Past exam of the quantum field theory course of the University of Oxford / 2023 / 2

For nonzero real $g$, the continuous internal symmetry of a classical field theory is the global U(1) symmetryThe kinetic terms and mass terms are invariant separately, while the phases in ${\varphi }^2{\Phi }^{†}$ cancel. Independent phase rotations would instead require $2{\alpha }_{\varphi }-{\alpha }_{\Phi }=0$ modulo $2\pi$, leaving one connected circle. Simultaneous complex conjugation is also a discrete symmetry, reversing the continuous charge; including it gives $U(1)⋊{\mathbb{Z}}_2\cong O(2)$ for generic parameters. At $g=0$ the continuous symmetry enlarges to $U(1)\times U(1)$.

The Noether current with particles assigned positive charges isFor example, put $A={\varphi }^2{\Phi }^{†}$ and $B={\varphi }^{†2}\Phi$. The Euler-Lagrange equations give ${\partial }_{\mu }{j}_{\varphi }^{\mu }=ig(A-B)$ and ${\partial }_{\mu }{j}_{\Phi }^{\mu }=ig(B-A)$, with the factor $2$ already included in $j_{\Phi }^{\mu }$. Their sum vanishes. The Noether charge $Q=\int d^{3}x{j}^0$ is conserved and, after normal ordering, countsThus a $\Phi$ particle carries two units of the same conserved charge carried by a $\varphi$ particle; their antiparticles carry the negatives. The interaction permits conversion of one $\Phi$ particle into two $\varphi$ particles when allowed by conservation of energy, while preserving this weighted number.

The two complex scalar fields have scalar propagatorscontractions of two unconjugated fields or two conjugated fields vanish. Draw an arrow on each line in the direction of positive conserved charge flow: a solid line carries one unit and a dashed line two units. A particle arrow follows its physical momentum; an antiparticle arrow opposes it.

There are two conjugate three-leg interaction vertices, both with Feynman rule $-ig$: one has a charge-two arrow entering and two charge-one arrows leaving, and the other reverses every arrow. These describe the two conjugate terms ${\varphi }^2{\Phi }^{†}$ and ${\varphi }^{†2}\Phi$. The two identical $\varphi$ attachments cancel the $1/2$ in the Lagrangian density. All crossed assignments are obtained by moving external lines between initial and final states while maintaining the arrows. There is no interaction vertex with three equal-sign charges.

Each interaction vertex conserves four-momentum, supplying $(2\pi {)}^4{\delta }^4(\sum p_j)$ with all momenta entering. Integrate each independent loop momentum with $d^{4}k/(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor. No minus sign is assigned to a closed scalar loop. These rules are for $i\mathcal{M}$ after amputating external lines and removing the overall four-momentum conservation factor.

Label the incoming four-momenta $p_1,p_2$ and the outgoing ones $p_3,p_4$. Define the Mandelstam variables $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$. Two incoming $\varphi$ particles have conserved charge $+2$, so the only tree-level Feynman diagram is an $s$-channel $\Phi$ exchange, shown in panel (b) of b's figure. Each end converts two charge-one lines into one charge-two line. There are no $t$- or $u$-channel exchanges: an incoming and an outgoing particle at such an end would transfer zero charge, for which this theory has no cubic interaction vertex.The single Feynman diagram is already symmetric under exchanging the two outgoing particles. This does not add another copy of the scattering amplitude; the identical final-state symmetry factor enters the integrated scattering cross-section instead.

Take $\varphi (p_1)+\bar{\varphi }(p_2)\to \varphi (p_3)+\bar{\varphi }(p_4)$ with the Mandelstam variables of i. The allowed tree-level Feynman diagram is the crossed $u$-channel $\Phi$ exchange in panel (c) of b's figure. At one end the incoming particle and outgoing antiparticle transfer charge $+2$; the other end absorbs that charge. The zero-charge $s$- and $t$-channel assignments are forbidden by the interaction vertices. ThusIn the center-of-mass frame with $|\mathbf{p}|\ll m$, $s=4m^2+O(|\mathbf{p}{|}^2)$ and $u=O(|\mathbf{p}{|}^2)$. For $M\gg m$, both scattering amplitudes approach $g^2/M^2$. The elastic two-body Lorentz-invariant phase space and incident flux agree for the two experiments. Usingfor distinguishable final particles, and an extra factor $1/2!$ when integrating the identical-particle final state over the whole sphere, givesThere is no additional division for identical incoming particles in the definition of a beam scattering cross-section. Before taking $M/m\to \infty$, the zero-momentum ratio is $\frac{1}{2}[M^2/(M^2-4m^2){]}^2$, provided the threshold is away from the heavy pole.

Near $s=M^2$, the bare $s$-channel scalar propagator makes fixed-order perturbation theory fail: repeated self-energy insertions contain powers of ${\Sigma }_{\Phi }(s)/(s-M^2)$. The $\Phi \to \varphi \varphi$ channel is open for $M>2m$, giving the leading decay widthResumming the self-energy gives a finite Breit-Wigner resonance cross section, with denominator $s-M_{\mathrm{pole}}^2+i{M}_{\mathrm{pole}}{\Gamma }_{\Phi }$ near the pole, rather than an infinite scattering cross-section. Also, $s\simeq M^2$ with $M\gg m$ lies outside the assumed nonrelativistic incoming kinematics; the ratio $1/2$ is not a resonance prediction.

Four external $\Phi$ lines require four cubic interaction vertices, since each such interaction vertex contains only one heavy field. Their other eight endpoints must form four light internal lines. A connected Feynman diagram therefore has $V=I=4$ and $L=1$: the leading contribution is a one-loop box diagram, of order $g^4$, with four $\varphi$ scalar propagators. Charge-two incoming and outgoing heavy lines attach alternately around the box, as shown in panel (d) of b's figure. Each incoming heavy line supplies the two outgoing light-charge arrows at its corner, and each outgoing heavy line absorbs them. With external labels fixed there is one such box and its symmetry factor is one.

For $|\mathbf{p}|\ll M\ll m$, the external energies are also small compared with $m$, so expand the loop integrand in all external four-momenta. Its leading term isA Wick rotation and the Euclidean massive loop integral giveHence, more precisely than just its requested scaling,The mass dimension $[g]=1$ makes $g^4/m^4$ dimensionless, as a four-dimensional scattering amplitude should be. The leading amplitude has no extra identical-particle multiplicity.

Cross the second incoming heavy particle and one outgoing heavy particle in i to obtain $\Phi \bar{\Phi }\to \Phi \bar{\Phi }$. Panel (e) of b's figure shows the resulting box diagram. Its four light-charge arrows again alternate source and sink corners; it is the only connected order-$g^4$ topology. There is no tree exchange because every cubic interaction vertex has only one heavy leg. With $|\mathbf{p}|\ll M\ll m$, crossing does not affect the zero-external-momentum loop integral, soAn independent check is to integrate out the complex scalar field $\varphi$ at constant $\Phi$. Its two real-component squared masses are $m^2\pm g|\Phi |$. The one-loop quantum effective potential is thereforeExpanding the logarithms to fourth order in $\Phi$ givesThus the effective Lagrangian has the positive local coefficient $g^4/(384{\pi }^2{m}^4)$. Its $2!2!$ external attachments give the common scattering amplitude $g^4/(96{\pi }^2{m}^4)$ in both channels. This also checks the box multiplicity and sign.