 Solution

Assume $m_{\psi }>0$, as required to have a rest frame. Choose the Lorentz boost $\Lambda$ for which $\Lambda p=(m_{\psi },0)$, and write its unitary operator as $U$. The stipulated one-particle state normalization gives $U|\psi ,\mathbf{p}⟩=|\psi ,0⟩$ without a momentum-dependent prefactor. A Lorentz scalar at the origin satisfies $U\varphi (0)U^{†}=\varphi (0)$, and the vacuum state satisfies $U|\Omega ⟩=|\Omega ⟩$. Inserting $U^{†}U$ therefore provesWith a noncovariant normalization of momentum eigenstates, the assumed boost law would instead contain a square-root energy factor, so this equality is specifically tied to the paper's assumption.

Let $q^{\mu }=(E_{\mathbf{q}}^{\chi },\mathbf{q})$ denote the four-momentum of the other energy eigenstate. Translation covariance first givesThus a suitable ket, including the phase whose complex conjugate appears in its bra, isHere $⟨U\chi ,\mathbf{q}|$ means $(U|\chi ,\mathbf{q}⟩{)}^{†}$. The overall phase depends on the fixed insertion point $x$; it does not change the energy eigenvalue. It cannot simply be dropped if the equality of complex matrix elements is to hold exactly.

The transformed state has four-momentum $q^{\prime }=\Lambda q$. Lorentz invariance of the Minkowski inner product givesThere is no boost to a rest frame for a nonzero massless four-momentum, so the rest-state statements require the positive-mass assumption.