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Choose the zero of energy so that the translation-invariant vacuum state has $P^{\mu }|\Omega ⟩=0$. A one-particle state of rest mass $m_{\psi }$ obeys the relativistic energy-momentum relationThe translation covariance of a quantum field in c remains valid for the full interacting four-momentum operator. Acting its exponential functions on the two momentum eigenstates givesHere $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. A different additive vacuum energy must be subtracted consistently from both energy eigenvalues before identifying the physical $p^0$.

For $p^2=m^2>0$, let a unitary operator $U$ implement a Lorentz boost taking $p$ to $(m,0)$. For a Lorentz scalar field, $⟨q|\varphi (x)|p⟩=⟨\Psi |\varphi (0)|m,0⟩$ with $|\Psi ⟩=e^{-i(q-p)\cdot x}U|q⟩$. The phase follows from translation covariance of a quantum field; its complex conjugate enters the bra. The transformed energy eigenvalue is $(\Lambda q{)}^0=p\cdot q/m$ by invariance of the Minkowski inner product. The reduction requires a massive initial state.

For a massive spin-zero one-particle state with covariant normalization and a Lorentz scalar field, Lorentz invariance makes $F=⟨\Omega |\varphi (0)|p⟩$ independent of the on-shell four-momentum within a fixed species. A Lorentz boost to the rest frame, invariance of the vacuum state and scalar transformation at the origin prove this. Translation covariance of a quantum field then supplies the plane-wave factor. With noncovariant normalization, the boost law and the displayed momentum independence acquire energy factors.