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.