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

The following claim is made:

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 $t-t_0\rightarrow t_0$.


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