= Solution
Choose the zero of <energy> so that the translation-invariant <vacuum state> has $P^\mu|\Omega\rangle=0$. A <one-particle state> of <rest mass> $m_\psi$ obeys the <relativistic energy-momentum relation>
$$
\boxed{p^\mu=(E_{\mathbf p},\mathbf p),\qquad E_{\mathbf p}=\sqrt{m_\psi^2+|\mathbf p|^2},\qquad p^2=m_\psi^2.}
$$
The <translation covariance of a quantum field> in <1/c> remains valid for the full interacting <four-momentum operator>. Acting its <exponential functions> on the two <momentum eigenstates> gives
$$
\begin{aligned}
\langle\Omega|\phi(x)|\psi,\mathbf p\rangle
&=\langle\Omega|e^{iP\cdot x}\phi(0)e^{-iP\cdot x}|\psi,\mathbf p\rangle\\
&=\boxed{e^{-ip\cdot x}\langle\Omega|\phi(0)|\psi,\mathbf p\rangle.}
\end{aligned}
$$
Here $p\cdot x=E_{\mathbf p}t-\mathbf p\cdot\mathbf x$. A different additive <vacuum energy> must be subtracted consistently from both <energy eigenvalues> before identifying the physical $p^0$.
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