In QFT, the Klein-Gordon equation is solved with the field operator $\hat \psi(x)$/$\hat \psi^\dagger(x)$ in the Heisenberg picture, and (as I understand it) gives the evolution of a single on-mass-shell particle/antiparticle.
My question is, could we solve it in the Schrodinger picture of the field configurations instead? What would this look like? Would it be unhelpful? How would it deal with positive frequency solutions?
In this case, the state changes with time as
$$\Psi\left[\phi_{2}, t_{2}\right]=\int \mathcal{D} \phi_{1} \mathcal{S}\left[\phi_{2}, t_{2} ; \phi_{1}, t_{1}\right] \Psi\left[\phi_{1}, t_{1}\right]$$ here $$\mathcal{S}\left[\phi_{2}, t_{2} ; \phi_{1}, t_{1}\right]=\left\langle\phi_{2}\left|e^{-i H\left(t_{2}-t_{1}\right) / \hbar}\right| \phi_{1}\right\rangle$$
where $H$ is constant. Since $H$ is constant in this picture we don't need to think about the time ordering in the unitary operator.
And the field operator is constant in time defined by (Note instead of 4 momentum and 4 position vector we have 3d vectors)
$$\psi (x)=\int \left.{\frac {\mathrm {d} ^{3}p}{(2\pi )^{3}}}{\frac {1}{2E(\mathbf {p} )}}\left(A(\mathbf {p} )e^{i\vec{p}\cdot \vec{x}}+B(\mathbf {p} )e^{-i\vec{p}\cdot \vec{x}}\right)\right|_{p^{0}=+E(\mathbf {p} )}$$
The S matrix of both pictures is equal.
$$\langle f|S| i\rangle_{\text {H}}=\langle f ; \infty \mid i ;-\infty\rangle_{\text {S}}$$
Read about Schrödinger functional for more information.
