Standard quantum mechanics postulates that, for an isolated system, time evolution is ruled by unitary operators, then one can prove Schrodinger equation (SE), which is not Lorentz invariant.
If we want a quantum relativistic theory we need, for some reasons, a field viewpoint and this leads to QFT.
My question is: can we still say that in QFT time evolution is, by postulate, unitary?
A fundamental postulate of QFT establishes that the theory admits a strongly continuous representation of (orthochronous proper) Poincaré group $\cal P$. A certain one-parameter subgroup of $\cal P$ describes time evolution (with respect to an inertial reference frame) which, as a consequence, turns out to be unitary since it is part of a larger unitary representation.
Alternatively, you can exploit Wigner theorem. Time is homogeneous in inertial systems both in classical and relativistic physics, so time evolution must preserve probability transitions for isolated physical systems. Wigner theorem implies that time evolution is represented by unitary or antiunitary transformations $U_t$. Assuming that $U_t$ is a representation of $\mathbb R$ (the axis of time), i.e. $U_tU_s = U_{t+s}$, one sees that each $U_t$ must be unitary, because $U_t = U_{t/2}U_{t/2}$ is unitary nomatter if $U_{t/2}$ is unitary or antiunitary.
Under some further quite mild requirements (separability of the Hilbert space and the fact that the complex valued maps $\mathbb R \ni t \mapsto \langle \psi | U_t \phi \rangle$ are measurable for every choice of the vectors) a theorem due to von Neumann proves that $\mathbb R \ni t \to U_t$ is strongly continuous and thus admits a unique self-adjoint generator (Stone theorem), that is the Hamiltonian operator, by definition.
