Past exam of the theoretical physics major option course of the University of Oxford / 2015 / 6

Evaluate the two-body Lorentz-invariant phase space in the decaying particle's rest frame, $p=(m,0)$. The spatial delta function sets ${\mathbf{q}}_2=-{\mathbf{q}}_1$. Both massless particles have energy $q=|{\mathbf{q}}_1|$, leavingHere the factor $1/2$ in the second equality is the Jacobian of the energy delta function. Integrating the solid angle gives $\int d{\Pi }_2=1/(8\pi )$. If the decay is isotropic, the squared amplitude is angle independent, soThis uses distinguishable final particles or an amplitude/phase-space convention with any required identical-particle symmetry factor already included. Two genuinely identical final bosons require an additional $1/2!$ if the full angular domain is integrated without such a factor. The complex particle and antiparticle in the following parts are distinct, so no such division is required there.

Define the free Feynman propagators byThe interaction preserves the global phase symmetry $\phi \mapsto e^{i\alpha }\phi$, with the real $\varphi$ neutral. Therefore the charge-two correlator vanishes in a symmetry-invariant perturbative vacuum:In Wick contractions, a complex field contracts only with its conjugate; one vertex has one of each and cannot absorb both external $\phi$ fields. The symmetry shows this vanishing holds to all perturbative orders, not only first order.

The neutral connected correlator has one allowed contraction to each external point at order $g$:The propagators are even functions of their separation; their orientations encode the conjugate-field contractions. Equivalently, with every external momentum Fourier transformed as incoming,No factor of two accompanies the vertex: $\phi$ and ${\phi }^{†}$ are distinct fields.

For the full correlator as literally written, a disconnected scalar three-point tadpole is also possible. Contract the two external complex fields with each other and contract the complex pair at the vertex into a loop. Unless normal ordering is imposed on the interaction or a tadpole counterterm fixes $⟨\varphi ⟩=0$, addThe coincident propagator is regulator dependent; in dimensional regularization the massless scaleless tadpole diagram vanishes. More generally this is $⟨T{\phi }^{†}(x)\phi (y){⟩}_0⟨\varphi (z)⟩$ to this order and is removed by taking the connected correlator or imposing the stated vacuum-renormalization condition. Normalizing by vacuum bubbles alone does not remove an external-leg disconnected tadpole diagram. The charged correlator remains zero in every one of these conventions.^[ref]

At a single interaction insertion, the Dyson series supplies $i{\mathcal{L}}_{\mathrm{int}}=-ig{\phi }^{†}\phi \varphi$. Amputating the three external propagators in b gives the momentum-space ruleIf the momentum-conservation factor is included in the vertex convention, append $(2\pi {)}^4{\delta }^{(4)}(p_{\varphi }+p_{\phi }+p_{{\phi }^{†}})$, with all momenta incoming. There is one charged line entering and one leaving the vertex, and a neutral real-scalar line. The absence of a factorial in the interaction is appropriate because its three fields are distinct. The amputated decay amplitude has magnitude $|g|$ with the usual convention that a graph computes $i\mathcal{A}$.

The leading decay is into one complex particle and its antiparticle, which are distinguishable. Using $|\mathcal{A}{|}^2=g^2$ in a givesIn four space-time dimensions a canonically normalized scalar has mass dimension one, so $g$ has mass dimension one. The amplitude starts at order $g$, a rate is quadratic in that amplitude, and the only mass scale in the massless-daughter decay is $m$. Those facts already require $\Gamma \propto g^2/m$, leaving only a dimensionless numerical factor to calculate.

For low-energy charged-scalar scattering through a neutral scalar, label the incoming particle and antiparticle as $1,2$ and their outgoing counterparts as $3,4$. Charge flow allows $s$-channel annihilation and $t=(p_1-p_3{)}^2$ exchange of the real massive scalar. It forbids a separate $u$-channel graph for this fixed assignment of distinguishable final charges. The tree amplitude isAt low energy it is approximately angle independent. The massless two-body scattering formula then givesIn particular the requested scale estimate is $\sigma \sim g^4/(m^{4}s)$, not $g^4/m^6$ with the incoming-energy dependence omitted. Integrating out $\varphi$ gives ${\mathcal{L}}_{\mathrm{eff}}=g^2({\phi }^{†}\phi {)}^2/(2m^2)$ at this order, whose complex-scalar four-leg combinatorics reproduce the same $2g^2/m^2$ amplitude. These are weak-coupling perturbative results about the specified vacuum; possible ultraviolet stabilizing interactions are not determined by the cubic term alone.