Past exam of the quantum field theory course of the University of Oxford / 2024 / 1

Write ${\int }_{\mathbf{p}}=\int d^3\mathbf{p}/(2\pi {)}^3$. The canonical commutation relations and the product rule for a commutator giveThe other-species terms vanish. Integrating the Dirac delta distributions in the Hamiltonian therefore giveswhere $E_{-p}=E_p$. Consequently the complex scalar field satisfiesDirect differentiation of the time factors gives the same expression, proving the Heisenberg equation of motion

Both a particle and its antiparticle carry the momentum of their mode label. Thus the normal-ordered total momentum operator isThe canonical commutation relations giveThe minus sign in the antiparticle mode label is essential: both terms in the printed mode expansion of a free field contain $e^{i\mathbf{p}\cdot \mathbf{x}}$. Hence

For a small real time displacement $ϵ$, the unitary time evolution operator is $T_t(ϵ)=1-iϵH+O({ϵ}^2)$. The corresponding Heisenberg picture field transformation is conjugation by its inverse:Here a gives $[H,\varphi ]=-i{\partial }_t\varphi$.

Likewise let $T_s(\mathbf{a})=1+i\mathbf{a}\cdot \mathbf{P}+O(|\mathbf{a}{|}^2)$, with the paper's sign for its spatial translation operator. From b,These are precisely the first-order Taylor expansions. Since $H$ and $\mathbf{P}$ commute in this translation-invariant theory, the finite result is

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 $p^0$.

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.