Does Heisenberg picture only work for time-dependent Schrödinger equation not Klein-Gordon equation? 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.
 A: The Klein-Gordon equation does not replace the Schrödinger equation. The former is an equation for the field operators, the latter an equation governing the time evolution of the state.
See for example this, this or this.
A: In normal quantum mechanics the Heisenberg picture is defined as
$${\displaystyle |\psi _{\rm {H}}\rangle=e^{iH_{\rm {S}}~t/\hbar }\displaystyle |\psi _{\rm {S}}(t)\rangle}$$
for states and
$${\displaystyle A_{\rm {H}}(t)=e^{iH_{\rm {S}}~t/\hbar }A_{\rm {S}}e^{-iH_{\rm {S}}~t/\hbar }}$$
for operators
In QFT instead of the state, we will mainly deal with an operator field which in Heisenberg's picture is defined similarly to above
$${\displaystyle \phi_{\rm {H}}(\vec{x},t)=e^{iH_{\rm {S}}~t/\hbar }\phi_{\rm {S}}e^{-iH_{\rm {S}}~t/\hbar }}$$
Here $\phi_{\rm {S}}$ is independent of time since in Schrödinger picture operators are constant.
The definition is not different. The difference is in QM we are not dealing with an operator but in QFT we deal with an operator.
Similarly in QFT the states in the Heisenberg picture are defined as
$${\displaystyle |a_{\rm {H}}\rangle=e^{iH_{\rm {S}}~t/\hbar }\displaystyle |a_{\rm {S}}(t)\rangle}$$
here the state is constant. In QFT Heisenberg picture is more useful when there are no interactions. In QM Schrödinger picture is more useful when there are no interactions. When interactions are there Dirac picture is better in both cases.
