Past exam of the theoretical physics major option course of the University of Oxford / 2018 / 3

The action is dimensionless and $[d^{4}x]=-4$, so $[\mathcal{L}]=4$. A canonical scalar kinetic term then requires $2+2[\varphi ]=4$, and the electromagnetic kinetic term requires $2[F_{\mu \nu }]=4$. ThusThe brackets here denote mass dimension. The Lagrangian has no explicit coordinate dependence, and all indices are contracted as Lorentz scalars. Its continuous spacetime symmetry is the Poincaré group. Noether's theorem gives conserved four-momentum from translations and conserved angular momentum, including the boost generators, from Lorentz transformations. The nonzero mass $M$ and dimensionful couplings prevent a dilation symmetry.

Sign qualification: the printed electromagnetic kinetic term has the opposite healthy relative sign to the scalar kinetic terms. With the stated mostly-minus Minkowski metric, a physical Maxwell Lagrangian is $-F_{\mu \nu }{F}^{\mu \nu }/4$. The printed plus sign instead gives photon Hamiltonian density $-({\mathbf{E}}^2+{\mathbf{B}}^2)/2$; a nonzero purely electric configuration already has negative energy, unbounded below as its amplitude grows. The physical decay widths below use the healthy Maxwell sign with the given interaction normalization. The dimensional and spacetime-symmetry conclusions themselves do not depend on this sign correction.

Counting the mass dimensions of each interaction operator givesIndeed $\Phi (\partial a{)}^2$ and $\Phi F^2$ have dimension five, whereas $\Phi b^2(\partial b{)}^2$ has dimension seven. These are interactions of an effective field theory with dimensionful couplings.

The direct one-vertex decay channels are $\Phi \to aa$, $\Phi \to bbbb$ and $\Phi \to \gamma \gamma$. Each leading scattering amplitude is linear in its corresponding coupling, and a decay width is quadratic in the amplitude. Therefore, holding the parent mass fixed,There are no tree-level interference terms between these different final states.

All daughter particles are massless, so the parent mass $M$ supplies the only kinematic scale at tree level. Each vertex has two derivatives, making its scattering amplitude scale as ${\lambda }_i{M}^2$. For the four-$b$ vertex, symmetrizing over the derivative-bearing legs yields a sum of pairwise four-momentum products, which is proportional to $M^2$ by four-momentum conservation.

For $n$ massless daughters, the Lorentz invariance Lorentz-invariant phase-space measure scales as $M^{2n-4}$: each $d^{3}q/(2E)$ contributes two powers and the four-dimensional Dirac delta removes four powers. Including the parent factor $1/(2M)$ gives $\Gamma \sim |{\lambda }_i{|}^2{M}^{2n-1}$. ThusThe unspecified factors are dimensionless angular integrals and identical-particle factors. Each result has mass dimension one, as a decay width must in natural units.

Every interaction contains an even number of $b$ fields, so the exact discrete symmetry $b\mapsto -b$ forbids $\Phi \to bbb$ to every perturbative order. There is no such prohibition for $\Phi \to bb$.

The latter decay channel is absent at tree level in the displayed interactions, but contracting two $b$ legs at a $\Phi b^2(\partial b{)}^2$ vertex produces a tadpole diagram with two external $b$ legs. For example, contracting the undifferentiated pair leaves the local operator $\Phi (\partial b{)}^2$ multiplied by an integral proportional to $\int d^4\ell /({\ell }^2+i0)$. Contracting the differentiated pair instead gives a local $\Phi b^2$ contribution. Both are allowed effective field theory operators. With a ultraviolet cutoff, these local terms generically carry quadratic or quartic cutoff dependence and require matching coefficients or counterterms.

There is a relevant qualification to a claim that the one-loop decay width must be nonzero. Since $b$ is exactly massless here, the one-vertex tadpole integrals have no external scale and are scaleless integrals. They vanish in dimensional regularization. Thus the correct conclusion is allowed, and generically generated by quantum corrections or matching, not a regulator-independent nonzero one-loop prediction from ${\lambda }_b$ alone. No exact symmetry protects the two-$b$ channel; its numerical rate requires the renormalized coefficients of the allowed operators.

Work in the parent rest frame, $p=(M,0)$. For massless daughters, the spatial Dirac delta fixes ${\mathbf{q}}_2=-{\mathbf{q}}_1$ and $E_1=E_2=q$. The Lorentz invariance two-body Lorentz-invariant phase-space measure becomesThe factor $1/2$ in the last step is the Jacobian of the energy-conservation Dirac delta. Consequently the labelled two-body decay width isFor two identical final particles, the full angular integral counts each unordered final state twice, so the physical decay width includes an additional factor $1/2!$. This factor is needed for both channels in part e; it is separate from any vertex combinatorics already included in the Feynman rule.

The massless two-body kinematics gives $q_1\cdot q_2=M^2/2$. The scalar scattering amplitude therefore has constant magnitude $|{\mathcal{A}}_{aa}|=|{\lambda }_a|M^2/2$. Use $\int d\Omega =4\pi$ and the identical-particle factor from part d:For the photons, the two-photon vertex of a scalar field-strength interaction following from $\Phi F_{\mu \nu }{F}^{\mu \nu }$ is, up to an irrelevant overall phase,Vertex qualification: the source prints same-index polarization contractions in the second product, whereas the contractions must be crossed as above. The printed expression fails the Ward identity. For a concrete counterexample take $q_1=(E,0,0,E)$, $q_2=(E,0,0,-E)$ and ${ϵ}_2=(1,1,0,-1)$. This is a transverse unit spacelike photon polarization vector, since $q_2\cdot {ϵ}_2=0$ and ${ϵ}_2^2=-1$. Replacing ${ϵ}_1$ by $q_1$ in the printed expression leaves $4{\lambda }_{\gamma }{E}^3$ instead of zero. The corrected crossed expression vanishes under either such replacement.

Choose physical spatial transverse photon polarizations along $x$ and $y$ in the parent rest frame. Then both crossed products vanish; the two equal-polarization combinations have $|{ϵ}_1\cdot {ϵ}_2{|}^2=1$ and the two orthogonal combinations give zero. There is no initial-spin average for a spin-zero parent. HenceThe calculation in this spatial transverse gauge also yields the intended result from the misprinted expression, but only the corrected vertex makes its gauge invariance explicit. For equal-magnitude couplings, the photon decay width is twice the two-scalar decay width because of its two nonzero polarization combinations.