For a Klein-Gordon field, our QFT lecture notes say we use the following relationship to define the Heisenberg picture.
$$i \frac{dQ}{dt} = [Q,H]$$
which leads to
$$Q(t) = e^{iHT}Q(0)e^{-iHt}$$
However, for a Klein-Gordon field, shouldn't the Klein-Gordon equation replace the time-dependent Schrödinger equation, and therefore (since there is a double time derivative now), shouldn't the general solution not just be in the form $|\psi(t)\rangle = e^{-iHt} |\psi(0)\rangle$ (since $|\psi(t)\rangle = e^{+iHt} |\psi(0)\rangle$ would also be a valid solution)?
Since all the rest of QFT seems to use this Heisenberg picture for their operators, this seems like an important point for me to understand.