# Expanding field operators at a fixed time $t_0$ (from Peskin/Schroeder)

This relates to the bottom of page 83 in Peskin and Schroeder.

At any fixed time $$t_0$$ we can of course expand $$\phi$$ in terms of ladder operators $$\phi(\textbf{x},t_0)=\displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}(a_pe^{i\textbf{p.x}}+a_p^\dagger e^{-i\textbf{p.x}}). \tag{i}$$ Then to obtain $$\phi(\textbf{x},t)$$ for $$t\neq t_0$$ we just switch to the Heisenberg picture $$\phi(\textbf{x},t)=e^{iH(t-t_0)}\phi(\textbf{x},t_0)e^{-iH(t-t_0)}.\tag{ii}$$
The usage of "at any fixed time $$t_0$$" is odd to me. It appears we can obtain, given the field operator on any timeslice $$\phi(\vec x,t_0)$$, the field on a different timeslice through (ii). Essentially given the "initial condition" $$\phi(\vec x,t_0)$$ we can obtain the entire field, noting that $$H$$ in the above is an interacting Hamiltonian.
However, since $$t_0$$ is supposedly arbitrary, if we use (ii) to obtain the field at some time $$t-t_0$$, we could re-choose our "fixed time" to be $$t-t_0$$, and simply expand the field as in (i). This seems odd since (i) is the free field expansion, which the above appears to imply is valid at all $$t$$, by a trivial relabelling $$t-t_0\rightarrow t_0$$.