRev.139.B684).
I would very much appreciate it if you could let me know if $U = \exp[-iH(t_1 – t_0)]$ is indeed the time evolution operator in RQFT, and if the answer is yes, then why the state vector $|\psi (t)\rangle$ is never defined and calculated for a finite time $t$. This question is also important for the path-integral quantization, since in order to construct a path integral one uses the expression $U = \exp[-iH(t_1 – t_0)]$ for the time evolution operator, and also state vectors at finite time. Otherwise, the path integral approach cannot be constructed and the path-integral formulas must be postulated! Are they merely postulated?
EDIT 1: There seems to be a lot of confusion in the comments and answers between the Schrodinger picture and the Schrodinger representation (in QFT). As in all the textbooks on QM there are, by the Schrodinger picture I understand a description of a quantum system by states represented by abstract time-dependent state vectors in some Hilbert space, and whose observables are represented by time-independent operators on that Hilbert space. There is nothing unclear here, and there is no question of any ill-defined "functionals" since no representation has been introduced yet, i.e., no particular basis of the Hilbert space has been chosen in order to pass from abstract vectors to wave functions (or wave functionals) by projecting the abstract vectors on that basis.
As a matter of fact, Dirac in his paper uses abstract vectors in the formalism of second quantization. What is really puzzling, and this is the main cause of my confusion, is that the time-evolution operator should exist in the Schrodinger picture (i.e., there must be a unitary operator relating any two state vectors that have different time arguments), based on the Poincare invariance imposed on any relativistic quantum system. Yet, as Dirac shows the math just doesn't make any sense (and this is not because of some "functionals") and it seems that the Schrodinger picture must be banned from QFT, and this, in turn, comes against Poincare invariance!
EDIT 2: It's interesting to note that $U(t)$ doesn't exist even in the Heisenberg picture and not even for a free field. V. Moretti's construction is faulty since his $U(t)$ doesn't have a domain of definition. Indeed, $ \psi(x) \propto a_p + a_p^{\dagger}$. Therefore, $(\psi(x))^2 \propto (a_{p})^2 + (a_p^{\dagger})^2 + …$, and hence $H \propto \int (\psi(x))^2 d^3x \propto \sum_p \{(a_{p})^2 + (a_p^{\dagger})^2 + …\}.$ Hence, $U(t) \propto t^2 \sum_p (a_{p})^{2}(a_p^{\dagger})^{2}$ already at second order in $t$. Hence, acting on the vacum state $U(t)$ gives $\infty$ (for finite $t$!), as Dirac has shown in his paper. Therefore, $U(t)$ doesn't have a domain of definition and it cannot exist, even in the Heisenberg picture!
EDIT 3: @Valter Moretti You claim in your answer that you work entirely in the Heisenberg picture. Therefore, the field depends on both space $\bf{x}$ and time $t$. However, when you construct the Hamiltonian you "smear" the field with a test function, but integrate only over the space, i.e. over $\bf{x}$, but not over time $t$. Therefore, your Hamiltonian operator should depend on time $t$. You cannot simply eliminate $t$ unless you also integrate over it! There is no such construction in the literature! If it's not so, please edit your answer by showing as to how you get rid of the time dependence in the Hamiltonian.
Best Answer
Consider for the shake of simplicity a free neutral scalar field $\phi$. Passing to the second quantization picture, it is a operator valued distribution $$C_0^\infty(M;\mathbb R) \ni f \mapsto \phi(f)$$ where $M$ is Minkowski spacetime and $\phi(f)$ is a densely defined symmetric operator on the Hilbert space $$F_+(\cal H) = \mathbb C \oplus \cal H \oplus(\cal H \otimes \cal H)_S \oplus (\cal H \otimes \cal H \otimes \cal H)_S \oplus \cdots $$ Obviously $F_+(\cal H)$ is the symmetrized ($(\cdots)_S$) Fock space generated by the one-particle Hilbert space $\cal H$. And $$\phi(f) = a_{Pf} + a^\dagger_{Pf} \tag{1}$$ Above $P$ extract the positive frequency part of the $4D$ Fourier transform of $f$ and $a^\dagger$ and $a$ are the standard creation and annihilation operators. Formally $$\phi(f) = \int_{\mathbb R^4} f(x) \phi(x) d^4x\:,\tag{2}$$ The domain of definition of $\phi(f)$ is the dense invariant span $D$ of vectors with finite number of particles associated with all possible $f$ as above. With this definition it is possible to give a rigorous meaning to objects (densely defined symmetric operators) of the form $$\int_{\mathbb R^3} f(\vec{x}) \phi(t,\vec{x})^2 d^3x \:\quad \mbox{and}\quad \int_{\mathbb R^3} g(\vec{x}) \partial_a \phi(t,\vec{x}) \partial_b \phi(t,\vec{x}) d^3x$$ where $\mathbb R^3$ is the rest space of an inertial reference frame referred to Minkowskian coordinates. The procedure to do it is usually called normal ordering. Also the limit for $f \to 1$ constantly on $M$ is well defined producing essentially self adjoint operators. In particular integrating the component $T_{00}$ of the stress energy tensor over the rest space $\mathbb R^3$ you obtain the Hamiltonian operator, $H$, of the field as the unique self-adjont extension of the integrated operator. It is finally possible to prove that $$H = \oplus_{k=0}^{+\infty} H_k\:,$$ where $H_0 =0$, $H_1 = \sqrt{m^2+ \vec{P}^2} : \cal H \to \cal H$ is the standard relativistic Hamiltonian operator of one particle associated with the quantumfield $\phi$, $P_\mu$ being the standard four momentum operator of one particle in $\cal H = L^2(\mathbb R^3, d\vec{p}/p^0)$, and $$H_n = H+ \cdots + H \quad \mbox{(n times)}\quad : \cal H \oplus \cdots \oplus \cal H \to \cal H \oplus \cdots \oplus \cal H$$ is the Hamiltonian of a system of $n$ particles. The time evolutor is finally well defined, as $H$ is self-adjoint and reads $$U_t := e^{-itH}$$ With this definition, it is possible to prove that with an interpretation of $\phi$ as the one given in (2), $$U_\tau \phi(t,x) U_\tau^{-1} = \phi(t+\tau, x)$$ which is the integrated version of Heisenberg equation of motion. This identity can be written in a rigorous form using the rigorous interpretation of $\phi$ provided in (1).